← 学习库 OpenStax Calculus Vol 3 目录

4 Differentiation of Functions of Several Variables

> 来源: OpenStax《Calculus Volume 3》| 原页: https://openstax.org/books/calculus-volume-3/pages/4-differentiation-of-functions-of-several-variables

（该页为章节总览/导航页，无独立正文；本章内容请见其下各小节。）

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Introduction

> 来源: OpenStax《Calculus Volume 3》| 原页: https://openstax.org/books/calculus-volume-3/pages/4-introduction

Chapter Outline

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4.1 Functions of Several Variables

> 来源: OpenStax《Calculus Volume 3》| 原页: https://openstax.org/books/calculus-volume-3/pages/4-1-functions-of-several-variables

4.1 Functions of Several Variables

Our first step is to explain what a function of more than one variable is, starting with functions of two independent variables. This step includes identifying the domain and range of such functions and learning how to graph them. We also examine ways to relate the graphs of functions in three dimensions to graphs of more familiar planar functions.

Functions of Two Variables

The definition of a function of two variables is very similar to the definition for a function of one variable. The main difference is that, instead of mapping values of one variable to values of another variable, we map ordered pairs of variables to another variable.

A function of two variables $z = {f\left( {x,y} \right)}$ maps each ordered pair $\left( {x,y} \right)$ in a subset $D$ of the real plane $\mathbb{R}^{2}$ to a unique real number $z.$ The set $D$ is called the *domain* of the function. The *range* of $f$ is the set of all real numbers $z$ that has at least one ordered pair $\left( {x,y} \right) \in D$ such that $f\left( {x,y} \right) = z$ as shown in the following figure.

Determining the domain of a function of two variables involves taking into account any domain restrictions that may exist. Let’s take a look.

Domains and Ranges for Functions of Two Variables

Find the domain and range of each of the following functions:

1. $f\left( {x,y} \right) = 3x + 5y + 2$

2. $g\left( {x,y} \right) = \sqrt{9 - x^{2} - y^{2}}$

Solution

1. This is an example of a linear function in two variables. There are no values or combinations of $x$ and $y$ that cause $f\left( {x,y} \right)$ to be undefined, so the domain of $f$ is $\mathbb{R}^{2}.$ To determine the range, first pick a value for $z.$ We need to find a solution to the equation $f\left( {x,y} \right) = z,$ or $3x + 5y + 2 = z.$ One such solution can be obtained by first setting $y = 0,$ which yields the equation $3x + 2 = z.$ The solution to this equation is $x = \frac{z - 2}{3},$ which gives the ordered pair $\left( {\frac{z - 2}{3},0} \right)$ as a solution to the equation $f\left( {x,y} \right) = z$ for any value of $z.$ Therefore, the range of the function is all real numbers, or $\mathbb{R}.$

2. For the function $g\left( {x,y} \right)$ to have a real value, the quantity under the square root must be nonnegative:

$$9 - x^{2} - y^{2} \geq 0.$$

This inequality can be written in the form

$$x^{2} + y^{2} \leq 9.$$

Therefore, the domain of $g\left( {x,y} \right)$ is $\left\{ {\left( {x,y} \right) \in \left. \mathbb{R}^{2} \right|x^{2} + y^{2} \leq 9} \right\}.$ The graph of this set of points can be described as a disk of radius $3$ centered at the origin. The domain includes the boundary circle as shown in the following graph.

To determine the range of $g\left( {x,y} \right) = \sqrt{9 - x^{2} - y^{2}}$ we start with a point $\left( {x_{0},y_{0}} \right)$ on the boundary of the domain, which is defined by the relation $x^{2} + y^{2} = 9.$ It follows that $x_{0}^{2} + y_{0}^{2} = 9$ and

$$g\left( {x_{0},y_{0}} \right) = \sqrt{9 - x_{0}^{2} - y_{0}^{2}} = \sqrt{9 - \left( {x_{0}^{2} + y_{0}^{2}} \right)} = \sqrt{9 - 9} = 0.$$

If $x_{0}^{2} + y_{0}^{2} = 0$ (in other words, $x_{0} = y_{0} = 0),$ then

$$g\left( {x_{0},y_{0}} \right) = \sqrt{9 - x_{0}^{2} - y_{0}^{2}} = \sqrt{9 - \left( {x_{0}^{2} + y_{0}^{2}} \right)} = \sqrt{9 - 0} = 3.$$

This is the maximum value of the function. Given any value *c* between $0\ \text{and}\ 3,$ we can find an entire set of points inside the domain of $g$ such that $g\left( {x,y} \right) = c\text{:}$

$$\begin{array}{rll}

\sqrt{9 - x^{2} - y^{2}} & = & c \\

{9 - x^{2} - y^{2}} & = & c^{2} \\

{x^{2} + y^{2}} & = & {9 - c^{2}.}

\end{array}$$

Since $9 - c^{2} > 0,$ this describes a circle of radius $\sqrt{9 - c^{2}}$ centered at the origin. Any point on this circle satisfies the equation $g\left( {x,y} \right) = c.$ Therefore, the range of this function can be written in interval notation as $\left\lbrack {0,3} \right\rbrack.$

Find the domain and range of the function $f\left( {x,y} \right) = \sqrt{36 - 9x^{2} - 9y^{2}}.$

Graphing Functions of Two Variables

Suppose we wish to graph the function $z = f\left( {x,y} \right).$ This function has two independent variables $\left( {x\ \text{and}\ y} \right)$ and one dependent variable $(z).$ When graphing a function $y = f(x)$ of one variable, we use the Cartesian plane. We are able to graph any ordered pair $\left( {x,y} \right)$ in the plane, and every point in the plane has an ordered pair $\left( {x,y} \right)$ associated with it. With a function of two variables, each ordered pair $\left( {x,y} \right)$ in the domain of the function is mapped to a real number $z.$ Therefore, the graph of the function $f$ consists of ordered triples $\left( {x,y,z} \right).$ The graph of a function $z = \left( {x,y} \right)$ of two variables is called a surface.

To understand more completely the concept of plotting a set of ordered triples to obtain a surface in three-dimensional space, imagine the $\left( {x,y} \right)$ coordinate system laying flat. Then, every point in the domain of the function $f$ has a unique $z\text{-value}$ associated with it. If $z$ is positive, then the graphed point is located above the $\textit{xy}\text{-plane,}$ if $z$ is negative, then the graphed point is located below the $\textit{xy}\text{-plane}.$ The set of all the graphed points becomes the two-dimensional surface that is the graph of the function $f.$

Graphing Functions of Two Variables

Create a graph of each of the following functions:

1. $g\left( {x,y} \right) = \sqrt{9 - x^{2} - y^{2}}$

2. $f\left( {x,y} \right) = x^{2} + y^{2}$

Solution

1. In Example 4.1, we determined that the domain of $g\left( {x,y} \right) = \sqrt{9 - x^{2} - y^{2}}$ is $\left\{ {\left( {x,y} \right) \in \left. \mathbb{R}^{2} \right|x^{2} + y^{2} \leq 9} \right\}$ and the range is $\left\{ {z \in \left. \mathbb{R} \right|0 \leq z \leq 3} \right\}.$ When $x^{2} + y^{2} = 9$ we have $g\left( {x,y} \right) = 0.$ Therefore any point on the circle of radius $3$ centered at the origin in the $x,y\text{-plane}$ maps to $z = 0$ in $\mathbb{R}^{3}.$ If $x^{2} + y^{2} = 8,$ then $g\left( {x,y} \right) = 1,$ so any point on the circle of radius $2\sqrt{2}$ centered at the origin in the $x,y\text{-plane}$ maps to $z = 1$ in $\mathbb{R}^{3}.$ As $x^{2} + y^{2}$ gets closer to zero, the value of *z* approaches 3. When $x^{2} + y^{2} = 0,$ then $g\left( {x,y} \right) = 3.$ This is the origin in the $x,y\text{-plane}.$ If $x^{2} + y^{2}$ is equal to any other value between $0\ \text{and}\ 9,$ then $g\left( {x,y} \right)$ equals some other constant between $0\ \text{and}\ 3.$ The surface described by this function is a hemisphere centered at the origin with radius $3$ as shown in the following graph.

2. This function also contains the expression $x^{2} + y^{2}.$ Setting this expression equal to various values starting at zero, we obtain circles of increasing radius. The minimum value of $f(x,y) = x^{2} + y^{2}$ is zero (attained when $x = y = 0.).$ When $x = 0,$ the function becomes $z = y^{2},$ and when $y = 0,$ then the function becomes $z = x^{2}.$ These are cross-sections of the graph, and are parabolas. Recall from Introduction to Vectors in Space that the name of the graph of $f(x,y) = x^{2} + y^{2}$ is a *paraboloid*. The graph of $f$ appears in the following graph.

Nuts and Bolts

A profit function for a hardware manufacturer is given by

$$f(x,y) = 16 - \left( {x - 3} \right)^{2} - \left( {y - 2} \right)^{2},$$

where $x$ is the number of nuts sold per month (measured in thousands) and $y$ represents the number of bolts sold per month (measured in thousands). Profit is measured in thousands of dollars. Sketch a graph of this function.

Solution

This function is a polynomial function in two variables. The domain of $f$ consists of $\left( {x,y} \right)$ coordinate pairs that yield a nonnegative profit:

$$\begin{array}{l}

{16 - \left( {x - 3} \right)^{2} - \left( {y - 2} \right)^{2} \geq 0} \\

{\left( {x - 3} \right)^{2} + \left( {y - 2} \right)^{2} \leq 16.}

\end{array}$$

This is a disk of radius $4$ centered at $\left( {3,2} \right).$ A further restriction is that both $x\ \text{and}\ y$ must be nonnegative. When $x = 3$ and $y = 2,$ $f\left( {x,y} \right) = 16.$ Note that it is possible for either value to be a noninteger; for example, it is possible to sell $2.5$ thousand nuts in a month. The domain, therefore, contains thousands of points, so we can consider all points within the disk. For any $z < 16,$ we can solve the equation $f\left( {x,y} \right) = z\text{:}$

$$\begin{array}{rll}

& & \\

{16 - \left( {x - 3} \right)^{2} - \left( {y - 2} \right)^{2}} & = & z \\

{\left( {x - 3} \right)^{2} + \left( {y - 2} \right)^{2}} & = & {16 - z.}

\end{array}$$

Since $z < 16,$ we know that $16 - z > 0,$ so the previous equation describes a circle with radius $\sqrt{16 - z}$ centered at the point $\left( {3,2} \right).$ Therefore. the range of $f\left( {x,y} \right)$ is $\left\{ {z \in \left. \mathbb{R} \right|z \leq 16} \right\}.$ The graph of $f\left( {x,y} \right)$ is also a paraboloid, and this paraboloid points downward as shown.

Level Curves

If hikers walk along rugged trails, they might use a topographical map that shows how steeply the trails change. A topographical map contains curved lines called *contour lines*. Each contour line corresponds to the points on the map that have equal elevation (Figure 4.7). A level curve of a function of two variables $f\left( {x,y} \right)$ is completely analogous to a contour line on a topographical map.

Given a function $f\left( {x,y} \right)$ and a number $c$ in the range of $f,\ \text{a}$ level curve of a function of two variables for the value $c$ is defined to be the set of points satisfying the equation $f\left( {x,y} \right) = c.$

Returning to the function $g\left( {x,y} \right) = \sqrt{9 - x^{2} - y^{2}},$ we can determine the level curves of this function. The range of $g$ is the closed interval $\left\lbrack {0,3} \right\rbrack.$ First, we choose any number in this closed interval—say, $c = 2.$ The level curve corresponding to $c = 2$ is described by the equation

$$\sqrt{9 - x^{2} - y^{2}} = 2.$$

To simplify, square both sides of this equation:

$$9 - x^{2} - y^{2} = 4.$$

Now, multiply both sides of the equation by $-1$ and add $9$ to each side:

$$x^{2} + y^{2} = 5.$$

This equation describes a circle centered at the origin with radius $\sqrt{5}.$ Using values of $c$ between $0\ \text{and}\ 3$ yields other circles also centered at the origin. If $c = 3,$ then the circle has radius $0,$ so it consists solely of the origin. Figure 4.8 is a graph of the level curves of this function corresponding to $c = 0,1,2,\ \text{and}\ 3.$ Note that in the previous derivation it may be possible that we introduced extra solutions by squaring both sides. This is not the case here because the range of the square root function is nonnegative.

A graph of the various level curves of a function is called a contour map.

Making a Contour Map

Given the function $f\left( {x,y} \right) = \sqrt{8 + 8x - 4y - 4x^{2} - y^{2}},$ find the level curve corresponding to $c = 0.$ Then create a contour map for this function. What are the domain and range of $f?$

Solution

To find the level curve for $c = 0,$ we set $f\left( {x,y} \right) = 0$ and solve. This gives

$$0 = \sqrt{8 + 8x - 4y - 4x^{2} - y^{2}}.$$

We then square both sides and multiply both sides of the equation by $-1\text{:}$

$$4x^{2} + y^{2} - 8x + 4y - 8 = 0.$$

Now, we rearrange the terms, putting the $x$ terms together and the $y$ terms together, and add $8$ to each side:

$$4x^{2} - 8x + y^{2} + 4y = 8.$$

Next, we group the pairs of terms containing the same variable in parentheses, and factor $4$ from the first pair:

$$4\left( {x^{2} - 2x} \right) + \left( {y^{2} + 4y} \right) = 8.$$

Then we complete the square in each pair of parentheses and add the correct value to the right-hand side:

$$4\left( {x^{2} - 2x + 1} \right) + \left( {y^{2} + 4y + 4} \right) = 8 + 4(1) + 4.$$

Next, we factor the left-hand side and simplify the right-hand side:

$$4\left( {x - 1} \right)^{2} + \left( {y + 2} \right)^{2} = 16.$$

Last, we divide both sides by $16\text{:}$

$$\frac{\left( {x - 1} \right)^{2}}{4} + \frac{\left( {y + 2} \right)^{2}}{16} = 1.$$ (4.1)

This equation describes an ellipse centered at $\left( {1,-2} \right).$ The graph of this ellipse appears in the following graph.

We can repeat the same derivation for values of $c$ less than $4.$ Then, Equation 4.1 becomes

$$\frac{4\left( {x - 1} \right)^{2}}{16 - c^{2}} + \frac{\left( {y + 2} \right)^{2}}{16 - c^{2}} = 1$$

for an arbitrary value of $c.$ Figure 4.10 shows a contour map for $f\left( {x,y} \right)$ using the values $c = 0,1,2,\ \text{and}\ 3.$ When $c = 4,$ the level curve is the point $\left( {1,-2} \right).$

$\begin{array}{l}

{\text{Domain:}\left( {x,y} \right)~\text{inside~the~ellipse}~\frac{(x - 1)^{2}}{4} + \frac{(y + 2)^{2}}{16} = 1} \\

{\text{Range:}~\lbrack 0,4)}

\end{array}$

Find and graph the level curve of the function $g\left( {x,y} \right) = x^{2} + y^{2} - 6x + 2y$ corresponding to $c = 15.$

Another useful tool for understanding the graph of a function of two variables is called a vertical trace. Level curves are always graphed in the $xy\text{-plane,}$ but as their name implies, vertical traces are graphed in the $xz$- or $yz\text{-planes.}$

Consider a function $z = f\left( {x,y} \right)$ with domain $D \subseteq \mathbb{R}^{2}.$ A vertical trace of the function can be either the set of points that solves the equation $f\left( {a,y} \right) = z$ for a given constant $x = a$ or $f\left( {x,b} \right) = z$ for a given constant $y = b.$

Finding Vertical Traces

Find vertical traces for the function $f\left( {x,y} \right) = \text{sin}\ x\ \text{cos}\ y$ corresponding to $x = - \frac{\pi}{4},0,\ \text{and}\ \frac{\pi}{4},$ and $y = - \frac{\pi}{4},0,\ \text{and}\ \frac{\pi}{4}.$

Solution

First set $x = - \frac{\pi}{4}$ in the equation $z = \text{sin}\ x\ \text{cos}\ y\text{:}$

$$z = \text{sin}\left( {- \frac{\pi}{4}} \right)\text{cos}\ y = - \frac{\sqrt{2}\ \text{cos}\ y}{2} \approx -0.7071\ \text{cos}\ y.$$

This describes a cosine graph in the plane $x = - \frac{\pi}{4}.$ The other values of $z$ appear in the following table.

| $c$ | *Vertical Trace for* $x = c$ |

|-------------------|-------------------------------------------|

| $- \frac{\pi}{4}$ | $z = - \frac{\sqrt{2}\ \text{cos}\ y}{2}$ |

| $0$ | $z = 0$ |

| $\frac{\pi}{4}$ | $z = \frac{\sqrt{2}\ \text{cos}\ y}{2}$ |

Table 4.1 Vertical Traces Parallel to the $xz\text{-Plane}$ for the Function $f\left( {x,y} \right) = \text{sin}\ x\ \text{cos}\ y$

In a similar fashion, we can substitute the $y\text{-values}$ in the equation $f(x,y)$ to obtain the traces in the $yz\text{-plane,}$ as listed in the following table.

| $d$ | *Vertical Trace for* $y = d$ |

|-------------------|-----------------------------------------|

| $- \frac{\pi}{4}$ | $z = \frac{\sqrt{2}\ \text{sin}\ x}{2}$ |

| $0$ | $z = \text{sin}\ x$ |

| $\frac{\pi}{4}$ | $z = \frac{\sqrt{2}\ \text{sin}\ x}{2}$ |

Table 4.2 Vertical Traces Parallel to the $yz\text{-Plane}$ for the Function $f\left( {x,y} \right) = \text{sin}\ x\ \text{cos}\ y$

The three traces in the $xz\text{-plane}$ are cosine functions; the three traces in the $yz\text{-plane}$ are sine functions. These curves appear in the intersections of the surface with the planes $x = - \frac{\pi}{4},x = 0,x = \frac{\pi}{4}$ and $y = - \frac{\pi}{4},y = 0,y = \frac{\pi}{4}$ as shown in the following figure.

Determine the equation of the vertical trace of the function $g\left( {x,y} \right) = \text{−}x^{2} - y^{2} + 2x + 4y - 1$ corresponding to $y = 3,$ and describe its graph.

Functions of two variables can produce some striking-looking surfaces. The following figure shows two examples.

Functions of More Than Two Variables

So far, we have examined only functions of two variables. However, it is useful to take a brief look at functions of more than two variables. Two such examples are

$$f\left( {x,y,z} \right) = x^{2} - 2xy + y^{2} + 3yz - z^{2} + 4x - 2y + 3x - 6\ \text{(a polynomial in three variables)}$$

and

$$g\left( {x,y,t} \right) = \left( {x^{2} - 4xy + y^{2}} \right)\text{sin}\ t - \left( {3x + 5y} \right)\text{cos}\ t.$$

In the first function, $\left( {x,y,z} \right)$ represents a point in space, and the function $f$ maps each point in space to a fourth quantity, such as temperature or wind speed. In the second function, $\left( {x,y} \right)$ can represent a point in the plane, and $t$ can represent time. The function might map a point in the plane to a third quantity (for example, pressure) at a given time $t.$ The method for finding the domain of a function of more than two variables is analogous to the method for functions of one or two variables.

Domains for Functions of Three Variables

Find the domain of each of the following functions:

1. $f\left( {x,y,z} \right) = \frac{3x - 4y + 2z}{\sqrt{9 - x^{2} - y^{2} - z^{2}}}$

2. $g\left( {x,y,t} \right) = \frac{\sqrt{2t - 4}}{x^{2} - y^{2}}$

Solution

1. For the function $f\left( {x,y,z} \right) = \frac{3x - 4y + 2z}{\sqrt{9 - x^{2} - y^{2} - z^{2}}}$ to be defined (and be a real value), two conditions must hold:

1. The denominator cannot be zero.

2. The radicand cannot be negative.

Combining these conditions leads to the inequality

$$9 - x^{2} - y^{2} - z^{2} > 0.$$

Moving the variables to the other side and reversing the inequality gives the domain as

$$\text{domain}(f) = \left\{ {\left( {x,y,z} \right) \in \left. \mathbb{R}^{3} \right|x^{2} + y^{2} + z^{2} < 9} \right\},$$

which describes a ball of radius $3$ centered at the origin. (*Note*: The surface of the ball is not included in this domain.)

2. For the function $g\left( {x,y,t} \right) = \frac{\sqrt{2t - 4}}{x^{2} - y^{2}}$ to be defined (and be a real value), two conditions must hold:

1. The radicand cannot be negative.

2. The denominator cannot be zero.

Since the radicand cannot be negative, this implies $2t - 4 \geq 0,$ and therefore that $t \geq 2.$ Since the denominator cannot be zero, $x^{2} - y^{2} \neq 0,$ or $x^{2} \neq y^{2},$ Which can be rewritten as $y \neq \text{±}x$, which are the equations of two lines passing through the origin. Therefore, the domain of $g$ is

$$\text{domain}(g) = \left\{ \left. \left( {x,y,t} \right) \right|y \neq \text{±}x,t \geq 2 \right\}.$$

Find the domain of the function $h\left( {x,y,t} \right) = \left( {3t - 6} \right)\sqrt{y - 4x^{2} + 4}.$

Functions of two variables have level curves, which are shown as curves in the $xy\text{-plane.}$ However, when the function has three variables, the curves become surfaces, so we can define level surfaces for functions of three variables.

Given a function $f\left( {x,y,z} \right)$ and a number $c$ in the range of $f,$ a level surface of a function of three variables is defined to be the set of points satisfying the equation $f\left( {x,y,z} \right) = c.$

Finding a Level Surface

Find the level surface for the function $f\left( {x,y,z} \right) = 4x^{2} + 9y^{2} - z^{2}$ corresponding to $c = 1.$

Solution

The level surface is defined by the equation $4x^{2} + 9y^{2} - z^{2} = 1.$ This equation describes a hyperboloid of one sheet as shown in the following figure.

Find an equation of the level surface of the function

$$g\left( {x,y,z} \right) = x^{2} + y^{2} + z^{2} - 2x + 4y - 6z$$

corresponding to $c = 2,$ and describe the surface, if possible.

Section 4.1 Exercises

For the following exercises, evaluate each function at the indicated values.

1.

$W(x,y) = 4x^{2} + y^{2}.$ Find $W(2,-1),$ $W(-3,6).$

2\.

$W(x,y) = 4x^{2} + y^{2}.$ Find $W(2 + h,3 + h).$

3.

The volume of a right circular cylinder is calculated by a function of two variables, $V(x,y) = \pi x^{2}y,$ where $x$ is the radius of the right circular cylinder and $y$ represents the height of the cylinder. Evaluate $V(2,5)$ and explain what this means.

4\.

An oxygen tank is constructed of a right cylinder of height $y$ and radius $x$ with two hemispheres of radius $x$ mounted on the top and bottom of the cylinder. Express the volume of the tank as a function of two variables, $x\ \text{and}\ y,$ find $V(10,2),$ and explain what this means.

For the following exercises, find the domain of the function.

5.

$V(x,y) = 4x^{2} + y^{2}$

6\.

$f(x,y) = \sqrt{x^{2} + y^{2} - 4}$

7.

$f(x,y) = 4\ \text{ln}(y^{2} - x)$

8\.

$g(x,y) = \sqrt{16 - 4x^{2} - y^{2}}$

9.

$z(x,y) = y^{2} - x^{2}$

10\.

$f(x,y) = \frac{y + 2}{x^{2}}$

Find the range of the functions.

11.

$g(x,y) = \sqrt{16 - 4x^{2} - y^{2}}$

12\.

$V(x,y) = 4x^{2} + y^{2}$

13.

$z = y^{2} - x^{2}$

For the following exercises, find the level curves of each function at the indicated value of $c$ to visualize the given function.

14\.

$z(x,y) = y^{2} - x^{2},$ $c = 1$

15.

$z(x,y) = y^{2} - x^{2},$ $c = 4$

16\.

$g(x,y) = x^{2} + y^{2};c = 4,c = 9$

17.

$g(x,y) = 4 - x - y;c = 0,4$

18\.

$f(x,y) = xy;c = 1;c = -1$

19.

$h(x,y) = 2x - y;c = 0,-2,2$

20\.

$f(x,y) = x^{2} - y;c = 1,2$

21.

$g(x,y) = \frac{x}{x + y};c = -1,0,2$

22\.

$g(x,y) = x^{3} - y;c = -1,0,2$

23.

$g(x,y) = e_{}^{xy};c = \frac{1}{2},3$

24\.

$f(x,y) = x^{2};c = 4,9$

25.

$f(x,y) = xy - x;c = -2,0,2$

26\.

$h(x,y) = \text{ln}(x^{2} + y^{2});c = -1,0,1$

27.

$g(x,y) = \text{ln}\left( \frac{y}{x^{2}} \right);c = -2,0,2$

28\.

$z = f(x,y) = \sqrt{x^{2} + y^{2}},$ $c = 3$

29.

$f(x,y) = \frac{y + 2}{x^{2}},$ $c =$ any constant

For the following exercises, find the vertical traces of the functions at the indicated values of $x$ and *y*, and plot the traces.

30\.

$z = 4 - x - y;x = 2$

31.

$f(x,y) = 3x + y^{3},x = 1$

32\.

$z = \text{cos}\sqrt{x^{2} + y^{2}}$ $x = 1$

Find the domain of the following functions.

33.

$z = \sqrt{100 - 4x^{2} - 25y^{2}}$

34\.

$z = \text{ln}\left( {x - y^{2}} \right)$

35.

$f(x,y,z) = \frac{1}{\sqrt{36 - 4x^{2} - 9y^{2} - z^{2}}}$

36\.

$f(x,y,z) = \sqrt{49 - x^{2} - y^{2} - z^{2}}$

37.

$f(x,y,z) = \sqrt[3]{16 - x^{2} - y^{2} - z^{2}}$

38\.

$f(x,y) = \text{cos}\sqrt{x^{2} + y^{2}}$

For the following exercises, plot a graph of the function.

39.

$z = f(x,y) = \sqrt{x^{2} + y^{2}}$

40\.

$z = x^{2} + y^{2}$

41.

Use technology to graph $z = x^{2}y.$

Sketch the following by finding the level curves. Verify the graph using technology.

42\.

$f(x,y) = \sqrt{4 - x^{2} - y^{2}}$

43.

$f(x,y) = 2 - \sqrt{x^{2} + y^{2}}$

44\.

$z = 1 + e^{\text{−}x^{2} - y^{2}}$

45.

$z = \text{cos}\sqrt{x^{2} + y^{2}}$

46\.

$z = y^{2} - x^{2}$

47.

Describe the contour lines for several values of $c$ for $z = x^{2} + y^{2} - 2x - 2y.$

Find the level surface for the functions of three variables and describe it.

48\.

$w(x,y,z) = x - 2y + z,c = 4$

49.

$w(x,y,z) = x^{2} + y^{2} + z^{2},c = 9$

50\.

$w(x,y,z) = x^{2} + y^{2} - z^{2},c = -4$

51.

$w(x,y,z) = x^{2} + y^{2} - z^{2},c = 4$

52\.

$w(x,y,z) = 9x^{2} - 4y^{2} + 36z^{2},c = 0$

For the following exercises, find an equation of the level curve of $f$ that contains the point $P.$

53.

$f(x,y) = 1 - 4x^{2} - y^{2},P(0,1)$

54\.

$g(x,y) = y^{2}\text{arctan}\ x,P(1,2)$

55.

$g(x,y) = e^{xy}(x^{2} + y^{2}),P(1,0)$

56\.

The strength $E$ of an electric field at point $\left( {x,y,z} \right)$ resulting from an infinitely long charged wire lying along the $z\text{-axis}$ is given by $E(x,y,z) = {k\text{/}\sqrt{x^{2} + y^{2}}},$ where $k$ is a positive constant. For simplicity, let $k = 1$ and find the equations of the level surfaces for $E = 10\ \text{and}\ E = 100.$

57.

A thin plate made of iron is located in the $xy\text{-plane.}$ The temperature $T$ in degrees Celsius at a point $P\left( {x,y} \right)$ is inversely proportional to the square of its distance from the origin. Express $T$ as a function of $x\ \text{and}\ y.$

58\.

Refer to the preceding problem. Using the temperature function found there, determine the proportionality constant if the temperature at point $P\left( {1,2} \right)\ is\ 50\text{°}\text{C}.$ Use this constant to determine the temperature at point $Q\left( {3,4} \right).$

59.

Refer to the preceding problem. Find the level curves for $T = 40\text{°}\text{C and}\ T = 100\text{°}\text{C},$ and describe what the level curves represent.

---

4.2 Limits and Continuity

> 来源: OpenStax《Calculus Volume 3》| 原页: https://openstax.org/books/calculus-volume-3/pages/4-2-limits-and-continuity

4.2 Limits and Continuity

We have now examined functions of more than one variable and seen how to graph them. In this section, we see how to take the limit of a function of more than one variable, and what it means for a function of more than one variable to be continuous at a point in its domain. It turns out these concepts have aspects that just don’t occur with functions of one variable.

Limit of a Function of Two Variables

Recall from The Limit of a Function the definition of a limit of a function of one variable:

Let $f(x)$ be defined for all $x \neq a$ in an open interval containing $a.$ Let $L$ be a real number. Then

$$\underset{x\rightarrow a}{\text{lim}}f(x) = L$$

if for every $\varepsilon > 0,$ there exists a $\delta > 0,$ such that if $0 < \left| {x - a} \right| < \delta$ for all $x$ in the domain of $f,$ then

$$\left| {f(x) - L} \right| < \varepsilon.$$

Before we can adapt this definition to define a limit of a function of two variables, we first need to see how to extend the idea of an open interval in one variable to an open interval in two variables.

Consider a point $(a,b) \in \mathbb{R}^{2}.$ A $\delta$ disk centered at point $(a,b)$ is defined to be an open disk of radius $\delta$ centered at point $\left( {a,b} \right)$—that is,

$$\left\{ (x,y) \in \mathbb{R}^{2} \middle| {(x - a)}^{2} + {(y - b)}^{2} < \delta^{2} \right\}$$

as shown in the following graph.

The idea of a $\delta$ disk appears in the definition of the limit of a function of two variables. If $\delta$ is small, then all the points $\left( {x,y} \right)$ in the $\delta$ disk are close to $\left( {a,b} \right).$ This is completely analogous to $x$ being close to $a$ in the definition of a limit of a function of one variable. In one dimension, we express this restriction as

$$a - \delta < x < a + \delta.$$

In more than one dimension, we use a $\delta$ disk.

Let $f$ be a function of two variables, $x$ and $y.$ The limit of $f\left( {x,y} \right)$ as $\left( {x,y} \right)$ approaches $\left( {a,b} \right)$ is $L,$ written

$$\underset{{({x,y})}\rightarrow{({a,b})}}{\text{lim}}f(x,y) = L$$

if for each $\varepsilon > 0$ there exists a small enough $\delta > 0$ such that for all points $\left( {x,y} \right)$ in a $\delta$ disk around $\left( {a,b} \right),$ except possibly for $\left( {a,b} \right)$ itself, the value of $f\left( {x,y} \right)$ is no more than $\varepsilon$ away from $L$ (Figure 4.15). Using symbols, we write the following: For any $\varepsilon > 0,$ there exists a number $\delta > 0$ such that

$$\left| {f\left( {x,y} \right) - L} \right| < \varepsilon\ \text{whenever}\ 0 < \sqrt{\left( {x - a} \right)^{2} + \left( {y - b} \right)^{2}} < \delta.$$

Proving that a limit exists using the definition of a limit of a function of two variables can be challenging. Instead, we use the following theorem, which gives us shortcuts to finding limits. The formulas in this theorem are an extension of the formulas in the limit laws theorem in The Limit Laws.

Limit laws for functions of two variables

Let $f\left( {x,y} \right)$ and $g\left( {x,y} \right)$ be defined for all $\left( {x,y} \right) \neq \left( {a,b} \right)$ in a neighborhood around $\left( {a,b} \right),$ and assume the neighborhood is contained completely inside the domain of $f.$ Assume that $L$ and $M$ are real numbers such that $\underset{(x,y)\rightarrow(a,b)}{\text{lim}}f(x,y) = L$ and $\underset{(x,y)\rightarrow(a,b)}{\text{lim}}g(x,y) = M,$ and let $c$ be a constant. Then each of the following statements holds:

Constant Law:

$$\underset{(x,y)\rightarrow(a,b)}{\text{lim}}c = c$$ (4.2)

Identity Laws:

$$\underset{(x,y)\rightarrow(a,b)}{\text{lim}}x = a$$ (4.3) $$\underset{(x,y)\rightarrow(a,b)}{\text{lim}}y = b$$ (4.4)

Sum Law:

$$\underset{(x,y)\rightarrow(a,b)}{\text{lim}}\left( {f\left( {x,y} \right) + g\left( {x,y} \right)} \right) = L + M$$ (4.5)

Difference Law:

$$\underset{(x,y)\rightarrow(a,b)}{\text{lim}}\left( {f\left( {x,y} \right) - g\left( {x,y} \right)} \right) = L - M$$ (4.6)

Constant Multiple Law:

$$\underset{(x,y)\rightarrow(a,b)}{\text{lim}}\left( {cf\left( {x,y} \right)} \right) = cL$$ (4.7)

Product Law:

$$\underset{(x,y)\rightarrow(a,b)}{\text{lim}}\left( {f\left( {x,y} \right)g\left( {x,y} \right)} \right) = LM$$ (4.8)

Quotient Law:

$$\underset{(x,y)\rightarrow(a,b)}{\text{lim}}\frac{f\left( {x,y} \right)}{g\left( {x,y} \right)} = \frac{L}{M}\ \text{for}\ M \neq 0$$ (4.9)

Power Law:

$$\underset{(x,y)\rightarrow(a,b)}{\text{lim}}\left( {f\left( {x,y} \right)} \right)^{n} = L^{n}$$ (4.10)

for any positive integer $n.$

Root Law:

$$\underset{(x,y)\rightarrow(a,b)}{\text{lim}}\sqrt[n]{f\left( {x,y} \right)} = \sqrt[n]{L}$$ (4.11)

for all $L$ if $n$ is odd and positive, and for $L \geq 0$ if $n$ is even and positive provided that $f\left( x,\operatorname{}y \right)\operatorname{} \geq \operatorname{}0$ for all $\left( x,\operatorname{}y \right)\operatorname{} \neq \operatorname{}\left( a,\operatorname{}b \right)$ in neighborhood of $\left( a,\operatorname{}b \right)$.

The proofs of these properties are similar to those for the limits of functions of one variable. We can apply these laws to finding limits of various functions.

Finding the Limit of a Function of Two Variables

Find each of the following limits:

1. $\underset{{({x,y})}\rightarrow{({2,-1})}}{\text{lim}}\left( {x^{2} - 2xy + 3y^{2} - 4x + 3y - 6} \right)$

2. $\underset{{({x,y})}\rightarrow{({2,-1})}}{\text{lim}}\frac{2x + 3y}{4x - 3y}$

Solution

1. First use the sum and difference laws to separate the terms:

$$\begin{array}{l}

\\

{\underset{{({x,y})}\rightarrow{({2,-1})}}{\text{lim}}\left( {x^{2} - 2xy + 3y^{2} - 4x + 3y - 6} \right)} \\

{= \left( {\underset{{({x,y})}\rightarrow{({2,-1})}}{\text{lim}}x^{2}} \right) - \left( {\underset{{({x,y})}\rightarrow{({2,-1})}}{\text{lim}}2xy} \right) + \left( {\underset{{({x,y})}\rightarrow{({2,-1})}}{\text{lim}}3y^{2}} \right) - \left( {\underset{{({x,y})}\rightarrow{({2,-1})}}{\text{lim}}4x} \right)} \\

{\mspace{14mu}{+ \left( {\underset{{({x,y})}\rightarrow{({2,-1})}}{\text{lim}}3y} \right)} - \left( {\underset{{({x,y})}\rightarrow{({2,-1})}}{\text{lim}}6} \right).}

\end{array}$$

Next, use the constant multiple law on the second, third, fourth, and fifth limits:

$$\begin{array}{l}

\\

\\

{= \left( {\underset{{({x,y})}\rightarrow{({2,-1})}}{\text{lim}}x^{2}} \right) - 2\left( {\underset{{({x,y})}\rightarrow{({2,-1})}}{\text{lim}}xy} \right) + 3\left( {\underset{{({x,y})}\rightarrow{({2,-1})}}{\text{lim}}y^{2}} \right) - 4\left( {\underset{{({x,y})}\rightarrow{({2,-1})}}{\text{lim}}x} \right)} \\

{\mspace{9mu} + 3\left( {\underset{{({x,y})}\rightarrow{({2,-1})}}{\text{lim}}y} \right) - \underset{{({x,y})}\rightarrow{({2,-1})}}{\text{lim}}6.}

\end{array}$$

Now, use the power law on the first and third limits, and the product law on the second limit:

$$\begin{array}{l}

\\

\\

\\

{= \left( {\underset{{({x,y})}\rightarrow{({2,-1})}}{\text{lim}}x} \right)^{2} - 2\left( {\underset{{({x,y})}\rightarrow{({2,-1})}}{\text{lim}}x} \right)\left( {\underset{{({x,y})}\rightarrow{({2,-1})}}{\text{lim}}y} \right) + 3\left( {\underset{{({x,y})}\rightarrow{({2,-1})}}{\text{lim}}y} \right)^{2}} \\

{\mspace{9mu} - 4\left( {\underset{{({x,y})}\rightarrow{({2,-1})}}{\text{lim}}x} \right) + 3\left( {\underset{{({x,y})}\rightarrow{({2,-1})}}{\text{lim}}y} \right) - \underset{{({x,y})}\rightarrow{({2,-1})}}{\text{lim}}6.}

\end{array}$$

Last, use the identity laws on the first six limits and the constant law on the last limit:

$\begin{array}{cl}

{\underset{{({x,y})}\rightarrow{({2,-1})}}{\text{lim}}\left( {x^{2} - 2xy + 3y^{2} - 4x + 3y - 6} \right)} & {= (2)^{2} - 2(2)(-1) + 3(-1)^{2} - 4(2) + 3(-1) - 6} \\

& {= -6.}

\end{array}$

2. Before applying the quotient law, we need to verify that the limit of the denominator is nonzero. Using the difference law, constant multiple law, and identity law,

$$\begin{array}{cl}

{\underset{{({x,y})}\rightarrow{({2,-1})}}{\text{lim}}\left( {4x - 3y} \right)} & {= \underset{{({x,y})}\rightarrow{({2,-1})}}{\text{lim}}4x - \underset{{({x,y})}\rightarrow{({2,-1})}}{\text{lim}}3y} \\

& {= 4\left( {\underset{{({x,y})}\rightarrow{({2,-1})}}{\text{lim}}x} \right) - 3\left( {\underset{{({x,y})}\rightarrow{({2,-1})}}{\text{lim}}y} \right)} \\

& {= 4(2) - 3(-1) = 11.}

\end{array}$$

Since the limit of the denominator is nonzero, the quotient law applies. We now calculate the limit of the numerator using the difference law, constant multiple law, and identity law:

$$\begin{array}{cl}

{\underset{{({x,y})}\rightarrow{({2,-1})}}{\text{lim}}\left( {2x + 3y} \right)} & {= \underset{{({x,y})}\rightarrow{({2,-1})}}{\text{lim}}2x + \underset{{({x,y})}\rightarrow{({2,-1})}}{\text{lim}}3y} \\

& {= 2\left( {\underset{{({x,y})}\rightarrow{({2,-1})}}{\text{lim}}x} \right) + 3\left( {\underset{{({x,y})}\rightarrow{({2,-1})}}{\text{lim}}y} \right)} \\

& {= 2(2) + 3(-1)} \\

& {= 1.}

\end{array}$$

Therefore, according to the quotient law we have

$$\underset{{({x,y})}\rightarrow{({2,-1})}}{\text{lim}}\frac{2x + 3y}{4x - 3y} = \frac{\underset{{({x,y})}\rightarrow{({2,-1})}}{\text{lim}}\left( {2x + 3y} \right)}{\underset{{({x,y})}\rightarrow{({2,-1})}}{\text{lim}}\left( {4x - 3y} \right)} = \frac{1}{11}.$$

Evaluate the following limit:

$$\underset{(x,y)\rightarrow(5,-2)}{\text{lim}}\sqrt[3]{\frac{x^{2} - y}{y^{2} + x - 1}}.$$

Since we are taking the limit of a function of two variables, the point $(a,b)$ is in $\mathbb{R}^{2},$ and it is possible to approach this point from an infinite number of directions. Sometimes when calculating a limit, the answer varies depending on the path taken toward $(a,b).$ If this is the case, then the limit fails to exist. In other words, the limit must be unique, regardless of path taken.

Limits That Fail to Exist

Show that neither of the following limits exist:

1. $\underset{(x,y)\rightarrow(0,0)}{\text{lim}}\frac{2xy}{3x^{2} + y^{2}}$

2. $\underset{(x,y)\rightarrow(0,0)}{\text{lim}}\frac{4xy^{2}}{x^{2} + 3y^{4}}$

Solution

1. The domain of the function $f(x,y) = \frac{2xy}{3x^{2} + y^{2}}$ consists of all points in the $xy\text{-plane}$ except for the point $\left( {0,0} \right)$ (Figure 4.16). To show that the limit does not exist as $\left( {x,y} \right)$ approaches $\left( {0,0} \right),$ we note that it is impossible to satisfy the definition of a limit of a function of two variables because of the fact that the function takes different values along different lines passing through point $\left( {0,0} \right).$ First, consider the line $y = 0$ in the $xy\text{-plane.}$ Substituting $y = 0$ into $f\left( {x,y} \right)$ gives

$$f\left( {x,0} \right) = \frac{2x(0)}{3x^{2} + 0^{2}} = 0$$

for any value of $x.$ Therefore the value of $f$ remains constant for any point on the $x\text{-axis,}$ and as $y$ approaches zero, the function remains fixed at zero.

Next, consider the line $y = x.$ Substituting $y = x$ into $f\left( {x,y} \right)$ gives

$$f\left( {x,x} \right) = \frac{2x(x)}{3x^{2} + x^{2}} = \frac{2x^{2}}{4x^{2}} = \frac{1}{2}.$$

This is true for any point on the line $y = x.$ If we let $x$ approach zero while staying on this line, the value of the function remains fixed at $\frac{1}{2},$ regardless of how small $x$ is.

Choose a value for $\varepsilon$ that is less than $1\text{/}2$—say, $1\text{/}4.$ Then, no matter how small a $\delta$ disk we draw around $\left( {0,0} \right),$ the values of $f\left( {x,y} \right)$ for points inside that $\delta$ disk will include both $0$ and $\frac{1}{2}.$ Therefore, the definition of limit at a point is never satisfied and the limit fails to exist.

In a similar fashion to a., we can approach the origin along any straight line passing through the origin. If we try the $x\text{-axis}$ (i.e., $y = 0),$ then the function remains fixed at zero. The same is true for the $y\text{-axis.}$ Suppose we approach the origin along a straight line of slope $k.$ The equation of this line is $y = kx.$ Then the limit becomes

$$\begin{array}{cl}

{\underset{(x,y)\rightarrow(0,0)}{\text{lim}}\frac{4xy^{2}}{x^{2} + 3y^{4}}} & {= \underset{(x,y)\rightarrow(0,0)}{\text{lim}}\frac{4x\left( {kx} \right)^{2}}{x^{2} + 3\left( {kx} \right)^{4}}} \\

& {= \underset{(x,y)\rightarrow(0,0)}{\text{lim}}\frac{4k^{2}x^{3}}{x^{2} + 3k^{4}x^{4}}} \\

& {= \underset{(x,y)\rightarrow(0,0)}{\text{lim}}\frac{4k^{2}x}{1 + 3k^{4}x^{2}}} \\

& {= \frac{\underset{(x,y)\rightarrow(0,0)}{\text{lim}}\left( {4k^{2}x} \right)}{\underset{(x,y)\rightarrow(0,0)}{\text{lim}}\left( {1 + 3k^{4}x^{2}} \right)}} \\

& {= 0}

\end{array}$$

regardless of the value of $k.$ It would seem that the limit is equal to zero. What if we chose a curve passing through the origin instead? For example, we can consider the parabola given by the equation $x = y^{2}.$ Substituting $y^{2}$ in place of $x$ in $f\left( {x,y} \right)$ gives

$$\begin{array}{cl}

{\underset{{({x,y})}\rightarrow{({0,0})}}{\text{lim}}\frac{4xy^{2}}{x^{2} + 3y^{4}}} & {= \underset{{({x,y})}\rightarrow{({0,0})}}{\text{lim}}\frac{4\left( y^{2} \right)y^{2}}{\left( y^{2} \right)^{2} + 3y^{4}}} \\

& {= \underset{{({x,y})}\rightarrow{({0,0})}}{\text{lim}}\frac{4y^{4}}{y^{4} + 3y^{4}}} \\

& {= \underset{{({x,y})}\rightarrow{({0,0})}}{\text{lim}}1} \\

& {= 1.}

\end{array}$$

By the same logic in a., it is impossible to find a $\delta$ disk around the origin that satisfies the definition of the limit for any value of $\varepsilon < 1.$ Therefore, $\underset{{({x,y})}\rightarrow{({0,0})}}{\text{lim}}\frac{4xy^{2}}{x^{2} + 3y^{4}}$ does not exist.

Show that

$$\underset{{({x,y})}\rightarrow{({2,1})}}{\text{lim}}\frac{\left( {x - 2} \right)\left( {y - 1} \right)}{\left( {x - 2} \right)^{2} + \left( {y - 1} \right)^{2}}$$

does not exist.

Interior Points and Boundary Points

To study continuity and differentiability of a function of two or more variables, we first need to learn some new terminology.

Let *S* be a subset of $\mathbb{R}^{2}$ (Figure 4.17).

1. A point $P_{0}$ is called an interior point of $S$ if there is a $\delta$ disk centered around $P_{0}$ contained completely in $S.$

2. A point $P_{0}$ is called a boundary point of $S$ if every $\delta$ disk centered around $P_{0}$ contains points both inside and outside $S.$

Let *S* be a subset of $\mathbb{R}^{2}$ (Figure 4.17).

1. $S$ is called an open set if every point of $S$ is an interior point.

2. $S$ is called a closed set if it contains all its boundary points.

An example of an open set is a $\delta$ disk. If we include the boundary of the disk, then it becomes a closed set. A set that contains some, but not all, of its boundary points is neither open nor closed. For example if we include half the boundary of a $\delta$ disk but not the other half, then the set is neither open nor closed.

Let *S* be a subset of $\mathbb{R}^{2}$ (Figure 4.17).

1. An open set $S$ is a connected set if it cannot be represented as the union of two or more disjoint, nonempty open subsets.

2. A set $S$ is a region if it is open, connected, and nonempty.

The definition of a limit of a function of two variables requires the $\delta$ disk to be contained inside the domain of the function. However, if we wish to find the limit of a function at a boundary point of the domain, the $\delta\ \text{disk}$ is not contained inside the domain. By definition, some of the points of the $\delta\ \text{disk}$ are inside the domain and some are outside. Therefore, we need only consider points that are inside both the $\delta$ disk and the domain of the function. This leads to the definition of the limit of a function at a boundary point.

Let $f$ be a function of two variables, $x$ and $y,$ and suppose $\left( {a,b} \right)$ is on the boundary of the domain of $f.$ Then, the limit of $f\left( {x,y} \right)$ as $\left( {x,y} \right)$ approaches $\left( {a,b} \right)$ is $L,$ written

$$\underset{{({x,y})}\rightarrow{({a,b})}}{\text{lim}}f\left( {x,y} \right) = L,$$

if for any $\varepsilon > 0,$ there exists a number $\delta > 0$ such that for any point $\left( {x,y} \right)$ inside the domain of $f$ and within a suitably small distance positive $\delta$ of $\left( {a,b} \right),$ the value of $f\left( {x,y} \right)$ is no more than $\varepsilon$ away from $L$ (Figure 4.15). Using symbols, we can write: For any $\varepsilon > 0,$ there exists a number $\delta > 0$ such that

$$\left| {f\left( {x,y} \right) - L} \right| < \varepsilon\ \text{whenever}\ 0 < \sqrt{\left( {x - a} \right)^{2} + \left( {y - b} \right)^{2}} < \delta.$$

Limit of a Function at a Boundary Point

Prove $\underset{{({x,y})}\rightarrow{({4,3})}}{\text{lim}}\sqrt{25 - x^{2} - y^{2}} = 0.$

Solution

The domain of the function $f\left( {x,y} \right) = \sqrt{25 - x^{2} - y^{2}}$ is $\left\{ {(x,y) \in \mathbb{R}^{2}\left| {x^{2} + y^{2} \leq 25} \right.} \right\},$ which is a circle of radius $5$ centered at the origin, along with its interior as shown in the following graph.

We can use the limit laws, which apply to limits at the boundary of domains as well as interior points:

$$\begin{array}{cl}

{\underset{{({x,y})}\rightarrow{({4,3})}}{\text{lim}}\sqrt{25 - x^{2} - y^{2}}} & {= \sqrt{\underset{{({x,y})}\rightarrow{({4,3})}}{\text{lim}}\left( {25 - x^{2} - y^{2}} \right)}} \\

& {= \sqrt{\underset{{({x,y})}\rightarrow{({4,3})}}{\text{lim}}25 - \underset{{({x,y})}\rightarrow{({4,3})}}{\text{lim}}x^{2} - \underset{{({x,y})}\rightarrow{({4,3})}}{\text{lim}}y^{2}}} \\

& {= \sqrt{25 - 4^{2} - 3^{2}}} \\

& {= 0.}

\end{array}$$

See the following graph.

Evaluate the following limit:

$$\underset{{({x,y})}\rightarrow{({5,-2})}}{\text{lim}}\sqrt{29 - x^{2} - y^{2}}.$$

Continuity of Functions of Two Variables

In Continuity, we defined the continuity of a function of one variable and saw how it relied on the limit of a function of one variable. In particular, three conditions are necessary for $f(x)$ to be continuous at point $x = a\text{:}$

1. $f(a)$ exists.

2. $\underset{x\rightarrow a}{\text{lim}}f(x)$ exists.

3. $\underset{x\rightarrow a}{\text{lim}}f(x) = f(a).$

These three conditions are necessary for continuity of a function of two variables as well.

A function $f(x,y)$ is continuous at a point $\left( {a,b} \right)$ in its domain if the following conditions are satisfied:

1. $f\left( {a,b} \right)$ exists.

2. $\underset{{({x,y})}\rightarrow{({a,b})}}{\text{lim}}f\left( {x,y} \right)$ exists.

3. $\underset{{({x,y})}\rightarrow{({a,b})}}{\text{lim}}f\left( {x,y} \right) = f\left( {a,b} \right).$

Demonstrating Continuity for a Function of Two Variables

Show that the function $f(x,y) = \frac{3x + 2y}{x + y + 1}$ is continuous at point $\left( {5,-3} \right).$

Solution

There are three conditions to be satisfied, per the definition of continuity. In this example, $a = 5$ and $b = -3.$

1. $f\left( {a,b} \right)$ exists. This is true because the domain of the function $f$ consists of those ordered pairs for which the denominator is nonzero (i.e., $x + y + 1 \neq 0).$ Point $\left( {5,-3} \right)$ satisfies this condition. Furthermore,

$$f\left( {a,b} \right) = f\left( {5,-3} \right) = \frac{3(5) + 2(-3)}{5 + (-3) + 1} = \frac{15 - 6}{2 + 1} = 3.$$

2. $\underset{{({x,y})}\rightarrow{({a,b})}}{\text{lim}}f\left( {x,y} \right)$ exists. This is also true:

$$\begin{array}{cl}

{\underset{{({x,y})}\rightarrow{({a,b})}}{\text{lim}}f\left( {x,y} \right)} & {= \underset{{({x,y})}\rightarrow{({5,-3})}}{\text{lim}}\frac{3x + 2y}{x + y + 1}} \\

& {= \frac{\underset{{({x,y})}\rightarrow{({5,-3})}}{\text{lim}}\left( {3x + 2y} \right)}{\underset{{({x,y})}\rightarrow{({5,-3})}}{\text{lim}}\left( {x + y + 1} \right)}} \\

& {= \frac{15 - 6}{5 - 3 + 1}} \\

& {= 3.}

\end{array}$$

3. $\underset{{({x,y})}\rightarrow{({a,b})}}{\text{lim}}f\left( {x,y} \right) = f\left( {a,b} \right).$ This is true because we have just shown that both sides of this equation equal three.

Show that the function $f\left( {x,y} \right) = \sqrt{26 - 2x^{2} - y^{2}}$ is continuous at point $\left( {2,-3} \right).$

Continuity of a function of any number of variables can also be defined in terms of delta and epsilon. A function of two variables is continuous at a point $\left( {x_{0},y_{0}} \right)$ in its domain if for every $\varepsilon > 0$ there exists a $\delta > 0$ such that, whenever $\sqrt{\left( {x - x_{0}} \right)^{2} + \left( {y - y_{0}} \right)^{2}} < \delta$ it is true, $\left| {f\left( {x,y} \right) - f\left( {a,b} \right)} \right| < \varepsilon.$ This definition can be combined with the formal definition (that is, the *epsilon–delta definition*) of continuity of a function of one variable to prove the following theorems:

The Sum of Continuous Functions Is Continuous

If $f\left( {x,y} \right)$ is continuous at $\left( {x_{0},y_{0}} \right),$ and $g\left( {x,y} \right)$ is continuous at $\left( {x_{0},y_{0}} \right),$ then $f\left( {x,y} \right) + g\left( {x,y} \right)$ is continuous at $\left( {x_{0},y_{0}} \right).$

The Product of Continuous Functions Is Continuous

If $g(x)$ is continuous at $x_{0}$ and $h(y)$ is continuous at $y_{0},$ then $f\left( {x,y} \right) = g(x)h(y)$ is continuous at $\left( {x_{0},y_{0}} \right).$

The Composition of Continuous Functions Is Continuous

Let $g$ be a function of two variables from a domain $D \subseteq \mathbb{R}^{2}$ to a range $R \subseteq \mathbb{R}.$ Suppose $g$ is continuous at some point $\left( {x_{0},y_{0}} \right) \in D$ and define $z_{0} = g\left( {x_{0},y_{0}} \right).$ Let $f$ be a function that maps $\mathbb{R}$ to $\mathbb{R}$ such that $z_{0}$ is in the domain of $f.$ Last, assume $f$ is continuous at $z_{0}.$ Then $f \circ g$ is continuous at $\left( {x_{0},y_{0}} \right)$ as shown in the following figure.

Let’s now use the previous theorems to show continuity of functions in the following examples.

More Examples of Continuity of a Function of Two Variables

Show that the functions $f\left( {x,y} \right) = 4x^{3}y^{2}$ and $g\left( {x,y} \right) = \text{cos}\left( {4x^{3}y^{2}} \right)$ are continuous everywhere.

Solution

The polynomials $g(x) = 4x^{3}$ and $h(y) = y^{2}$ are continuous at every real number, and therefore by the product of continuous functions theorem, $f\left( {x,y} \right) = 4x^{3}y^{2}$ is continuous at every point $\left( {x,y} \right)$ in the $xy\text{-plane.}$ Since $f\left( {x,y} \right) = 4x^{3}y^{2}$ is continuous at every point $\left( {x,y} \right)$ in the $xy\text{-plane}$ and $g(x) = \text{cos}\ x$ is continuous at every real number $x,$ the continuity of the composition of functions tells us that $g\left( {x,y} \right) = \text{cos}\left( {4x^{3}y^{2}} \right)$ is continuous at every point $\left( {x,y} \right)$ in the $xy\text{-plane.}$

Show that the functions $f\left( {x,y} \right) = 2x^{2}y^{3} + 3$ and $g\left( {x,y} \right) = \left( {2x^{2}y^{3} + 3} \right)^{4}$ are continuous everywhere.

Functions of Three or More Variables

The limit of a function of three or more variables occurs readily in applications. For example, suppose we have a function $f\left( {x,y,z} \right)$ that gives the temperature at a physical location $\left( {x,y,z} \right)$ in three dimensions. Or perhaps a function $g\left( {x,y,z,t} \right)$ can indicate air pressure at a location $\left( {x,y,z} \right)$ at time $t.$ How can we take a limit at a point in $\mathbb{R}^{3}?$ What does it mean to be continuous at a point in four dimensions?

The answers to these questions rely on extending the concept of a $\delta$ disk into more than two dimensions. Then, the ideas of the limit of a function of three or more variables and the continuity of a function of three or more variables are very similar to the definitions given earlier for a function of two variables.

Let $\left( {x_{0},y_{0},z_{0}} \right)$ be a point in $\mathbb{R}^{3}.$ Then, a $\delta$ ball in three dimensions consists of all points in $\mathbb{R}^{3}$ lying at a distance of less than $\delta$ from $\left( {x_{0},y_{0},z_{0}} \right)$—that is,

$$\left\{ \left( {x,y,z} \right) \in \mathbb{R}^{3} \middle| \sqrt{\left( {x - x_{0}} \right)^{2} + \left( {y - y_{0}} \right)^{2} + \left( {z - z_{0}} \right)^{2}} < \delta \right\}.$$

To define a $\delta$ ball in higher dimensions, add additional terms under the radical to correspond to each additional dimension. For example, given a point $P = \left( {w_{0},x_{0},y_{0},z_{0}} \right)$ in $\mathbb{R}^{4},$ a $\delta$ ball around $P$ can be described by

$$\left\{ \left( {w,x,y,z} \right) \in \mathbb{R}^{4} \middle| \sqrt{\left( {w - w_{0}} \right)^{2} + \left( {x - x_{0}} \right)^{2} + \left( {y - y_{0}} \right)^{2} + \left( {z - z_{0}} \right)^{2}} < \delta \right\}.$$

To show that a limit of a function of three variables exists at a point $\left( {x_{0},y_{0},z_{0}} \right),$ it suffices to show that for any point in a $\delta$ ball centered at $\left( {x_{0},y_{0},z_{0}} \right),$ the value of the function at that point is arbitrarily close to a fixed value (the limit value). All the limit laws for functions of two variables hold for functions of more than two variables as well.

Finding the Limit of a Function of Three Variables

Find $\underset{{({x,y,z})}\rightarrow{({4,1,-3})}}{\text{lim}}\frac{x^{2}y - 3z}{2x + 5y - z}.$

Solution

Before we can apply the quotient law, we need to verify that the limit of the denominator is nonzero. Using the difference law, the identity law, and the constant law,

$$\begin{array}{cl}

{\underset{{({x,y,z})}\rightarrow{({4,1,-3})}}{\text{lim}}\left( {2x + 5y - z} \right)} & {= 2\left( {\underset{{({x,y,z})}\rightarrow{({4,1,-3})}}{\text{lim}}x} \right) + 5\left( {\underset{{({x,y,z})}\rightarrow{({4,1,-3})}}{\text{lim}}y} \right) - \left( {\underset{{({x,y,z})}\rightarrow{({4,1,-3})}}{\text{lim}}z} \right)} \\

& {= 2(4) + 5(1) - (-3)} \\

& {= 16.}

\end{array}$$

Since this is nonzero, we next find the limit of the numerator. Using the product law, difference law, constant multiple law, and identity law,

$$\begin{array}{cl}

{\underset{{({x,y,z})}\rightarrow{({4,1,-3})}}{\text{lim}}\left( {x^{2}y - 3z} \right)} & {= \left( {\underset{{({x,y,z})}\rightarrow{({4,1,-3})}}{\text{lim}}x} \right)^{2}\left( {\underset{{({x,y,z})}\rightarrow{({4,1,-3})}}{\text{lim}}y} \right) - 3\underset{{({x,y,z})}\rightarrow{({4,1,-3})}}{\text{lim}}z} \\

& {= \left( 4^{2} \right)(1) - 3(-3)} \\

& {= 16 + 9} \\

& {= 25.}

\end{array}$$

Last, applying the quotient law:

$$\begin{array}{cl}

{\underset{{({x,y,z})}\rightarrow{({4,1,-3})}}{\text{lim}}\frac{x^{2}y - 3z}{2x + 5y - z}} & {= \frac{\underset{{({x,y,z})}\rightarrow{({4,1,-3})}}{\text{lim}}\left( {x^{2}y - 3z} \right)}{\underset{{({x,y,z})}\rightarrow{({4,1,-3})}}{\text{lim}}\left( {2x + 5y - z} \right)}} \\

& {= \frac{25}{16}.}

\end{array}$$

Find $\underset{{({x,y,z})}\rightarrow{({4,-1,3})}}{\text{lim}}\sqrt{13 - x^{2} - 2y^{2} + z^{2}}.$

Section 4.2 Exercises

For the following exercises, find the limit of the function.

60\.

$\underset{(x,y)\rightarrow(1,2)}{\text{lim}}x$

61.

$\underset{(x,y)\rightarrow(1,2)}{\text{lim}}\frac{5x^{2}y}{x^{2} + y^{2}}$

62\.

Show that the limit $\underset{(x,y)\rightarrow(0,0)}{\text{lim}}\frac{5x^{2}y}{x^{2} + y^{2}}$ exists and is the same along the paths: $y\text{-axis}$ and $x\text{-axis,}$ and along $y = x.$

For the following exercises, evaluate the limits at the indicated values of $x\ \text{and}\ y.$ If the limit does not exist, state this and explain why the limit does not exist.

63.

$\underset{(x,y)\rightarrow(0,0)}{\text{lim}}\frac{4x^{2} + 10y^{2} + 4}{4x^{2} - 10y^{2} + 6}$

64\.

$\underset{(x,y)\rightarrow(11,13)}{\text{lim}}\sqrt{\frac{1}{xy}}$

65.

$\underset{(x,y)\rightarrow(0,1)}{\text{lim}}\frac{y^{2}\text{sin}\ x}{x}$

66\.

$\underset{(x,y)\rightarrow(0,0)}{\text{lim}}\text{sin}\left( \frac{x^{8} + y^{7}}{x - y + 10} \right)$

67.

$\underset{(x,y)\rightarrow(\pi\text{/}4,1)}{\text{lim}}\frac{y\ \text{tan}\ x}{y + 1}$

68\.

$\underset{(x,y)\rightarrow(0,\pi\text{/}4)}{\text{lim}}\frac{\text{sec}\ x + 2}{3x - \text{tan}\ y}$

69.

$\underset{(x,y)\rightarrow(2,5)}{\text{lim}}\left( {\frac{1}{x} - \frac{5}{y}} \right)$

70\.

$\underset{(x,y)\rightarrow(4,4)}{\text{lim}}x\ \text{ln}\ y$

71.

$\underset{(x,y)\rightarrow(4,4)}{\text{lim}}e^{\text{−}x^{2} - y^{2}}$

72\.

$\underset{(x,y)\rightarrow(0,0)}{\text{lim}}\sqrt{9 - x^{2} - y^{2}}$

73.

$\underset{(x,y)\rightarrow(1,2)}{\text{lim}}\left( {x^{2}y^{3} - x^{3}y^{2} + 3x + 2y} \right)$

74\.

$\underset{(x,y)\rightarrow(\pi,\pi)}{\text{lim}}x\ \text{sin}\left( \frac{x + y}{4} \right)$

75.

$\underset{(x,y)\rightarrow(0,0)}{\text{lim}}\frac{xy + 1}{x^{2} + y^{2} + 1}$

76\.

$\underset{(x,y)\rightarrow(0,0)}{\text{lim}}\frac{x^{2} + y^{2}}{\sqrt{x^{2} + y^{2} + 1} - 1}$

77.

$\underset{(x,y)\rightarrow(0,0)}{\text{lim}}\text{ln}\left( {x^{2} + y^{2}} \right)$

For the following exercises, complete the statement.

78\.

A point $\left( {x_{0},y_{0}} \right)$ in a plane region $R$ is an interior point of $R$ if \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_.

79.

A point $\left( {x_{0},y_{0}} \right)$ in a plane region $R$ is called a boundary point of $R$ if \_\_\_\_\_\_\_\_\_\_\_.

For the following exercises, use algebraic techniques to evaluate the limit.

80\.

$\underset{(x,y)\rightarrow(2,1)}{\text{lim}}\frac{x - y - 1}{\sqrt{x - y} - 1}$

81.

$\underset{(x,y)\rightarrow(0,0)}{\text{lim}}\frac{x^{4} - 4y^{4}}{x^{2} + 2y^{2}}$

82\.

$\underset{(x,y)\rightarrow(0,0)}{\text{lim}}\frac{x^{3} - y^{3}}{x - y}$

83.

$\underset{(x,y)\rightarrow(0,0)}{\text{lim}}\frac{x^{2} - xy}{\sqrt{x} - \sqrt{y}}$

For the following exercises, evaluate the limits of the functions of three variables.

84\.

$\underset{(x,y,z)\rightarrow(1,2,3)}{\text{lim}}\frac{xz^{2} - y^{2}z}{xyz - 1}$

85.

$\underset{(x,y,z)\rightarrow(0,0,0)}{\text{lim}}\frac{x^{2} - y^{2} - z^{2}}{x^{2} + y^{2} - z^{2}}$

For the following exercises, evaluate the limit of the function by determining the value the function approaches along the indicated paths. If the limit does not exist, explain why not.

86\.

$\underset{(x,y)\rightarrow(0,0)}{\text{lim}}\frac{xy + y^{3}}{x^{2} + y^{2}}$

1. Along the $x\text{-axis}$ $(y = 0)$

2. Along the $y\text{-axis}$ $(x = 0)$

3. Along the path $y = 2x$

87.

Evaluate $\underset{(x,y)\rightarrow(0,0)}{\text{lim}}\frac{xy + y^{3}}{x^{2} + y^{2}}$ using the results of previous problem.

88\.

$\underset{(x,y)\rightarrow(0,0)}{\text{lim}}\frac{x^{2}y}{x^{4} + y^{2}}$

1. Along the *x*-axis $(y = 0)$

2. Along the *y*-axis $(x = 0)$

3. Along the path $y = x^{2}$

89.

Evaluate $\underset{(x,y)\rightarrow(0,0)}{\text{lim}}\frac{x^{2}y}{x^{4} + y^{2}}$ using the results of previous problem.

Discuss the continuity of the following functions. Find the largest region in the $xy\text{-plane}$ in which the following functions are continuous.

90\.

$f(x,y) = \text{sin}(xy)$

91.

$f(x,y) = \text{ln}(x + y)$

92\.

$f(x,y) = e^{3xy}$

93.

$f(x,y) = \frac{1}{xy}$

For the following two exercises, determine the region in which the function is continuous. Explain your answer.

94\.

$f(x,y) = \frac{x^{2}y}{x^{2} + y^{2}}$

95.

$f(x,y) = \frac{\text{sin}(x^{2} + y^{2})}{x^{2} + y^{2}}$

96\.

Determine whether $f\left( {x,\operatorname{}y} \right)$ is continuous at $\left( {0,\operatorname{}0} \right)$. $f(x,y) = \left\{ \begin{array}{ll}

\frac{x^{2}y}{x^{2} + y^{2}} & {\text{if}\ (x,y) \neq (0,0)} \\

0 & {\text{if}\ (x,y) = (0,0)}

\end{array} \right\}$

97.

Determine whether $g(x,y) = \frac{x^{2} - y^{2}}{x^{2} + y^{2}}$ is continuous at $\left( {0,0} \right).$

98\.

Create a plot using graphing software to determine where the limit does not exist. Find where in the coordinate plane $f(x,y) = \frac{1}{x^{2} - y}$ is continuous.

99.

Determine the region of the $xy\text{-plane}$ in which the function $g(x,y) = \text{arctan}\left( \frac{xy^{2}}{x + y} \right)$ is continuous. Use technology to support your conclusion.

100\.

Determine the region of the $xy\text{-plane}$ in which $f(x,y) = \text{ln}(x^{2} + y^{2} - 1)$ is continuous. Use technology to support your conclusion. (*Hint*: Choose the range of values for $x\ \text{and}\ y$ carefully!)

101.

At what points in space is $g(x,y,z) = x^{2} + y^{2} - 2z^{2}$ continuous?

102\.

At what points in space is $g(x,y,z) = \frac{1}{x^{2} + z^{2} - 1}$ continuous?

103.

Show that $\underset{(x,y)\rightarrow(0,0)}{\text{lim}}\frac{1}{x^{2} + y^{2}}$ does not exist at $\left( {0,0} \right)$ by plotting the graph of the function.

104\.

\[T\] Evaluate $\underset{(x,y)\rightarrow(0,0)}{\text{lim}}\frac{\text{−}xy^{2}}{x^{2} + y^{4}}$ by plotting the function using a CAS. Determine analytically the limit along the path $x = y^{2}.$

105.

\[T\]

1. Use a CAS to draw a contour map of $z = \sqrt{9 - x^{2} - y^{2}}.$

2. What is the name of the geometric shape of the level curves?

3. Give the general equation of the level curves.

4. What is the maximum value of $z?$

5. What is the domain of the function?

6. What is the range of the function?

106\.

*True or False*: If we evaluate $\underset{(x,y)\rightarrow(0,0)}{\text{lim}}f(x)$ along several paths and each time the limit is $1,$ we can conclude that $\underset{(x,y)\rightarrow(0,0)}{\text{lim}}f(x) = 1.$

107.

Use polar coordinates to find $\underset{(x,y)\rightarrow(0,0)}{\text{lim}}\frac{\text{sin}\sqrt{x^{2} + y^{2}}}{\sqrt{x^{2} + y^{2}}}.$ You can also find the limit using L’Hôpital’s rule.

108\.

Use polar coordinates to find $\underset{(x,y)\rightarrow(0,0)}{\text{lim}}\text{cos}\left( {x^{2} + y^{2}} \right).$

109.

Discuss the continuity of $f(g(x,y))$ where $f(t) = {1\text{/}t}$ and $g(x,y) = 2x - 5y.$

110\.

Given $f(x,y) = x^{2} - 4y,$ find $\underset{h\rightarrow 0}{\text{lim}}\frac{f(x + h,y) - f(x,y)}{h}.$

111.

Given $f(x,y) = x^{2} - 4y,$ find $\underset{h\rightarrow 0}{\text{lim}}\frac{f(1 + h,y) - f(1,y)}{h}.$

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4.3 Partial Derivatives

> 来源: OpenStax《Calculus Volume 3》| 原页: https://openstax.org/books/calculus-volume-3/pages/4-3-partial-derivatives

4.3 Partial Derivatives

Now that we have examined limits and continuity of functions of two variables, we can proceed to study derivatives. Finding derivatives of functions of two variables is the key concept in this chapter, with as many applications in mathematics, science, and engineering as differentiation of single-variable functions. However, we have already seen that limits and continuity of multivariable functions have new issues and require new terminology and ideas to deal with them. This carries over into differentiation as well.

Derivatives of a Function of Two Variables

When studying derivatives of functions of one variable, we found that one interpretation of the derivative is an instantaneous rate of change of $y$ as a function of $x.$ Leibniz notation for the derivative is ${{dy}\text{/}{dx}},$ which implies that $y$ is the dependent variable and $x$ is the independent variable. For a function $z = f(x,y)$ of two variables, $x$ and $y$ are the independent variables and $z$ is the dependent variable. This raises two questions right away: How do we adapt Leibniz notation for functions of two variables? Also, what is an interpretation of the derivative? The answer lies in partial derivatives.

Let $f\left( {x,y} \right)$ be a function of two variables. Then the partial derivative of $f$ with respect to $x,$ written as ${{\partial f}\text{/}{\partial x}},$ or $f_{x},$ is defined as

$$\frac{\partial f}{\partial x} = \underset{h\rightarrow 0}{\text{lim}}\frac{f\left( {x + h,y} \right) - f\left( {x,y} \right)}{h}.$$ (4.12)

The partial derivative of $f$ with respect to $y,$ written as ${{\partial f}\text{/}{\partial y}},$ or $f_{y},$ is defined as

$$\frac{\partial f}{\partial y} = \underset{k\rightarrow 0}{\text{lim}}\frac{f\left( {x,y + k} \right) - f\left( {x,y} \right)}{k}.$$ (4.13)

This definition shows two differences already. First, the notation changes, in the sense that we still use a version of Leibniz notation, but the $d$ in the original notation is replaced with the symbol $\partial.$ (This rounded $\text{“d”}$ is usually called “partial,” so ${\partial f}\text{/}{\partial x}$ is spoken as the “partial of $f$ with respect to $\left. x\text{.”} \right)$ This is the first hint that we are dealing with partial derivatives. Second, we now have two different derivatives we can take, since there are two different independent variables. Depending on which variable we choose, we can come up with different partial derivatives altogether, and often do.

Calculating Partial Derivatives from the Definition

Use the definition of the partial derivative as a limit to calculate ${\partial f}\text{/}{\partial x}$ and ${\partial f}\text{/}{\partial y}$ for the function

$$f\left( {x,y} \right) = x^{2} - 3xy + 2y^{2} - 4x + 5y - 12.$$

Solution

First, calculate $f\left( {x + h,y} \right).$

$$\begin{array}{cl}

{f\left( {x + h,y} \right)} & {= \left( {x + h} \right)^{2} - 3\left( {x + h} \right)y + 2y^{2} - 4\left( {x + h} \right) + 5y - 12} \\

& {= x^{2} + 2xh + h^{2} - 3xy - 3hy + 2y^{2} - 4x - 4h + 5y - 12.}

\end{array}$$

Next, substitute this into Equation 4.12 and simplify:

$$\begin{array}{cl}

\frac{\partial f}{\partial x} & {= \underset{h\rightarrow 0}{\text{lim}}\frac{f\left( {x + h,y} \right) - f\left( {x,y} \right)}{h}} \\

& {= \underset{h\rightarrow 0}{\text{lim}}\frac{\left( {x^{2} + 2xh + h^{2} - 3xy - 3hy + 2y^{2} - 4x - 4h + 5y - 12} \right) - \left( {x^{2} - 3xy + 2y^{2} - 4x + 5y - 12} \right)}{h}} \\

& {= \underset{h\rightarrow 0}{\text{lim}}\frac{x^{2} + 2xh + h^{2} - 3xy - 3hy + 2y^{2} - 4x - 4h + 5y - 12 - x^{2} + 3xy - 2y^{2} + 4x - 5y + 12}{h}} \\

& {= \underset{h\rightarrow 0}{\text{lim}}\frac{2xh + h^{2} - 3hy - 4h}{h}} \\

& {= \underset{h\rightarrow 0}{\text{lim}}\frac{h\left( {2x + h - 3y - 4} \right)}{h}} \\

& {= \underset{h\rightarrow 0}{\text{lim}}\left( {2x + h - 3y - 4} \right)} \\

& {= 2x - 3y - 4.}

\end{array}$$

To calculate $\frac{\partial f}{\partial y},$ first calculate $f\left( {x,y + k} \right)\text{:}$

$$\begin{matrix}

{f(x,y + k)} & {= x^{2} - 3x(y + k) + 2(y + k)^{2} - 4x + 5(y + k) - 12} \\

& {= x^{2} - 3xy - 3xk + 2y^{2} + 4yk + 2k^{2} - 4x + 5y + 5k - 12.}

\end{matrix}$$

Next, substitute this into Equation 4.13 and simplify:

$$\begin{array}{cl}

\frac{\partial f}{\partial y} & {= \underset{k\rightarrow 0}{\text{lim}}\frac{f(x,y + k) - f(x,y)}{k}} \\

& {= \underset{k\rightarrow 0}{\text{lim}}\frac{\left( {x^{2} - 3xy - 3xk + 2y^{2} + 4yk + 2k^{2} - 4x + 5y + 5k - 12} \right) - \left( {x^{2} - 3xy + 2y^{2} - 4x + 5y - 12} \right)}{k}} \\

& {= \underset{k\rightarrow 0}{\text{lim}}\frac{x^{2} - 3xy - 3xk + 2y^{2} + 4yk + 2k^{2} - 4x + 5y + 5k - 12 - x^{2} + 3xy - 2y^{2} + 4x - 5y + 12}{k}} \\

& {= \underset{k\rightarrow 0}{\text{lim}}\frac{-3xk + 4yk + 2k^{2} + 5k}{k}} \\

& {= \underset{k\rightarrow 0}{\text{lim}}\frac{h\left( {-3x + 4y + 2k + 5} \right)}{k}} \\

& {= \underset{k\rightarrow 0}{\text{lim}}\left( {-3x + 4y + 2k + 5} \right)} \\

& {= -3x + 4y + 5.}

\end{array}$$

Use the definition of the partial derivative as a limit to calculate ${\partial f}\text{/}{\partial x}$ and ${\partial f}\text{/}{\partial y}$ for the function

$$f\left( {x,y} \right) = 4x^{2} + 2xy - y^{2} + 3x - 2y + 5.$$

The idea to keep in mind when calculating partial derivatives is to treat all independent variables, other than the variable with respect to which we are differentiating, as constants. Then proceed to differentiate as with a function of a single variable. To see why this is true, first fix $y$ and define $g(x) = f\left( {x,y} \right)$ as a function of $x.$ Then

$$g^{\prime}(x) = \underset{h\rightarrow 0}{\text{lim}}\frac{g\left( {x + h} \right) - g(x)}{h} = \underset{h\rightarrow 0}{\text{lim}}\frac{f\left( {x + h,y} \right) - f\left( {x,y} \right)}{h} = \frac{\partial f}{\partial x}.$$

The same is true for calculating the partial derivative of $f$ with respect to $y.$ This time, fix $x$ and define $h(y) = f\left( {x,y} \right)$ as a function of $y.$ Then

$$h^{\prime}(x) = \underset{k\rightarrow 0}{\text{lim}}\frac{h\left( {x + k} \right) - h(x)}{k} = \underset{k\rightarrow 0}{\text{lim}}\frac{f\left( {x,y + k} \right) - f\left( {x,y} \right)}{k} = \frac{\partial f}{\partial y}.$$

All differentiation rules from Introduction to Derivatives apply.

Calculating Partial Derivatives

Calculate both partial derivatives for the following functions by holding the opposite variable constant then differentiating:

1. $f\left( {x,y} \right) = x^{2} - 3xy + 2y^{2} - 4x + 5y - 12$

2. $g\left( {x,y} \right) = \text{sin}\left( {x^{2}y - 2x + 4} \right)$

Solution

1. To calculate ${{\partial f}\text{/}{\partial x}},$ treat the variable $y$ as a constant. Then differentiate $f\left( {x,y} \right)$ with respect to $x$ using the sum, difference, and power rules:

$$\begin{array}{cl}

\frac{\partial f}{\partial x} & {= \frac{\partial}{\partial x}\left\lbrack {x^{2} - 3xy + 2y^{2} - 4x + 5y - 12} \right\rbrack} \\

& {= \frac{\partial}{\partial x}\left\lbrack x^{2} \right\rbrack - \frac{\partial}{\partial x}\left\lbrack {3xy} \right\rbrack + \frac{\partial}{\partial x}\left\lbrack {2y^{2}} \right\rbrack - \frac{\partial}{\partial x}\left\lbrack {4x} \right\rbrack + \frac{\partial}{\partial x}\left\lbrack {5y} \right\rbrack - \frac{\partial}{\partial x}\lbrack 12\rbrack} \\

& {= 2x - 3y + 0 - 4 + 0 - 0} \\

& {= 2x - 3y - 4.}

\end{array}$$

The derivatives of the third, fifth, and sixth terms are all zero because they do not contain the variable $x,$ so they are treated as constant terms. The derivative of the second term is equal to the coefficient of $x,$ which is $-3y.$ Calculating ${{\partial f}\text{/}{\partial y}}\text{:}$

$$\begin{array}{cl}

\frac{\partial f}{\partial y} & {= \frac{\partial}{\partial y}\left\lbrack {x^{2} - 3xy + 2y^{2} - 4x + 5y - 12} \right\rbrack} \\

& {= \frac{\partial}{\partial y}\left\lbrack x^{2} \right\rbrack - \frac{\partial}{\partial y}\left\lbrack {3xy} \right\rbrack + \frac{\partial}{\partial y}\left\lbrack {2y^{2}} \right\rbrack - \frac{\partial}{\partial y}\left\lbrack {4x} \right\rbrack + \frac{\partial}{\partial y}\left\lbrack {5y} \right\rbrack - \frac{\partial}{\partial y}\lbrack 12\rbrack} \\

& {= -3x + 4y - 0 + 5 - 0} \\

& {= -3x + 4y + 5.}

\end{array}$$

These are the same answers obtained in Example 4.14.

2. To calculate ${{\partial g}\text{/}{\partial x}},$ treat the variable *y* as a constant. Then differentiate $g\left( {x,y} \right)$ with respect to *x* using the chain rule and power rule:

$$\begin{array}{cl}

\frac{\partial g}{\partial x} & {= \frac{\partial}{\partial x}\left\lbrack {\text{sin}\left( {x^{2}y - 2x + 4} \right)} \right\rbrack} \\

& {= \text{cos}\left( {x^{2}y - 2x + 4} \right)\frac{\partial}{\partial x}\left\lbrack {x^{2}y - 2x + 4} \right\rbrack} \\

& {= \left( {2xy - 2} \right)\text{cos}\left( {x^{2}y - 2x + 4} \right).}

\end{array}$$

To calculate ${{\partial g}\text{/}{\partial y}},$ treat the variable $x$ as a constant. Then differentiate $g\left( {x,y} \right)$ with respect to $y$ using the chain rule and power rule:

$$\begin{array}{cl}

\frac{\partial g}{\partial y} & {= \frac{\partial}{\partial y}\left\lbrack {\text{sin}\left( {x^{2}y - 2x + 4} \right)} \right\rbrack} \\

& {= \text{cos}\left( {x^{2}y - 2x + 4} \right)\frac{\partial}{\partial y}\left\lbrack {x^{2}y - 2x + 4} \right\rbrack} \\

& {= x^{2}\text{cos}\left( {x^{2}y - 2x + 4} \right).}

\end{array}$$

Calculate ${\partial f}\text{/}{\partial x}$ and ${\partial f}\text{/}{\partial y}$ for the function $f\left( {x,y} \right) = \text{tan}\left( {x^{3} - 3x^{2}y^{2} + 2y^{4}} \right)$ by holding the opposite variable constant, then differentiating.

How can we interpret these partial derivatives? Recall that the graph of a function of two variables is a surface in $\mathbb{R}^{3}.$ If we remove the limit from the definition of the partial derivative with respect to $x,$ the difference quotient remains:

$$\frac{f\left( {x + h,y} \right) - f\left( {x,y} \right)}{h}.$$

This resembles the difference quotient for the derivative of a function of one variable, except for the presence of the $y$ variable. Figure 4.21 illustrates a surface described by an arbitrary function $z = f\left( {x,y} \right).$

In Figure 4.21, the value of $h$ is positive. If we graph $f\left( {x,y} \right)$ and $f\left( {x + h,y} \right)$ for an arbitrary point $\left( {x,y} \right),$ then the slope of the secant line passing through these two points is given by

$$\frac{f\left( {x + h,y} \right) - f\left( {x,y} \right)}{h}.$$

This line is parallel to the $x–z~\text{plane.}$ Therefore, the slope of the secant line represents an average rate of change of the function $f$ as we travel parallel to the $x\text{-axis.}$ As $h$ approaches zero, the slope of the secant line approaches the slope of the tangent line.

If we choose to change $y$ instead of $x$ by the same incremental value $h,$ then the secant line is parallel to the $y\text{-axis}$ and so is the tangent line. Therefore, ${\partial f}\text{/}{\partial x}$ represents the slope of the tangent line passing through the point $\left( {x,y,f\left( {x,y} \right)} \right)$ parallel to the $x\text{-axis}$ and ${\partial f}\text{/}{\partial y}$ represents the slope of the tangent line passing through the point $\left( {x,y,f\left( {x,y} \right)} \right)$ parallel to the $y\text{-axis}.$ If we wish to find the slope of a tangent line passing through the same point in any other direction, then we need what are called *directional derivatives*, which we discuss in Directional Derivatives and the Gradient.

We now return to the idea of contour maps, which we introduced in Functions of Several Variables. We can use a contour map to estimate partial derivatives of a function $g\left( {x,y} \right).$

Partial Derivatives from a Contour Map

Use a contour map to estimate ${\partial g}\text{/}{\partial x}$ at the point $\left( {\sqrt{5},0} \right)$ for the function $g\left( {x,y} \right) = \sqrt{9 - x^{2} - y^{2}}.$

Solution

The following graph represents a contour map for the function $g\left( {x,y} \right) = \sqrt{9 - x^{2} - y^{2}}.$

The inner circle on the contour map corresponds to $c = 2$ and the next circle out corresponds to $c = 1.$ The first circle is given by the equation $2 = \sqrt{9 - x^{2} - y^{2}};$ the second circle is given by the equation $1 = \sqrt{9 - x^{2} - y^{2}}.$ The first equation simplifies to $x^{2} + y^{2} = 5$ and the second equation simplifies to $x^{2} + y^{2} = 8.$ The $x\text{-intercept}$ of the first circle is $\left( {\sqrt{5},0} \right)$ and the $x\text{-intercept}$ of the second circle is $\left( {2\sqrt{2},0} \right).$ We can estimate the value of ${\partial g}\text{/}{\partial x}$ evaluated at the point $\left( {\sqrt{5},0} \right)$ using the slope formula:

$$\left. \frac{\partial g}{\partial x} \right|_{{({x,y})} = {({\sqrt{5},0})}} \approx \frac{g\left( {\sqrt{5},0} \right) - g\left( {2\sqrt{2},0} \right)}{\sqrt{5} - 2\sqrt{2}} = \frac{2 - 1}{\sqrt{5} - 2\sqrt{2}} = \frac{1}{\sqrt{5} - 2\sqrt{2}} \approx -1.688.$$

To calculate the exact value of ${\partial g}\text{/}{\partial x}$ evaluated at the point $\left( {\sqrt{5},0} \right),$ we start by finding ${\partial g}\text{/}{\partial x}$ using the chain rule. First, we rewrite the function as $g\left( {x,y} \right) = \sqrt{9 - x^{2} - y^{2}} = \left( {9 - x^{2} - y^{2}} \right)^{1\text{/}2}$ and then differentiate with respect to $x$ while holding $y$ constant:

$$\frac{\partial g}{\partial x} = \frac{1}{2}\left( {9 - x^{2} - y^{2}} \right)^{-1\text{/}2}\left( {-2x} \right) = - \frac{x}{\sqrt{9 - x^{2} - y^{2}}}.$$

Next, we evaluate this expression using $x = \sqrt{5}$ and $y = 0\text{:}$

$$\left. \frac{\partial g}{\partial x} \right|_{{({x,y})} = {({\sqrt{5},0})}} = - \frac{\sqrt{5}}{\sqrt{9 - \left( \sqrt{5} \right)^{2} - (0)^{2}}} = - \frac{\sqrt{5}}{\sqrt{4}} = - \frac{\sqrt{5}}{2} \approx -1.118.$$

The estimate for the partial derivative corresponds to the slope of the secant line passing through the points $\left( {\sqrt{5},0,g\left( {\sqrt{5},0} \right)} \right)$ and $\left( {2\sqrt{2},0,g\left( {2\sqrt{2},0} \right)} \right).$ It represents an approximation to the slope of the tangent line to the surface through the point $\left( {\sqrt{5},0,g\left( {\sqrt{5},0} \right)} \right),$ which is parallel to the $x\text{-axis}.$

Use a contour map to estimate ${\partial f}\text{/}{\partial y}$ at point $\left( {0,\sqrt{2}} \right)$ for the function

$$f\left( {x,y} \right) = x^{2} - y^{2}.$$

Compare this with the exact answer.

Functions of More Than Two Variables

Suppose we have a function of three variables, such as $w = f\left( {x,y,z} \right).$ We can calculate partial derivatives of $w$ with respect to any of the independent variables, simply as extensions of the definitions for partial derivatives of functions of two variables.

Let $f\left( {x,y,z} \right)$ be a function of three variables. Then, the *partial derivative of $f$ with respect to x,* written as ${{\partial f}\text{/}{\partial x}},$ or $f_{x},$ is defined to be

$$\frac{\partial f}{\partial x} = \underset{h\rightarrow 0}{\text{lim}}\frac{f\left( {x + h,y,z} \right) - f\left( {x,y,z} \right)}{h}.$$ (4.14)

The *partial derivative of* $f$ *with respect to* $y,$ written as ${{\partial f}\text{/}{\partial y}},$ or $f_{y},$ is defined to be

$$\frac{\partial f}{\partial y} = \underset{k\rightarrow 0}{\text{lim}}\frac{f\left( {x,y + k,z} \right) - f\left( {x,y,z} \right)}{k}.$$ (4.15)

The *partial derivative of* $f$ *with respect to* $z,$ written as ${{\partial f}\text{/}{\partial z}},$ or $f_{z},$ is defined to be

$$\frac{\partial f}{\partial z} = \underset{m\rightarrow 0}{\text{lim}}\frac{f\left( {x,y,z + m} \right) - f\left( {x,y,z} \right)}{m}.$$ (4.16)

We can calculate a partial derivative of a function of three variables using the same idea we used for a function of two variables. For example, if we have a function $f$ of $x,y,\text{and}\ z,$ and we wish to calculate ${{\partial f}\text{/}{\partial x}},$ then we treat the other two independent variables as if they are constants, then differentiate with respect to $x.$

Calculating Partial Derivatives for a Function of Three Variables

Use the limit definition of partial derivatives to calculate ${\partial f}\text{/}{\partial x}$ for the function

$$f\left( {x,y,z} \right) = x^{2} - 3xy + 2y^{2} - 4xz + 5yz^{2} - 12x + 4y - 3z.$$

Then, find ${\partial f}\text{/}{\partial y}$ and ${\partial f}\text{/}{\partial z}$ by setting the other two variables constant and differentiating accordingly.

Solution

We first calculate ${\partial f}\text{/}{\partial x}$ using Equation 4.14, then we calculate the other two partial derivatives by holding the remaining variables constant. To use the equation to find ${{\partial f}\text{/}{\partial x}},$ we first need to calculate $f\left( {x + h,y,z} \right)\text{:}$

$$\begin{array}{cl}

{f\left( {x + h,y,z} \right)} & {= \left( {x + h} \right)^{2} - 3\left( {x + h} \right)y + 2y^{2} - 4\left( {x + h} \right)z + 5yz^{2} - 12\left( {x + h} \right) + 4y - 3z} \\

& {= x^{2} + 2xh + h^{2} - 3xy - 3xh + 2y^{2} - 4xz - 4hz + 5yz^{2} - 12x - 12h + 4y - 3z}

\end{array}$$

and recall that $f\left( {x,y,z} \right) = x^{2} - 3xy + 2y^{2} - 4zx + 5yz^{2} - 12x + 4y - 3z.$ Next, we substitute these two expressions into the equation:

$$\begin{array}{cl}

\frac{\partial f}{\partial x} & {= \underset{h\rightarrow 0}{\text{lim}}\left\lbrack \frac{x^{2} + 2xh + h^{2} - 3xy - 3hy + 2y^{2} - 4xz - 4hz + 5yz^{2} - 12x - 12h + 4y - 3z}{h} \right.} \\

& \left. \mspace{9mu} - \frac{x^{2} - 3xy + 2y^{2} - 4xz + 5yz^{2} - 12x + 4y - 3z}{h} \right\rbrack \\

& {= \underset{h\rightarrow 0}{\text{lim}}\left\lbrack \frac{2xh + h^{2} - 3hy - 4hz - 12h}{h} \right\rbrack} \\

& {= \underset{h\rightarrow 0}{\text{lim}}\left\lbrack \frac{h\left( {2x + h - 3y - 4z - 12} \right)}{h} \right\rbrack} \\

& {= \underset{h\rightarrow 0}{\text{lim}}\left( {2x + h - 3y - 4z - 12} \right)} \\

& {= 2x - 3y - 4z - 12.}

\end{array}$$

Then we find ${\partial f}\text{/}{\partial y}$ by holding $x\ \text{and}\ z$ constant. Therefore, any term that does not include the variable $y$ is constant, and its derivative is zero. We can apply the sum, difference, and power rules for functions of one variable:

$$\begin{array}{l}

{\frac{\partial}{\partial y}\left\lbrack {x^{2} - 3xy + 2y^{2} - 4xz + 5yz^{2} - 12x + 4y - 3z} \right\rbrack} \\

{= \frac{\partial}{\partial y}\left\lbrack x^{2} \right\rbrack - \frac{\partial}{\partial y}\left\lbrack {3xy} \right\rbrack + \frac{\partial}{\partial y}\left\lbrack {2y^{2}} \right\rbrack - \frac{\partial}{\partial y}\left\lbrack {4xz} \right\rbrack + \frac{\partial}{\partial y}\left\lbrack {5yz^{2}} \right\rbrack - \frac{\partial}{\partial y}\left\lbrack {12x} \right\rbrack + \frac{\partial}{\partial y}\left\lbrack {4y} \right\rbrack - \frac{\partial}{\partial y}\left\lbrack {3z} \right\rbrack} \\

{= 0 - 3x + 4y - 0 + 5z^{2} - 0 + 4 - 0} \\

{= -3x + 4y + 5z^{2} + 4.}

\end{array}$$

To calculate ${{\partial f}\text{/}{\partial z}},$ we hold *x* and *y* constant and apply the sum, difference, and power rules for functions of one variable:

$$\begin{array}{l}

{\frac{\partial}{\partial z}\left\lbrack {x^{2} - 3xy + 2y^{2} - 4xz + 5yz^{2} - 12x + 4y - 3z} \right\rbrack} \\

{= \frac{\partial}{\partial z}\left\lbrack x^{2} \right\rbrack - \frac{\partial}{\partial z}\left\lbrack {3xy} \right\rbrack + \frac{\partial}{\partial z}\left\lbrack {2y^{2}} \right\rbrack - \frac{\partial}{\partial z}\left\lbrack {4xz} \right\rbrack + \frac{\partial}{\partial z}\left\lbrack {5yz^{2}} \right\rbrack - \frac{\partial}{\partial z}\left\lbrack {12x} \right\rbrack + \frac{\partial}{\partial z}\left\lbrack {4y} \right\rbrack - \frac{\partial}{\partial z}\left\lbrack {3z} \right\rbrack} \\

{= 0 - 0 + 0 - 4x + 10yz - 0 + 0 - 3} \\

{= -4x + 10yz - 3.}

\end{array}$$

Use the limit definition of partial derivatives to calculate ${\partial f}\text{/}{\partial x}$ for the function

$$f\left( {x,y,z} \right) = 2x^{2} - 4x^{2}y + 2y^{2} + 5xz^{2} - 6x + 3z - 8.$$

Then find ${\partial f}\text{/}{\partial y}$ and ${\partial f}\text{/}{\partial z}$ by setting the other two variables constant and differentiating accordingly.

Calculating Partial Derivatives for a Function of Three Variables

Calculate the three partial derivatives of the following functions.

1. $f\left( {x,y,z} \right) = \frac{x^{2}y - 4xz + y^{2}}{x - 3yz}$

2. $g\left( {x,y,z} \right) = \text{sin}\left( {x^{2}y - z} \right) + \text{cos}\left( {x^{2} - yz} \right)$

Solution

In each case, treat all variables as constants except the one whose partial derivative you are calculating.

1. $\begin{array}{cl}

\frac{\partial f}{\partial x} & {= \frac{\partial}{\partial x}\left\lbrack \frac{x^{2}y - 4xz + y^{2}}{x - 3yz} \right\rbrack} \\

& {= \frac{\frac{\partial}{\partial x}\left( {x^{2}y - 4xz + y^{2}} \right)\left( {x - 3yz} \right) - \left( {x^{2}y - 4xz + y^{2}} \right)\frac{\partial}{\partial x}\left( {x - 3yz} \right)}{\left( {x - 3yz} \right)^{2}}} \\

& {= \frac{\left( {2xy - 4z} \right)\left( {x - 3yz} \right) - \left( {x^{2}y - 4xz + y^{2}} \right)(1)}{\left( {x - 3yz} \right)^{2}}} \\

& {= \frac{2x^{2}y - 6xy^{2}z - 4xz + 12yz^{2} - x^{2}y + 4xz - y^{2}}{\left( {x - 3yz} \right)^{2}}} \\

& {= \frac{x^{2}y - 6xy^{2}z - 4xz + 12yz^{2} + 4xz - y^{2}}{\left( {x - 3yz} \right)^{2}}}

\end{array}$

$\begin{array}{cl}

\frac{\partial f}{\partial y} & {= \frac{\partial}{\partial y}\left\lbrack \frac{x^{2}y - 4xz + y^{2}}{x - 3yz} \right\rbrack} \\

& {= \frac{\frac{\partial}{\partial y}\left( {x^{2}y - 4xz + y^{2}} \right)\left( {x - 3yz} \right) - \left( {x^{2}y - 4xz + y^{2}} \right)\frac{\partial}{\partial y}\left( {x - 3yz} \right)}{\left( {x - 3yz} \right)^{2}}} \\

& {= \frac{\left( {x^{2} + 2y} \right)\left( {x - 3yz} \right) - \left( {x^{2}y - 4xz + y^{2}} \right)\left( {-3z} \right)}{\left( {x - 3yz} \right)^{2}}} \\

& {= \frac{x^{3} - 3x^{2}yz + 2xy - 6y^{2}z + 3x^{2}yz - 12xz^{2} + 3y^{2}z}{\left( {x - 3yz} \right)^{2}}} \\

& {= \frac{x^{3} + 2xy - 3y^{2}z - 12xz^{2}}{\left( {x - 3yz} \right)^{2}}}

\end{array}$

$\begin{array}{cl}

\frac{\partial f}{\partial z} & {= \frac{\partial}{\partial z}\left\lbrack \frac{x^{2}y - 4xz + y^{2}}{x - 3yz} \right\rbrack} \\

& {= \frac{\frac{\partial}{\partial z}\left( {x^{2}y - 4xz + y^{2}} \right)\left( {x - 3yz} \right) - \left( {x^{2}y - 4xz + y^{2}} \right)\frac{\partial}{\partial z}\left( {x - 3yz} \right)}{\left( {x - 3yz} \right)^{2}}} \\

& {= \frac{\left( {-4x} \right)\left( {x - 3yz} \right) - \left( {x^{2}y - 4xz + y^{2}} \right)\left( {-3y} \right)}{\left( {x - 3yz} \right)^{2}}} \\

& {= \frac{-4x^{2} + 12xyz + 3x^{2}y^{2} - 12xyz + 3y^{3}}{\left( {x - 3yz} \right)^{2}}} \\

& {= \frac{-4x^{2} + 3x^{2}y^{2} + 3y^{3}}{\left( {x - 3yz} \right)^{2}}}

\end{array}$

2. $\begin{array}{cl}

\frac{\partial f}{\partial x} & {= \frac{\partial}{\partial x}\left\lbrack {\text{sin}\left( {x^{2}y - z} \right) + \text{cos}\left( {x^{2} - yz} \right)} \right\rbrack} \\

& {= \left( {\text{cos}\left( {x^{2}y - z} \right)} \right)\frac{\partial}{\partial x}\left( {x^{2}y - z} \right) - \left( {\text{sin}\left( {x^{2} - yz} \right)} \right)\frac{\partial}{\partial x}\left( {x^{2} - yz} \right)} \\

& {= 2xy\ \text{cos}\left( {x^{2}y - z} \right) - 2x\ \text{sin}\left( {x^{2} - yz} \right)} \\

\frac{\partial f}{\partial y} & {= \frac{\partial}{\partial y}\left\lbrack {\text{sin}\left( {x^{2}y - z} \right) + \text{cos}\left( {x^{2} - yz} \right)} \right\rbrack} \\

& {= \left( {\text{cos}\left( {x^{2}y - z} \right)} \right)\frac{\partial}{\partial y}\left( {x^{2}y - z} \right) - \left( {\text{sin}\left( {x^{2} - yz} \right)} \right)\frac{\partial}{\partial y}\left( {x^{2} - yz} \right)} \\

& {= x^{2}\text{cos}\left( {x^{2}y - z} \right) + z\ \text{sin}\left( {x^{2} - yz} \right)} \\

\frac{\partial f}{\partial z} & {= \frac{\partial}{\partial z}\left\lbrack {\text{sin}\left( {x^{2}y - z} \right) + \text{cos}\left( {x^{2} - yz} \right)} \right\rbrack} \\

& {= \left( {\text{cos}\left( {x^{2}y - z} \right)} \right)\frac{\partial}{\partial z}\left( {x^{2}y - z} \right) - \left( {\text{sin}\left( {x^{2} - yz} \right)} \right)\frac{\partial}{\partial z}\left( {x^{2} - yz} \right)} \\

& {= \text{−}\text{cos}\left( {x^{2}y - z} \right) + y\ \text{sin}\left( {x^{2} - yz} \right)}

\end{array}$

Calculate ${{\partial f}\text{/}{\partial x}},$ ${{\partial f}\text{/}{\partial y}},$ and ${\partial f}\text{/}{\partial z}$ for the function $f\left( {x,y,z} \right) = \text{sec}\left( {x^{2}y} \right) - \text{tan}\left( {x^{3}yz^{2}} \right).$

Higher-Order Partial Derivatives

Consider the function

$$f(x,y) = 2x^{3} - 4xy^{2} + 5y^{3} - 6xy + 5x - 4y + 12.$$

Its partial derivatives are

$$\frac{\partial f}{\partial x} = 6x^{2} - 4y^{2} - 6y + 5\ \text{and}\ \frac{\partial f}{\partial y} = -8xy + 15y^{2} - 6x - 4.$$

Each of these partial derivatives is a function of two variables, so we can calculate partial derivatives of these functions. Just as with derivatives of single-variable functions, we can call these *second-order derivatives, third-order derivatives*, and so on. In general, they are referred to as higher-order partial derivatives. There are four second-order partial derivatives for any function (provided they all exist):

$$\frac{\partial^{2}f}{\partial x^{2}} = \frac{\partial}{\partial x}\left\lbrack \frac{\partial f}{\partial x} \right\rbrack,\quad\frac{\partial^{2}f}{\partial x\partial y} = \frac{\partial}{\partial x}\left\lbrack \frac{\partial f}{\partial y} \right\rbrack,\quad\frac{\partial^{2}f}{\partial y\partial x} = \frac{\partial}{\partial y}\left\lbrack \frac{\partial f}{\partial x} \right\rbrack,\quad\frac{\partial^{2}f}{\partial y^{2}} = \frac{\partial}{\partial y}\left\lbrack \frac{\partial f}{\partial y} \right\rbrack.$$

An alternative notation for each is $f_{xx},f_{yx},f_{xy},$ and $f_{yy},$ respectively. Higher-order partial derivatives calculated with respect to different variables, such as $f_{xy}$ and $f_{yx},$ are commonly called mixed partial derivatives.

Calculating Second Partial Derivatives

Calculate all four second partial derivatives for the function

$$f\left( {x,y} \right) = xe^{-3y} + \text{sin}\left( {2x - 5y} \right).$$

Solution

To calculate ${\partial^{2}f}\text{/}{dx^{2}}$ and ${{\partial^{2}f}\text{/}{\partial y\partial x}},$ we first calculate ${{\partial f}\text{/}{\partial x}}\text{:}$

$$\frac{\partial f}{\partial x} = e^{-3y} + 2\ \text{cos}\left( {2x - 5y} \right).$$

To calculate ${{\partial^{2}f}\text{/}{dx^{2}}},$ differentiate ${\partial f}\text{/}{\partial x}$ with respect to $x\text{:}$

$$\begin{array}{cl}

\frac{\partial^{2}f}{\partial x^{2}} & {= \frac{\partial}{\partial x}\left\lbrack \frac{\partial f}{\partial x} \right\rbrack} \\

& {= \frac{\partial}{\partial x}\left\lbrack {e^{-3y} + 2\ \text{cos}\left( {2x - 5y} \right)} \right\rbrack} \\

& {= -4\ \text{sin}\left( {2x - 5y} \right).}

\end{array}$$

To calculate ${{\partial^{2}f}\text{/}{\partial y\partial x}},$ differentiate ${\partial f}\text{/}{\partial x}$ with respect to $y\text{:}$

$$\begin{array}{cl}

\frac{\partial^{2}f}{\partial y\partial x} & {= \frac{\partial}{\partial y}\left\lbrack \frac{\partial f}{\partial x} \right\rbrack} \\

& {= \frac{\partial}{\partial y}\left\lbrack {e^{-3y} + 2\ \text{cos}\left( {2x - 5y} \right)} \right\rbrack} \\

& {= -3e^{-3y} + 10\ \text{sin}\left( {2x - 5y} \right).}

\end{array}$$

To calculate ${\partial^{2}f}\text{/}{\partial x\partial y}$ and ${{\partial^{2}f}\text{/}{dy^{2}}},$ first calculate ${{\partial f}\text{/}{\partial y}}\text{:}$

$$\frac{\partial f}{\partial y} = -3xe^{-3y} - 5\ \text{cos}\left( {2x - 5y} \right).$$

To calculate ${{\partial^{2}f}\text{/}{\partial x\partial y}},$ differentiate ${\partial f}\text{/}{\partial y}$ with respect to $x\text{:}$

$$\begin{array}{cl}

\frac{\partial^{2}f}{\partial x\partial y} & {= \frac{\partial}{\partial x}\left\lbrack \frac{\partial f}{\partial y} \right\rbrack} \\

& {= \frac{\partial}{\partial x}\left\lbrack {-3xe^{-3y} - 5\ \text{cos}\left( {2x - 5y} \right)} \right\rbrack} \\

& {= -3e^{-3y} + 10\ \text{sin}\left( {2x - 5y} \right).}

\end{array}$$

To calculate ${{\partial^{2}f}\text{/}{\partial y^{2}}},$ differentiate ${\partial f}\text{/}{\partial y}$ with respect to $y\text{:}$

$$\begin{array}{cl}

\frac{\partial^{2}f}{\partial y^{2}} & {= \frac{\partial}{\partial y}\left\lbrack \frac{\partial f}{\partial y} \right\rbrack} \\

& {= \frac{\partial}{\partial y}\left\lbrack {-3xe^{-3y} - 5\ \text{cos}\left( {2x - 5y} \right)} \right\rbrack} \\

& {= 9xe^{-3y} - 25\ \text{sin}\left( {2x - 5y} \right).}

\end{array}$$

Calculate all four second partial derivatives for the function

$$f\left( {x,y} \right) = \text{sin}\left( {3x - 2y} \right) + \text{cos}\left( {x + 4y} \right).$$

At this point we should notice that, in both Example 4.19 and the checkpoint, it was true that ${{\partial^{2}f}\text{/}{\partial x\partial y}} = {{\partial^{2}f}\text{/}{\partial y\partial x}}.$ Under certain conditions, this is always true. In fact, it is a direct consequence of the following theorem.

Equality of Mixed Partial Derivatives (Clairaut’s Theorem)

Suppose that $f\left( {x,y} \right)$ is defined on an open disk $D$ that contains the point $\left( {a,b} \right).$ If the functions $f_{xy}$ and $f_{yx}$ are continuous on $D,$ then $f_{xy} = f_{yx}.$

Clairaut’s theorem guarantees that as long as mixed second-order derivatives are continuous, the order in which we choose to differentiate the functions (i.e., which variable goes first, then second, and so on) does not matter. It can be extended to higher-order derivatives as well. The proof of Clairaut’s theorem can be found in most advanced calculus books.

Two other second-order partial derivatives can be calculated for any function $f\left( {x,y} \right).$ The partial derivative $f_{xx}$ is equal to the partial derivative of $f_{x}$ with respect to $x,$ and $f_{yy}$ is equal to the partial derivative of $f_{y}$ with respect to $y.$

Partial Differential Equations

In Introduction to Differential Equations, we studied differential equations in which the unknown function had one independent variable. A partial differential equation is an equation that involves an unknown function of more than one independent variable and one or more of its partial derivatives. Examples of partial differential equations are

$$u_{t} = c^{2}\left( {u_{xx} + u_{yy}} \right)$$ (4.17)

(heat equation in two dimensions)

$$u_{tt} = c^{2}\left( {u_{xx} + u_{yy}} \right)$$ (4.18)

(wave equation in two dimensions)

$$u_{xx} + u_{yy} = 0$$ (4.19)

(Laplace’s equation in two dimensions)

In the first two equations, the unknown function $u$ has three independent variables—$t,x,\ \text{and}\ y$—and $c$ is an arbitrary constant. The independent variables $x\ \text{and}\ y$ are considered to be spatial variables, and the variable $t$ represents time. In Laplace’s equation, the unknown function $u$ has two independent variables $x\ \text{and}\ y.$

A Solution to the Wave Equation

Verify that

$$u\left( {x,y,t} \right) = 5\ \text{sin}\left( {3\pi x} \right)\text{sin}\left( {4\pi y} \right)\text{cos}\left( {10\pi t} \right)$$

is a solution to the wave equation

$$u_{tt} = 4\left( {u_{xx} + u_{yy}} \right).$$ (4.20)

Solution

First, we calculate $u_{tt},u_{xx},$ and $u_{yy}\text{:}$

$$\begin{array}{cll}

u_{tt} & = & {\frac{\partial}{\partial t}\left\lbrack \frac{\partial u}{\partial t} \right\rbrack} \\

& = & {\frac{\partial}{\partial t}\left\lbrack {5\ \text{sin}\left( {3\pi x} \right)\text{sin}\left( {4\pi y} \right)\left( {-10\pi\ \text{sin}\left( {10\pi t} \right)} \right)} \right\rbrack} \\

& = & {\frac{\partial}{\partial t}\left\lbrack {-50\pi\ \text{sin}\left( {3\pi x} \right)\text{sin}\left( {4\pi y} \right)\text{sin}\left( {10\pi t} \right)} \right\rbrack} \\

& = & {-500\pi^{2}\text{sin}\left( {3\pi x} \right)\text{sin}\left( {4\pi y} \right)\text{cos}\left( {10\pi t} \right)} \\

u_{xx} & = & {\frac{\partial}{\partial x}\left\lbrack \frac{\partial u}{\partial x} \right\rbrack} \\

& = & {\frac{\partial}{\partial x}\left\lbrack {15\pi\ \text{cos}\left( {3\pi x} \right)\text{sin}\left( {4\pi y} \right)\text{cos}\left( {10\pi t} \right)} \right\rbrack} \\

& = & {-45\pi^{2}\text{sin}\left( {3\pi x} \right)\text{sin}\left( {4\pi y} \right)\text{cos}\left( {10\pi t} \right)} \\

u_{yy} & = & {\frac{\partial}{\partial y}\left\lbrack \frac{\partial u}{\partial y} \right\rbrack} \\

& = & {\frac{\partial}{\partial y}\left\lbrack {5\ \text{sin}\left( {3\pi x} \right)\left( {4\pi\ \text{cos}\left( {4\pi y} \right)} \right)\text{cos}\left( {10\pi t} \right)} \right\rbrack} \\

& = & {\frac{\partial}{\partial y}\left\lbrack {20\pi\ \text{sin}\left( {3\pi x} \right)\text{cos}\left( {4\pi y} \right)\text{cos}\left( {10\pi t} \right)} \right\rbrack} \\

& = & {-80\pi^{2}\text{sin}\left( {3\pi x} \right)\text{sin}\left( {4\pi y} \right)\text{cos}\left( {10\pi t} \right).}

\end{array}$$

Next, we substitute each of these into the right-hand side of Equation 4.20 and simplify:

$$\begin{array}{cl}

{4\left( {u_{xx} + u_{yy}} \right)} & {= 4\left( {-45\pi^{2}\text{sin}\left( {3\pi x} \right)\text{sin}\left( {4\pi y} \right)\text{cos}\left( {10\pi t} \right) + - 80\pi^{2}\text{sin}\left( {3\pi x} \right)\text{sin}\left( {4\pi y} \right)\text{cos}\left( {10\pi t} \right)} \right)} \\

& {= 4\left( {-125\pi^{2}\text{sin}\left( {3\pi x} \right)\text{sin}\left( {4\pi y} \right)\text{cos}\left( {10\pi t} \right)} \right)} \\

& {= -500\pi^{2}\text{sin}\left( {3\pi x} \right)\text{sin}\left( {4\pi y} \right)\text{cos}\left( {10\pi t} \right)} \\

& {= u_{tt}.}

\end{array}$$

This verifies the solution.

Verify that $u\left( {x,y,t} \right) = 2\ \text{sin}\left( \frac{x}{3} \right)\text{sin}\left( \frac{y}{4} \right)e^{-25t\text{/}16}$ is a solution to the heat equation

$$u_{t} = 9\left( {u_{xx} + u_{yy}} \right).$$ (4.21)

Since the solution to the two-dimensional heat equation is a function of three variables, it is not easy to create a visual representation of the solution. We can graph the solution for fixed values of *t*, which amounts to snapshots of the heat distributions at fixed times. These snapshots show how the heat is distributed over a two-dimensional surface as time progresses. The graph of the preceding solution at time $t = 0$ appears in the following figure. As time progresses, the extremes level out, approaching zero as *t* approaches infinity.

If we consider the heat equation in one dimension, then it is possible to graph the solution over time. The heat equation in one dimension becomes

$$u_{t} = c^{2}u_{xx},$$

where $c^{2}$ represents the thermal diffusivity of the material in question. A solution of this differential equation can be written in the form

$$u_{m}\left( {x,t} \right) = e^{\text{−}\pi^{2}m^{2}c^{2}t}\text{sin}\left( {m\pi x} \right)$$ (4.22)

where $m$ is any positive integer. A graph of this solution using $m = 1$ appears in Figure 4.24, where the initial temperature distribution over a wire of length $1$ is given by $u\left( {x,0} \right) = \text{sin}\ \pi x.$ Notice that as time progresses, the wire cools off. This is seen because, from left to right, the highest temperature (which occurs in the middle of the wire) decreases and changes color from red to blue.

Lord Kelvin and the Age of Earth

During the late 1800s, the scientists of the new field of geology were coming to the conclusion that Earth must be “millions and millions” of years old. At about the same time, Charles Darwin had published his treatise on evolution. Darwin’s view was that evolution needed many millions of years to take place, and he made a bold claim that the Weald chalk fields, where important fossils were found, were the result of $300$ million years of erosion.

At that time, eminent physicist William Thomson (Lord Kelvin) used an important partial differential equation, known as the *heat diffusion equation*, to estimate the age of Earth by determining how long it would take Earth to cool from molten rock to what we had at that time. His conclusion was a range of $20\ \text{to}\ 400$ million years, but most likely about $50$ million years. For many decades, the proclamations of this irrefutable icon of science did not sit well with geologists or with Darwin.

Read Kelvin’s paper on estimating the age of the Earth.

Kelvin made reasonable assumptions based on what was known in his time, but he also made several assumptions that turned out to be wrong. One incorrect assumption was that Earth is solid and that the cooling was therefore via conduction only, hence justifying the use of the diffusion equation. But the most serious error was a forgivable one—omission of the fact that Earth contains radioactive elements that continually supply heat beneath Earth’s mantle. The discovery of radioactivity came near the end of Kelvin’s life and he acknowledged that his calculation would have to be modified.

Kelvin used the simple one-dimensional model applied only to Earth’s outer shell, and derived the age from graphs and the roughly known temperature gradient near Earth’s surface. Let’s take a look at a more appropriate version of the diffusion equation in radial coordinates, which has the form

$$\frac{\partial T}{\partial t} = K\left\lbrack {\frac{\partial^{2}T}{\partial^{2}r} + \frac{2}{r}\ \frac{\partial T}{\partial r}} \right\rbrack.$$ (4.23)

Here, $T\left( {r,t} \right)$ is temperature as a function of $r$ (measured from the center of Earth) and time $t.$ $K$ is the heat conductivity—for molten rock, in this case. The standard method of solving such a partial differential equation is by separation of variables, where we express the solution as the product of functions containing each variable separately. In this case, we would write the temperature as

$$T\left( {r,t} \right) = R(r)f(t).$$

1. Substitute this form into Equation 4.13 and, noting that $f(t)$ is constant with respect to distance $(r)$ and $R(r)$ is constant with respect to time $(t),$ show that

$$\frac{1}{f}\ \frac{\partial f}{\partial t} = \frac{K}{R}\left\lbrack {\frac{\partial^{2}R}{\partial r^{2}} + \frac{2}{r}\ \frac{\partial R}{\partial r}} \right\rbrack.$$

2. This equation represents the separation of variables we want. The left-hand side is only a function of $t$ and the right-hand side is only a function of $r,$ and they must be equal for all values of $r\ \text{and}\ t.$ Therefore, they both must be equal to a constant. Let’s call that constant $\text{−}\lambda^{2}.$ (The convenience of this choice is seen on substitution.) So, we have

$$\frac{1}{f}\ \frac{\partial f}{\partial t} = \text{−}\lambda^{2}\quad\text{and}\quad\frac{K}{R}\left\lbrack {\frac{\partial^{2}R}{\partial r^{2}} + \frac{2}{r}\ \frac{\partial R}{\partial r}} \right\rbrack = \text{−}\lambda^{2}.$$

Now, we can verify through direct substitution for each equation that the solutions are $f(t) = Ae^{\text{−}\lambda^{2}t}$ and $R(r) = B\left( \frac{\text{sin}\ \alpha r}{r} \right) + C\left( \frac{\text{cos}\ \alpha r}{r} \right),$ where $\alpha = {\lambda\text{/}{\sqrt{K}.}}$ Note that $f(t) = Ae^{+ \lambda n^{2}t}$ is also a valid solution, so we could have chosen $+ \lambda^{2}$ for our constant. Can you see why it would not be valid for this case as time increases?

3. Let’s now apply boundary conditions.

1. The temperature must be finite at the center of Earth, $r = 0.$ Which of the two constants, $B$ or $C,$ must therefore be zero to keep $R$ finite at $r = 0?$ (Recall that $\text{sin}(\alpha r)\text{/}r\rightarrow\alpha =$ as $r\rightarrow 0,$ but $\text{cos}(\alpha r)\text{/}r$ behaves very differently.)

2. Kelvin argued that when magma reaches Earth’s surface, it cools very rapidly. A person can often touch the surface within weeks of the flow. Therefore, the surface reached a moderate temperature very early and remained nearly constant at a surface temperature $T_{s}.$ For simplicity, let’s set $T = 0\ \text{at}\ r = R_{E}$ and find $\alpha$ such that this is the temperature there for all time $t.$ (Kelvin took the value to be $300\ \text{K} \approx 80\text{°}\text{F}.$ We can add this $300\ \text{K}$ constant to our solution later.) For this to be true, the sine argument must be zero at $r = R_{E}.$ Note that $\alpha$ has an infinite series of values that satisfies this condition. Each value of $\alpha$ represents a valid solution (each with its own value for $A).$ The total or general solution is the sum of all these solutions.

3. At $t = 0,$ we assume that all of Earth was at an initial hot temperature $T_{0}$ (Kelvin took this to be about $7000\ \text{K}.)$ The application of this boundary condition involves the more advanced application of Fourier coefficients. As noted in part b. each value of $\alpha_{n}$ represents a valid solution, and the general solution is a sum of all these solutions. This results in a series solution:

$$T(r,t) = \left( \frac{T_{0}R_{\text{E}}}{\pi} \right){\sum\limits_{n}{\frac{{(-1)}^{n - 1}}{n}e^{\text{−}\lambda n^{2}t}\frac{\text{sin}(\alpha_{n}r)}{r},}}\text{where}\ \alpha_{n} = {{n\pi}\text{/}R_{\text{E}}}.$$

Note how the values of $\alpha_{n}$ come from the boundary condition applied in part b. The term $\frac{\text{−}1^{n - 1}}{n}$ is the constant $A_{n}$ for each term in the series, determined from applying the Fourier method. Letting $\beta = \frac{\pi}{R_{\text{E}}},$ examine the first few terms of this solution shown here and note how $\lambda^{2}$ in the exponential causes the higher terms to decrease quickly as time progresses:

$$T(r,t) = \frac{T_{0}R_{\text{E}}}{\pi r}\begin{pmatrix}

{e^{\text{−}K\beta^{2}t}\left( {\text{sin}\ \beta r} \right) - \frac{1}{2}e^{-4K\beta^{2}t}\left( {\text{sin}\ 2\beta r} \right) + \frac{1}{3}e^{-9K\beta^{2}t}\left( {\text{sin}\ 3\beta r} \right)} \\

{- \frac{1}{4}e^{-16K\beta^{2}t}\left( {\text{sin}\ 4\beta r} \right) + \frac{1}{5}e^{-25K\beta^{2}t}\left( {\text{sin}\ 5\beta r} \right)...}

\end{pmatrix}.$$

Near time $t = 0,$ many terms of the solution are needed for accuracy. Inserting values for the conductivity $K$ and $\beta = {\pi\text{/}R_{\text{E}}}$ for time approaching merely thousands of years, only the first few terms make a significant contribution. Kelvin only needed to look at the solution near Earth’s surface (Figure 4.26) and, after a long time, determine what time best yielded the estimated temperature gradient known during his era $(1\text{°}\text{F}$ increase per $50\ \text{ft}).$ He simply chose a range of times with a gradient close to this value. In Figure 4.26, the solutions are plotted and scaled, with the $300 - \text{K}$ surface temperature added. Note that the center of Earth would be relatively cool. At the time, it was thought Earth must be solid.

Epilog

On May $20,1904,$ physicist Ernest Rutherford spoke at the Royal Institution to announce a revised calculation that included the contribution of radioactivity as a source of Earth’s heat. In Rutherford’s own words:

“I came into the room, which was half-dark, and presently spotted Lord Kelvin in the audience, and realised that I was in for trouble at the last part of my speech dealing with the age of the Earth, where my views conflicted with his. To my relief, Kelvin fell fast asleep, but as I came to the important point, I saw the old bird sit up, open an eye and cock a baleful glance at me.

Then a sudden inspiration came, and I said Lord Kelvin had limited the age of the Earth, *provided no new source* \[*of heat*\] *was discovered*. That prophetic utterance referred to what we are now considering tonight, radium! Behold! The old boy beamed upon me.”

Rutherford calculated an age for Earth of about $500$ million years. Today’s accepted value of Earth’s age is about $4.6$ billion years.

Section 4.3 Exercises

For the following exercises, calculate the partial derivative using the limit definitions only.

112\.

$\frac{\partial z}{\partial x}$ for $z = x^{2} - 3xy + y^{2}$

113.

$\frac{\partial z}{\partial y}$ for $z = x^{2} - 3xy + y^{2}$

For the following exercises, calculate the sign of the partial derivative using the graph of the surface.

114\.

$f_{x}(1,1)$

115.

$f_{x}(-1,1)$

116\.

$f_{y}(1,1)$

117.

$f_{x}(0,0)$

For the following exercises, calculate the partial derivatives.

118\.

$\frac{\partial z}{\partial x}$ for $z = \text{sin}(3x)\text{cos}(3y)$

119.

$\frac{\partial z}{\partial y}$ for $z = \text{sin}(3x)\text{cos}(3y)$

120\.

$\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ for $z = x^{8}e^{3y}$

121.

$\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ for $z = \text{ln}\left( {x^{6} + y^{4}} \right)$

122\.

Find $f_{y}(x,y)$ for $f(x,y) = e^{xy}\text{cos}(x)\text{sin}(y).$

123.

Let $z = e^{xy}.$ Find $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}.$

124\.

Let $z = \text{ln}\left( \frac{x}{y} \right).$ Find $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}.$

125.

Let $z = \text{tan}(2x - y).$ Find $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}.$

126\.

Let $z = \text{sinh}\left( {2x + 3y} \right).$ Find $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}.$

127.

Let $f(x,y) = \text{arctan}\left( \frac{y}{x} \right).$ Evaluate $f_{x}(2,-2)$ and $f_{y}(2,-2).$

128\.

Let $f(x,y) = \frac{xy}{x - y}.$ Find $f_{x}(2,-2)$ and $f_{y}(2,-2).$

129.

Find $\frac{\partial z}{\partial x}$ at $\left( {0,1} \right)$ for $z = e^{\text{−}x}\text{cos}(y).$ Evaluate the partial derivatives at point $P(0,1).$

130\.

Given $f(x,y,z) = x^{3}yz^{2},$ find $\frac{\partial^{2}f}{\partial x\partial y}$ and $f_{z}(1,1,1).$

131.

Given $f(x,y,z) = 2\ \text{sin}\left( {x + y} \right),$ find $f_{x}\left( {0,\frac{\pi}{2},-4} \right),$ $f_{y}\left( {0,\frac{\pi}{2},-4} \right),$ and $f_{z}\left( {0,\frac{\pi}{2},-4} \right).$

132\.

The area of a parallelogram with adjacent side lengths that are $a\ \text{and}\ b,$ and in which the angle between these two sides is $\theta,$ is given by the function $A(a,b,\theta) = ba\ \text{sin}(\theta).$ Find the rate of change of the area of the parallelogram with respect to the following:

1. Side *a*

2. Side *b*

3. $\text{Angle}\ \theta$

133.

Express the volume of a right circular cylinder as a function of two variables:

1. its radius $r$ and its height $h.$

2. Show that the rate of change of the volume of the cylinder with respect to its radius is the product of its circumference multiplied by its height.

3. Show that the rate of change of the volume of the cylinder with respect to its height is equal to the area of the circular base.

134\.

Calculate $\frac{\partial w}{\partial z}$ for $w = z\ \text{sin}(xy^{2} + 2z).$

Find the indicated higher-order partial derivatives.

135.

$f_{xy}$ for $z = \text{ln}(x - y)$

136\.

$f_{yx}$ for $z = \text{ln}(x - y)$

137.

Let $z = x^{2} + 3xy + 2y^{2}.$ Find $\frac{\partial^{2}z}{\partial x^{2}}$ and $\frac{\partial^{2}z}{\partial y^{2}}.$

138\.

Given $z = e^{x}\text{tan}\ y,$ find $\frac{\partial^{2}z}{\partial x\partial y}$ and $\frac{\partial^{2}z}{\partial y\partial x}.$

139.

Given $f(x,y,z) = xyz,$ find $f_{xyy},f_{yxy},$ and $f_{yyx}.$

140\.

Given $f(x,y,z) = e^{-2x}\text{sin}\left( {z^{2}y} \right),$ show that $f_{xyy} = f_{yxy}.$

141.

Show that $z = \frac{1}{2}\left( {e^{y} - e^{\text{−}y}} \right)\text{sin}\ x$ is a solution of the differential equation $\frac{\partial^{2}z}{\partial x^{2}} + \frac{\partial^{2}z}{\partial y^{2}} = 0.$

142\.

Find $f_{xx}(x,y)$ for $f(x,y) = \frac{4x^{2}}{y} + \frac{y^{2}}{2x}.$

143.

Let $f(x,y,z) = x^{2}y^{3}z - 3xy^{2}z^{3} + 5x^{2}z - y^{3}z.$ Find $f_{xyz}.$

144\.

Let $F(x,y,z) = x^{3}yz^{2} - 2x^{2}yz + 3xz - 2y^{3}z.$ Find $F_{xyz}.$

145.

Given $f(x,y) = x^{2} + x - 3xy + y^{3} - 5,$ find all points at which $f_{x} = f_{y} = 0$ simultaneously.

146\.

Given $f(x,y) = 2x^{2} + 2xy + y^{2} + 2x - 3,$ find all points at which $\frac{\partial f}{\partial x} = 0$ and $\frac{\partial f}{\partial y} = 0$ simultaneously.

147.

Given $f(x,y) = y^{3} - 3yx^{2} - 3y^{2} - 3x^{2} + 1,$ find all points on $f$ at which $f_{x} = f_{y} = 0$ simultaneously.

148\.

Given $f(x,y) = 15x^{3} - 3xy + 15y^{3},$ find all points at which $f_{x}(x,y) = f_{y}(x,y) = 0$ simultaneously.

149.

Show that $z = e^{x}\text{sin}\ y$ satisfies the equation $\frac{\partial^{2}z}{\partial x^{2}} + \frac{\partial^{2}z}{\partial y^{2}} = 0.$

150\.

Show that $f(x,y) = \text{ln}\left( {x^{2} + y^{2}} \right)$ solves Laplace’s equation $\frac{\partial^{2}z}{\partial x^{2}} + \frac{\partial^{2}z}{\partial y^{2}} = 0.$

151.

Show that $z = e^{\text{−}t}\text{cos}\left( \frac{x}{c} \right)$ satisfies the heat equation $\frac{\partial z}{\partial t} = c^{2}\left( \frac{\partial^{2}z}{\partial x^{2}} \right)$

152\.

Find $\underset{\text{Δ}x\rightarrow 0}{\text{lim}}\frac{f(x + \text{Δ}x\text{,}y) - f(x,y)}{\text{Δ}x}$ for $f(x,y) = -7x - 2xy + 7y.$

153.

Find $\underset{\text{Δ}y\rightarrow 0}{\text{lim}}\frac{f(x,y + \text{Δ}y) - f(x,y)}{\text{Δ}y}$ for $f(x,y) = -7x - 2xy + 7y.$

154\.

Find $\underset{\text{Δ}x\rightarrow 0}{\text{lim}}\frac{\text{Δ}f}{\text{Δ}x} = \underset{\text{Δ}x\rightarrow 0}{\text{lim}}\frac{f(x + \text{Δ}x,y) - f(x,y)}{\text{Δ}x}$ for $f(x,y) = x^{2}y^{2} + xy + y.$

155.

Find $\underset{\text{Δ}x\rightarrow 0}{\text{lim}}\frac{\text{Δ}f}{\text{Δ}x} = \underset{\text{Δ}x\rightarrow 0}{\text{lim}}\frac{f(x + \text{Δ}x,y) - f(x,y)}{\text{Δ}x}$ for $f(x,y) = \text{sin}(xy).$

156\.

The function $P(T,V) = \frac{nRT}{V}$ gives the pressure at a point in a gas as a function of temperature $T$ and volume $V.$ The letters $n\ \text{and}\ R$ are constants. Find $\frac{\partial P}{\partial V}$ and $\frac{\partial P}{\partial T},$ and explain what these quantities represent.

157\.

The equation for heat flow in the $xy\text{-plane}$ is $\frac{\partial f}{\partial t} = \frac{\partial^{2}f}{\partial x^{2}} + \frac{\partial^{2}f}{\partial y^{2}}.$ Show that $f(x,y,t) = e^{-2t}\text{sin}\ x\ \text{sin}\ y$ is a solution.

158\.

The basic wave equation is $f_{tt} = f_{xx}.$ Verify that $f(x,t) = \text{sin}(x + t)$ and $f(x,t) = \text{sin}(x - t)$ are solutions.

159.

The law of cosines can be thought of as a function of three variables. Let $x,y,$ and $\theta$ be two sides of any triangle where the angle $\theta$ is the included angle between the two sides. Then, $F(x,y,\theta) = x^{2} + y^{2} - 2xy\ \text{cos}\ \theta$ gives the square of the third side of the triangle. Find $\frac{\partial F}{\partial\theta}$ and $\frac{\partial F}{\partial x}$ when $x = 2,y = 3,$ and $\theta = \frac{\pi}{6}.$

160\.

Suppose the sides of a rectangle are changing with respect to time. The first side is changing at a rate of $2$ in./sec whereas the second side is changing at the rate of $4$ in/sec. How fast is the diagonal of the rectangle changing when the first side measures $16$ in. and the second side measures $20$ in.? (Round answer to three decimal places.)

161.

A Cobb-Douglas production function is $f(x,y) = 200x^{0.7}y^{0.3},$ where $x\ \text{and}\ y$ represent the amount of labor and capital available. Let $x = 500$ and $y = 1000.$ Find $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ at these values, which represent the marginal productivity of labor and capital, respectively.

162\.

The apparent temperature index is a measure of how the temperature feels, and it is based on two variables: $h,$ which is relative humidity, and $t,$ which is the air temperature.

$A = 0.885t - 22.4h + 1.20th - 0.544.$ Find $\frac{\partial A}{\partial t}$ and $\frac{\partial A}{\partial h}$ when $t = 20\text{°}\text{F}$ and $h = 0.90.$

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4.4 Tangent Planes and Linear Approximations

> 来源: OpenStax《Calculus Volume 3》| 原页: https://openstax.org/books/calculus-volume-3/pages/4-4-tangent-planes-and-linear-approximations

4.4 Tangent Planes and Linear Approximations

In this section, we consider the problem of finding the tangent plane to a surface, which is analogous to finding the equation of a tangent line to a curve when the curve is defined by the graph of a function of one variable, $y = f(x).$ The slope of the tangent line at the point $x = a$ is given by $m = f\prime(a);$ what is the slope of a tangent plane? We learned about the equation of a plane in Equations of Lines and Planes in Space; in this section, we see how it can be applied to the problem at hand.

Tangent Planes

Intuitively, it seems clear that, in a plane, only one line can be tangent to a curve at a point. However, in three-dimensional space, many lines can be tangent to a given point. If these lines lie in the same plane, they determine the tangent plane at that point. A tangent plane at a regular point contains all of the lines tangent to that point. A more intuitive way to think of a tangent plane is to assume the surface is smooth at that point (no corners). Then, a tangent line to the surface at that point in any direction does not have any abrupt changes in slope because the direction changes smoothly.

Let $P_{0} = \left( {x_{0},y_{0},z_{0}} \right)$ be a point on a surface $S,$ and let $C$ be any curve passing through $P_{0}$ and lying entirely in $S.$ If the tangent lines to all such curves $C$ at $P_{0}$ lie in the same plane, then this plane is called the tangent plane to $S$ at $P_{0}$ (Figure 4.27).

For a tangent plane to a surface to exist at a point on that surface, it is sufficient for the function that defines the surface to be differentiable at that point, defined later in this section. We define the term tangent plane here and then explore the idea intuitively.

Let $S$ be a surface defined by a differentiable function $z = f\left( {x,y} \right),$ and let $P_{0} = \left( {x_{0},y_{0}} \right)$ be a point in the domain of $f.$ Then, the equation of the tangent plane to $S$ at $P_{0}$ is given by

$$z = f\left( {x_{0},y_{0}} \right) + f_{x}\left( {x_{0},y_{0}} \right)\left( {x - x_{0}} \right) + f_{y}\left( {x_{0},y_{0}} \right)\left( {y - y_{0}} \right).$$ (4.24)

To see why this formula is correct, let’s first find two tangent lines to the surface $S.$ The equation of the tangent line to the curve that is represented by the intersection of $S$ with the vertical trace given by $x = x_{0}$ is $z = f\left( {x_{0},y_{0}} \right) + f_{y}\left( {x_{0},y_{0}} \right)\left( {y - y_{0}} \right).$ Similarly, the equation of the tangent line to the curve that is represented by the intersection of $S$ with the vertical trace given by $y = y_{0}$ is $z = f\left( {x_{0},y_{0}} \right) + f_{x}\left( {x_{0},y_{0}} \right)\left( {x - x_{0}} \right).$ A parallel vector to the first tangent line is $\mathbf{a} = \mathbf{j} + f_{y}\left( {x_{0},y_{0}} \right)\mathbf{k;}$ a parallel vector to the second tangent line is $\mathbf{b} = \mathbf{i} + f_{x}\left( {x_{0},y_{0}} \right)\mathbf{k}.$ We can take the cross product of these two vectors:

$$\begin{array}{cl}

{\mathbf{a}\ \times \ \mathbf{b}} & {= \left( {\mathbf{j} + f_{y}\left( {x_{0},y_{0}} \right)\mathbf{k}} \right)\ \times \ \left( {\mathbf{i} + f_{x}\left( {x_{0},y_{0}} \right)\mathbf{k}} \right)} \\

& {= \left| \begin{array}{lll}

\mathbf{i} & \mathbf{j} & \mathbf{k} \\

0 & 1 & {f_{y}\left( {x_{0},y_{0}} \right)} \\

1 & 0 & {f_{x}\left( {x_{0},y_{0}} \right)}

\end{array} \right|} \\

& {= f_{x}\left( {x_{0},y_{0}} \right)\mathbf{i} + f_{y}\left( {x_{0},y_{0}} \right)\mathbf{j} - \mathbf{k}.}

\end{array}$$

This vector is perpendicular to both lines and is therefore perpendicular to the tangent plane. We can use this vector as a normal vector to the tangent plane, along with the point $P_{0} = \left( {x_{0},y_{0},f\left( {x_{0},y_{0}} \right)} \right)$ in the equation for a plane:

$$\begin{array}{rll}

{\mathbf{n} \cdot \left( {\left( {x - x_{0}} \right)\mathbf{i} + \left( {y - y_{0}} \right)\mathbf{j} + \left( {z - f\left( {x_{0},y_{0}} \right)} \right)\mathbf{k}} \right)} & = & 0 \\

{\left( {f_{x}\left( {x_{0},y_{0}} \right)\mathbf{i} + f_{y}\left( {x_{0},y_{0}} \right)\mathbf{j}\text{-}\mathbf{k}} \right) \cdot \left( {\left( {x - x_{0}} \right)\mathbf{i} + \left( {y - y_{0}} \right)\mathbf{j} + \left( {z - f\left( {x_{0},y_{0}} \right)} \right)\mathbf{k}} \right)} & = & 0 \\

{f_{x}\left( {x_{0},y_{0}} \right)\left( {x - x_{0}} \right) + f_{y}\left( {x_{0},y_{0}} \right)\left( {y - y_{0}} \right) - \left( {z - f\left( {x_{0},y_{0}} \right)} \right)} & = & {0.}

\end{array}$$

Solving this equation for $z$ gives Equation 4.24.

Finding a Tangent Plane

Find an equation of the tangent plane to the surface defined by the function $f\left( {x,y} \right) = 2x^{2} - 3xy + 8y^{2} + 2x - 4y + 4$ at point $\left( {2,-1} \right).$

Solution

First, we must calculate $f_{x}\left( {x,y} \right)$ and $f_{y}\left( {x,y} \right),$ then use Equation 4.24 with $x_{0} = 2$ and $y_{0} = -1\text{:}$

$$\begin{array}{rll}

{f_{x}\left( {x,y} \right)} & = & {4x - 3y + 2} \\

{f_{y}\left( {x,y} \right)} & = & {-3x + 16y - 4} \\

{f\left( {2,-1} \right)} & = & {2(2)^{2} - 3(2)(-1) + 8(-1)^{2} + 2(2) - 4(-1) + 4 = 34.} \\

{f_{x}\left( {2,-1} \right)} & = & {4(2) - 3(-1) + 2 = 13} \\

{f_{y}\left( {2,-1} \right)} & = & {-3(2) + 16(-1) - 4 = -26.}

\end{array}$$

Then Equation 4.24 becomes

$$\begin{array}{l}

{z = f\left( {x_{0},y_{0}} \right) + f_{x}\left( {x_{0},y_{0}} \right)\left( {x - x_{0}} \right) + f_{y}\left( {x_{0},y_{0}} \right)\left( {y - y_{0}} \right)} \\

{z = 34 + 13\left( {x - 2} \right) - 26\left( {y - (-1)} \right)} \\

{z = 34 + 13x - 26 - 26y - 26} \\

{z = 13x - 26y - 18.}

\end{array}$$

(See the following figure).

Find an equation of the tangent plane to the surface defined by the function $f\left( {x,y} \right) = x^{3} - x^{2}y + y^{2} - 2x + 3y - 2$ at point $\left( {-1,3} \right).$

Finding Another Tangent Plane

Find an equation of the tangent plane to the surface defined by the function $f\left( {x,y} \right) = \text{sin}\left( {2x} \right)\text{cos}\left( {3y} \right)$ at the point $\left( {{\pi\text{/}3},{\pi\text{/}4}} \right).$

Solution

First, calculate $f_{x}\left( {x,y} \right)$ and $f_{y}\left( {x,y} \right),$ then use Equation 4.24 with $x_{0} = {\pi\text{/}3}$ and $y_{0} = {\pi\text{/}4}\text{:}$

$$\begin{array}{rll}

{f_{x}\left( {x,y} \right)} & = & {2\ \text{cos}\left( {2x} \right)\text{cos}\left( {3y} \right)} \\

{f_{y}\left( {x,y} \right)} & = & {-3\ \text{sin}\left( {2x} \right)\text{sin}\left( {3y} \right)} \\

{f\left( {\frac{\pi}{3},\frac{\pi}{4}} \right)} & = & {\text{sin}\left( {2\left( \frac{\pi}{3} \right)} \right)\text{cos}\left( {3\left( \frac{\pi}{4} \right)} \right) = \left( \frac{\sqrt{3}}{2} \right)\left( {- \frac{\sqrt{2}}{2}} \right) = - \frac{\sqrt{6}}{4}} \\

{f_{x}\left( {\frac{\pi}{3},\frac{\pi}{4}} \right)} & = & {2\ \text{cos}\left( {2\left( \frac{\pi}{3} \right)} \right)\text{cos}\left( {3\left( \frac{\pi}{4} \right)} \right) = 2\left( {- \frac{1}{2}} \right)\left( {- \frac{\sqrt{2}}{2}} \right) = \frac{\sqrt{2}}{2}} \\

{f_{y}\left( {\frac{\pi}{3},\frac{\pi}{4}} \right)} & = & {-3\ \text{sin}\left( {2\left( \frac{\pi}{3} \right)} \right)\text{sin}\left( {3\left( \frac{\pi}{4} \right)} \right) = -3\left( \frac{\sqrt{3}}{2} \right)\left( \frac{\sqrt{2}}{2} \right) = - \frac{3\sqrt{6}}{4}.}

\end{array}$$

Then Equation 4.24 becomes

$$\begin{array}{l}

\\

{z = f\left( {x_{0},y_{0}} \right) + f_{x}\left( {x_{0},y_{0}} \right)\left( {x - x_{0}} \right) + f_{y}\left( {x_{0},y_{0}} \right)\left( {y - y_{0}} \right)} \\

\\

{z = - \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{2}\left( {x - \frac{\pi}{3}} \right) - \frac{3\sqrt{6}}{4}\left( {y - \frac{\pi}{4}} \right)} \\

{z = \frac{\sqrt{2}}{2}x - \frac{3\sqrt{6}}{4}y - \frac{\sqrt{6}}{4} - \frac{\pi\sqrt{2}}{6} + \frac{3\pi\sqrt{6}}{16}.}

\end{array}$$

A tangent plane to a surface does not always exist at every point on the surface. Consider the function

$$f(x,y) = \begin{cases}

\frac{xy}{\sqrt{x^{2} + y^{2}}} & {\left( {x,y} \right) \neq \left( {0,0} \right)} \\

0 & {\left( {x,y} \right) = \left( {0,0} \right).}

\end{cases}$$

The graph of this function follows.

If either $x = 0$ or $y = 0,$ then $f\left( {x,y} \right) = 0,$ so the value of the function does not change on either the *x*- or *y*-axis. Therefore, $f_{x}\left( {x,0} \right) = f_{y}\left( {0,y} \right) = 0,$ so as either $x\ or\ y$ approach zero, these partial derivatives stay equal to zero. Substituting them into Equation 4.24 gives $z = 0$ as the equation of the tangent line. However, if we approach the origin from a different direction, we get a different story. For example, suppose we approach the origin along the line $y = x.$ If we put $y = x$ into the original function, it becomes

$$f\left( {x,x} \right) = \frac{x(x)}{\sqrt{x^{2} + (x)^{2}}} = \frac{x^{2}}{\sqrt{2x^{2}}} = \frac{|x|}{\sqrt{2}}.$$

When $x > 0,$ the slope of this curve is equal to ${\sqrt{2}\text{/}2};$ when $x < 0,$ the slope of this curve is equal to $\text{−}\left( {\sqrt{2}\text{/}2} \right).$ This presents a problem. In the definition of *tangent plane*, we presumed that all tangent lines through point $P$ (in this case, the origin) lay in the same plane. This is clearly not the case here. When we study differentiable functions, we will see that this function is not differentiable at the origin.

Linear Approximations

Recall from Linear Approximations and Differentials that the formula for the linear approximation of a function $f(x)$ at the point $x = a$ is given by

$$y \approx f(a) + f\prime(a)\left( {x - a} \right).$$

The diagram for the linear approximation of a function of one variable appears in the following graph.

The tangent line can be used as an approximation to the function $f(x)$ for values of $x$ reasonably close to $x = a.$ When working with a function of two variables, the tangent line is replaced by a tangent plane, but the approximation idea is much the same.

Given a function $z = f\left( {x,y} \right)$ with continuous partial derivatives that exist at the point $\left( {x_{0},y_{0}} \right),$ the linear approximation of $f$ at the point $\left( {x_{0},y_{0}} \right)$ is given by the equation

$$L\left( {x,y} \right) = f\left( {x_{0},y_{0}} \right) + f_{x}\left( {x_{0},y_{0}} \right)\left( {x - x_{0}} \right) + f_{y}\left( {x_{0},y_{0}} \right)\left( {y - y_{0}} \right).$$ (4.25)

Notice that this equation also represents the tangent plane to the surface defined by $z = f\left( {x,y} \right)$ at the point $\left( {x_{0},y_{0}} \right).$ The idea behind using a linear approximation is that, if there is a point $\left( {x_{0},y_{0}} \right)$ at which the precise value of $f\left( {x,y} \right)$ is known, then for values of $\left( {x,y} \right)$ reasonably close to $\left( {x_{0},y_{0}} \right),$ the linear approximation (i.e., tangent plane) yields a value that is also reasonably close to the exact value of $f\left( {x,y} \right)$ (Figure 4.31). Furthermore the plane that is used to find the linear approximation is also the tangent plane to the surface at the point $\left( {x_{0},y_{0}} \right).$

Using a Tangent Plane Approximation

Given the function $f\left( {x,y} \right) = \sqrt{41 - 4x^{2} - y^{2}},$ approximate $f\left( {2.1,2.9} \right)$ using point $\left( {2,3} \right)$ for $\left( {x_{0},y_{0}} \right).$ What is the approximate value of $f\left( {2.1,2.9} \right)$ to four decimal places?

Solution

To apply Equation 4.25, we first must calculate $f\left( {x_{0},y_{0}} \right),$ $f_{x}\left( {x_{0},y_{0}} \right),$ and $f_{y}\left( {x_{0},y_{0}} \right)$ using $x_{0} = 2$ and $y_{0} = 3\text{:}$

$$\begin{array}{rll}

{f\left( {x_{0},y_{0}} \right)} & = & {f\left( {2,3} \right) = \sqrt{41 - 4(2)^{2} - (3)^{2}} = \sqrt{41 - 16 - 9} = \sqrt{16} = 4} \\

{f_{x}\left( {x,y} \right)} & = & {- \frac{4x}{\sqrt{41 - 4x^{2} - y^{2}}}\ \text{so}\ f_{x}\left( {x_{0},y_{0}} \right) = - \frac{4(2)}{\sqrt{41 - 4(2)^{2} - (3)^{2}}} = -2} \\

{f_{y}\left( {x,y} \right)} & = & {- \frac{y}{\sqrt{41 - 4x^{2} - y^{2}}}\ \text{so}\ f_{y}\left( {x_{0},y_{0}} \right) = - \frac{3}{\sqrt{41 - 4(2)^{2} - (3)^{2}}} = - \frac{3}{4}.}

\end{array}$$

Now we substitute these values into Equation 4.25:

$$\begin{array}{cl}

{L\left( {x,y} \right)} & {= f\left( {x_{0},y_{0}} \right) + f_{x}\left( {x_{0},y_{0}} \right)\left( {x - x_{0}} \right) + f_{y}\left( {x_{0},y_{0}} \right)\left( {y - y_{0}} \right)} \\

& {= 4 - 2\left( {x - 2} \right) - \frac{3}{4}\left( {y - 3} \right)} \\

& {= \frac{41}{4} - 2x - \frac{3}{4}y.}

\end{array}$$

Last, we substitute $x = 2.1$ and $y = 2.9$ into $L\left( {x,y} \right)\text{:}$

$$L\left( {2.1,2.9} \right) = \frac{41}{4} - 2(2.1) - \frac{3}{4}(2.9) = 10.25 - 4.2 - 2.175 = 3.875.$$

The approximate value of $f\left( {2.1,2.9} \right)$ to four decimal places is

$$f\left( {2.1,2.9} \right) = \sqrt{41 - 4(2.1)^{2} - (2.9)^{2}} = \sqrt{14.95} \approx 3.8665,$$

which corresponds to a $0.2\text{\%}$ error in approximation.

Given the function $f\left( {x,y} \right) = e^{5 - 2x + 3y},$ approximate $f\left( {4.1,0.9} \right)$ using point $\left( {4,1} \right)$ for $\left( {x_{0},y_{0}} \right).$ What is the approximate value of $f\left( {4.1,0.9} \right)$ to four decimal places?

Differentiability

When working with a function $y = f(x)$ of one variable, the function is said to be differentiable at a point $x = a$ if $f^{\prime}(a)$ exists. Furthermore, if a function of one variable is differentiable at a point, the graph is “smooth” at that point (i.e., no corners exist) and a tangent line is well-defined at that point.

The idea behind differentiability of a function of two variables is connected to the idea of smoothness at that point. In this case, a surface is considered to be smooth at point $P$ if a tangent plane to the surface exists at that point. If a function is differentiable at a point, then a tangent plane to the surface exists at that point. Recall the formula for a tangent plane at a point $\left( {x_{0},y_{0}} \right)$ is given by

$$z = f\left( {x_{0},y_{0}} \right) + f_{x}\left( {x_{0},y_{0}} \right)\left( {x - x_{0}} \right) + f_{y}\left( {x_{0},y_{0}} \right)\left( {y - y_{0}} \right),$$

For a tangent plane to exist at the point $\left( {x_{0},y_{0}} \right),$ the partial derivatives must therefore exist at that point. However, this is not a sufficient condition for smoothness, as was illustrated in Figure 4.29. In that case, the partial derivatives existed at the origin, but the function also had a corner on the graph at the origin.

A function $f\left( {x,y} \right)$ is differentiable at a point $P\left( {x_{0},y_{0}} \right)$ if, for all points $\left( {x,y} \right)$ in a $\delta$ disk around $P,$ we can write

$$f\left( {x,y} \right) = f\left( {x_{0},y_{0}} \right) + f_{x}\left( {x_{0},y_{0}} \right)\left( {x - x_{0}} \right) + f_{y}\left( {x_{0},y_{0}} \right)\left( {y - y_{0}} \right) + E\left( {x,y} \right),$$ (4.26)

where the error term $E$ satisfies

$$\underset{{({x,y})}\rightarrow{({x_{0},y_{0}})}}{\text{lim}}\frac{E\left( {x,y} \right)}{\sqrt{\left( {x - x_{0}} \right)^{2} + \left( {y - y_{0}} \right)^{2}}} = 0.$$

The last term in Equation 4.26 is referred to as the *error term* and it represents how closely the tangent plane comes to the surface in a small neighborhood $(\delta$ disk) of point $P.$ For the function $f$ to be differentiable at $P,$ the function must be smooth—that is, the graph of $f$ must be close to the tangent plane for points near $P.$

Demonstrating Differentiability

Show that the function $f\left( {x,y} \right) = 2x^{2} - 4y$ is differentiable at point $\left( {2,-3} \right).$

Solution

First, we calculate $f\left( {x_{0},y_{0}} \right),f_{x}\left( {x_{0},y_{0}} \right),\ \text{and}\ f_{y}\left( {x_{0},y_{0}} \right)$ using $x_{0} = 2$ and $y_{0} = -3,$ then we use Equation 4.26:

$$\begin{array}{rll}

{f\left( {2,-3} \right)} & = & {2(2)^{2} - 4(-3) = 8 + 12 = 20} \\

{f_{x}\left( {2,-3} \right)} & = & {4(2) = 8} \\

{f_{y}\left( {2,-3} \right)} & = & -4.

\end{array}$$

Therefore $m_{1} = 8$ and $m_{2} = -4,$ and Equation 4.26 becomes

$$\begin{array}{rll}

{f\left( {x,y} \right)} & = & {f\left( {2,-3} \right) + f_{x}\left( {2,-3} \right)\left( {x - 2} \right) + f_{y}\left( {2,-3} \right)\left( {y + 3} \right) + E\left( {x,y} \right)} \\

{2x^{2} - 4y} & = & {20 + 8\left( {x - 2} \right) - 4\left( {y + 3} \right) + E\left( {x,y} \right)} \\

{2x^{2} - 4y} & = & {20 + 8x - 16 - 4y - 12 + E\left( {x,y} \right)} \\

{2x^{2} - 4y} & = & {8x - 4y - 8 + E\left( {x,y} \right)} \\

{E\left( {x,y} \right)} & = & {2x^{2} - 8x + 8.}

\end{array}$$

Next, we calculate $\underset{{({x,y})}\rightarrow{({x_{0},y_{0}})}}{\text{lim}}\frac{E\left( {x,y} \right)}{\sqrt{\left( {x - x_{0}} \right)^{2} + \left( {y - y_{0}} \right)^{2}}}\text{:}$

$$\begin{array}{cl}

{\underset{{({x,y})}\rightarrow{({x_{0},y_{0}})}}{\text{lim}}\frac{E\left( {x,y} \right)}{\sqrt{\left( {x - x_{0}} \right)^{2} + \left( {y - y_{0}} \right)^{2}}}} & {= \underset{{({x,y})}\rightarrow{({2,-3})}}{\text{lim}}\frac{2x^{2} - 8x + 8}{\sqrt{\left( {x - 2} \right)^{2} + \left( {y + 3} \right)^{2}}}} \\

& {= \underset{{({x,y})}\rightarrow{({2,-3})}}{\text{lim}}\frac{2\left( {x^{2} - 4x + 4} \right)}{\sqrt{\left( {x - 2} \right)^{2} + \left( {y + 3} \right)^{2}}}} \\

& {= \underset{{({x,y})}\rightarrow{({2,-3})}}{\text{lim}}\frac{2\left( {x - 2} \right)^{2}}{\sqrt{\left( {x - 2} \right)^{2} + \left( {y + 3} \right)^{2}}}} \\

& {\leq \underset{{({x,y})}\rightarrow{({2,-3})}}{\text{lim}}\frac{2\left( {\left( {x - 2} \right)^{2} + \left( {y + 3} \right)^{2}} \right)}{\sqrt{\left( {x - 2} \right)^{2} + \left( {y + 3} \right)^{2}}}} \\

& {= \underset{{({x,y})}\rightarrow{({2,-3})}}{\text{lim}}2\sqrt{\left( {x - 2} \right)^{2} + \left( {y + 3} \right)^{2}}} \\

& {= 0.}

\end{array}$$

Since $E\left( {x,y} \right) \geq 0$ for any value of $x\ \text{or}\ y,$ the original limit must be equal to zero. Therefore, $f\left( {x,y} \right) = 2x^{2} - 4y$ is differentiable at point $\left( {2,-3} \right).$

Show that the function $f\left( {x,y} \right) = 3x - 4y^{2}$ is differentiable at point $\left( {-1,2} \right).$

The function $f\left( {x,y} \right) = \begin{cases}

\frac{xy}{\sqrt{x^{2} + y^{2}}} & {\left( {x,y} \right) \neq \left( {0,0} \right)} \\

0 & {\left( {x,y} \right) = \left( {0,0} \right)}

\end{cases}$ is not differentiable at the origin. We can see this by calculating the partial derivatives. This function appeared earlier in the section, where we showed that $f_{x}\left( {0,0} \right) = f_{y}\left( {0,0} \right) = 0.$ Substituting this information into Equation 4.26 using $x_{0} = 0$ and $y_{0} = 0,$ we get

$$\begin{array}{rll}

{f\left( {x,y} \right)} & = & {f\left( {0,0} \right) + f_{x}\left( {0,0} \right)\left( {x - 0} \right) + f_{y}\left( {0,0} \right)\left( {y - 0} \right) + E\left( {x,y} \right)} \\

{E\left( {x,y} \right)} & = & {\frac{xy}{\sqrt{x^{2} + y^{2}}}.}

\end{array}$$

Calculating $\underset{{({x,y})}\rightarrow{({x_{0},y_{0}})}}{\text{lim}}\frac{E\left( {x,y} \right)}{\sqrt{\left( {x - x_{0}} \right)^{2} + \left( {y - y_{0}} \right)^{2}}}$ gives

$$\begin{array}{cl}

{\underset{{({x,y})}\rightarrow{({x_{0},y_{0}})}}{\text{lim}}\frac{E\left( {x,y} \right)}{\sqrt{\left( {x - x_{0}} \right)^{2} + \left( {y - y_{0}} \right)^{2}}}} & {= \underset{{({x,y})}\rightarrow{({0,0})}}{\text{lim}}\frac{\frac{xy}{\sqrt{x^{2} + y^{2}}}}{\sqrt{x^{2} + y^{2}}}} \\

& {= \underset{{({x,y})}\rightarrow{({0,0})}}{\text{lim}}\frac{xy}{x^{2} + y^{2}}.}

\end{array}$$

Depending on the path taken toward the origin, this limit takes different values. Therefore, the limit does not exist and the function $f$ is not differentiable at the origin as shown in the following figure.

Differentiability and continuity for functions of two or more variables are connected, the same as for functions of one variable. In fact, with some adjustments of notation, the basic theorem is the same.

Differentiability Implies Continuity

Let $z = f\left( {x,y} \right)$ be a function of two variables with $\left( {x_{0},y_{0}} \right)$ in the domain of $f.$ If $f\left( {x,y} \right)$ is differentiable at $\left( {x_{0},y_{0}} \right),$ then $f\left( {x,y} \right)$ is continuous at $\left( {x_{0},y_{0}} \right).$

Differentiability Implies Continuity shows that if a function is differentiable at a point, then it is continuous there. However, if a function is continuous at a point, then it is not necessarily differentiable at that point. For example,

$$f\left( {x,y} \right) = \begin{cases}

\frac{xy}{\sqrt{x^{2} + y^{2}}} & {\left( {x,y} \right) \neq \left( {0,0} \right)} \\

0 & {\left( {x,y} \right) = \left( {0,0} \right)}

\end{cases}$$

is continuous at the origin, but it is not differentiable at the origin. This observation is also similar to the situation in single-variable calculus.

Continuity of First Partials Implies Differentiability further explores the connection between continuity and differentiability at a point. This theorem says that if the function and its partial derivatives are continuous at a point, the function is differentiable.

Continuity of First Partials Implies Differentiability

Let $z = f\left( {x,y} \right)$ be a function of two variables with $\left( {x_{0},y_{0}} \right)$ in the domain of $f.$ If $f\left( {x,y} \right),$ $f_{x}\left( {x,y} \right),$ and $f_{y}\left( {x,y} \right)$ all exist in a neighborhood of $\left( {x_{0},y_{0}} \right)$ and are continuous at $\left( {x_{0},y_{0}} \right),$ then $f\left( {x,y} \right)$ is differentiable there.

Recall that earlier we showed that the function

$$f\left( {x,y} \right) = \begin{cases}

\frac{xy}{\sqrt{x^{2} + y^{2}}} & {\left( {x,y} \right) \neq \left( {0,0} \right)} \\

0 & {\left( {x,y} \right) = \left( {0,0} \right)}

\end{cases}$$

was not differentiable at the origin. Let’s calculate the partial derivatives $f_{x}$ and $f_{y}\text{:}$

$$\frac{\partial f}{\partial x} = \frac{y^{3}}{\left( {x^{2} + y^{2}} \right)^{3\text{/}2}}\quad\text{and}\quad\frac{\partial f}{\partial y} = \frac{x^{3}}{\left( {x^{2} + y^{2}} \right)^{3\text{/}2}}.$$

The contrapositive of the preceding theorem states that if a function is not differentiable, then at least one of the hypotheses must be false. Let’s explore the condition that $f_{x}\left( {0,0} \right)$ must be continuous. For this to be true, it must be true that $\underset{{({x,y})}\rightarrow{({0,0})}}{\text{lim}}f_{x}\left( {0,0} \right) = f_{x}\left( {0,0} \right)\text{:}$

$$\underset{{({x,y})}\rightarrow{({0,0})}}{\text{lim}}f_{x}\left( {x,y} \right) = \underset{{({x,y})}\rightarrow{({0,0})}}{\text{lim}}\frac{y^{3}}{\left( {x^{2} + y^{2}} \right)^{3\text{/}2}}.$$

Let $x = ky.$ Then

$$\begin{array}{cl}

{\underset{{({x,y})}\rightarrow{({0,0})}}{\text{lim}}\frac{y^{3}}{\left( {x^{2} + y^{2}} \right)^{3\text{/}2}}} & {= \underset{y\rightarrow 0}{\text{lim}}\frac{y^{3}}{\left( {\left( {ky} \right)^{2} + y^{2}} \right)^{3\text{/}2}}} \\

& {= \underset{y\rightarrow 0}{\text{lim}}\frac{y^{3}}{\left( {k^{2}y^{2} + y^{2}} \right)^{3\text{/}2}}} \\

& {= \underset{y\rightarrow 0}{\text{lim}}\frac{y^{3}}{|y|^{3}\left( {k^{2} + 1} \right)^{3\text{/}2}}} \\

& {= \frac{1}{\left( {k^{2} + 1} \right)^{3\text{/}2}}\underset{y\rightarrow 0}{\text{lim}}\frac{|y|}{y}.}

\end{array}$$

If $y > 0,$ then this expression equals ${1\text{/}\left( {k^{2} + 1} \right)^{3\text{/}2}};$ if $y < 0,$ then it equals $\text{−}\left( {1\text{/}\left( {k^{2} + 1} \right)^{3\text{/}2}} \right).$ In either case, the value depends on $k,$ so the limit fails to exist.

Differentials

In Linear Approximations and Differentials we first studied the concept of differentials. The differential of $y,$ written $dy,$ is defined as $f^{\prime}(x)dx.$ The differential is used to approximate $\text{Δ}y = f\left( {x + \text{Δ}x} \right) - f(x),$ where $\text{Δ}x = dx.$ Extending this idea to the linear approximation of a function of two variables at the point $\left( {x_{0},y_{0}} \right)$ yields the formula for the total differential for a function of two variables.

Let $z = f\left( {x,y} \right)$ be a function of two variables with $\left( {x_{0},y_{0}} \right)$ in the domain of $f,$ and let $\text{Δ}x$ and $\text{Δ}y$ be chosen so that $\left( {x_{0} + \text{Δ}x,y_{0} + \text{Δ}y} \right)$ is also in the domain of $f.$ If $f$ is differentiable at the point $\left( {x_{0},y_{0}} \right),$ then the differentials $dx$ and $dy$ are defined as

$$dx = \text{Δ}x\ \text{and}\ dy = \text{Δ}y.$$

The differential $dz,$ also called the total differential of $z = f\left( {x,y} \right)$ at $\left( {x_{0},y_{0}} \right),$ is defined as

$$dz = f_{x}\left( {x_{0},y_{0}} \right)dx + f_{y}\left( {x_{0},y_{0}} \right)dy.$$ (4.27)

Notice that the symbol $\partial$ is not used to denote the total differential; rather, $d$ appears in front of $z.$ Now, let’s define $\text{Δ}z = f\left( {x + \text{Δ}x,y + \text{Δ}y} \right) - f\left( {x,y} \right).$ We use $dz$ to approximate $\text{Δ}z,$ so

$$\text{Δ}z \approx dz = f_{x}\left( {x_{0},y_{0}} \right)dx + f_{y}\left( {x_{0},y_{0}} \right)dy.$$

Therefore, the differential is used to approximate the change in the function $z = f\left( {x_{0},y_{0}} \right)$ at the point $\left( {x_{0},y_{0}} \right)$ for given values of $\text{Δ}x$ and $\text{Δ}y.$ Since $\text{Δ}z = f\left( {x + \text{Δ}x,y + \text{Δ}y} \right) - f\left( {x,y} \right),$ this can be used further to approximate $f\left( {x + \text{Δ}x,y + \text{Δ}y} \right)\text{:}$

$$\begin{array}{cl}

{f\left( {x + \text{Δ}x,y + \text{Δ}y} \right)} & {= f\left( {x,y} \right) + \text{Δ}z} \\

& {\approx f\left( {x,y} \right) + f_{x}\left( {x_{0},y_{0}} \right)\text{Δ}x + f_{y}\left( {x_{0},y_{0}} \right)\text{Δ}y.}

\end{array}$$

See the following figure.

One such application of this idea is to determine error propagation. For example, if we are manufacturing a gadget and are off by a certain amount in measuring a given quantity, the differential can be used to estimate the error in the total volume of the gadget.

Approximation by Differentials

Find the differential $dz$ of the function $f\left( {x,y} \right) = 3x^{2} - 2xy + y^{2}$ and use it to approximate $\text{Δ}z$ at point $\left( {2,-3} \right).$ Use $\text{Δ}x = 0.1$ and $\text{Δ}y = -0.05.$ What is the exact value of $\text{Δ}z?$

Solution

First, we must calculate $f\left( {x_{0},y_{0}} \right),f_{x}\left( {x_{0},y_{0}} \right),\ \text{and}\ f_{y}\left( {x_{0},y_{0}} \right)$ using $x_{0} = 2$ and $y_{0} = -3\text{:}$

$$\begin{array}{rll}

{f\left( {x_{0},y_{0}} \right)} & = & {f\left( {2,-3} \right) = 3(2)^{2} - 2(2)(-3) + (-3)^{2} = 12 + 12 + 9 = 33} \\

{f_{x}\left( {x,y} \right)} & = & {6x - 2y} \\

{f_{y}\left( {x,y} \right)} & = & {-2x + 2y} \\

{f_{x}\left( {x_{0},y_{0}} \right)} & = & {f_{x}\left( {2,-3} \right) = 6(2) - 2(-3) = 12 + 6 = 18} \\

{f_{y}\left( {x_{0},y_{0}} \right)} & = & {f_{y}\left( {2,-3} \right) = -2(2) + 2(-3) = -4 - 6 = -10.}

\end{array}$$

Then, we substitute these quantities into Equation 4.27:

$$\begin{array}{l}

{dz = f_{x}\left( {x_{0},y_{0}} \right)dx + f_{y}\left( {x_{0},y_{0}} \right)dy} \\

{dz = 18(0.1) - 10(-0.05) = 1.8 + 0.5 = 2.3.}

\end{array}$$

This is the approximation to $\text{Δ}z = f\left( {x_{0} + \text{Δ}x,y_{0} + \text{Δ}y} \right) - f\left( {x_{0},y_{0}} \right).$ The exact value of $\text{Δ}z$ is given by

$$\begin{array}{cl}

{\text{Δ}z} & {= f\left( {x_{0} + \text{Δ}x,y_{0} + \text{Δ}y} \right) - f\left( {x_{0},y_{0}} \right)} \\

& {= f\left( {2 + 0.1,-3 - 0.05} \right) - f\left( {2,-3} \right)} \\

& {= f\left( {2.1,-3.05} \right) - f\left( {2,-3} \right)} \\

& {= 2.3425.}

\end{array}$$

Find the differential $dz$ of the function $f\left( {x,y} \right) = 4y^{2} + x^{2}y - 2xy$ and use it to approximate $\text{Δ}z$ at point $\left( {1,-1} \right).$ Use $\text{Δ}x = 0.03$ and $\text{Δ}y = -0.02.$ What is the exact value of $\text{Δ}z?$

Differentiability of a Function of Three Variables

All of the preceding results for differentiability of functions of two variables can be generalized to functions of three variables. First, the definition:

A function $f\left( {x,y,z} \right)$ is differentiable at a point $P\left( {x_{0},y_{0},z_{0}} \right)$ if for all points $\left( {x,y,z} \right)$ in a $\delta$ disk around $P$ we can write

$$\begin{array}{cl}

{f\left( {x,y,z} \right)} & {= f\left( {x_{0},y_{0},z_{0}} \right) + f_{x}\left( {x_{0},y_{0},z_{0}} \right)\left( {x - x_{0}} \right) + f_{y}\left( {x_{0},y_{0},z_{0}} \right)\left( {y - y_{0}} \right)} \\

& {\mspace{9mu} + f_{z}\left( {x_{0},y_{0},z_{0}} \right)\left( {z - z_{0}} \right) + E\left( {x,y,z} \right),}

\end{array}$$ (4.28)

where the error term *E* satisfies

$$\underset{{({x,y,z})}\rightarrow{({x_{0},y_{0},z_{0}})}}{\text{lim}}\frac{E\left( {x,y,z} \right)}{\sqrt{\left( {x - x_{0}} \right)^{2} + \left( {y - y_{0}} \right)^{2} + \left( {z - z_{0}} \right)^{2}}} = 0.$$

If a function of three variables is differentiable at a point $\left( {x_{0},y_{0},z_{0}} \right),$ then it is continuous there. Furthermore, continuity of first partial derivatives at that point guarantees differentiability.

Section 4.4 Exercises

For the following exercises, find a unit normal vector to the surface at the indicated point.

163.

$z = {f(x,y) = x^{3},(2,-1,8)}$

164\.

$\text{ln}\left( \frac{x}{y - z} \right) = 0$ when $x = y = 1$

For the following exercises, as a useful review for techniques used in this section, find a normal vector and a tangent vector at point $P.$

165.

$x^{2} + xy + y^{2} = 3,P(-1,-1)$

166\.

$\left( {x^{2} + y^{2}} \right)^{2} = 9\left( {x^{2} - y^{2}} \right),P(\sqrt{2},1)$

167.

$xy^{2} - 2x^{2} + y + 5x = 6,P(4,2)$

168\.

$2x^{3} - x^{2}y^{2} = 3x - y - 7,P(1,-2)$

169.

$ze^{x^{2} - y^{2}} - 3 = 0,$ $P(2,2,3)$

For the following exercises, find the equation for the tangent plane to the surface at the indicated point. (*Hint:* Solve for $z$ in terms of $x$ and $y.)$

170\.

$-8x - 3y - 7z = -19,P(1,-1,2)$

171.

$z = -9x^{2} - 3y^{2},P(2,1,-39)$

172\.

$x^{2} + 10xyz + y^{2} + 8z^{2} = 0,P(-1,-1,-1)$

173.

$z = \text{ln}(10x^{2} + 2y^{2} + 1),P(0,0,0)$

174\.

$z = e^{7x^{2} + 4y^{2}},$ $P(0,0,1)$

175.

$xy + yz + zx = 11,P(1,2,3)$

176\.

$x^{2} + 4y^{2} = z^{2},P(3,2,5)$

177.

$x^{3} + y^{3} = 3xyz,P\left( {1,2,\frac{3}{2}} \right)$

178\.

$z = axy,P\left( {1,\frac{1}{a},1} \right)$

179.

$z = \text{sin}\ x + \text{sin}\ y + \text{sin}(x + y),P(0,0,0)$

180\.

$z = {h(x,y) = \text{ln}\sqrt{x^{2} + y^{2}},P(3,4)}$

181.

$z = x^{2} - 2xy + y^{2},P(1,2,1)$

For the following exercises, find parametric equations for the normal line to the surface at the indicated point. (Recall that to find the equation of a line in space, you need a point on the line, $P_{0}\left( {x_{0,}y_{0},z_{0}} \right),$ and a vector $\mathbf{n} = \left\langle {a,b,c} \right\rangle$ that is parallel to the line. Then the equation of the line is $x - x_{0} = at,y - y_{0} = bt,z - z_{0} = ct.)$

182\.

$-3x + 9y + 4z = -4,P(1,-1,2)$

183.

$z = 5x^{2} - 2y^{2},P(2,1,18)$

184\.

$x^{2} - 8xyz + y^{2} + 6z^{2} = 0,P(1,1,1)$

185.

$z = \text{ln}\left( {3x^{2} + 7y^{2} + 1} \right),P(0,0,0)$

186\.

$z = e^{4x^{2} + 6y^{2}},P(0,0,1)$

187.

$z = x^{2} - 2xy + y^{2}$ at point $P(1,2,1)$

For the following exercises, use the figure shown here.

188\.

The length of line segment $AC$ is equal to what mathematical expression?

189.

The length of line segment $BC$ is equal to what mathematical expression?

190\.

Using the figure, explain what the length of line segment $AB$ represents.

For the following exercises, complete each task.

191.

Show that $f(x,y) = e^{xy}x$ is differentiable at point $(1,0).$

192\.

Find the total differential of the function $w = e^{y}\text{cos}(x) + z^{2}.$

193.

Show that $f(x,y) = x^{2} + 3y$ is differentiable at every point. In other words, show that $\text{Δ}z = f(x + \text{Δ}x,y + \text{Δ}y) - f(x,y) = f_{x}\text{Δ}x + f_{y}\text{Δ}y + \varepsilon_{1}\text{Δ}x + \varepsilon_{2}\text{Δ}y,$ where both $\varepsilon_{1}$ and $\varepsilon_{2}$ approach zero as $\left( {\text{Δ}x,\text{Δ}y} \right)$ approaches $\left( {0,0} \right).$

194\.

Find the total differential of the function $z = \frac{xy}{y + x}$ where $x$ changes from $10\ \text{to}\ 10.5$ and $y$ changes from $15\ \text{to}\ 13.$

195.

Let $z = f(x,y) = xe^{y}.$ Compute $\text{Δ}z$ from $P(1,2)$ to $Q(1.05,2.1)$ and then find the approximate change in $z$ from point $P$ to point $Q.$ Recall $\text{Δ}z = f(x + \text{Δ}x,y + \text{Δ}y) - f(x,y),$ and $dz$ and $\text{Δ}z$ are approximately equal.

196\.

The volume of a right circular cylinder is given by $V(r,h) = \pi r^{2}h.$ Find the differential $dV.$ Interpret the formula geometrically.

197.

See the preceding problem. Use differentials to estimate the volume of aluminum in an enclosed aluminum can with diameter $8.0\ \text{cm}$ and height $12\ \text{cm}$ if the aluminum is $0.04$ cm thick.

198\.

Use the differential $dz$ to approximate the change in $z = \sqrt{4 - x^{2} - y^{2}}$ as $(x,y)$ moves from point $(1,1)$ to point $(1.01,0.97).$ Compare this approximation with the actual change in the function.

199.

Let $z = f(x,y) = x^{2} + 3xy - y^{2}.$ Find the exact change in the function and the approximate change in the function as $x$ changes from $2.00\ \text{to}\ 2.05$ and $y$ changes from $3.00\ \text{to}\ 2.96.$

200\.

The centripetal acceleration of a particle moving in a circle is given by $a(r,v) = \frac{v^{2}}{r},$ where $v$ is the velocity and $r$ is the radius of the circle. Approximate the maximum percent error in measuring the acceleration resulting from errors of $3\text{\%}$ in $v$ and $2\text{\%}$ in $r.$ (Recall that the percentage error is the ratio of the amount of error over the original amount. So, in this case, the percentage error in $a$ is given by $\frac{da}{a}.)$

201.

The radius $r$ and height $h$ of a right circular cylinder are measured with possible errors of $4\text{\%}\ \text{and}\ 5\text{\%},$ respectively. Approximate the maximum possible percentage error in measuring the volume (Recall that the percentage error is the ratio of the amount of error over the original amount. So, in this case, the percentage error in $V$ is given by $\frac{dV}{V}.)$

202\.

The base radius and height of a right circular cone are measured as $10$ in. and $25$ in., respectively, with a possible error in measurement of as much as $0.1$ in. each. Use differentials to estimate the maximum error in the calculated volume of the cone.

203.

The electrical resistance $R$ produced by wiring resistors $R_{1}$ and $R_{2}$ in parallel can be calculated from the formula $\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}}.$ If $R_{1}$ and $R_{2}$ are measured to be $7\text{Ω}$ and $6\text{Ω},$ respectively, and if these measurements are accurate to within $0.05\text{Ω},$ estimate the maximum possible error in computing $R.$ (The symbol $\text{Ω}$ represents an ohm, the unit of electrical resistance.)

204\.

The area of an ellipse with axes of length $2a$ and $2b$ is given by the formula

$A = \pi ab.$ Approximate the percent change in the area when $a$ increases by $2\text{\%}$ and $b$ increases by $1.5\text{\%}.$

205.

The period $T$ of a simple pendulum with small oscillations is calculated from the formula $T = 2\pi\sqrt{\frac{L}{g}},$ where $L$ is the length of the pendulum and $g$ is the acceleration resulting from gravity. Suppose that $L$ and $g$ have errors of, at most, $0.5\text{\%}$ and $0.1\text{\%},$ respectively. Use differentials to approximate the maximum percentage error in the calculated value of $T.$

206\.

Electrical power $P$ is given by $P = \frac{V^{2}}{R},$ where $V$ is the voltage and $R$ is the resistance. Approximate the maximum percentage error in calculating power if $120$ $V$ is applied to a $2000 - \text{Ω}$ resistor and the possible percent errors in measuring $V$ and $R$ are $3\text{\%}$ and $4\text{\%},$ respectively.

For the following exercises, find the linear approximation of each function at the indicated point.

207.

$f(x,y) = x\sqrt{y},\quad P(1,4)$

208\.

$f(x,y) = e^{x}\text{cos}\ y;P(0,0)$

209.

$f(x,y) = \text{arctan}(x + 2y),P(1,0)$

210\.

$f(x,y) = \sqrt{20 - x^{2} - 7y^{2}},\quad P(2,1)$

211.

$f(x,y,z) = \sqrt{x^{2} + y^{2} + z^{2}},\quad P(3,2,6)$

212\.

\[T\] Find an equation of the tangent plane to the surface $f(x,y) = x^{2} + y^{2}$ at point $\left( {1,2,5} \right),$ and graph the surface and the tangent plane at the point.

213.

\[T\] Find the equation for the tangent plane to the surface at the indicated point, and graph the surface and the tangent plane: $z = \text{ln}(10x^{2} + 2y^{2} + 1),P(0,0,0).$

214\.

\[T\] Find an equation of the tangent plane to the surface $z = f(x,y) = \text{sin}(x + y^{2})$ at point $\left( {\frac{\pi}{4},0,\frac{\sqrt{2}}{2}} \right),$ and graph the surface and the tangent plane.

---

4.5 The Chain Rule

> 来源: OpenStax《Calculus Volume 3》| 原页: https://openstax.org/books/calculus-volume-3/pages/4-5-the-chain-rule

4.5 The Chain Rule

In single-variable calculus, we found that one of the most useful differentiation rules is the chain rule, which allows us to find the derivative of the composition of two functions. The same thing is true for multivariable calculus, but this time we have to deal with more than one form of the chain rule. In this section, we study extensions of the chain rule and learn how to take derivatives of compositions of functions of more than one variable.

Chain Rules for One or Two Independent Variables

Recall that the chain rule for the derivative of a composite of two functions can be written in the form

$$\frac{d}{dx}\left( {f\left( {g(x)} \right)} \right) = f\prime\left( {g(x)} \right)g\prime(x).$$

In this equation, both $f(x)$ and $g(x)$ are functions of one variable. Now suppose that $f$ is a function of two variables and $g$ is a function of one variable. Or perhaps they are both functions of two variables, or even more. How would we calculate the derivative in these cases? The following theorem gives us the answer for the case of one independent variable.

Chain Rule for One Independent Variable

Suppose that $x = g(t)$ and $y = h(t)$ are differentiable functions of $t$ and $z = f\left( {x,y} \right)$ is a differentiable function of $x\ \text{and}\ y.$ Then $z = f\left( {x(t),y(t)} \right)$ is a differentiable function of $t$ and

$$\frac{dz}{dt} = \frac{\partial z}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial z}{\partial y} \cdot \frac{dy}{dt},$$ (4.29)

where the ordinary derivatives are evaluated at $t$ and the partial derivatives are evaluated at $\left( {x,y} \right).$

Proof

The proof of this theorem uses the definition of differentiability of a function of two variables. Suppose that *f* is differentiable at the point $P\left( {x_{0},y_{0}} \right),$ where $x_{0} = g\left( t_{0} \right)$ and $y_{0} = h\left( t_{0} \right)$ for a fixed value of $t_{0}.$ We wish to prove that $z = f\left( {x(t),y(t)} \right)$ is differentiable at $t = t_{0}$ and that Equation 4.29 holds at that point as well.

Since $f$ is differentiable at $P,$ we know that

$$z(t) = f\left( {x,y} \right) = f\left( {x_{0},y_{0}} \right) + f_{x}\left( {x_{0},y_{0}} \right)\left( {x - x_{0}} \right) + f_{y}\left( {x_{0},y_{0}} \right)\left( {y - y_{0}} \right) + E\left( {x,y} \right),$$ (4.30)

where $\underset{{({x,y})}\rightarrow{({x_{0},y_{0}})}}{\text{lim}}\frac{E\left( {x,y} \right)}{\sqrt{\left( {x - x_{0}} \right)^{2} + \left( {y - y_{0}} \right)^{2}}} = 0.$ We then subtract $z_{0} = f\left( {x_{0},y_{0}} \right)$ from both sides of this equation:

$$\begin{array}{cl}

{z(t) - z\left( t_{0} \right)} & {= f\left( {x(t),y(t)} \right) - f\left( {x\left( t_{0} \right),y\left( t_{0} \right)} \right)} \\

& {= f_{x}\left( {x_{0},y_{0}} \right)\left( {x(t) - x\left( t_{0} \right)} \right) + f_{y}\left( {x_{0},y_{0}} \right)\left( {y(t) - y\left( t_{0} \right)} \right) + E\left( {x(t),y(t)} \right).}

\end{array}$$

Next, we divide both sides by $t - t_{0}\text{:}$

$$\frac{z(t) - z\left( t_{0} \right)}{t - t_{0}} = f_{x}\left( {x_{0},y_{0}} \right)\left( \frac{x(t) - x\left( t_{0} \right)}{t - t_{0}} \right) + f_{y}\left( {x_{0},y_{0}} \right)\left( \frac{y(t) - y\left( t_{0} \right)}{t - t_{0}} \right) + \frac{E\left( {x(t),y(t)} \right)}{t - t_{0}}.$$

Then we take the limit as $t$ approaches $t_{0}\text{:}$

$$\begin{array}{cl}

{\underset{t\rightarrow t_{0}}{\text{lim}}\frac{z(t) - z\left( t_{0} \right)}{t - t_{0}}} & {= f_{x}\left( {x_{0},y_{0}} \right)\underset{t\rightarrow t_{0}}{\text{lim}}\left( \frac{x(t) - x\left( t_{0} \right)}{t - t_{0}} \right) + f_{y}\left( {x_{0},y_{0}} \right)\underset{t\rightarrow t_{0}}{\text{lim}}\left( \frac{y(t) - y\left( t_{0} \right)}{t - t_{0}} \right)} \\

& {\mspace{9mu} + \underset{t\rightarrow t_{0}}{\text{lim}}\frac{E\left( {x(t),y(t)} \right)}{t - t_{0}}.}

\end{array}$$

The left-hand side of this equation is equal to ${{dz}\text{/}{dt}},$ which leads to

$$\frac{dz}{dt} = f_{x}\left( {x_{0},y_{0}} \right)\frac{dx}{dt} + f_{y}\left( {x_{0},y_{0}} \right)\frac{dy}{dt} + \underset{t\rightarrow t_{0}}{\text{lim}}\frac{E\left( {x(t),y(t)} \right)}{t - t_{0}}.$$

The last term can be rewritten as

$$\begin{array}{cl}

{\underset{t\rightarrow t_{0}}{\text{lim}}\frac{E\left( {x(t),y(t)} \right)}{t - t_{0}}} & {= \underset{t\rightarrow t_{0}}{\text{lim}}\left( {\frac{E\left( {x,y} \right)}{\sqrt{\left( {x - x_{0}} \right)^{2} + \left( {y - y_{0}} \right)^{2}}}\ \frac{\sqrt{\left( {x - x_{0}} \right)^{2} + \left( {y - y_{0}} \right)^{2}}}{t - t_{0}}} \right)} \\

& {= \underset{t\rightarrow t_{0}}{\text{lim}}\left( \frac{E\left( {x,y} \right)}{\sqrt{\left( {x - x_{0}} \right)^{2} + \left( {y - y_{0}} \right)^{2}}} \right)\underset{t\rightarrow t_{0}}{\text{lim}}\left( \frac{\sqrt{\left( {x - x_{0}} \right)^{2} + \left( {y - y_{0}} \right)^{2}}}{t - t_{0}} \right).}

\end{array}$$

As $t$ approaches $t_{0},$ $\left( {x(t),y(t)} \right)$ approaches $\left( {x\left( t_{0} \right),y\left( t_{0} \right)} \right),$ so we can rewrite the last product as

$$\underset{{({x,y})}\rightarrow{({x_{0},y_{0}})}}{\text{lim}}\left( \frac{E\left( {x,y} \right)}{\sqrt{\left( {x - x_{0}} \right)^{2} + \left( {y - y_{0}} \right)^{2}}} \right)\underset{{({x,y})}\rightarrow{({x_{0},y_{0}})}}{\text{lim}}\left( \frac{\sqrt{\left( {x - x_{0}} \right)^{2} + \left( {y - y_{0}} \right)^{2}}}{t - t_{0}} \right).$$

Since the first limit is equal to zero, we need only show that the second limit is finite:

$$\begin{array}{cl}

{\underset{{({x,y})}\rightarrow{({x_{0},y_{0}})}}{\text{lim}}\left( \frac{\sqrt{\left( {x - x_{0}} \right)^{2} + \left( {y - y_{0}} \right)^{2}}}{t - t_{0}} \right)} & {= \underset{{({x,y})}\rightarrow{({x_{0},y_{0}})}}{\text{lim}}\left( \sqrt{\frac{\left( {x - x_{0}} \right)^{2} + \left( {y - y_{0}} \right)^{2}}{\left( {t - t_{0}} \right)^{2}}} \right)} \\

& {= \underset{{({x,y})}\rightarrow{({x_{0},y_{0}})}}{\text{lim}}\left( \sqrt{\left( \frac{x - x_{0}}{t - t_{0}} \right)^{2} + \left( \frac{y - y_{0}}{t - t_{0}} \right)^{2}} \right)} \\

& {= \sqrt{\left( {\underset{{({x,y})}\rightarrow{({x_{0},y_{0}})}}{\text{lim}}\left( \frac{x - x_{0}}{t - t_{0}} \right)} \right)^{2} + \left( {\underset{{({x,y})}\rightarrow{({x_{0},y_{0}})}}{\text{lim}}\left( \frac{y - y_{0}}{t - t_{0}} \right)} \right)^{2}}.}

\end{array}$$

Since $x(t)$ and $y(t)$ are both differentiable functions of $t,$ both limits inside the last radical exist. Therefore, this value is finite. This proves the chain rule at $t = t_{0};$ the rest of the theorem follows from the assumption that all functions are differentiable over their entire domains.

□

Closer examination of Equation 4.29 reveals an interesting pattern. The first term in the equation is $\frac{\partial f}{\partial x} \cdot \frac{dx}{dt}$ and the second term is $\frac{\partial f}{\partial y} \cdot \frac{dy}{dt}.$ Recall that when multiplying fractions, cancelation can be used. If we treat these derivatives as fractions, then each product “simplifies” to something resembling ${{\partial f}\text{/}{dt}}.$ The variables $x\ \text{and}\ y$ that disappear in this simplification are often called intermediate variables: they are independent variables for the function $f,$ but are dependent variables for the variable $t.$ Two terms appear on the right-hand side of the formula, and $f$ is a function of two variables. This pattern works with functions of more than two variables as well, as we see later in this section.

Using the Chain Rule

Calculate ${dz}\text{/}{dt}$ for each of the following functions:

1. $z = f\left( {x,y} \right) = 4x^{2} + 3y^{2},x = x(t) = \text{sin}\ t,y = y(t) = \text{cos}\ t$

2. $z = f\left( {x,y} \right) = \sqrt{x^{2} - y^{2}},x = x(t) = e^{2t},y = y(t) = e^{\text{−}t}$

Solution

1. To use the chain rule, we need four quantities—${{\partial z}\text{/}{\partial x}},{{\partial z}\text{/}{\partial y}},{{dx}\text{/}{dt}},$ and ${{dy}\text{/}{dt}}\text{:}$

$$\begin{array}{lccl}

{\frac{\partial z}{\partial x} = 8x} & & & {\frac{\partial z}{\partial y} = 6y} \\

{\frac{dx}{dt} = \text{cos}\ t} & & & {\frac{dy}{dt} = \text{−}\text{sin}\ t}

\end{array}$$

Now, we substitute each of these into Equation 4.29:

$$\begin{array}{cl}

\frac{dz}{dt} & {= \frac{\partial z}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial z}{\partial y} \cdot \frac{dy}{dt}} \\

& {= \left( {8x} \right)\left( {\text{cos}\ t} \right) + \left( {6y} \right)\left( {\text{−}\text{sin}\ t} \right)} \\

& {= 8x\ \text{cos}\ t - 6y\ \text{sin}\ t.}

\end{array}$$

This answer has three variables in it. To reduce it to one variable, use the fact that $x(t) = \text{sin}\ t\ \text{and}\ y(t) = \text{cos}\ t.$ We obtain

$$\begin{array}{cl}

\frac{dz}{dt} & {= 8x\ \text{cos}\ t - 6y\ \text{sin}\ t} \\

& {= 8\left( {\text{sin}\ t} \right)\text{cos}\ t - 6\left( {\text{cos}\ t} \right)\text{sin}\ t} \\

& {= 2\ \text{sin}\ t\ \text{cos}\ t.}

\end{array}$$

This derivative can also be calculated by first substituting $x(t)$ and $y(t)$ into $f\left( {x,y} \right),$ then differentiating with respect to $t\text{:}$

$$\begin{array}{cl}

z & {= f\left( {x,y} \right)} \\

& {= f\left( {x(t),y(t)} \right)} \\

& {= 4\left( {x(t)} \right)^{2} + 3\left( {y(t)} \right)^{2}} \\

& {= 4\text{sin}^{2}t + 3\text{cos}^{2}t.}

\end{array}$$

Then

$$\begin{array}{cl}

\frac{dz}{dt} & {= 2(4\ \text{sin}\ t)(\text{cos}\ t) + 2(3\ \text{cos}\ t)(\text{−}\text{sin}\ t)} \\

& {= 8\ \text{sin}\ t\ \text{cos}\ t - 6\ \text{sin}\ t\ \text{cos}\ t} \\

& {= 2\ \text{sin}\ t\ \text{cos}\ t,}

\end{array}$$

which is the same solution. However, it may not always be this easy to differentiate in this form.

2. To use the chain rule, we again need four quantities—${{\partial z}\text{/}{\partial x}},{{\partial z}\text{/}{dy}},{{dx}\text{/}{dt}},$ and ${{dy}\text{/}{dt}}\text{:}$

$$\begin{matrix}

{\frac{\partial z}{\partial x} = \frac{x}{\sqrt{x^{2} - y^{2}}}} & & & {\frac{\partial z}{\partial y} = \frac{\text{−}y}{\sqrt{x^{2} - y^{2}}}} \\

{\frac{dx}{dt} = 2e^{2t}} & & & {\frac{dy}{dt} = \text{−}e^{\text{−}t}.}

\end{matrix}$$

We substitute each of these into Equation 4.29:

$$\begin{array}{cl}

\frac{dz}{dt} & {= \frac{\partial z}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial z}{\partial y} \cdot \frac{dy}{dt}} \\

& {= \left( \frac{x}{\sqrt{x^{2} - y^{2}}} \right)\left( {2e^{2t}} \right) + \left( \frac{\text{−}y}{\sqrt{x^{2} - y^{2}}} \right)\left( {\text{−}e^{\text{−}t}} \right)} \\

& {= \frac{2xe^{2t} + ye^{\text{−}t}}{\sqrt{x^{2} - y^{2}}}.}

\end{array}$$

To reduce this to one variable, we use the fact that $x(t) = e^{2t}$ and $y(t) = e^{\text{−}t}.$ Therefore,

$$\begin{array}{cl}

\frac{dz}{dt} & {= \frac{2xe^{2t} + ye^{\text{−}t}}{\sqrt{x^{2} - y^{2}}}} \\

& {= \frac{2\left( e^{2t} \right)e^{2t} + \left( e^{\text{−}t} \right)e^{\text{−}t}}{\sqrt{e^{4t} - e^{-2t}}}} \\

& {= \frac{2e^{4t} + e^{-2t}}{\sqrt{e^{4t} - e^{-2t}}}.}

\end{array}$$

To eliminate negative exponents, we multiply the top by $e^{2t}$ and the bottom by $\sqrt{e^{4t}}\text{:}$

$$\begin{array}{cl}

\frac{dz}{dt} & {= \frac{2e^{4t} + e^{-2t}}{\sqrt{e^{4t} - e^{-2t}}} \cdot \frac{e^{2t}}{\sqrt{e^{4t}}}} \\

& {= \frac{2e^{6t} + 1}{\sqrt{e^{8t} - e^{2t}}}} \\

& {= \frac{2e^{6t} + 1}{\sqrt{e^{2t}\left( {e^{6t} - 1} \right)}}} \\

& {= \frac{2e^{6t} + 1}{e^{t}\sqrt{e^{6t} - 1}}.}

\end{array}$$

Again, this derivative can also be calculated by first substituting $x(t)$ and $y(t)$ into $f\left( {x,y} \right),$ then differentiating with respect to $t\text{:}$

$$\begin{array}{cl}

z & {= f\left( {x,y} \right)} \\

& {= f\left( {x(t),y(t)} \right)} \\

& {= \sqrt{\left( {x(t)} \right)^{2} - \left( {y(t)} \right)^{2}}} \\

& {= \sqrt{e^{4t} - e^{-2t}}} \\

& {= \left( {e^{4t} - e^{-2t}} \right)^{1\text{/}2}.}

\end{array}$$

Then

$$\begin{array}{cl}

\frac{dz}{dt} & {= \frac{1}{2}\left( {e^{4t} - e^{-2t}} \right)^{\text{−}{1\text{/}2}}\left( {4e^{4t} + 2e^{-2t}} \right)} \\

& {= \frac{2e^{4t} + e^{-2t}}{\sqrt{e^{4t} - e^{-2t}}}.}

\end{array}$$

This is the same solution.

Calculate ${dz}\text{/}{dt}$ given the following functions. Express the final answer in terms of $t.$

$$z = f\left( {x,y} \right) = x^{2} - 3xy + 2y^{2},x = x(t) = 3\ \text{sin}\ 2t,y = y(t) = 4\ \text{cos}\ 2t$$

It is often useful to create a visual representation of Equation 4.29 for the chain rule. This is called a tree diagram for the chain rule for functions of one variable and it provides a way to remember the formula (Figure 4.34). This diagram can be expanded for functions of more than one variable, as we shall see very shortly.

In this diagram, the leftmost corner corresponds to $z = f\left( {x,y} \right).$ Since $f$ has two independent variables, there are two lines coming from this corner. The upper branch corresponds to the variable $x$ and the lower branch corresponds to the variable $y.$ Since each of these variables is then dependent on one variable $t,$ one branch then comes from $x$ and one branch comes from $y.$ Last, each of the branches on the far right has a label that represents the path traveled to reach that branch. The top branch is reached by following the $x$ branch, then the $t$ branch; therefore, it is labeled $\left( {{\partial z}\text{/}{\partial x}} \right)\ \times \ \left( {{dx}\text{/}{dt}} \right).$ The bottom branch is similar: first the $y$ branch, then the $t$ branch. This branch is labeled $\left( {{\partial z}\text{/}{\partial y}} \right)\ \times \ \left( {{dy}\text{/}{dt}} \right).$ To get the formula for ${{dz}\text{/}{dt}},$ add all the terms that appear on the rightmost side of the diagram. This gives us Equation 4.29.

In Chain Rule for Two Independent Variables, $z = f\left( {x,y} \right)$ is a function of $x\ \text{and}\ y,$ and both $x = g\left( {u,v} \right)$ and $y = h\left( {u,v} \right)$ are functions of the independent variables $u\ \text{and}\ v.$

Chain Rule for Two Independent Variables

Suppose $x = g\left( {u,v} \right)$ and $y = h\left( {u,v} \right)$ are differentiable functions of $u$ and $v,$ and $z = f\left( {x,y} \right)$ is a differentiable function of $x\ \text{and}\ y.$ Then, $z = f\left( {g\left( {u,v} \right),h\left( {u,v} \right)} \right)$ is a differentiable function of $u\ \text{and}\ v,$ and

$$\frac{\partial z}{\partial u} = \frac{\partial z}{\partial x}\ \frac{\partial x}{\partial u} + \frac{\partial z}{\partial y}\ \frac{\partial y}{\partial u}$$ (4.31)

and

$$\frac{\partial z}{\partial v} = \frac{\partial z}{\partial x}\ \frac{\partial x}{\partial v} + \frac{\partial z}{\partial y}\ \frac{\partial y}{\partial v}.$$ (4.32)

We can draw a tree diagram for each of these formulas as well as follows.

To derive the formula for ${{\partial z}\text{/}{\partial u}},$ start from the left side of the diagram, then follow only the branches that end with $u$ and add the terms that appear at the end of those branches. For the formula for ${{\partial z}\text{/}{\partial v}},$ follow only the branches that end with $v$ and add the terms that appear at the end of those branches.

There is an important difference between these two chain rule theorems. In Chain Rule for One Independent Variable, the left-hand side of the formula for the derivative is not a partial derivative, but in Chain Rule for Two Independent Variables it is. The reason is that, in Chain Rule for One Independent Variable, $z$ is ultimately a function of $t$ alone, whereas in Chain Rule for Two Independent Variables, $z$ is a function of both $u\ \text{and}\ v.$

Using the Chain Rule for Two Variables

Calculate ${\partial z}\text{/}{\partial u}$ and ${\partial z}\text{/}{\partial v}$ using the following functions:

$$z = f\left( {x,y} \right) = 3x^{2} - 2xy + y^{2},x = x\left( {u,v} \right) = 3u + 2v,y = y\left( {u,v} \right) = 4u - v.$$

Solution

To implement the chain rule for two variables, we need six partial derivatives—${{\partial z}\text{/}{\partial x}},{{\partial z}\text{/}{\partial y,{{\partial x}\text{/}{\partial u,{{\partial x}\text{/}{\partial v,{{\partial y}\text{/}{\partial u}}}}}}}},$ and ${{\partial y}\text{/}{\partial v}}\text{:}$

$$\begin{array}{lccl}

{\frac{\partial z}{\partial x} = 6x - 2y} & & & {\frac{\partial z}{\partial y} = -2x + 2y} \\

{\frac{\partial x}{\partial u} = 3} & & & {\frac{\partial x}{\partial v} = 2} \\

{\frac{\partial y}{\partial u} = 4} & & & {\frac{\partial y}{\partial v} = -1.}

\end{array}$$

To find ${{\partial z}\text{/}{\partial u}},$ we use Equation 4.31:

$$\begin{array}{cl}

\frac{\partial z}{\partial u} & {= \frac{\partial z}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial u}} \\

& {= 3\left( {6x - 2y} \right) + 4\left( {-2x + 2y} \right)} \\

& {= 10x + 2y.}

\end{array}$$

Next, we substitute $x\left( {u,v} \right) = 3u + 2v$ and $y\left( {u,v} \right) = 4u - v\text{:}$

$$\begin{array}{cl}

\frac{\partial z}{\partial u} & {= 10x + 2y} \\

& {= 10\left( {3u + 2v} \right) + 2\left( {4u - v} \right)} \\

& {= 38u + 18v.}

\end{array}$$

To find ${{\partial z}\text{/}{\partial v}},$ we use Equation 4.32:

$$\begin{array}{cl}

\frac{\partial z}{\partial v} & {= \frac{\partial z}{\partial x}\ \frac{\partial x}{\partial v} + \frac{\partial z}{\partial y}\ \frac{\partial y}{\partial v}} \\

& {= 2\left( {6x - 2y} \right) + (-1)\left( {-2x + 2y} \right)} \\

& {= 14x - 6y.}

\end{array}$$

Then we substitute $x\left( {u,v} \right) = 3u + 2v$ and $y\left( {u,v} \right) = 4u - v\text{:}$

$$\begin{array}{cl}

\frac{\partial z}{\partial v} & {= 14x - 6y} \\

& {= 14\left( {3u + 2v} \right) - 6\left( {4u - v} \right)} \\

& {= 18u + 34v.}

\end{array}$$

Calculate ${\partial z}\text{/}{\partial u}$ and ${\partial z}\text{/}{\partial v}$ given the following functions:

$$z = f\left( {x,y} \right) = \frac{2x - y}{x + 3y},x\left( {u,v} \right) = e^{2u}\text{cos}\ 3v,y\left( {u,v} \right) = e^{2u}\text{sin}\ 3v.$$

The Generalized Chain Rule

Now that we’ve see how to extend the original chain rule to functions of two variables, it is natural to ask: Can we extend the rule to more than two variables? The answer is yes, as the generalized chain rule states.

Generalized Chain Rule

Let $w = f\left( {x_{1},x_{2}\text{,…,}\ x_{m}} \right)$ be a differentiable function of $m$ independent variables, and for each $i \in \left\{ 1\text{,…,}\ m \right\},$ let $x_{i} = x_{i}(t_{1},t_{2}\text{,…,}\ t_{n})$ be a differentiable function of $n$ independent variables. Then

$$\frac{\partial w}{\partial t_{j}} = \frac{\partial w}{\partial x_{1}}\ \frac{\partial x_{1}}{\partial t_{j}} + \frac{\partial w}{\partial x_{2}}\ \frac{\partial x_{2}}{\partial t_{j}} + \text{⋯} + \frac{\partial w}{\partial x_{m}}\ \frac{\partial x_{m}}{\partial t_{j}}$$ (4.33)

for any $j \in \left\{ 1,2\text{,…,}\ n \right\}.$

In the next example we calculate the derivative of a function of three independent variables in which each of the three variables is dependent on two other variables.

Using the Generalized Chain Rule

Calculate ${\partial w}\text{/}{\partial u}$ and ${\partial w}\text{/}{\partial v}$ using the following functions:

$$\begin{array}{rll}

& & \\

w & = & {f\left( {x,y,z} \right) = 3x^{2} - 2xy + 4z^{2}} \\

x & = & {x\left( {u,v} \right) = e^{u}\text{sin}\ v} \\

y & = & {y\left( {u,v} \right) = e^{u}\text{cos}\ v} \\

z & = & {z\left( {u,v} \right) = e^{u}.}

\end{array}$$

Solution

The formulas for ${\partial w}\text{/}{\partial u}$ and ${\partial w}\text{/}{\partial v}$ are

$$\begin{array}{l}

{\frac{\partial w}{\partial u} = \frac{\partial w}{\partial x} \cdot \frac{\partial x}{\partial u} + \frac{\partial w}{\partial y} \cdot \frac{\partial y}{\partial u} + \frac{\partial w}{\partial z} \cdot \frac{\partial z}{\partial u}} \\

{\frac{\partial w}{\partial v} = \frac{\partial w}{\partial x} \cdot \frac{\partial x}{\partial v} + \frac{\partial w}{\partial y} \cdot \frac{\partial y}{\partial v} + \frac{\partial w}{\partial z} \cdot \frac{\partial z}{\partial v}.}

\end{array}$$

Therefore, there are nine different partial derivatives that need to be calculated and substituted. We need to calculate each of them:

$$\begin{array}{rllccrllccrll}

\frac{\partial w}{\partial x} & = & {6x - 2y} & & & \frac{\partial w}{\partial y} & = & {-2x} & & & \frac{\partial w}{\partial z} & = & {8z} \\

\frac{\partial x}{\partial u} & = & {e^{u}\text{sin}\ v} & & & \frac{\partial y}{\partial u} & = & {e^{u}\text{cos}\ v} & & & \frac{\partial z}{\partial u} & = & e^{u} \\

\frac{\partial x}{\partial v} & = & {e^{u}\text{cos}\ v} & & & \frac{\partial y}{\partial v} & = & {\text{−}e^{u}\text{sin}\ v} & & & \frac{\partial z}{\partial v} & = & {0.}

\end{array}$$

Now, we substitute each of them into the first formula to calculate $\partial w\text{/}\partial u\text{:}$

$$\begin{array}{cl}

\frac{\partial w}{\partial u} & {= \frac{\partial w}{\partial x} \cdot \frac{\partial x}{\partial u} + \frac{\partial w}{\partial y} \cdot \frac{\partial y}{\partial u} + \frac{\partial w}{\partial z} \cdot \frac{\partial z}{\partial u}} \\

& {= \left( {6x - 2y} \right)e^{u}\text{sin}\ v - 2xe^{u}\text{cos}\ v + 8ze^{u},}

\end{array}$$

then substitute $x(u,v) = e^{u}\text{sin}\ v,y(u,v) = e^{u}\text{cos}\ v,$ and $z(u,v) = e^{u}$ into this equation:

$$\begin{array}{cl}

\frac{\partial w}{\partial u} & {= \left( {6x - 2y} \right)e^{u}\text{sin}\ v - 2xe^{u}\text{cos}\ v + 8ze^{u}} \\

& {= \left( {6e^{u}\text{sin}\ v - 2e^{u}\text{cos}\ v} \right)e^{u}\text{sin}\ v - 2\left( {e^{u}\text{sin}\ v} \right)e^{u}\text{cos}\ v + 8e^{2u}} \\

& {= 6e^{2u}\text{sin}^{2}v - 4e^{2u}\text{sin}\ v\ \text{cos}\ v + 8e^{2u}} \\

& {= 2e^{2u}\left( {3\ \text{sin}^{2}v - 2\ \text{sin}\ v\ \text{cos}\ v + 4} \right).}

\end{array}$$

Next, we calculate ${{\partial w}\text{/}{\partial v}}\text{:}$

$$\begin{array}{cl}

\frac{\partial w}{\partial v} & {= \frac{\partial w}{\partial x} \cdot \frac{\partial x}{\partial v} + \frac{\partial w}{\partial y} \cdot \frac{\partial y}{\partial v} + \frac{\partial w}{\partial z} \cdot \frac{\partial z}{\partial v}} \\

& {= \left( {6x - 2y} \right)e^{u}\text{cos}\ v - 2x\left( {\text{−}e^{u}\text{sin}\ v} \right) + 8z(0),}

\end{array}$$

then we substitute $x(u,v) = e^{u}\text{sin}\ v,y(u,v) = e^{u}\text{cos}\ v,$ and $z(u,v) = e^{u}$ into this equation:

$$\begin{matrix}

\frac{\partial w}{\partial v} & {= (6x - 2y)e^{u}\text{cos}\ v - 2x\left( \text{−}e^{u}\text{sin}\ v \right)} \\

& {= \left( 6e^{u}\text{sin}\ v - 2e^{u}\text{cos}\ v \right)e^{u}\text{cos}\ v + 2\left( e^{u}\text{sin}\ v \right)\left( e^{u}\text{sin}\ v \right)} \\

& {= 2e^{2u}\text{sin}^{2}v + 6e^{2u}\text{sin}\ v\ \text{cos}\ v - 2e^{2u}\text{cos}^{2}v} \\

& {= 2e^{2u}\left( v + 3\text{sin}~\text{sin~}v~\text{cos}~\text{cos~}v - v \right).}

\end{matrix}$$

Calculate ${\partial w}\text{/}{\partial u}$ and ${\partial w}\text{/}{\partial v}$ given the following functions:

$$\begin{array}{rll}

& & \\

& & \\

w & = & {f\left( {x,y,z} \right) = \frac{x + 2y - 4z}{2x - y + 3z}} \\

x & = & {x\left( {u,v} \right) = e^{2u}\text{cos}\ 3v} \\

y & = & {y\left( {u,v} \right) = e^{2u}\text{sin}\ 3v} \\

z & = & {z\left( {u,v} \right) = e^{2u}.}

\end{array}$$

Drawing a Tree Diagram

Create a tree diagram for the case when

$$w = f\left( {x,y,z} \right),x = x\left( {t,u,v} \right),y = y\left( {t,u,v} \right),z = z\left( {t,u,v} \right)$$

and write out the formulas for the three partial derivatives of $w.$

Solution

Starting from the left, the function $f$ has three independent variables: $x,y,\text{and}\ z.$ Therefore, three branches must be emanating from the first node. Each of these three branches also has three branches, for each of the variables $t,u,\ \text{and}\ v.$

The three formulas are

$$\begin{array}{l}

{\frac{\partial w}{\partial t} = \frac{\partial w}{\partial x}\ \frac{\partial x}{\partial t} + \frac{\partial w}{\partial y}\ \frac{\partial y}{\partial t} + \frac{\partial w}{\partial z}\ \frac{\partial z}{\partial t}} \\

{\frac{\partial w}{\partial u} = \frac{\partial w}{\partial x}\ \frac{\partial x}{\partial u} + \frac{\partial w}{\partial y}\ \frac{\partial y}{\partial u} + \frac{\partial w}{\partial z}\ \frac{\partial z}{\partial u}} \\

{\frac{\partial w}{\partial v} = \frac{\partial w}{\partial x}\ \frac{\partial x}{\partial v} + \frac{\partial w}{\partial y}\ \frac{\partial y}{\partial v} + \frac{\partial w}{\partial z}\ \frac{\partial z}{\partial v}.}

\end{array}$$

Create a tree diagram for the case when

$$w = f\left( {x,y} \right),x = x\left( {t,u,v} \right),y = y\left( {t,u,v} \right)$$

and write out the formulas for the three partial derivatives of $w.$

Implicit Differentiation

Recall from Implicit Differentiation that implicit differentiation provides a method for finding ${dy}\text{/}{dx}$ when $y$ is defined implicitly as a function of $x.$ The method involves differentiating both sides of the equation defining the function with respect to $x,$ then solving for ${{dy}\text{/}{dx}}.$ Partial derivatives provide an alternative to this method.

Consider the ellipse defined by the equation $x^{2} + 3y^{2} + 4y - 4 = 0$ as follows.

This equation implicitly defines $y$ as a function of $x.$ As such, we can find the derivative ${dy}\text{/}{dx}$ using the method of implicit differentiation:

$$\begin{array}{rll}

{\frac{d}{dx}\left( {x^{2} + 3y^{2} + 4y - 4} \right)} & = & {\frac{d}{dx}(0)} \\

{2x + 6y\frac{dy}{dx} + 4\frac{dy}{dx}} & = & 0 \\

{\left( {6y + 4} \right)\frac{dy}{dx}} & = & {-2x} \\

\frac{dy}{dx} & = & {- \frac{x}{3y + 2}.}

\end{array}$$

We can also define a function $z = f\left( {x,y} \right)$ by using the left-hand side of the equation defining the ellipse. Then $f\left( {x,y} \right) = x^{2} + 3y^{2} + 4y - 4.$ The ellipse $x^{2} + 3y^{2} + 4y - 4 = 0$ can then be described by the equation $f\left( {x,y} \right) = 0.$ Using this function and the following theorem gives us an alternative approach to calculating ${{dy}\text{/}{dx}}.$

Implicit Differentiation of a Function of Two or More Variables

Suppose the function $z = f\left( {x,y} \right)$ defines $y$ implicitly as a function $y = g(x)$ of $x$ via the equation $f\left( {x,y} \right) = 0.$ Then

$$\frac{dy}{dx} = - \frac{{\partial f}\text{/}{\partial x}}{{\partial f}\text{/}{\partial y}}$$ (4.34)

provided $f_{y}\left( {x,y} \right) \neq 0.$

If the equation $f\left( {x,y,z} \right) = 0$ defines $z$ implicitly as a differentiable function of $x\ \text{and}\ y,$ then

$$\frac{\partial z}{\partial x} = - \frac{{\partial f}\text{/}{\partial x}}{{\partial f}\text{/}{\partial z}}\quad\text{and}\quad\frac{\partial z}{\partial y} = - \frac{{\partial f}\text{/}{\partial y}}{{\partial f}\text{/}{\partial z}}$$ (4.35)

as long as $f_{z}\left( {x,y,z} \right) \neq 0.$

Equation 4.34 is a direct consequence of Equation 4.31. In particular, if we assume that $y$ is defined implicitly as a function of $x$ via the equation $f\left( {x,y} \right) = 0,$ we can apply the chain rule to find ${{dy}\text{/}{dx}}\text{:}$

$$\begin{array}{rll}

{\frac{d}{dx}f\left( {x,y} \right)} & = & {\frac{d}{dx}(0)} \\

{\frac{\partial f}{\partial x} \cdot \frac{dx}{dx} + \frac{\partial f}{\partial y} \cdot \frac{dy}{dx}} & = & 0 \\

{\frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \cdot \frac{dy}{dx}} & = & 0.

\end{array}$$

Solving this equation for ${dy}\text{/}{dx}$ gives Equation 4.34. Equation 4.35 can be derived in a similar fashion.

Let’s now return to the problem that we started before the previous theorem. Using Implicit Differentiation of a Function of Two or More Variables and the function $f\left( {x,y} \right) = x^{2} + 3y^{2} + 4y - 4,$ we obtain

$$\begin{array}{l}

{\frac{\partial f}{\partial x} = 2x} \\

{\frac{\partial f}{\partial y} = 6y + 4.}

\end{array}$$

Then Equation 4.34 gives

$$\frac{dy}{dx} = - \frac{{\partial f}\text{/}{\partial x}}{{\partial f}\text{/}{\partial y}} = - \frac{2x}{6y + 4} = - \frac{x}{3y + 2},$$

which is the same result obtained by the earlier use of implicit differentiation.

Implicit Differentiation by Partial Derivatives

1. Calculate ${dy}\text{/}{dx}$ if $y$ is defined implicitly as a function of $x$ via the equation $3x^{2} - 2xy + y^{2} + 4x - 6y - 11 = 0.$ What is the equation of the tangent line to the graph of this curve at point $\left( {2,1} \right)?$

2. Calculate ${\partial z}\text{/}{\partial x}$ and ${{\partial z}\text{/}{\partial y}},$ given $x^{2}e^{y} - yze^{x} = 0.$

Solution

1. Set $f(x,y) = 3x^{2} - 2xy + y^{2} + 4x - 6y - 11 = 0,$ then calculate $f_{x}$ and $f_{y}\text{:}$ $\begin{array}{l}

{f_{x} = 6x - 2y + 4} \\

{f_{y} = -2x + 2y - 6.}

\end{array}$

The derivative is given by

$$\frac{dy}{dx} = - \frac{{\partial f}\text{/}{\partial x}}{{\partial f}\text{/}{\partial y}} = - \frac{6x - 2y + 4}{-2x + 2y - 6} = \frac{3x - y + 2}{x - y + 3}.$$

The slope of the tangent line at point $\left( {2,1} \right)$ is given by

$$\left. \frac{dy}{dx} \right|_{{({x,y})} = {({2,1})}} = \frac{3(2) - 1 + 2}{2 - 1 + 3} = \frac{7}{4}.$$

To find an equation of the tangent line, we use the point-slope form (Figure 4.38):

$$\begin{array}{rll}

{y - y_{0}} & = & {m\left( {x - x_{0}} \right)} \\

{y - 1} & = & {\frac{7}{4}\left( {x - 2} \right)} \\

y & = & {\frac{7}{4}x - \frac{7}{2} + 1} \\

y & = & {\frac{7}{4}x - \frac{5}{2}.}

\end{array}$$

2. We have $f\left( {x,y,z} \right) = x^{2}e^{y} - yze^{x}.$ Therefore,

$$\begin{array}{l}

{\frac{\partial f}{\partial x} = 2xe^{y} - yze^{x}} \\

{\frac{\partial f}{\partial y} = x^{2}e^{y} - ze^{x}} \\

{\frac{\partial f}{\partial z} = \text{−}ye^{x}.}

\end{array}$$

Using Equation 4.35,

$$\begin{matrix}

\begin{matrix}

\frac{\partial z}{\partial x} & {= - \frac{\partial f\text{/}\partial x}{\partial f\text{/}\partial z}} \\

& {= - \frac{2xe^{y} - yze^{x}}{\text{−}ye^{x}}} \\

& {= \frac{2xe^{y} - yze^{x}}{ye^{x}}}

\end{matrix} & & & \text{and} & & & \begin{matrix}

\frac{\partial z}{\partial y} & {= - \frac{\partial f\text{/}\partial y}{\partial f\text{/}\partial z}} \\

& {= - \frac{x^{2}e^{y} - ze^{x}}{\text{−}ye^{x}}} \\

& {= \frac{x^{2}e^{y} - ze^{x}}{ye^{x}}.}

\end{matrix}

\end{matrix}$$

Find ${dy}\text{/}{dx}$ if $y$ is defined implicitly as a function of $x$ by the equation $x^{2} + xy - y^{2} + 7x - 3y - 26 = 0.$ What is the equation of the tangent line to the graph of this curve at point $\left( {3,-2} \right)?$

Section 4.5 Exercises

For the following exercises, use the information provided to solve the problem.

215.

Let $w\left( {x,y,z} \right) = xy\ \text{cos}\ z,$ where $x = t,y = t^{2},$ and $z = \text{arcsin}\ t.$ Find $\frac{dw}{dt}.$

216\.

Let $w\left( {t,v} \right) = e^{tv}$ where $t = r + s$ and $v = rs.$ Find $\frac{\partial w}{\partial r}$ and $\frac{\partial w}{\partial s}.$

217.

If $w = 5x^{2} + 2y^{2},x = -3s + t,$ and $y = s - 4t,$ find $\frac{\partial w}{\partial s}$ and $\frac{\partial w}{\partial t}.$

218\.

If $w = xy^{2},x = 5\ \text{cos}(2t),$ and $y = 5\ \text{sin}(2t),$ find $\frac{dw}{dt}.$

219.

If $f(x,y) = xy,x = r\ \text{cos}\ \theta,$ and $y = r\ \text{sin}\ \theta,$ find $\frac{\partial f}{\partial r}$ and express the answer in terms of $r$ and $\theta.$

220\.

Suppose $f(x,y) = x + y,$ where $x = r\ \text{cos}\ \theta$ and $y = r\ \text{sin}\ \theta.$ Find $\frac{\partial f}{\partial\theta}.$

For the following exercises, find $\frac{df}{dt}$ using the chain rule and direct substitution.

221.

$f\left( {x,y} \right) = x^{2} + y^{2},$ $x = t,y = t^{2}$

222\.

$f\left( {x,y} \right) = \sqrt{x^{2} + y^{2}},y = t^{2},x = t$

223.

$f\left( {x,y} \right) = xy,x = 1 - \sqrt{t},y = 1 + \sqrt{t}$

224\.

$f\left( {x,y} \right) = \frac{x}{y},x = e^{t},y = 2e^{t}$

225.

$f\left( {x,y} \right) = \text{ln}\left( {x + y} \right),$ $x = e^{t},y = e^{t}$

226\.

$f\left( {x,y} \right) = x^{4},$ $x = t,y = t$

227.

Let $w\left( {x,y,z} \right) = x^{2} + y^{2} + z^{2},$ $x = \text{cos}\ t,y = \text{sin}\ t,$ and $z = e^{t}.$ Express $w$ as a function of $t$ and find $\frac{dw}{dt}$ directly. Then, find $\frac{dw}{dt}$ using the chain rule.

228\.

Let $z = x^{2}y,$ where $x = t^{2}$ and $y = t^{3}.$ Find $\frac{dz}{dt}.$

229.

Let $u = e^{x}\text{sin}\ y,$ where $x = \operatorname{-ln}2t$ and $y = \pi t.$ Find $\frac{du}{dt}$ when $x = \text{ln}\ 2$ and $y = \frac{\pi}{4}.$

For the following exercises, find $\frac{dy}{dx}$ using partial derivatives.

230\.

$\text{sin}\left( {6x} \right) + \text{tan}\left( {8y} \right) + 5 = 0$

231.

$x^{3} + y^{2}x - 3 = 0$

232\.

$\text{sin}\left( {x + y} \right) + \text{cos}\left( {x - y} \right) = 4$

233.

$x^{2} - 2xy + y^{4} = 4$

234\.

$xe^{y} + ye^{x} - 2x^{2}y = 0$

235.

$x^{2\text{/}3} + y^{2\text{/}3} = a^{2\text{/}3}$

236\.

$x\ \text{cos}(xy) + y\ \text{cos}\ x = 2$

237.

$e^{xy} + ye^{y} = 1$

238\.

$x^{2}y^{3} + \text{cos}\ y = 0$

239.

Find $\frac{dz}{dt}$ using the chain rule where $z = 3x^{2}y^{3},x = t^{4},$ and $y = t^{2}.$

240\.

Let $z = 3\ \text{cos}\ x - \text{sin}(xy),x = \frac{1}{t},$ and $y = 3t.$ Find $\frac{dz}{dt}.$

241.

Let $z = e^{1 - xy},x = t^{1\text{/}3},$ and $y = t^{3}.$ Find $\frac{dz}{dt}.$

242\.

Find $\frac{dz}{dt}$ by the chain rule where $z = \text{cosh}^{2}(xy),x = \frac{1}{2}t,$ and $y = e^{t}.$

243.

Let $z = \frac{x}{y},x = 2\ \text{cos}\ u,$ and $y = 3\ \text{sin}\ v.$ Find $\frac{\partial z}{\partial u}$ and $\frac{\partial z}{\partial v}.$

244\.

Let $z = e^{x^{2}y},$ where $x = \sqrt{uv}$ and $y = \frac{1}{v}.$ Find $\frac{\partial z}{\partial u}$ and $\frac{\partial z}{\partial v}.$

245.

If $z = xye^{x\text{/}y},$ $x = r\ \text{cos}\ \theta,$ and $y = r\ \text{sin}\ \theta,$ find $\frac{\partial z}{\partial r}$ and $\frac{\partial z}{\partial\theta}$ when $r = 2$ and $\theta = \frac{\pi}{6}.$

246\.

Find $\frac{\partial w}{\partial s}$ if $w = 4x + y^{2} + z^{3},x = e^{rs^{2}},y = \text{ln}\left( \frac{r + s}{t} \right),$ and $z = rst^{2}.$

247.

If $w = \text{sin}(xyz),x = 1 - 3t,y = e^{1 - t},$ and $z = 4t,$ find $\frac{\partial w}{\partial t}.$

For the following exercises, use this information: A function $f(x,y)$ is said to be homogeneous of degree $n$ if $f(tx,ty) = t^{n}f(x,y).$ For all homogeneous functions of degree $n,$ the following equation is true: $x\frac{\partial f}{\partial x} + y\frac{\partial f}{\partial y} = nf(x,y).$ Show that the given function is homogeneous and verify that $x\frac{\partial f}{\partial x} + y\frac{\partial f}{\partial y} = nf(x,y).$

248\.

$f(x,y) = 3x^{2} + y^{2}$

249.

$f(x,y) = \sqrt{x^{2} + y^{2}}$

250\.

$f(x,y) = x^{2}y - 2y^{3}$

251.

The volume of a right circular cylinder is given by $V(x,y) = \pi x^{2}y,$ where $x$ is the radius of the cylinder and *y* is the cylinder height. Suppose $x$ and $y$ are functions of $t$ given by $x = \frac{1}{2}t$ and $y = \frac{1}{3}t$ so that $x\ \text{and}\ y$ are both increasing with time. How fast is the volume increasing when $x = 2$ and $y = \frac{4}{3}?$

252\.

The pressure $P$ of a gas is related to the volume and temperature by the formula $PV = kT,$ where temperature is expressed in kelvins. Express the pressure of the gas as a function of both $V$ and $T.$ Find $\frac{dP}{dt}$ when $k = 1,$ $\frac{dV}{dt} = 2$ cm3/min, $\frac{dT}{dt} = \frac{1}{2}$ K/min, $V = 20$ cm3, and $T = 20\text{°}\text{F}.$

253.

The radius of a right circular cone is increasing at $3$ cm/min whereas the height of the cone is decreasing at $2$ cm/min. Find the rate of change of the volume of the cone when the radius is $13$ cm and the height is $18$ cm.

254\.

The volume of a frustum of a cone is given by the formula $V = \frac{1}{3}\pi z\left( {x^{2} + y^{2} + xy} \right),$ where $x$ is the radius of the smaller circle, $y$ is the radius of the larger circle, and $z$ is the height of the frustum (see figure). Find the rate of change of the volume of this frustum when $x = 10\ \text{in}\text{.,}\ y = 12\ \text{in.,}\ \text{and}\ z = 18\ \text{in}.$ if $\frac{dz}{dt} = - 5,\frac{dx}{dt} = 1,\frac{dy}{dt} = 1$ (all in/min).

255.

A closed box is in the shape of a rectangular solid with dimensions $x,y,\ \text{and}\ z.$ (Dimensions are in inches.) Suppose each dimension is changing at the rate of $0.5$ in./min. Find the rate of change of the total surface area of the box when $x = 2\ \text{in}\text{.,}\ y = 3\ \text{in.,}\ \text{and}\ z = 1\ \text{in}.$

256\.

The total resistance in a circuit that has three individual resistances represented by $x,y,$ and $z$ is given by the formula $R(x,y,z) = \frac{xyz}{yz + xz + xy}.$ Suppose at a given time the $x$ resistance is $100\text{Ω},$ the *y* resistance is $200\text{Ω},$ and the $z$ resistance is $300\text{Ω}.$ Also, suppose the $x$ resistance is changing at a rate of $2\text{Ω}\text{/}\text{min},$ the $y$ resistance is changing at the rate of $1\text{Ω}\text{/}\text{min},$ and the $z$ resistance has no change. Find the rate of change of the total resistance in this circuit at this time.

257.

The temperature $T$ at a point $(x,y)$ is $T(x,y)$ and is measured using the Celsius scale. A fly crawls so that its position after $t$ seconds is given by $x = \sqrt{1 + t}$ and $y = 2 + \frac{1}{3}t,$ where $x\ \text{and}\ y$ are measured in centimeters. The temperature function satisfies $T_{x}\left( {2,3} \right) = 4$ and $T_{y}\left( {2,3} \right) = 3.$ How fast is the temperature increasing on the fly’s path after $3$ sec?

258\.

The $x\ \text{and}\ y$ components of a fluid moving in two dimensions are given by the following functions: $u(x,y) = 2y$ and $v(x,y) = -2x;$ $x \geq 0;y \geq 0.$ The speed of the fluid at the point $(x,y)$ is $s(x,y) = \sqrt{u{(x,y)}^{2} + v{(x,y)}^{2}}.$ Find $\frac{\partial s}{\partial x}$ and $\frac{\partial s}{\partial y}$ using the chain rule.

259.

Let $u = u\left( {x,y,z} \right),$ where $x = x(w,t),y = y(w,t),z = z(w,t),w = w(r,s),\text{and}\ t = t(r,s).$ Use a tree diagram and the chain rule to find an expression for $\frac{\partial u}{\partial r}.$

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4.6 Directional Derivatives and the Gradient

> 来源: OpenStax《Calculus Volume 3》| 原页: https://openstax.org/books/calculus-volume-3/pages/4-6-directional-derivatives-and-the-gradient

4.6 Directional Derivatives and the Gradient

In Partial Derivatives we introduced the partial derivative. A function $z = f\left( {x,y} \right)$ has two partial derivatives: ${\partial z}\text{/}{\partial x}$ and ${{\partial z}\text{/}{\partial y}}.$ These derivatives correspond to each of the independent variables and can be interpreted as instantaneous rates of change (that is, as slopes of a tangent line). For example, ${\partial z}\text{/}{\partial x}$ represents the slope of a tangent line passing through a given point on the surface defined by $z = f\left( {x,y} \right),$ assuming the tangent line is parallel to the *x*-axis. Similarly, ${\partial z}\text{/}{\partial y}$ represents the slope of the tangent line parallel to the $y\text{-axis.}$ Now we consider the possibility of a tangent line parallel to neither axis.

Directional Derivatives

We start with the graph of a surface defined by the equation $z = f\left( {x,y} \right).$ Given a point $\left( {a,b} \right)$ in the domain of $f,$ we choose a direction to travel from that point. We measure the direction using an angle $\theta,$ which is measured counterclockwise in the *x*, *y*-plane, starting at zero from the positive *x*-axis (Figure 4.39). The distance we travel is $h$ and the direction we travel is given by the unit vector $\mathbf{u} = \left( {\text{cos}\ \theta} \right)\mathbf{i} + \left( {\text{sin}\ \theta} \right)\mathbf{j}.$ Therefore, the *z*-coordinate of the second point on the graph is given by $z = f\left( {a + h\ \text{cos}\ \theta,b + h\ \text{sin}\ \theta} \right).$

We can calculate the slope of the secant line by dividing the difference in $z\text{-values}$ by the length of the line segment connecting the two points in the domain. The length of the line segment is $h.$ Therefore, the slope of the secant line is

$$m_{\text{sec}} = \frac{f\left( {a + h\ \text{cos}\ \theta,b + h\ \text{sin}\ \theta} \right) - f\left( {a,b} \right)}{h}.$$

To find the slope of the tangent line in the same direction, we take the limit as $h$ approaches zero.

Suppose $z = f\left( {x,y} \right)$ is a function of two variables with a domain of $D.$ Let $\left( {a,b} \right) \in D$ and define $\text{u} = \text{cos}\ \theta\mathbf{i} + \text{sin}\ \theta\mathbf{j}.$ Then the directional derivative of $f$ in the direction of $\mathbf{u}$ is given by

$$D_{\mathbf{u}}f\left( {a,b} \right) = \underset{h\rightarrow 0}{\text{lim}}\frac{f\left( {a + h\ \text{cos}\ \theta,b + h\ \text{sin}\ \theta} \right) - f\left( {a,b} \right)}{h},$$ (4.36)

provided the limit exists.

Equation 4.36 provides a formal definition of the directional derivative that can be used in many cases to calculate a directional derivative.

Finding a Directional Derivative from the Definition

Let $\theta = \text{arccos}\left( {3\text{/}5} \right).$ Find the directional derivative $D_{\mathbf{u}}f\left( {x,y} \right)$ of $f\left( {x,y} \right) = x^{2} - xy + 3y^{2}$ in the direction of $\text{u} = \left( {\text{cos}\ \theta} \right)\mathbf{i} + \left( {\text{sin}\ \theta} \right)\mathbf{j}.$ What is $D_{\mathbf{u}}f\left( {-1,2} \right)?$

Solution

First of all, since $\text{cos}\ \theta = {3\text{/}5}$ and $\theta$ is acute, this implies

$$\text{sin}\ \theta = \sqrt{1 - \left( \frac{3}{5} \right)^{2}} = \sqrt{\frac{16}{25}} = \frac{4}{5}.$$

Using $f\left( {x,y} \right) = x^{2} - xy + 3y^{2},$ we first calculate $f\left( {x + h\ \text{cos}\ \theta,y + h\ \text{sin}\ \theta} \right)\text{:}$

$$\begin{array}{cl}

{f\left( {x + h\ \text{cos}\ \theta,y + h\ \text{sin}\ \theta} \right)} & {= \left( {x + h\ \text{cos}\ \theta} \right)^{2} - \left( {x + h\ \text{cos}\ \theta} \right)\left( {y + h\ \text{sin}\ \theta} \right) + 3\left( {y + h\ \text{sin}\ \theta} \right)^{2}} \\

& {= x^{2} + 2xh\ \text{cos}\ \theta + h^{2}\text{cos}^{2}\theta - xy - xh\ \text{sin}\ \theta - yh\ \text{cos}\ \theta} \\

& {\mspace{9mu}\textit{−h}^{2}\text{sin}\ \theta\ \text{cos}\ \theta + 3y^{2} + 6yh\ \text{sin}\ \theta + 3h^{2}\text{sin}^{2}\theta} \\

& {= x^{2} + 2xh\left( \frac{3}{5} \right) + \frac{9h^{2}}{25} - xy - \frac{4xh}{5} - \frac{3yh}{5} - \frac{12h^{2}}{25} + 3y^{2}} \\

& {\mspace{9mu} + 6yh\left( \frac{4}{5} \right) + 3h^{2}\left( \frac{16}{25} \right)} \\

& {= x^{2} - xy + 3y^{2} + \frac{2xh}{5} + \frac{9h^{2}}{5} + \frac{21yh}{5}.}

\end{array}$$

We substitute this expression into Equation 4.36:

$$\begin{array}{cl}

{D_{\mathbf{u}}f\left( {a,b} \right)} & {= \underset{h\rightarrow 0}{\text{lim}}\frac{f\left( {a + h\ \text{cos}\ \theta,b + h\ \text{sin}\ \theta} \right) - f\left( {a,b} \right)}{h}} \\

& {= \underset{h\rightarrow 0}{\text{lim}}\frac{\left( {x^{2} - xy + 3y^{2} + \frac{2xh}{5} + \frac{9h^{2}}{5} + \frac{21yh}{5}} \right) - \left( {x^{2} - xy + 3y^{2}} \right)}{h}} \\

& {= \underset{h\rightarrow 0}{\text{lim}}\frac{\frac{2xh}{5} + \frac{9h^{2}}{5} + \frac{21yh}{5}}{h}} \\

& {= \underset{h\rightarrow 0}{\text{lim}}\frac{2x}{5} + \frac{9h}{5} + \frac{21y}{5}} \\

& {= \frac{2x + 21y}{5}.}

\end{array}$$

To calculate $D_{\mathbf{u}}f\left( {-1,2} \right),$ we substitute $x = -1$ and $y = 2$ into this answer:

$$\begin{array}{cl}

{D_{\mathbf{u}}f\left( {-1,2} \right)} & {= \frac{2(-1) + 21(2)}{5}} \\

& {= \frac{-2 + 42}{5}} \\

& {= 8.}

\end{array}$$

(See the following figure.)

Another approach to calculating a directional derivative involves partial derivatives, as outlined in the following theorem.

Directional Derivative of a Function of Two Variables

Let $z = f\left( {x,y} \right)$ be a function of two variables $x\ \text{and}\ y,$ and assume that $f_{x}$ and $f_{y}$ exist and $f(x,~y)$ is differentiable everywhere. Then the directional derivative of $f$ in the direction of $\textbf{u} = \text{cos}\ \theta\textbf{i} + \text{sin}\ \theta\textbf{j}$ is given by

$$D_{\mathbf{u}}f\left( {x,y} \right) = f_{x}\left( {x,y} \right)\text{cos}\ \theta + f_{y}\left( {x,y} \right)\text{sin}\ \theta.$$ (4.37)

Proof

Equation 4.36 states that the directional derivative of *f* in the direction of $\mathbf{u} = \text{cos}\ \theta\mathbf{i} + \text{sin}\ \theta\mathbf{j}$ is given by

$$D_{\mathbf{u}}f\left( {a,b} \right) = \underset{t\rightarrow 0}{\text{lim}}\frac{f\left( {a + t\ \text{cos}\ \theta,b + t\ \text{sin}\ \theta} \right) - f\left( {a,b} \right)}{t}.$$

Let $x = a + t\ \text{cos}\ \theta$ and $y = b + t\ \text{sin}\ \theta,$ and define $g(t) = f\left( {x,y} \right).$ Since $f_{x}$ and $f_{y}$ both exist, and therefore $f$ is differentiable, we can use the chain rule for functions of two variables to calculate $g^{\prime}(t)\text{:}$

$$\begin{array}{cl}

{g^{\prime}(t)} & {= \frac{\partial\textit{f}}{\partial\textit{x}}\mspace{2mu}\frac{dx}{dt} + \frac{\partial\textit{f}}{\partial\textit{y}}\mspace{2mu}\frac{dy}{dt}} \\

& {= f_{x}\left( {x,y} \right)\text{cos}\ \theta + f_{y}\left( {x,y} \right)\text{sin}\ \theta.}

\end{array}$$

If $t = 0,$ then $x = x_{0}( = a)$ and $y = y_{0}( = b),$ so

$$g^{\prime}(0) = f_{x}\left( {x_{0},y_{0}} \right)\text{cos}\ \theta + f_{y}\left( {x_{0},y_{0}} \right)\text{sin}\ \theta.$$

By the definition of $g^{\prime}(t),$ it is also true that

$$\begin{array}{cl}

{g^{\prime}(0)} & {= \underset{t\rightarrow 0}{\text{lim}}\frac{g(t) - g(0)}{t}} \\

& {= \underset{t\rightarrow 0}{\text{lim}}\frac{f\left( {x_{0} + t\ \text{cos}\ \theta,y_{0} + t\ \text{sin}\ \theta} \right) - f\left( {x_{0},y_{0}} \right)}{t}.}

\end{array}$$

Therefore, $D_{\mathbf{u}}f\left( {x_{0},y_{0}} \right) = f_{x}\left( {x,y} \right)\text{cos}\ \theta + f_{y}\left( {x,y} \right)\text{sin}\ \theta.$

□

Finding a Directional Derivative: Alternative Method

Let $\theta = \text{arccos}\left( {3\text{/}5} \right).$ Find the directional derivative $D_{\mathbf{u}}f\left( {x,y} \right)$ of $f\left( {x,y} \right) = x^{2} - xy + 3y^{2}$ in the direction of $\mathbf{u} = \left( {\text{cos}\ \theta} \right)\mathbf{i} + \left( {\text{sin}\ \theta} \right)\mathbf{j}.$ What is $D_{\mathbf{u}}f\left( {-1,2} \right)?$

Solution

First, we must calculate the partial derivatives of $f\text{:}$

$$\begin{array}{l}

{f_{x} = 2x - y} \\

{f_{y} = \text{−}x + 6y,}

\end{array}$$

Then we use Equation 4.37 with $\theta = \text{arccos}\left( {3\text{/}5} \right)\text{:}$

$$\begin{array}{cl}

{D_{\mathbf{u}}f\left( {x,y} \right)} & {= f_{x}\left( {x,y} \right)\text{cos}\ \theta + f_{y}\left( {x,y} \right)\text{sin}\ \theta} \\

& {= \left( {2x - y} \right)\frac{3}{5} + \left( {\text{−}x + 6y} \right)\frac{4}{5}} \\

& {= \frac{6x}{5} - \frac{3y}{5} - \frac{4x}{5} + \frac{24y}{5}} \\

& {= \frac{2x + 21y}{5}.}

\end{array}$$

To calculate $D_{\mathbf{u}}f\left( {-1,2} \right),$ let $x = -1$ and $y = 2\text{:}$

$$D_{\mathbf{u}}f\left( {-1,2} \right) = \frac{2(-1) + 21(2)}{5} = \frac{-2 + 42}{5} = 8.$$

This is the same answer obtained in Example 4.31.

Gradient

The right-hand side of Equation 4.37 is equal to $f_{x}\left( {x,y} \right)\text{cos}\ \theta + f_{y}\left( {x,y} \right)\text{sin}\ \theta,$ which can be written as the dot product of two vectors. Define the first vector as $\nabla f\left( {x,y} \right) = f_{x}\left( {x,y} \right)\text{i} + f_{y}\left( {x,y} \right)\text{j}$ and the second vector as $\mathbf{u} = \left( {\text{cos}\ \theta} \right)\mathbf{i} + \left( {\text{sin}\ \theta} \right)\mathbf{j}.$ Then the right-hand side of the equation can be written as the dot product of these two vectors:

$$D_{\mathbf{u}}f\left( {x,y} \right) = \nabla f\left( {x,y} \right) \cdot \mathbf{u}.$$ (4.38)

The first vector in Equation 4.38 has a special name: the gradient of the function $f.$ The symbol $\nabla$ is called *nabla* and the vector $\nabla f$ is read $\text{“del}\ f\text{.”}$

Let $z = f\left( {x,y} \right)$ be a function of $x\ \text{and}\ y$ such that $f_{x}$ and $f_{y}$ exist. The vector $\nabla f\left( {x,y} \right)$ is called the gradient of $f$ and is defined as

$$\nabla f\left( {x,y} \right) = f_{x}\left( {x,y} \right)\mathbf{i} + f_{y}\left( {x,y} \right)\mathbf{j}.$$ (4.39)

The vector $\nabla f\left( {x,y} \right)$ is also written as $\text{“grad}\ f\text{.”}$

Find the directional derivative $D_{\mathbf{u}}f\left( {x,y} \right)$ of $f\left( {x,y} \right) = 3x^{2}y - 4xy^{3} + 3y^{2} - 4x$ in the direction of $\mathbf{u} = \left( {\text{cos}\ \frac{\pi}{3}} \right)\mathbf{i} + \left( {\text{sin}\ \frac{\pi}{3}} \right)\mathbf{j}$ using Equation 4.37. What is $D_{\mathbf{u}}f\left( {3,4} \right)?$

If the vector that is given for the direction of the derivative is not a unit vector, then it is only necessary to divide by the norm of the vector. For example, if we wished to find the directional derivative of the function in Example 4.32 in the direction of the vector $\left\langle {-5,12} \right\rangle,$ we would first divide by its magnitude to get $\mathbf{u}.$ This gives us $\mathbf{u} = \left\langle {\text{−}\left( {5\text{/}13} \right),{12\text{/}13}} \right\rangle.$ Then

$$\begin{array}{cl}

{D_{\mathbf{u}}f\left( {x,y} \right)} & {= \nabla f\left( {x,y} \right) \cdot \mathbf{u}} \\

& {= - \frac{5}{13}\left( {2x - \frac{77}{12}y} \right) + \frac{12}{13}\left( {\text{−}x + 6y} \right)} \\

& {= - \frac{22}{13}x + \frac{17}{13}y.}

\end{array}$$

Finding Gradients

Find the gradient $\nabla f\left( {x,y} \right)$ of each of the following functions:

1. $f\left( {x,y} \right) = x^{2} - xy + 3y^{2}$

2. $f\left( {x,y} \right) = \text{sin}\ 3x\ \text{cos}\ 3y$

Solution

For both parts a. and b., we first calculate the partial derivatives $f_{x}$ and $f_{y},$ then use Equation 4.39.

1.

$\begin{array}{cll}

{f_{x}\left( {x,y} \right)} & = & {2x - y\ \text{and}\ f_{y}\left( {x,y} \right) = \text{−}x + 6y,\ \text{so}} \\

{\nabla f\left( {x,y} \right)} & = & {f_{x}\left( {x,y} \right)\mathbf{i} + f_{y}\left( {x,y} \right)\mathbf{j}} \\

& = & {\left( {2x - y} \right)\mathbf{i} + \left( {\text{−}x + 6y} \right)\mathbf{j.}}

\end{array}$

2.

$\begin{array}{cll}

{f_{x}\left( {x,y} \right)} & = & {3\ \text{cos}\ 3x\ \text{cos}\ 3y\ \text{and}\ f_{y}\left( {x,y} \right) = -3\ \text{sin}\ 3x\ \text{sin}\ 3y,\ \text{so}} \\

{\nabla f\left( {x,y} \right)} & = & {f_{x}\left( {x,y} \right)\mathbf{i} + f_{y}\left( {x,y} \right)\mathbf{j}} \\

& = & {\left( {3\ \text{cos}\ 3x\ \text{cos}\ 3y} \right)\mathbf{i} - \left( {3\ \text{sin}\ 3x\ \text{sin}\ 3y} \right)\mathbf{j}.}

\end{array}$

Find the gradient $\nabla f\left( {x,y} \right)$ of $f(x,y) = {\left( {x^{2} - 3y^{2}} \right)\text{/}{\left( {2x + y} \right).}}$

The gradient has some important properties. We have already seen one formula that uses the gradient: the formula for the directional derivative. Recall from The Dot Product that if the angle between two vectors $\mathbf{a}$ and $\mathbf{b}$ is $\varphi,$ then $\mathbf{a} \cdot \text{b} = \left\| \mathbf{a} \right\|\left\| \text{b} \right\|\text{cos}\ \varphi.$ Therefore, if the angle between $\nabla f\left( {x_{0},y_{0}} \right)$ and $\mathbf{u} = \left( {\text{cos}\ \theta} \right)\mathbf{i} + \left( {\text{sin}\ \theta} \right)\mathbf{j}$ is $\varphi,$ we have

$$D_{\mathbf{u}}f\left( {x_{0},y_{0}} \right) = \nabla f\left( {x_{0},y_{0}} \right) \cdot \mathbf{u} = \left\| {\nabla f\left( {x_{0},y_{0}} \right)} \right\|\left\| \mathbf{u} \right\|\text{cos}\ \varphi = \left\| {\nabla f\left( {x_{0},y_{0}} \right)} \right\|\text{cos}\ \varphi.$$

The $\left\| \mathbf{u} \right\|$ disappears because $\mathbf{u}$ is a unit vector. Therefore, the directional derivative is equal to the magnitude of the gradient evaluated at $\left( {x_{0},y_{0}} \right)$ multiplied by $\text{cos}\ \varphi.$ Recall that $\text{cos}\ \varphi$ ranges from $-1$ to $1.$ If $\varphi = 0,$ then $\text{cos}\ \varphi = 1$ and $\nabla f\left( {x_{0},y_{0}} \right)$ and $\mathbf{u}$ both point in the same direction. If $\varphi = \pi,$ then $\text{cos}\ \varphi = -1$ and $\nabla f\left( {x_{0},y_{0}} \right)$ and $\mathbf{u}$ point in opposite directions. In the first case, the value of $D_{\text{u}}f\left( {x_{0},y_{0}} \right)$ is maximized; in the second case, the value of $D_{\text{u}}f\left( {x_{0},y_{0}} \right)$ is minimized. If $\nabla f\left( {x_{0},y_{0}} \right) = 0,$ then $D_{\mathbf{u}}f\left( {x_{0},y_{0}} \right) = \nabla f\left( {x_{0},y_{0}} \right) \cdot \mathbf{u} = 0$ for any vector $\mathbf{u}.$ These three cases are outlined in the following theorem.

Properties of the Gradient

Suppose the function $z = f\left( {x,y} \right)$ is differentiable at $\left( {x_{0},y_{0}} \right)$ (Figure 4.41).

1. If $\nabla f\left( {x_{0},y_{0}} \right) = \mathbf{0},$ then $D_{\mathbf{u}}f\left( {x_{0},y_{0}} \right) = 0$ for any unit vector $\mathbf{u}.$

2. If $\nabla f\left( {x_{0},y_{0}} \right) \neq \mathbf{0},$ then $D_{\mathbf{u}}f\left( {x_{0},y_{0}} \right)$ is maximized when $\mathbf{u}$ points in the same direction as $\nabla f\left( {x_{0},y_{0}} \right).$ The maximum value of $D_{\mathbf{u}}f\left( {x_{0},y_{0}} \right)$ is $\left\| {\nabla f\left( {x_{0},y_{0}} \right)} \right\|\text{.}$

3. If $\nabla f\left( {x_{0},y_{0}} \right) \neq \mathbf{0},$ then $D_{\mathbf{u}}f\left( {x_{0},y_{0}} \right)$ is minimized when $\mathbf{u}$ points in the opposite direction from $\nabla f\left( {x_{0},y_{0}} \right).$ The minimum value of $D_{\mathbf{u}}f\left( {x_{0},y_{0}} \right)$ is $\text{−}\left\| {\nabla f\left( {x_{0},y_{0}} \right)} \right\|\text{.}$

Finding a Maximum Directional Derivative

Find the direction for which the directional derivative of $f\left( {x,y} \right) = 3x^{2} - 4xy + 2y^{2}$ at $\left( {-2,3} \right)$ is a maximum. What is the maximum value?

Solution

The maximum value of the directional derivative occurs when $\nabla f$ and the unit vector point in the same direction. Therefore, we start by calculating $\nabla f\left( {x,y} \right)\text{:}$

$$\begin{array}{rll}

{f_{x}\left( {x,y} \right)} & = & {6x - 4y\ \text{and}\ f_{y}\left( {x,y} \right) = -4x + 4y,\ \text{so}} \\

{\nabla f\left( {x,y} \right)} & = & {f_{x}\left( {x,y} \right)\mathbf{i} + f_{y}\left( {x,y} \right)\mathbf{j} = \left( {6x - 4y} \right)\mathbf{i} + \left( {-4x + 4y} \right)\mathbf{j}.}

\end{array}$$

Next, we evaluate the gradient at $\left( {-2,3} \right)\text{:}$

$$\nabla f\left( {-2,3} \right) = \left( {6(-2) - 4(3)} \right)\mathbf{i} + \left( {-4(-2) + 4(3)} \right)\mathbf{j} = -24\mathbf{i} + 20\mathbf{j}.$$

We need to find a unit vector that points in the same direction as $\nabla f\left( {-2,3} \right),$ so the next step is to divide $\nabla f\left( {-2,3} \right)$ by its magnitude, which is $\sqrt{(-24)^{2} + (20)^{2}} = \sqrt{976} = 4\sqrt{61}.$ Therefore,

$$\frac{\nabla f\left( {-2,3} \right)}{\left\| {\nabla f\left( {-2,3} \right)} \right\|} = \frac{-24}{4\sqrt{61}}\mathbf{i} + \frac{20}{4\sqrt{61}}\mathbf{j} = \frac{-6\sqrt{61}}{61}\mathbf{i} + \frac{5\sqrt{61}}{61}\mathbf{j}.$$

This is the unit vector that points in the same direction as $\nabla f\left( {-2,3} \right).$ To find the angle corresponding to this unit vector, we solve the equations

$$\text{cos}\ \theta = \frac{-6\sqrt{61}}{61}\ \text{and}\ \text{sin}\ \theta = \frac{5\sqrt{61}}{61}$$

for $\theta.$ Since cosine is negative and sine is positive, the angle must be in the second quadrant. Therefore, $\theta = \pi - \text{arcsin}\left( {\left( {5\sqrt{61}} \right)\text{/}61} \right) \approx 2.45\ \text{rad.}$

The maximum value of the directional derivative at $\left( {-2,3} \right)$ is $\left\| {\nabla f\left( {-2,3} \right)} \right\| = 4\sqrt{61}$ (see the following figure).

Find the direction for which the directional derivative of $g\left( {x,y} \right) = 4x - xy + 2y^{2}$ at $\left( {-2,3} \right)$ is a maximum. What is the maximum value?

Figure 4.43 shows a portion of the graph of the function $f\left( {x,y} \right) = 3 + \text{sin}\ x\ \text{sin}\ y.$ Given a point $\left( {a,b} \right)$ in the domain of $f,$ the maximum value of the gradient at that point is given by $\left\| {\nabla f\left( {a,b} \right)} \right\|.$ This would equal the rate of greatest ascent if the surface represented a topographical map. If we went in the opposite direction, it would be the rate of greatest descent.

When using a topographical map, the steepest slope is always in the direction where the contour lines are closest together (see Figure 4.44). This is analogous to the contour map of a function, assuming the level curves are obtained for equally spaced values throughout the range of that function.

Gradients and Level Curves

Recall that if a curve is defined parametrically by the function pair $\left( {x(t),y(t)} \right),$ then the vector $x^{\prime}(t)\mathbf{i} + y^{\prime}(t)\mathbf{j}$ is tangent to the curve for every value of $t$ in the domain. Now let’s assume $z = f\left( {x,y} \right)$ is a differentiable function of $x\ \text{and}\ y,$ and $\left( {x_{0},y_{0}} \right)$ is in its domain. Let’s suppose further that $x_{0} = x\left( t_{0} \right)$ and $y_{0} = y\left( t_{0} \right)$ for some value of $t,$ and consider the level curve $f\left( {x,y} \right) = k.$ Define $g(t) = f\left( {x(t),y(t)} \right)$ and calculate $g^{\prime}(t)$ on the level curve. By the chain Rule,

$$g^{\prime}(t) = f_{x}\left( {x(t),y(t)} \right)x^{\prime}(t) + f_{y}\left( {x(t),y(t)} \right)y^{\prime}(t).$$

But $g^{\prime}(t) = 0$ because $g(t) = k$ for all $t.$ Therefore, on the one hand,

$$f_{x}\left( {x(t),y(t)} \right)x^{\prime}(t) + f_{y}\left( {x(t),y(t)} \right)y^{\prime}(t) = 0;$$

on the other hand,

$$f_{x}\left( {x(t),y(t)} \right)x^{\prime}(t) + f_{y}\left( {x(t),y(t)} \right)y^{\prime}(t) = \nabla f\left( {x,y} \right) \cdot \left\langle {x^{\prime}(t),y^{\prime}(t)} \right\rangle\text{.}$$

Therefore,

$$\nabla f\left( {x,y} \right) \cdot \left\langle {x^{\prime}(t),y^{\prime}(t)} \right\rangle = 0.$$

Thus, the dot product of these vectors is equal to zero, which implies they are orthogonal. However, the second vector is tangent to the level curve, which implies the gradient must be normal to the level curve, which gives rise to the following theorem.

Gradient Is Normal to the Level Curve

Suppose the function $z = f\left( {x,y} \right)$ has continuous first-order partial derivatives in an open disk centered at a point $\left( {x_{0},y_{0}} \right).$ If $\nabla f\left( {x_{0},y_{0}} \right) \neq \mathbf{0},$ then $\nabla f\left( {x_{0},y_{0}} \right)$ is normal to the level curve of $f$ at $\left( {x_{0},y_{0}} \right).$

We can use this theorem to find tangent and normal vectors to level curves of a function.

Finding Tangents to Level Curves

For the function $f\left( {x,y} \right) = 2x^{2} - 3xy + 8y^{2} + 2x - 4y + 4,$ find a tangent vector to the level curve at point $\left( {-2,1} \right).$ Graph the level curve corresponding to $f\left( {x,y} \right) = 18$ and draw in $\nabla f\left( {-2,1} \right)$ and a tangent vector.

Solution

First, we must calculate $\nabla f\left( {x,y} \right)\text{:}$

$$f_{x}\left( {x,y} \right) = 4x - 3y + 2\ \text{and}\ f_{y} = -3x + 16y - 4\ \text{so}\ \nabla f\left( {x,y} \right) = \left( {4x - 3y + 2} \right)\mathbf{i} + \left( {-3x + 16y - 4} \right)\mathbf{j}.$$

Next, we evaluate $\nabla f\left( {x,y} \right)$ at $\left( {-2,1} \right)\text{:}$

$$\nabla f\left( {-2,1} \right) = \left( {4(-2) - 3(1) + 2} \right)\mathbf{i} + \left( {-3(-2) + 16(1) - 4} \right)\mathbf{j} = -9\mathbf{i} + 18\mathbf{j}.$$

This vector is orthogonal to the curve at point $\left( {-2,1} \right).$ We can obtain a tangent vector by reversing the components and multiplying either one by $-1.$ Thus, for example, $-18\mathbf{i} - 9\mathbf{j}$ is a tangent vector (see the following graph).

For the function $f\left( {x,y} \right) = x^{2} - 2xy + 5y^{2} + 3x - 2y + 4,$ find the tangent to the level curve at point $\left( {1,1} \right).$ Draw the graph of the level curve corresponding to $f\left( {x,y} \right) = 9$ and draw $\nabla f\left( {1,1} \right)$ and a tangent vector.

Three-Dimensional Gradients and Directional Derivatives

The definition of a gradient can be extended to functions of more than two variables.

Let $w = f\left( {x,y,z} \right)$ be a function of three variables such that $f_{x},f_{y},\text{and}\ f_{z}$ exist. The vector $\nabla f\left( {x,y,z} \right)$ is called the gradient of $f$ and is defined as

$$\nabla f\left( {x,y,z} \right) = f_{x}\left( {x,y,z} \right)\mathbf{i} + f_{y}\left( {x,y,z} \right)\mathbf{j} + f_{z}\left( {x,y,z} \right)\mathbf{k}.$$ (4.40)

$\nabla f\left( {x,y,z} \right)$ can also be written as $\text{grad}\ f\left( {x,y,z} \right).$

Calculating the gradient of a function in three variables is very similar to calculating the gradient of a function in two variables. First, we calculate the partial derivatives $f_{x},f_{y},$ and $f_{z},$ and then we use Equation 4.40.

Finding Gradients in Three Dimensions

Find the gradient $\nabla f\left( {x,y,z} \right)$ of each of the following functions:

1. $f\left( {x,y,z} \right) = 5x^{2} - 2xy + y^{2} - 4yz + z^{2} + 3xz$

2. $f\left( {x,y,z} \right) = e^{-2z}\text{sin}\ 2x\ \text{cos}\ 2y$

Solution

For both parts a. and b., we first calculate the partial derivatives $f_{x},f_{y},$ and $f_{z},$ then use Equation 4.40.

1.

$\begin{matrix}

{f_{x}(x,y,z)} & = & {10x - 2y + 3z,\ f_{y}(x,y,z) = -2x + 2y - 4z\ \text{and}\ f_{z}(x,y,z) = 3x - 4y + 2z,\ \text{so}} \\

{\nabla f(x,y,z)} & = & {f_{x}(x,y,z)\mathbf{i} + f_{y}(x,y,z)\mathbf{j} + f_{z}(x,y,z)\mathbf{k}} \\

& = & {(10x - 2y + 3z)\mathbf{i} + (-2x + 2y - 4z)\mathbf{j} + (3x - 4y + 2z)\mathbf{k}.}

\end{matrix}$

2.

$\begin{matrix}

{f_{x}(x,y,z)} & = & {2e^{-2z}\text{cos}\ 2x\ \text{cos}\ 2y,\ f_{y}(x,y,z) = -2e^{-2z}\text{sin}\ 2x\ \text{sin}\ 2y\ \text{and}} \\

{f_{z}(x,y,z)} & = & {-2e^{-2z}\text{sin}\ 2x\ \text{cos}\ 2y,\ \text{so}} \\

{\nabla f(x,y,z)} & = & {f_{x}(x,y,z)\mathbf{i} + f_{y}(x,y,z)\mathbf{j} + f_{z}(x,y,z)\mathbf{k}} \\

& = & {\left( 2e^{-2z}\text{cos}\ 2x\ \text{cos}\ 2y \right)\mathbf{i} + \left( - 2e^{- 2z}\sin~2x~\sin~2y \right)\mathbf{j} + \left( - 2e^{- 2z}\sin~2x~\cos~2y \right)\mathbf{k}} \\

& = & {2e^{-2z}\left( \text{cos}\ 2x\ \text{cos}\ 2y\ \text{i} - \text{sin}\ 2x\ \text{sin}\ 2y\ \text{j} - \text{sin}\ 2x\ \text{cos}\ 2y\ \text{k} \right).}

\end{matrix}$

Find the gradient $\nabla f\left( {x,y,z} \right)$ of $f(x,y,z) = \frac{x^{2} - 3y^{2} + z^{2}}{2x + y - 4z}.$

The directional derivative can also be generalized to functions of three variables. To determine a direction in three dimensions, a vector with three components is needed. This vector is a unit vector, and the components of the unit vector are called *directional cosines*. Given a three-dimensional unit vector $\mathbf{u}$ in standard form (i.e., the initial point is at the origin), this vector forms three different angles with the positive $x - ,y - ,$ and *z-*axes. Let’s call these angles $\alpha,\beta,$ and $\gamma.$ Then the directional cosines are given by $\text{cos}\ \alpha,\text{cos}\ \beta,$ and $\text{cos}\ \gamma.$ These are the components of the unit vector $\mathbf{u};$ since $\mathbf{u}$ is a unit vector, it is true that $\text{cos}^{2}\alpha + \text{cos}^{2}\beta + \text{cos}^{2}\gamma = 1.$

Suppose $w = f\left( {x,y,z} \right)$ is a function of three variables with a domain of $D.$ Let $\left( {x_{0},y_{0},z_{0}} \right) \in D$ and let $\text{u} = \text{cos}\ \alpha\mathbf{i} + \text{cos}\ \beta\mathbf{j} + \text{cos}\ \gamma\mathbf{k}$ be a unit vector. Then, the directional derivative of $f$ in the direction of $u$ is given by

$$D_{\mathbf{u}}f\left( {x_{0},y_{0},z_{0}} \right) = \underset{t\rightarrow 0}{\text{lim}}\frac{f\left( {x_{0} + t\ \text{cos}\ \alpha,y_{0} + t\ \text{cos}\ \beta,z_{0} + t\ \text{cos}\ \gamma} \right) - f\left( {x_{0},y_{0},z_{0}} \right)}{t},$$ (4.41)

provided the limit exists.

We can calculate the directional derivative of a function of three variables by using the gradient, leading to a formula that is analogous to Equation 4.38.

Directional Derivative of a Function of Three Variables

Let $f\left( {x,y,z} \right)$ be a differentiable function of three variables and let $\mathbf{u} = \text{cos}\ \alpha\mathbf{i} + \text{cos}\ \beta\mathbf{j} + \text{cos}\ \gamma\mathbf{k}$ be a unit vector. Then, the directional derivative of $f$ in the direction of $\mathbf{u}$ is given by

$$\begin{array}{cl}

{D_{\mathbf{u}}f\left( {x,y,z} \right)} & {= \nabla f\left( {x,y,z} \right) \cdot \mathbf{u}} \\

& {= f_{x}\left( {x,y,z} \right)\text{cos}\ \alpha + f_{y}\left( {x,y,z} \right)\text{cos}\ \beta + f_{z}\left( {x,y,z} \right)\text{cos}\ \gamma.}

\end{array}$$ (4.42)

The three angles $\alpha,\beta,\ \text{and}\ \gamma$ determine the unit vector $\mathbf{u}.$ In practice, we can use an arbitrary (nonunit) vector, then divide by its magnitude to obtain a unit vector in the desired direction.

Finding a Directional Derivative in Three Dimensions

Calculate $D_{\mathbf{u}}f\left( {1,-2,3} \right)$ in the direction of $\text{v} = \text{−}\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}$ for the function

$$f\left( {x,y,z} \right) = 5x^{2} - 2xy + y^{2} - 4yz + z^{2} + 3xz.$$

Solution

First, we find the magnitude of $\mathbf{v}\text{:}$

$$\left\| \mathbf{v} \right\| = \sqrt{(-1)^{2} + (2)^{2} + (2)^{2}} = 3.$$

Therefore, $\frac{\mathbf{v}}{\left\| \mathbf{v} \right\|} = \frac{\text{−}\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}}{3} = - \frac{1}{3}\mathbf{i} + \frac{2}{3}\mathbf{j} + \frac{2}{3}\mathbf{k}$ is a unit vector in the direction of $\mathbf{v},$ so $\text{cos}\ \alpha = - \frac{1}{3},\text{cos}\ \beta = \frac{2}{3},\ \text{and}\ \text{cos}\ \gamma = \frac{2}{3}.$ Next, we calculate the partial derivatives of $f\text{:}$

$$\begin{array}{rll}

{f_{x}\left( {x,y,z} \right)} & = & {10x - 2y + 3z} \\

{f_{y}\left( {x,y,z} \right)} & = & {-2x + 2y - 4z} \\

{f_{z}\left( {x,y,z} \right)} & = & {-4y + 2z + 3x,}

\end{array}$$

then substitute them into Equation 4.42:

$$\begin{array}{cl}

{D_{\mathbf{u}}f\left( {x,y,z} \right)} & {= f_{x}\left( {x,y,z} \right)\text{cos}\ \alpha + f_{y}\left( {x,y,z} \right)\text{cos}\ \beta + f_{z}\left( {x,y,z} \right)\text{cos}\ \gamma} \\

& {= \left( {10x - 2y + 3z} \right)\left( {- \frac{1}{3}} \right) + \left( {-2x + 2y - 4z} \right)\left( \frac{2}{3} \right) + \left( {-4y + 2z + 3x} \right)\left( \frac{2}{3} \right)} \\

& {= - \frac{10x}{3} + \frac{2y}{3} - \frac{3z}{3} - \frac{4x}{3} + \frac{4y}{3} - \frac{8z}{3} - \frac{8y}{3} + \frac{4z}{3} + \frac{6x}{3}} \\

& {= - \frac{8x}{3} - \frac{2y}{3} - \frac{7z}{3}.}

\end{array}$$

Last, to find $D_{\mathbf{u}}f\left( {1,-2,3} \right),$ we substitute $x = 1,y = -2,\ \text{and}\ z = 3\text{:}$

$$\begin{array}{cl}

{D_{\mathbf{u}}f\left( {1,-2,3} \right)} & {= - \frac{8(1)}{3} - \frac{2(-2)}{3} - \frac{7(3)}{3}} \\

& {= - \frac{8}{3} + \frac{4}{3} - \frac{21}{3}} \\

& {= - \frac{25}{3}.}

\end{array}$$

Calculate $D_{\mathbf{u}}f\left( {x,y,z} \right)$ and $D_{\mathbf{u}}f\left( {0,-2,5} \right)$ in the direction of $\text{v} = -3\mathbf{i} + 12\mathbf{j} - 4\mathbf{k}$ for the function $f\left( {x,y,z} \right) = 3x^{2} + xy - 2y^{2} + 4yz - z^{2} + 2xz.$

Section 4.6 Exercises

For the following exercises, find the directional derivative using the limit definition only.

260\.

$f\left( {x,y} \right) = 5 - 2x^{2} - \frac{1}{2}y^{2}$ at point $P\left( {3,4} \right)$ in the direction of $\text{u} = \left( {\text{cos}\ \frac{\pi}{4}} \right)\mathbf{i} + \left( {\text{sin}\ \frac{\pi}{4}} \right)\mathbf{j}$

261.

$f\left( {x,y} \right) = y^{2}\text{cos}\left( {2x} \right)$ at point $P\left( {\frac{\pi}{3},2} \right)$ in the direction of $\text{u} = \left( {\text{cos}\ \frac{\pi}{4}} \right)\mathbf{i} + \left( {\text{sin}\ \frac{\pi}{4}} \right)\mathbf{j}$

262\.

Find the directional derivative of $f\left( {x,y} \right) = y^{2}\text{sin}\left( {2x} \right)$ at point $P\left( {\frac{\pi}{4},2} \right)$ in the direction of $\mathbf{u} = 5\mathbf{i} + 12\mathbf{j}.$

For the following exercises, find the directional derivative of the function at point $P$ in the direction of $\mathbf{u}$ or $\mathbf{v}$ as appropriate.

263.

$f\left( {x,y} \right) = xy,$ $P\left( {0,-2} \right),$ $\mathbf{v} = \frac{1}{2}\mathbf{i} + \frac{\sqrt{3}}{2}\mathbf{j}$

264\.

$h\left( {x,y} \right) = e^{x}\text{sin}\ y,P\left( {1,\frac{\pi}{2}} \right),\mathbf{v} = \text{−}\mathbf{i}$

265.

$h\left( {x,y,z} \right) = xyz,P\left( {2,1,1} \right),\mathbf{v} = 2\mathbf{i} + \mathbf{j} - \mathbf{k}$

266\.

$f(x,y) = xy,P(1,1),\mathbf{u} = \left\langle {\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}} \right\rangle$

267.

$f(x,y) = x^{2} - y^{2},\begin{array}{ll}

{\mathbf{u} = \left\langle {\frac{\sqrt{3}}{2},\frac{1}{2}} \right\rangle,} & {P(1,0)}

\end{array}$

268\.

$f(x,y) = 3x + 4y + 7,\begin{array}{ll}

{\mathbf{u} = \left\langle {\frac{3}{5},\frac{4}{5}} \right\rangle,} & {P\left( {0,\frac{\pi}{2}} \right)}

\end{array}$

269.

$\begin{array}{lll}

{f(x,y) = e^{x}\text{cos}\ y,} & {\mathbf{u} = \left\langle {0,1} \right\rangle,} & {P = \left( {0,\frac{\pi}{2}} \right)}

\end{array}$

270\.

$\begin{array}{lll}

{f(x,y) = y^{10},} & {\mathbf{u} = \left\langle {0,-1} \right\rangle,} & {P = (1,-1)}

\end{array}$

271.

$f(x,y) = \text{ln}(x^{2} + y^{2}),\begin{array}{ll}

{\mathbf{u} = \left\langle {\frac{3}{5},\frac{4}{5}} \right\rangle,} & {P\left( {1,2} \right)}

\end{array}$

272\.

$f(x,y) = x^{2}y,\begin{array}{ll}

{P(-5,5),} & {\mathbf{v} = 3\mathbf{i} - 4\mathbf{j}}

\end{array}$

273.

$f(x,y,z) = y^{2} + xz,\begin{array}{ll}

{P(1,2,2),} & {\mathbf{v} = \left\langle {2,-1,2} \right\rangle}

\end{array}$

For the following exercises, find the directional derivative of the function in the direction of the unit vector $\mathbf{u} = \text{cos}\ \theta\mathbf{i} + \text{sin}\ \theta\mathbf{j}.$

274\.

$f\left( {x,y} \right) = x^{2} + 2y^{2},\theta = \frac{\pi}{6}$

275.

$f\left( {x,y} \right) = \frac{y}{x + 2y},\theta = - \frac{\pi}{4}$

276\.

$f\left( {x,y} \right) = \text{cos}\left( {3x + y} \right),\theta = \frac{\pi}{4}$

277.

$w\left( {x,y} \right) = ye^{x},\theta = \frac{\pi}{3}$

278\.

$\begin{array}{ll}

{f\left( {x,y} \right) = x\ \text{arctan}(y),} & {\theta = \frac{\pi}{2}}

\end{array}$

279.

$\begin{array}{ll}

{f\left( {x,y} \right) = \text{ln}(x + 2y),} & {\theta = \frac{\pi}{3}}

\end{array}$

For the following exercises, find the gradient.

280\.

Find the gradient of $f(x,y) = \frac{14 - x^{2} - y^{2}}{3}.$ Then, find the gradient at point $P\left( {1,2} \right).$

281.

Find the gradient of $f(x,y,z) = xy + yz + xz$ at point $P\left( {1,2,3} \right).$

282\.

Find the gradient of $f(x,y,z)$ at $P$ and the directional derivative in the direction of $\mathbf{u}\text{:}$ $f(x,y,z) = \text{ln}(x^{2} + 2y^{2} + 3z^{2}),\begin{array}{ll}

{P(2,1,4),} & {\mathbf{u} = \frac{-3}{13}\mathbf{i} - \frac{4}{13}\mathbf{j} - \frac{12}{13}\mathbf{k}}

\end{array}.$

283.

$f(x,y,z) = 4x^{5}y^{2}z^{3},\begin{array}{ll}

{P(2,-1,1),} & {\mathbf{u} = \frac{1}{3}\mathbf{i} + \frac{2}{3}\mathbf{j} - \frac{2}{3}\mathbf{k}}

\end{array}$

For the following exercises, find the directional derivative of the function at point $P$ in the direction of $Q.$

284\.

$f(x,y) = x^{2} + 3y^{2},\begin{array}{ll}

{P(1,1),} & {Q(4,5)}

\end{array}$

285.

$f(x,y,z) = \frac{y}{x + z},\begin{array}{ll}

{P(2,1,-1),} & {Q(-1,2,0)}

\end{array}$

For the following exercises, find the derivative of the function at $P$ in the direction of $\mathbf{u}.$

286\.

$f(x,y) = -7x + 2y,\begin{array}{ll}

{P(2,-4),} & {\mathbf{u} = 4\mathbf{i} - 3\mathbf{j}}

\end{array}$

287.

$f(x,y) = \text{ln}(5x + 4y),\begin{array}{ll}

{P(3,9),} & {\mathbf{u} = 6\mathbf{i} + 8\mathbf{j}}

\end{array}$

288\.

\[T\] Use technology to sketch the level curve of $f(x,y) = 4x - 2y + 3$ that passes through $P(1,2)$ and draw the gradient vector at $P.$

289.

\[T\] Use technology to sketch the level curve of $f(x,y) = x^{2} + 4y^{2}$ that passes through $P(-2,0)$ and draw the gradient vector at $P.$

For the following exercises, find the gradient vector at the indicated point.

290\.

$f(x,y) = xy^{2} - yx^{2},P(-1,1)$

291.

$f(x,y) = xe^{y} - \text{ln}(x),P\left( {3,0} \right)$

292\.

$f(x,y,z) = xy - \text{ln}(z),P(2,-2,2)$

293.

$f(x,y,z) = x\sqrt{y^{2} + z^{2}},P(-2,-1,-1)$

For the following exercises, find the derivative of the function.

294\.

$f(x,y) = x^{2} + xy + y^{2}$ at point $\left( {-5,-4} \right)$ in the direction the function increases most rapidly

295.

$f(x,y) = e^{xy}$ at point $\left( {6,7} \right)$ in the direction the function increases most rapidly

296\.

$f(x,y) = \text{arctan}\left( \frac{y}{x} \right)$ at point $\left( {-9,9} \right)$ in the direction the function increases most rapidly

297.

$f(x,y,z) = \text{ln}(xy + yz + zx)$ at point $\left( {-9,-18,-27} \right)$ in the direction the function increases most rapidly

298\.

$f(x,y,z) = \frac{x}{y} + \frac{y}{z} + \frac{z}{x}$ at point $\left( {5,-5,5} \right)$ in the direction the function increases most rapidly

For the following exercises, find the maximum rate of change of $f$ at the given point and the direction in which it occurs.

299.

$f(x,y) = xe^{\text{−}y},$ $\left( {1,0} \right)$

300\.

$f(x,y) = \sqrt{x^{2} + 2y},$ $\left( {4,10} \right)$

301.

$f(x,y) = \text{cos}(3x + 2y),\left( {\frac{\pi}{6}, - \frac{\pi}{8}} \right)$

For the following exercises, find equations of

1. the tangent plane and

2. the normal line to the given surface at the given point.

302\.

The level surface $f(x,y,z) = 12$ for $f(x,y,z) = 4x^{2} - 2y^{2} + z^{2}$ at point $\left( {2,2,2} \right).$

303.

$f(x,y,z) = xy + yz + xz = 3$ at point $\left( {1,1,1} \right)$

304\.

$f(x,y,z) = xyz = 6$ at point $\left( {1,2,3} \right)$

305.

$f(x,y,z) = xe^{y}\text{cos}\ z - z = 1$ at point $\left( {1,0,0} \right)$

For the following exercises, solve the problem.

306\.

The temperature $T$ in a metal sphere is inversely proportional to the distance from the center of the sphere (the origin: $\left( {0,0,0} \right)).$ The temperature at point $\left( {1,2,2} \right)$ is $120\text{°}\text{C}.$

1. Find the rate of change of the temperature at point $\left( {1,2,2} \right)$ in the direction toward point $\left( {2,1,3} \right).$

2. Show that, at any point in the sphere, the direction of greatest increase in temperature is given by a vector that points toward the origin.

307.

The electrical potential (voltage) in a certain region of space is given by the function $V(x,y,z) = 5x^{2} - 3xy + xyz.$

1. Find the rate of change of the voltage at point $\left( {3,4,5} \right)$ in the direction of the vector $\left\langle {1,1,-1} \right\rangle.$

2. In which direction does the voltage change most rapidly at point $\left( {3,4,5} \right)?$

3. What is the maximum rate of change of the voltage at point $\left( {3,4,5} \right)?$

308\.

If the electric potential at a point $\left( {x,y} \right)$ in the *xy*-plane is $V(x,y) = e^{-2x}\text{cos}(2y),$ then the electric intensity vector at $\left( {x,y} \right)$ is $\mathbf{E} = \text{−}\nabla V(x,y).$

1. Find the electric intensity vector at $\left( {\frac{\pi}{4},0} \right).$

2. Show that, at each point in the plane, the electric potential decreases most rapidly in the direction of the vector $\mathbf{E}.$

309.

In two dimensions, the motion of an ideal fluid is governed by a velocity potential $\varphi.$ The velocity components of the fluid $u$ in the *x-*direction and $v$ in the *y*-direction, are given by $\left\langle {u,v} \right\rangle = \nabla\varphi.$ Find the velocity components associated with the velocity potential $\varphi(x,y) = \text{sin}\ \pi x\ \text{sin}\ 2\pi y.$

---

4.7 Maxima/Minima Problems

> 来源: OpenStax《Calculus Volume 3》| 原页: https://openstax.org/books/calculus-volume-3/pages/4-7-maxima-minima-problems

4.7 Maxima/Minima Problems

One of the most useful applications for derivatives of a function of one variable is the determination of maximum and/or minimum values. This application is also important for functions of two or more variables, but as we have seen in earlier sections of this chapter, the introduction of more independent variables leads to more possible outcomes for the calculations. The main ideas of finding critical points and using derivative tests are still valid, but new wrinkles appear when assessing the results.

Critical Points

For functions of a single variable, we defined critical points as the values of the function when the derivative equals zero or does not exist. For functions of two or more variables, the concept is essentially the same, except for the fact that we are now working with partial derivatives.

Let $z = f\left( {x,y} \right)$ be a function of two variables that is defined on an open set containing the point $\left( {x_{0},y_{0}} \right).$ The point $\left( {x_{0},y_{0}} \right)$ is called a critical point of a function of two variables $f$ if one of the two following conditions holds:

1. $f_{x}\left( {x_{0},y_{0}} \right) = f_{y}\left( {x_{0},y_{0}} \right) = 0$

2. Either $f_{x}\left( {x_{0},y_{0}} \right)\ \text{or}\ f_{y}\left( {x_{0},y_{0}} \right)$ does not exist.

Finding Critical Points

Find the critical points of each of the following functions:

1. $f\left( {x,y} \right) = \sqrt{4y^{2} - 9x^{2} + 24y + 36x + 36}$

2. $g\left( {x,y} \right) = x^{2} + 2xy - 4y^{2} + 4x - 6y + 4$

Solution

1. First, we calculate $f_{x}\left( {x,y} \right)\ \text{and}\ f_{y}\left( {x,y} \right)\text{:}$

$$\begin{array}{cll}

{f_{x}\left( {x,y} \right)} & = & {\frac{1}{2}\left( {-18x + 36} \right)\left( {4y^{2} - 9x^{2} + 24y + 36x + 36} \right)^{-1\text{/}2}} \\

& = & \frac{-9x + 18}{\sqrt{4y^{2} - 9x^{2} + 24y + 36x + 36}} \\

{f_{y}\left( {x,y} \right)} & = & {\frac{1}{2}\left( {8y + 24} \right)\left( {4y^{2} - 9x^{2} + 24y + 36x + 36} \right)^{-1\text{/}2}} \\

& = & {\frac{4y + 12}{\sqrt{4y^{2} - 9x^{2} + 24y + 36x + 36}}.}

\end{array}$$

Next, we set each of these expressions equal to zero:

$$\begin{array}{rll}

\frac{-9x + 18}{\sqrt{4y^{2} - 9x^{2} + 24y + 36x + 36}} & = & 0 \\

\frac{4y + 12}{\sqrt{4y^{2} - 9x^{2} + 24y + 36x + 36}} & = & {0.}

\end{array}$$

Then, multiply each equation by its common denominator:

$$\begin{array}{rll}

{- 9x + 18} & = & 0 \\

{4y + 12} & = & {0.}

\end{array}$$

Therefore, $x = 2$ and $y = -3,$ so $\left( {2,-3} \right)$ is a critical point of $f.$

We must also check for the possibility that the denominator of each partial derivative can equal zero, thus causing the partial derivative not to exist. Since the denominator is the same in each partial derivative, we need only do this once:

$$4y^{2} - 9x^{2} + 24y + 36x + 36 = 0.$$

This equation represents a hyperbola. We should also note that the domain of $f$ consists of points satisfying the inequality

$$4y^{2} - 9x^{2} + 24y + 36x + 36 \geq 0.$$

Therefore, any points on the hyperbola are not only critical points, they are also on the boundary of the domain. To put the hyperbola in standard form, we use the method of completing the square:

$$\begin{array}{rll}

{4y^{2} - 9x^{2} + 24y + 36x + 36} & = & 0 \\

{4y^{2} - 9x^{2} + 24y + 36x} & = & -36 \\

{4y^{2} + 24y - 9x^{2} + 36x} & = & -36 \\

{4\left( {y^{2} + 6y} \right) - 9\left( {x^{2} - 4x} \right)} & = & -36 \\

{4\left( {y^{2} + 6y + 9} \right) - 9\left( {x^{2} - 4x + 4} \right)} & = & {-36 + 36 - 36} \\

{4\left( {y + 3} \right)^{2} - 9\left( {x - 2} \right)^{2}} & = & {-36.}

\end{array}$$

Dividing both sides by $-36$ puts the equation in standard form:

$$\begin{array}{rll}

{\frac{4\left( {y + 3} \right)^{2}}{-36} - \frac{9\left( {x - 2} \right)^{2}}{-36}} & = & 1 \\

{\frac{\left( {x - 2} \right)^{2}}{4} - \frac{\left( {y + 3} \right)^{2}}{9}} & = & {1.}

\end{array}$$

Notice that point $\left( {2,-3} \right)$ is the center of the hyperbola.

2. First, we calculate $g_{x}\left( {x,y} \right)\ \text{and}\ g_{y}\left( {x,y} \right)\text{:}$

$$\begin{array}{rll}

{g_{x}\left( {x,y} \right)} & = & {2x + 2y + 4} \\

{g_{y}\left( {x,y} \right)} & = & {2x - 8y - 6.}

\end{array}$$

Next, we set each of these expressions equal to zero, which gives a system of equations in $x\ \text{and}\ y\text{:}$

$$\begin{array}{rll}

{2x + 2y + 4} & = & 0 \\

{2x - 8y - 6} & = & {0.}

\end{array}$$

Subtracting the second equation from the first gives $10y + 10 = 0,\ \text{so}\ y = -1.$ Substituting this into the first equation gives $2x + 2(-1) + 4 = 0,$ so $x = -1.$ Therefore $\left( {-1,-1} \right)$ is a critical point of $g$ (Figure 4.46). There are no points in $\mathbb{R}^{2}$ that make either partial derivative not exist.

Find the critical point of the function $f\left( {x,y} \right) = x^{3} + 2xy - 2x - 4y.$

The main purpose for determining critical points is to locate relative maxima and minima, as in single-variable calculus. When working with a function of one variable, the definition of a local extremum involves finding an interval around the critical point such that the function value is either greater than or less than all the other function values in that interval. When working with a function of two or more variables, we work with an open disk around the point.

Let $z = f\left( {x,y} \right)$ be a function of two variables that is defined and continuous on an open set containing the point $\left( {x_{0},y_{0}} \right).$ Then *f* has a *local maximum* at $\left( {x_{0},y_{0}} \right)$ if

$$f\left( {x_{0},y_{0}} \right) \geq f\left( {x,y} \right)$$

for all points $\left( {x,y} \right)$ within some disk centered at $\left( {x_{0},y_{0}} \right).$ The number $f\left( {x_{0},y_{0}} \right)$ is called a *local maximum value*. If the preceding inequality holds for every point $\left( {x,y} \right)$ in the domain of $f,$ then $f$ has a *global maximum* (also called an *absolute maximum*) at $\left( {x_{0},y_{0}} \right).$

The function $f$ has a *local minimum* at $\left( {x_{0},y_{0}} \right)$ if

$$f\left( {x_{0},y_{0}} \right) \leq f\left( {x,y} \right)$$

for all points $\left( {x,y} \right)$ within some disk centered at $\left( {x_{0},y_{0}} \right).$ The number $f\left( {x_{0},y_{0}} \right)$ is called a *local minimum value*. If the preceding inequality holds for every point $\left( {x,y} \right)$ in the domain of $f,$ then $f$ has a *global minimum* (also called an *absolute minimum*) at $\left( {x_{0},y_{0}} \right).$

If $f\left( {x_{0},y_{0}} \right)$ is either a local maximum or local minimum value, then it is called a *local extremum* (see the following figure).

In Maxima and Minima, we showed that extrema of functions of one variable occur at critical points. The same is true for functions of more than one variable, as stated in the following theorem.

Fermat’s Theorem for Functions of Two Variables

Let $z = f\left( {x,y} \right)$ be a function of two variables that is defined and continuous on an open set containing the point $(x_{0},y_{0}).$ Suppose $f_{x}$ and $f_{y}$ each exists at $\left( {x_{0},y_{0}} \right).$ If $f$ has a local extremum at $\left( {x_{0},y_{0}} \right),$ then $\left( {x_{0},y_{0}} \right)$ is a critical point of $f.$

Second Derivative Test

Consider the function $f(x) = x^{3}.$ This function has a critical point at $x = 0,$ since $f\prime(0) = 3(0)^{2} = 0.$ However, $f$ does not have an extreme value at $x = 0.$ Therefore, the existence of a critical value at $x = x_{0}$ does not guarantee a local extremum at $x = x_{0}.$ The same is true for a function of two or more variables. One way this can happen is at a saddle point. An example of a saddle point appears in the following figure.

In this graph, the origin is a saddle point. This is because the first partial derivatives of $f\left( {x,y} \right) = x^{2} - y^{2}$ are both equal to zero at this point, but it is neither a maximum nor a minimum for the function. Furthermore the vertical trace corresponding to $y = 0$ is $z = x^{2}$ (a parabola opening upward), but the vertical trace corresponding to $x = 0$ is $z = \text{−}y^{2}$ (a parabola opening downward). Therefore, it is both a global maximum for one trace and a global minimum for another.

Given the function $z = f\left( {x,y} \right),$ the point $\left( {x_{0},y_{0},f\left( {x_{0},y_{0}} \right)} \right)$ is a saddle point if both $f_{x}\left( {x_{0},y_{0}} \right) = 0$ and $f_{y}\left( {x_{0},y_{0}} \right) = 0,$ but $f$ does not have a local extremum at $\left( {x_{0},y_{0}} \right).$

The second derivative test for a function of one variable provides a method for determining whether an extremum occurs at a critical point of a function. When extending this result to a function of two variables, an issue arises related to the fact that there are, in fact, four different second-order partial derivatives, although equality of mixed partials reduces this to three. The second derivative test for a function of two variables, stated in the following theorem, uses a discriminant $D$ that replaces $f^{''}\left( x_{0} \right)$ in the second derivative test for a function of one variable.

Second Derivative Test

Let $z = f\left( {x,y} \right)$ be a function of two variables for which the first- and second-order partial derivatives are continuous on some disk containing the point $\left( {x_{0},y_{0}} \right).$ Suppose $f_{x}\left( {x_{0},y_{0}} \right) = 0$ and $f_{y}\left( {x_{0},y_{0}} \right) = 0.$ Define the quantity

$$D = f_{xx}\left( {x_{0},y_{0}} \right)f_{yy}\left( {x_{0},y_{0}} \right) - \left( {f_{xy}\left( {x_{0},y_{0}} \right)} \right)^{2}.$$ (4.43)

1. If $D > 0$ and $f_{xx}\left( {x_{0},y_{0}} \right) > 0,$ then $f$ has a local minimum at $\left( {x_{0},y_{0}} \right).$

2. If $D > 0$ and $f_{xx}\left( {x_{0},y_{0}} \right) < 0,$ then $f$ has a local maximum at $\left( {x_{0},y_{0}} \right).$

3. If $D < 0,$ then $f$ has a saddle point at $\left( {x_{0},y_{0}} \right).$

4. If $D = 0,$ then the test is inconclusive.

See Figure 4.49.

To apply the second derivative test, it is necessary that we first find the critical points of the function. There are several steps involved in the entire procedure, which are outlined in a problem-solving strategy.

Using the Second Derivative Test for Functions of Two Variables

Let $z = f\left( {x,y} \right)$ be a function of two variables for which the first- and second-order partial derivatives are continuous on some disk containing the point $\left( {x_{0},y_{0}} \right).$ To apply the second derivative test to find local extrema, use the following steps:

1. Determine the critical points $\left( {x_{0},y_{0}} \right)$ of the function $f$ where $f_{x}\left( {x_{0},y_{0}} \right) = f_{y}\left( {x_{0},y_{0}} \right) = 0.$ Discard any points where at least one of the partial derivatives does not exist.

2. Calculate the discriminant $D = f_{xx}\left( {x_{0},y_{0}} \right)f_{yy}\left( {x_{0},y_{0}} \right) - \left( {f_{xy}\left( {x_{0},y_{0}} \right)} \right)^{2}$ for each critical point of $f.$

3. Apply Second Derivative Test to determine whether each critical point is a local maximum, local minimum, or saddle point, or whether the theorem is inconclusive.

Using the Second Derivative Test

Find the critical points for each of the following functions, and use the second derivative test to find the local extrema:

1. $f\left( {x,y} \right) = 4x^{2} + 9y^{2} + 8x - 36y + 24$

2. $g\left( {x,y} \right) = \frac{1}{3}x^{3} + y^{2} + 2xy - 6x - 3y + 4$

Solution

1. Step $1$ of the problem-solving strategy involves finding the critical points of $f.$ To do this, we first calculate $f_{x}\left( {x,y} \right)$ and $f_{y}\left( {x,y} \right),$ then set each of them equal to zero:

$$\begin{array}{rll}

{f_{x}\left( {x,y} \right)} & = & {8x + 8} \\

{f_{y}\left( {x,y} \right)} & = & {18y - 36.}

\end{array}$$

Setting them equal to zero yields the system of equations

$$\begin{array}{rll}

{8x + 8} & = & 0 \\

{18y - 36} & = & {0.}

\end{array}$$

The solution to this system is $x = -1$ and $y = 2.$ Therefore $\left( {-1,2} \right)$ is a critical point of $f.$

Step 2 of the problem-solving strategy involves calculating $D.$ To do this, we first calculate the second partial derivatives of $f\text{:}$

$$\begin{array}{rll}

{f_{xx}\left( {x,y} \right)} & = & 8 \\

{f_{xy}\left( {x,y} \right)} & = & 0 \\

{f_{yy}\left( {x,y} \right)} & = & {18.}

\end{array}$$

Therefore, $D = f_{xx}\left( {-1,2} \right)f_{yy}\left( {-1,2} \right) - \left( {f_{xy}\left( {-1,2} \right)} \right)^{2} = (8)(18) - (0)^{2} = 144.$

Step 3 states to check the Second Derivative Test for Functions of Two Variables.. Since $D > 0$ and $f_{xx}\left( {-1,2} \right) > 0,$ this corresponds to case 1. Therefore, $f$ has a local minimum at $\left( {-1,2} \right)$ as shown in the following figure.

2. For step 1, we first calculate $g_{x}\left( {x,y} \right)$ and $g_{y}\left( {x,y} \right),$ then set each of them equal to zero:

$$\begin{array}{rll}

{g_{x}\left( {x,y} \right)} & = & {x^{2} + 2y - 6} \\

{g_{y}\left( {x,y} \right)} & = & {2y + 2x - 3.}

\end{array}$$

Setting them equal to zero yields the system of equations

$$\begin{array}{rll}

{x^{2} + 2y - 6} & = & 0 \\

{2y + 2x - 3} & = & {0.}

\end{array}$$

To solve this system, first solve the second equation for *y.* This gives $y = \frac{3 - 2x}{2}.$ Substituting this into the first equation gives

$$\begin{array}{rll}

{x^{2} + 3 - 2x - 6} & = & 0 \\

{x^{2} - 2x - 3} & = & 0 \\

{\left( {x - 3} \right)\left( {x + 1} \right)} & = & {0.}

\end{array}$$

Therefore, $x = -1$ or $x = 3.$ Substituting these values into the equation $y = \frac{3 - 2x}{2}$ yields the critical points $\left( {-1,\frac{5}{2}} \right)$ and $\left( {3, - \frac{3}{2}} \right).$

Step 2 involves calculating the second partial derivatives of $g\text{:}$

$$\begin{array}{rll}

{g_{xx}\left( {x,y} \right)} & = & {2x} \\

{g_{xy}\left( {x,y} \right)} & = & 2 \\

{g_{yy}\left( {x,y} \right)} & = & {2.}

\end{array}$$

Then, we find a general formula for $D\text{:}$

$$\begin{array}{cl}

D & {= g_{xx}\left( {x_{0},y_{0}} \right)g_{yy}\left( {x_{0},y_{0}} \right) - \left( {g_{xy}\left( {x_{0},y_{0}} \right)} \right)^{2}} \\

& {= \left( {2x_{0}} \right)(2) - 2^{2}} \\

& {= 4x_{0} - 4.}

\end{array}$$

Next, we substitute each critical point into this formula:

$$\begin{array}{rll}

{D\left( {-1,\frac{5}{2}} \right)} & = & {\left( {2(-1)} \right)(2) - (2)^{2} = -4 - 4 = -8} \\

{D\left( {3, - \frac{3}{2}} \right)} & = & {\left( {2(3)} \right)(2) - (2)^{2} = 12 - 4 = 8.}

\end{array}$$

In step 3, we note that, applying the Second Derivative Test for Functions of Two Variables. to point $\left( {-1,\frac{5}{2}} \right)$ leads to case $3,$ which means that $\left( {-1,\frac{5}{2}} \right)$ is a saddle point. Applying the theorem to point $\left( {3, - \frac{3}{2}} \right)$ leads to case 1, which means that $\left( {3, - \frac{3}{2}} \right)$ corresponds to a local minimum as shown in the following figure.

Use the second derivative to find the local extrema of the function

$$f\left( {x,y} \right) = x^{3} + 2xy - 6x - 4y^{2}.$$

Absolute Maxima and Minima

When finding global extrema of functions of one variable on a closed interval, we start by checking the critical values over that interval and then evaluate the function at the endpoints of the interval. When working with a function of two variables, the closed interval is replaced by a closed, bounded set. A set is *bounded* if all the points in that set can be contained within a ball (or disk) of finite radius. First, we need to find the critical points inside the set and calculate the corresponding critical values. Then, it is necessary to find the maximum and minimum value of the function on the boundary of the set. When we have all these values, the largest function value corresponds to the global maximum and the smallest function value corresponds to the absolute minimum. First, however, we need to be assured that such values exist. The following theorem does this.

Extreme Value Theorem

A continuous function $f(x,y)$ on a closed and bounded set $D$ in the plane attains an absolute maximum value at some point of $D$ and an absolute minimum value at some point of $D.$

Now that we know any continuous function $f$ defined on a closed, bounded set attains its extreme values, we need to know how to find them.

Finding Extreme Values of a Function of Two Variables

Assume $z = f\left( {x,y} \right)$ is a differentiable function of two variables defined on a closed, bounded set $D.$ Then $f$ will attain the absolute maximum value and the absolute minimum value, which are, respectively, the largest and smallest values found among the following:

1. The values of $f$ at the critical points of $f$ in $D.$

2. The values of $f$ on the boundary of $D.$

The proof of this theorem is a direct consequence of the extreme value theorem and Fermat’s theorem. In particular, if either extremum is not located on the boundary of $D,$ then it is located at an interior point of $D.$ But an interior point $\left( {x_{0},y_{0}} \right)$ of $D$ that’s an absolute extremum is also a local extremum; hence, $\left( {x_{0},y_{0}} \right)$ is a critical point of $f$ by Fermat’s theorem. Therefore the only possible values for the global extrema of $f$ on $D$ are the extreme values of $f$ on the interior or boundary of $D.$

Finding Absolute Maximum and Minimum Values

Let $z = f\left( {x,y} \right)$ be a continuous function of two variables defined on a closed, bounded set $D,$ and assume $f$ is differentiable on $D.$ To find the absolute maximum and minimum values of $f$ on $D,$ do the following:

1. Determine the critical points of $f$ in $D.$

2. Calculate $f$ at each of these critical points.

3. Determine the maximum and minimum values of $f$ on the boundary of its domain.

4. The maximum and minimum values of $f$ will occur at one of the values obtained in steps $2\ \text{and}\ 3.$

Finding the maximum and minimum values of $f$ on the boundary of $D$ can be challenging. If the boundary is a rectangle or set of straight lines, then it is possible to parameterize the line segments and determine the maxima on each of these segments, as seen in Example 4.40. The same approach can be used for other shapes such as circles and ellipses.

If the boundary of the set $D$ is a more complicated curve defined by a function $g\left( {x,y} \right) = c$ for some constant $c,$ and the first-order partial derivatives of $g$ exist, then the method of Lagrange multipliers can prove useful for determining the extrema of $f$ on the boundary. The method of Lagrange multipliers is introduced in Lagrange Multipliers.

Finding Absolute Extrema

Use the problem-solving strategy for finding absolute extrema of a function to determine the absolute extrema of each of the following functions:

1. $f\left( {x,y} \right) = x^{2} - 2xy + 4y^{2} - 4x - 2y + 24$ on the domain defined by $0 \leq x \leq 4$ and $0 \leq y \leq 2$

2. $g\left( {x,y} \right) = x^{2} + y^{2} + 4x - 6y$ on the domain defined by $x^{2} + y^{2} \leq 16$

Solution

1. Using the problem-solving strategy, step $1$ involves finding the critical points of $f$ on its domain. Therefore, we first calculate $f_{x}\left( {x,y} \right)$ and $f_{y}\left( {x,y} \right),$ then set them each equal to zero:

$$\begin{array}{rll}

{f_{x}\left( {x,y} \right)} & = & {2x - 2y - 4} \\

{f_{y}\left( {x,y} \right)} & = & {-2x + 8y - 2.}

\end{array}$$

Setting them equal to zero yields the system of equations

$$\begin{array}{rll}

{2x - 2y - 4} & = & 0 \\

{- 2x + 8y - 2} & = & {0.}

\end{array}$$

The solution to this system is $x = 3$ and $y = 1.$ Therefore $\left( {3,1} \right)$ is a critical point of $f.$ Calculating $f\left( {3,1} \right)$ gives $f\left( {3,1} \right) = 17.$

The next step involves finding the extrema of $f$ on the boundary of its domain. The boundary of its domain consists of four line segments as shown in the following graph:

$L_{1}$ is the line segment connecting $\left( {0,0} \right)$ and $\left( {4,0} \right),$ and it can be parameterized by the equations $x(t) = t,y(t) = 0$ for $0 \leq t \leq 4.$ Define $g(t) = f\left( {x(t),y(t)} \right).$ This gives $g(t) = t^{2} - 4t + 24.$ Differentiating *g* leads to $g^{\prime}(t) = 2t - 4.$ Therefore, $g$ has a critical value at $t = 2,$ which corresponds to the point $\left( {2,0} \right).$ Calculating $f\left( {2,0} \right)$ gives the *z-*value $20.$

$L_{2}$ is the line segment connecting $\left( {4,0} \right)$ and $\left( {4,2} \right),$ and it can be parameterized by the equations $x(t) = 4,y(t) = t$ for $0 \leq t \leq 2.$ Again, define $g(t) = f\left( {x(t),y(t)} \right).$ This gives $g(t) = 4t^{2} - 10t + 24.$ Then, $g^{\prime}(t) = 8t - 10.$ $g$ has a critical value at $t = \frac{5}{4},$ which corresponds to the point $\left( {4,\frac{5}{4}} \right).$ Calculating $f\left( {4,\frac{5}{4}} \right)$ gives the *z-*value $17.75.$

$L_{3}$ is the line segment connecting $\left( {0,2} \right)$ and $\left( {4,2} \right),$ and it can be parameterized by the equations $x(t) = t,y(t) = 2$ for $0 \leq t \leq 4.$ Again, define $g(t) = f\left( {x(t),y(t)} \right).$ This gives $g(t) = t^{2} - 8t + 36.$ The critical value corresponds to the point $\left( {4,2} \right).$ So, calculating $f\left( {4,2} \right)$ gives the *z-*value $20.$

$L_{4}$ is the line segment connecting $\left( {0,0} \right)$ and $\left( {0,2} \right),$ and it can be parameterized by the equations $x(t) = 0,y(t) = t$ for $0 \leq t \leq 2.$ This time, $g(t) = 4t^{2} - 2t + 24$ and the critical value $t = \frac{1}{4}$ correspond to the point $\left( {0,\frac{1}{4}} \right).$ Calculating $f\left( {0,\frac{1}{4}} \right)$ gives the *z-*value $23.75.$

We also need to find the values of $f\left( {x,y} \right)$ at the corners of its domain. These corners are located at $\left( {0,0} \right),\left( {4,0} \right),\left( {4,2} \right)\ \text{and}\ \left( {0,2} \right)\text{:}$

$$\begin{array}{rllll}

& & & & \\

{f\left( {0,0} \right)} & = & {(0)^{2} - 2(0)(0) + 4(0)^{2} - 4(0) - 2(0) + 24} & = & 24 \\

{f\left( {4,0} \right)} & = & {(4)^{2} - 2(4)(0) + 4(0)^{2} - 4(4) - 2(0) + 24} & = & 24 \\

{f\left( {4,2} \right)} & = & {(4)^{2} - 2(4)(2) + 4(2)^{2} - 4(4) - 2(2) + 24} & = & 20 \\

{f\left( {0,2} \right)} & = & {(0)^{2} - 2(0)(2) + 4(2)^{2} - 4(0) - 2(2) + 24} & = & {36.}

\end{array}$$

The absolute maximum value is $36,$ which occurs at $\left( {0,2} \right),$ and the global minimum value is $17,$ which occurs at $\left( {3,1} \right)$ as shown in the following figure.

2. Using the problem-solving strategy, step $1$ involves finding the critical points of $g$ on its domain. Therefore, we first calculate $g_{x}\left( {x,y} \right)$ and $g_{y}\left( {x,y} \right),$ then set them each equal to zero:

$$\begin{array}{rll}

{g_{x}\left( {x,y} \right)} & = & {2x + 4} \\

{g_{y}\left( {x,y} \right)} & = & {2y - 6.}

\end{array}$$

Setting them equal to zero yields the system of equations

$$\begin{array}{rll}

{2x + 4} & = & 0 \\

{2y - 6} & = & {0.}

\end{array}$$

The solution to this system is $x = -2$ and $y = 3.$ Therefore, $\left( {-2,3} \right)$ is a critical point of $g.$ Calculating $g\left( {-2,3} \right),$ we get

$$g\left( {-2,3} \right) = (-2)^{2} + 3^{2} + 4(-2) - 6(3) = 4 + 9 - 8 - 18 = -13.$$

The next step involves finding the extrema of *g* on the boundary of its domain. The boundary of its domain consists of a circle of radius $4$ centered at the origin as shown in the following graph.

The boundary of the domain of $g$ can be parameterized using the functions $x(t) = 4\ \text{cos}\ t,y(t) = 4\ \text{sin}\ t$ for $0 \leq t \leq 2\pi.$ Define $h(t) = g\left( {x(t),y(t)} \right)\text{:}$

$$\begin{array}{cl}

{h(t)} & {= g\left( {x(t),y(t)} \right)} \\

& {= \left( {4\ \text{cos}\ t} \right)^{2} + \left( {4\ \text{sin}\ t} \right)^{2} + 4\left( {4\ \text{cos}\ t} \right) - 6\left( {4\ \text{sin}\ t} \right)} \\

& {= 16\text{cos}^{2}\mspace{2mu} t + 16\text{sin}^{2}\mspace{2mu} t + 16\ \text{cos}\ t - 24\ \text{sin}\ t} \\

& {= 16 + 16\ \text{cos}\ t - 24\ \text{sin}\ t.}

\end{array}$$

Setting $h^{\prime}(t) = 0$ leads to

$$\begin{array}{rll}

{- 16\ \text{sin}\ t - 24\ \text{cos}\ t} & = & 0 \\

{- 16\ \text{sin}\ t} & = & {24\ \text{cos}\ t} \\

\frac{-16\ \text{sin}\ t}{-16\ \text{cos}\ t} & = & \frac{24\ \text{cos}\ t}{-16\ \text{cos}\ t} \\

{\text{tan}\ t} & = & {- \frac{3}{2}.}

\end{array}$$

This equation has two solutions over the interval $0 \leq t \leq 2\pi.$ One is $t = \pi - \text{arctan}\left( \frac{3}{2} \right)$ and the other is $t = 2\pi - \text{arctan}\left( \frac{3}{2} \right).$ For the first angle,

$$\begin{array}{rll}

{\text{sin}\ t} & = & {\text{sin}\left( {\pi - \text{arctan}\left( \frac{3}{2} \right)} \right) = \text{sin}\left( {\text{arctan}\left( \frac{3}{2} \right)} \right) = \frac{3\sqrt{13}}{13}} \\

{\text{cos}\ t} & = & {\text{cos}\left( {\pi - \text{arctan}\left( \frac{3}{2} \right)} \right) = \text{−}\text{cos}\left( {\text{arctan}\left( \frac{3}{2} \right)} \right) = - \frac{2\sqrt{13}}{13}.}

\end{array}$$

Therefore, $x(t) = 4\ \text{cos}\ t = - \frac{8\sqrt{13}}{13}$ and $y(t) = 4\ \text{sin}\ t = \frac{12\sqrt{13}}{13},$ so $\left( {- \frac{8\sqrt{13}}{13},\frac{12\sqrt{13}}{13}} \right)$ is a critical point on the boundary and

$$\begin{array}{cl}

{g\left( {- \frac{8\sqrt{13}}{13},\frac{12\sqrt{13}}{13}} \right)} & {= \left( {- \frac{8\sqrt{13}}{13}} \right)^{2} + \left( \frac{12\sqrt{13}}{13} \right)^{2} + 4\left( {- \frac{8\sqrt{13}}{13}} \right) - 6\left( \frac{12\sqrt{13}}{13} \right)} \\

& {= \frac{144}{13} + \frac{64}{13} - \frac{32\sqrt{13}}{13} - \frac{72\sqrt{13}}{13}} \\

& {= \frac{208 - 104\sqrt{13}}{13} \approx -12.844.}

\end{array}$$

For the second angle,

$$\begin{array}{rll}

{\text{sin}\ t} & = & {\text{sin}\left( {2\pi - \text{arctan}\left( \frac{3}{2} \right)} \right) = \text{−}\text{sin}\left( {\text{arctan}\left( \frac{3}{2} \right)} \right) = - \frac{3\sqrt{13}}{13}} \\

{\text{cos}\ t} & = & {\text{cos}\left( {2\pi - \text{arctan}\left( \frac{3}{2} \right)} \right) = \text{cos}\left( {\text{arctan}\left( \frac{3}{2} \right)} \right) = \frac{2\sqrt{13}}{13}.}

\end{array}$$

Therefore, $x(t) = 4\ \text{cos}\ t = \frac{8\sqrt{13}}{13}$ and $y(t) = 4\ \text{sin}\ t = - \frac{12\sqrt{13}}{13},$ so $\left( {\frac{8\sqrt{13}}{13}, - \frac{12\sqrt{13}}{13}} \right)$ is a critical point on the boundary and

$$\begin{array}{cl}

{g\left( {\frac{8\sqrt{13}}{13}, - \frac{12\sqrt{13}}{13}} \right)} & {= \left( \frac{8\sqrt{13}}{13} \right)^{2} + \left( {- \frac{12\sqrt{13}}{13}} \right)^{2} + 4\left( \frac{8\sqrt{13}}{13} \right) - 6\left( {- \frac{12\sqrt{13}}{13}} \right)} \\

& {= \frac{144}{13} + \frac{64}{13} + \frac{32\sqrt{13}}{13} + \frac{72\sqrt{13}}{13}} \\

& {= \frac{208 + 104\sqrt{13}}{13} \approx 44.844.}

\end{array}$$

The absolute minimum of *g* is $-13,$ which is attained at the point $\left( {-2,3} \right),$ which is an interior point of *D*. The absolute maximum of *g* is approximately equal to 44.844, which is attained at the boundary point $\left( {\frac{8\sqrt{13}}{13}, - \frac{12\sqrt{13}}{13}} \right).$ These are the absolute extrema of *g* on *D* as shown in the following figure.

Use the problem-solving strategy for finding absolute extrema of a function to find the absolute extrema of the function

$$f\left( {x,y} \right) = 4x^{2} - 2xy + 6y^{2} - 8x + 2y + 3$$

on the domain defined by $0 \leq x \leq 2$ and $-1 \leq y \leq 3.$

Chapter Opener: Profitable Golf Balls

Pro-$\text{T}$ company has developed a profit model that depends on the number *x* of golf balls sold per month (measured in thousands), and the number of hours per month of advertising *y*, according to the function

$$z = f\left( {x,y} \right) = 48x + 96y - x^{2} - 2xy - 9y^{2},$$

where $z$ is measured in thousands of dollars. The maximum number of golf balls that can be produced and sold is $50,000,$ and the maximum number of hours of advertising that can be purchased is $25.$ Find the values of $x$ and $y$ that maximize profit, and find the maximum profit.

Solution

Using the problem-solving strategy, step $1$ involves finding the critical points of $f$ on its domain. Therefore, we first calculate $f_{x}\left( {x,y} \right)$ and $f_{y}\left( {x,y} \right),$ then set them each equal to zero:

$$\begin{array}{rll}

{f_{x}\left( {x,y} \right)} & = & {48 - 2x - 2y} \\

{f_{y}\left( {x,y} \right)} & = & {96 - 2x - 18y.}

\end{array}$$

Setting them equal to zero yields the system of equations

$$\begin{array}{rll}

{48 - 2x - 2y} & = & 0 \\

{96 - 2x - 18y} & = & {0.}

\end{array}$$

The solution to this system is $x = 21$ and $y = 3.$ Therefore $\left( {21,3} \right)$ is a critical point of $f.$ Calculating $f\left( {21,3} \right)$ gives $f\left( {21,3} \right) = 48(21) + 96(3) - 21^{2} - 2(21)(3) - 9(3)^{2} = 648.$

The domain of this function is $0 \leq x \leq 50$ and $0 \leq y \leq 25$ as shown in the following graph.

$L_{1}$ is the line segment connecting $\left( {0,0} \right)$ and $\left( {50,0} \right),$ and it can be parameterized by the equations $x(t) = t,y(t) = 0$ for $0 \leq t \leq 50.$ We then define $g(t) = f\left( {x(t),y(t)} \right)\text{:}$

$$\begin{array}{cl}

{g(t)} & {= f\left( {x(t),y(t)} \right)} \\

& {= f\left( {t,0} \right)} \\

& {= 48t + 96(0) - t^{2} - 2(t)(0) - 9(0)^{2}} \\

& {= 48t - t^{2}.}

\end{array}$$

Setting $g^{\prime}(t) = 0$ yields the critical point $t = 24,$ which corresponds to the point $\left( {24,0} \right)$ in the domain of $f.$ Calculating $f\left( {24,0} \right)$ gives $576.$

$L_{2}$ is the line segment connecting and $\left( {50,25} \right),$ and it can be parameterized by the equations $x(t) = 50,y(t) = t$ for $0 \leq t \leq 25.$ Once again, we define $g(t) = f\left( {x(t),y(t)} \right)\text{:}$

$$\begin{array}{cl}

{g(t)} & {= f\left( {x(t),y(t)} \right)} \\

& {= f\left( {50,t} \right)} \\

& {= 48(50) + 96t - 50^{2} - 2(50)t - 9t^{2}} \\

& {= -9t^{2} - 4t - 100.}

\end{array}$$

This function has a critical point at $t = - \frac{2}{9},$ which corresponds to the point $\left( {50, - \frac{2}{9}} \right).$ This point is not in the domain of $f.$

$L_{3}$ is the line segment connecting $\left( {0,25} \right)\ \text{and}\ \left( {50,25} \right),$ and it can be parameterized by the equations $x(t) = t,y(t) = 25$ for $0 \leq t \leq 50.$ We define $g(t) = f\left( {x(t),y(t)} \right)\text{:}$

$$\begin{array}{cl}

{g(t)} & {= f\left( {x(t),y(t)} \right)} \\

& {= f\left( {t,25} \right)} \\

& {= 48t + 96(25) - t^{2} - 2t(25) - 9\left( 25^{2} \right)} \\

& {= \text{−}t^{2} - 2t - 3225.}

\end{array}$$

This function has a critical point at $t = -1,$ which corresponds to the point $\left( {-1,25} \right),$ which is not in the domain.

$L_{4}$ is the line segment connecting $\left( {0,0} \right)\ \text{to}\ \left( {0,25} \right),$ and it can be parameterized by the equations $x(t) = 0,y(t) = t$ for $0 \leq t \leq 25.$ We define $g(t) = f\left( {x(t),y(t)} \right)\text{:}$

$$\begin{array}{cl}

{g(t)} & {= f\left( {x(t),y(t)} \right)} \\

& {= f\left( {0,t} \right)} \\

& {= 48(0) + 96t - (0)^{2} - 2(0)t - 9t^{2}} \\

& {= 96t - t^{2}.}

\end{array}$$

This function has a critical point at $t = \frac{16}{3},$ which corresponds to the point $\left( {0,\frac{16}{3}} \right),$ which is on the boundary of the domain. Calculating $f\left( {0,\frac{16}{3}} \right)$ gives $256.$

We also need to find the values of $f\left( {x,y} \right)$ at the corners of its domain. These corners are located at $\left( {0,0} \right),\left( {50,0} \right),\left( {50,25} \right)\ \text{and}\ \left( {0,25} \right)\text{:}$

$$\begin{array}{rll}

{f\left( {0,0} \right)} & = & {48(0) + 96(0) - (0)^{2} - 2(0)(0) - 9(0)^{2} = 0} \\

{f\left( {50,0} \right)} & = & {48(50) + 96(0) - (50)^{2} - 2(50)(0) - 9(0)^{2} = -100} \\

{f\left( {50,25} \right)} & = & {48(50) + 96(25) - (50)^{2} - 2(50)(25) - 9(25)^{2} = -5825} \\

{f\left( {0,25} \right)} & = & {48(0) + 96(25) - (0)^{2} - 2(0)(25) - 9(25)^{2} = -3225.}

\end{array}$$

The maximum critical value is $648,$ which occurs at $\left( {21,3} \right).$ Therefore, a maximum profit of $\text{\$}648,000$ is realized when $21,000$ golf balls are sold and $3$ hours of advertising are purchased per month as shown in the following figure.

Section 4.7 Exercises

For the following exercises, find all critical points.

310\.

$f(x,y) = 1 + x^{2} + y^{2}$

311.

$f(x,y) = {(3x - 2)}^{2} + {(y - 4)}^{2}$

312\.

$f(x,y) = x^{4} + y^{4} - 16xy$

313.

$f(x,y) = 15x^{3} - 3xy + 15y^{3}$

For the following exercises, find the critical points of the function by using algebraic techniques (completing the square) or by examining the form of the equation. Verify your results using the partial derivatives test.

314\.

$f(x,y) = \sqrt{x^{2} + y^{2} + 1}$

315.

$f(x,y) = \text{−}x^{2} - 5y^{2} + 8x - 10y - 13$

316\.

$f(x,y) = x^{2} + y^{2} + 2x - 6y + 6$

317.

$f(x,y) = \sqrt{x^{2} + y^{2}} + 1$

For the following exercises, use the second derivative test to classify any critical points and determine whether each critical point is a maximum, minimum, saddle point, or none of these.

318\.

$f(x,y) = \text{−}x^{3} + 4xy - 2y^{2} + 1$

319.

$f(x,y) = x^{2}y^{2}$

320\.

$f(x,y) = x^{2} - 6x + y^{2} + 4y - 8$

321.

$f(x,y) = 2xy + 3x + 4y$

322\.

$f(x,y) = 8xy(x + y) + 7$

323.

$f(x,y) = x^{2} + 4xy + y^{2}$

324\.

$f(x,y) = x^{3} + y^{3} - 300x - 75y - 3$

325.

$f(x,y) = 9 - x^{4}y^{4}$

326\.

$f(x,y) = 7x^{2}y + 9xy^{2}$

327.

$f(x,y) = 3x^{2} - 2xy + y^{2} - 8y$

328\.

$f(x,y) = 3x^{2} + 2xy + y^{2}$

329.

$f(x,y) = y^{2} + xy + 3y + 2x + 3$

330\.

$f(x,y) = x^{2} + xy + y^{2} - 3x$

331.

$f(x,y) = x^{2} + 2y^{2} - x^{2}y$

332\.

$f(x,y) = x^{2} + y - e^{y}$

333.

$f(x,y) = e^{\text{−}(x^{2} + y^{2} + 2x)}$

334\.

$f(x,y) = x^{2} + xy + y^{2} - x - y + 1$

335.

$f(x,y) = x^{2} + 10xy + y^{2}$

336\.

$f(x,y) = \text{−}x^{2} - 5y^{2} + 10x - 30y - 62$

337.

$f(x,y) = 120x + 120y - xy - x^{2} - y^{2}$

338\.

$f(x,y) = 2x^{2} + 2xy + y^{2} + 2x - 3$

339.

$f(x,y) = x^{2} + x - 3xy + y^{3} - 5$

340\.

$f(x,y) = 2xye^{\text{−}x^{2} - y^{2}}$

For the following exercises, determine the extreme values and the saddle points. Use a CAS to graph the function.

341.

\[T\] $f(x,y) = ye^{x} - e^{y}$

342\.

\[T\] $f(x,y) = x\ \text{sin}(y)$

343.

\[T\] $f(x,y) = \text{sin}(x)\text{sin}(y),x \in \left( {0,2\pi} \right),y \in \left( {0,2\pi} \right)$

Find the absolute extrema of the given function on the indicated closed and bounded set $R.$

344\.

$f(x,y) = xy - x - 3y;$ $R$ is the triangular region with vertices $\left( {0,0} \right),\left( {0,4} \right),\ \text{and}\ \left( {5,0} \right).$

345.

Find the absolute maximum and minimum values of $f(x,y) = x^{2} + y^{2} - 2y + 1$ on the region $R = \left\{ {(x,y)\left| {x^{2} + y^{2} \leq 4} \right.} \right\}.$

346\.

$f(x,y) = x^{3} - 3xy - y^{3}$ on $R = \left\{ (x,y)\text{:}\ -2 \leq x \leq 2,-2 \leq y \leq 2 \right\}$

347.

$f(x,y) = \frac{-2y}{x^{2} + y^{2} + 1}$ on $R = \left\{ {(x,y)\text{:}\ x^{2} + y^{2} \leq 4} \right\}$

348\.

Find three positive numbers the sum of which is $27,$ such that the sum of their squares is as small as possible.

349.

Find the points on the surface $x^{2} - yz = 5$ that are closest to the origin.

350\.

Find the maximum volume of a rectangular box with three faces in the coordinate planes and a vertex in the first octant on the plane $x + y + z = 1.$

351.

The sum of the length and the girth (perimeter of a cross-section) of a package carried by a delivery service cannot exceed $108$ in. Find the dimensions of the rectangular package of largest volume that can be sent.

352\.

A cardboard box without a lid is to be made with a volume of $4$ ft3. Find the dimensions of the box that requires the least amount of cardboard.

353.

Find the point on the surface $f(x,y) = x^{2} + y^{2} + 10$ nearest the plane $x + 2y - z = 0.$ Identify the point on the plane.

354\.

Find the point in the plane $2x - y + 2z = 16$ that is closest to the origin.

355.

A company manufactures two types of athletic shoes: jogging shoes and cross-trainers. The total revenue from $x$ jogging shoes and $y$ cross-trainers is given by $R(x,y) = -5x^{2} - 8y^{2} - 2xy + 42x + 102y,$ where $x$ and $y$ are in thousands of units. Find the values of *x* and *y* to maximize the total revenue.

356\.

A shipping company handles rectangular boxes provided the sum of the length, width, and height of the box does not exceed $96$ in. Find the dimensions of the box that meets this condition and has the largest volume.

357.

Find the maximum volume of a cylindrical soda can such that the sum of its height and circumference is $120$ cm.

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4.8 Lagrange Multipliers

> 来源: OpenStax《Calculus Volume 3》| 原页: https://openstax.org/books/calculus-volume-3/pages/4-8-lagrange-multipliers

4.8 Lagrange Multipliers

Solving optimization problems for functions of two or more variables can be similar to solving such problems in single-variable calculus. However, techniques for dealing with multiple variables allow us to solve more varied optimization problems for which we need to deal with additional conditions or constraints. In this section, we examine one of the more common and useful methods for solving optimization problems with constraints.

Lagrange Multipliers

Example 4.41 was an applied situation involving maximizing a profit function, subject to certain constraints. In that example, the constraints involved a maximum number of golf balls that could be produced and sold in $1$ month $(x),$ and a maximum number of advertising hours that could be purchased per month $(y).$ Suppose these were combined into a budgetary constraint, such as $20x + 4y \leq 216,$ that took into account the cost of producing the golf balls and the number of advertising hours purchased per month. The goal is, still, to maximize profit, but now there is a different type of constraint on the values of $x$ and $y.$ This constraint, when combined with the profit function $f\left( {x,y} \right) = 48x + 96y - x^{2} - 2xy - 9y^{2},$ is an example of an optimization problem, and the function $f\left( {x,y} \right)$ is called the objective function. A graph of various level curves of the function $f\left( {x,y} \right)$ follows.

In Figure 4.59, the value $c$ represents different profit levels (i.e., values of the function $f).$ As the value of $c$ increases, the curve shifts to the right. Since our goal is to maximize profit, we want to choose a curve as far to the right as possible. If there was no restriction on the number of golf balls the company could produce, or the number of units of advertising available, then we could produce as many golf balls as we want, and advertise as much as we want, and there would not be a maximum profit for the company. Unfortunately, we have a budgetary constraint that is modeled by the inequality $20x + 4y \leq 216.$ To see how this constraint interacts with the profit function, Figure 4.60 shows the graph of the line $20x + 4y = 216$ superimposed on the previous graph.

As mentioned previously, the maximum profit occurs when the level curve is as far to the right as possible. However, the level of production corresponding to this maximum profit must also satisfy the budgetary constraint, so the point at which this profit occurs must also lie on (or to the left of) the red line in Figure 4.60. Inspection of this graph reveals that this point exists where the line is tangent to the level curve of $f.$ Trial and error reveals that this profit level seems to be around $395,$ when $x$ and $y$ are both just less than $5.$ We return to the solution of this problem later in this section. From a theoretical standpoint, at the point where the profit curve is tangent to the constraint line, the gradient of both of the functions evaluated at that point must point in the same (or opposite) direction. Recall that the gradient of a function of more than one variable is a vector. If two vectors point in the same (or opposite) directions, then one must be a constant multiple of the other. This idea is the basis of the method of Lagrange multipliers.

Method of Lagrange Multipliers: One Constraint

Let $f$ and $g$ be functions of two variables with continuous partial derivatives at every point of some open set containing the smooth curve $g\left( {x,y} \right) = 0.$ Suppose that $f,$ when restricted to points on the curve $g\left( {x,y} \right) = 0,$ has a local extremum at the point $\left( {x_{0},y_{0}} \right)$ and that $\nabla g\left( {x_{0},y_{0}} \right) \neq 0.$ Then there is a number $\lambda$ called a Lagrange multiplier, for which

$$\nabla f\left( {x_{0},y_{0}} \right) = \lambda\nabla g\left( {x_{0},y_{0}} \right).$$

Proof

Assume that a constrained extremum occurs at the point $\left( {x_{0},y_{0}} \right).$ Furthermore, we assume that the equation $g\left( {x,y} \right) = 0$ can be smoothly parameterized as

$$x = x(s)\ \text{and}\ y = y(s)$$

where *s* is an arc length parameter with reference point $\left( {x_{0},y_{0}} \right)$ at $s = 0.$ Therefore, the quantity $z = f\left( {x(s),y(s)} \right)$ has a relative maximum or relative minimum at $s = 0,$ and this implies that $\frac{dz}{ds} = 0$ at that point. From the chain rule,

$$\frac{dz}{ds} = \frac{\partial f}{\partial x} \cdot \frac{dx}{ds} + \frac{\partial f}{\partial y} \cdot \frac{dy}{ds} = \left( \frac{\partial f}{\partial x}\hat{\mathbf{i}} + \frac{\partial f}{\partial y}\hat{\mathbf{j}} \right) \cdot \left( \frac{dx}{ds}\hat{\mathbf{i}} + \frac{dy}{ds}\hat{\mathbf{j}} \right) = 0,$$

where the derivatives are all evaluated at $s = 0.$ However, the first factor in the dot product is the gradient of $f,$ and the second factor is the unit tangent vector $\text{T}(0)$ to the constraint curve. Since the point $\left( {x_{0},y_{0}} \right)$ corresponds to $s = 0,$ it follows from this equation that

$$\nabla f\left( {x_{0},y_{0}} \right) \cdot \text{T}(0) = 0,$$

which implies that the gradient is either $\mathbf{0}$ or is normal to the constraint curve at a constrained relative extremum. However, the constraint curve $g\left( {x,y} \right) = 0$ is a level curve for the function $g\left( {x,y} \right)$ so that if $\nabla g\left( {x_{0},y_{0}} \right) \neq 0$ then $\nabla g\left( {x_{0},y_{0}} \right)$ is normal to this curve at $\left( {x_{0},y_{0}} \right)$ It follows, then, that there is some scalar $\lambda$ such that

$$\nabla f\left( {x_{0},y_{0}} \right) = \lambda\nabla g\left( {x_{0},y_{0}} \right)$$

□

To apply Method of Lagrange Multipliers: One Constraint to an optimization problem similar to that for the golf ball manufacturer, we need a problem-solving strategy.

Steps for Using Lagrange Multipliers

1. Determine the objective function $f\left( {x,y} \right)$ and the constraint function $g\left( {x,y} \right).$ Does the optimization problem involve maximizing or minimizing the objective function?

2. Set up a system of equations using the following template:

$$\begin{array}{rll}

{\nabla f\left( {x_{0},y_{0}} \right)} & = & {\lambda\nabla g\left( {x_{0},y_{0}} \right)} \\

{g\left( {x_{0},y_{0}} \right)} & = & {0.}

\end{array}$$

3. Solve for $x_{0}$ and $y_{0}.$

4. The largest of the values of $f$ at the solutions found in step $3$ maximizes $f;$ the smallest of those values minimizes $f.$

Using Lagrange Multipliers

Use the method of Lagrange multipliers to find the minimum value of $f\left( {x,y} \right) = x^{2} + 4y^{2} - 2x + 8y$ subject to the constraint $x + 2y = 7.$

Solution

Let’s follow the problem-solving strategy:

1. The optimization function is $f\left( {x,y} \right) = x^{2} + 4y^{2} - 2x + 8y.$ To determine the constraint function, we must first subtract $7$ from both sides of the constraint. This gives $x + 2y - 7 = 0.$ The constraint function is equal to the left-hand side, so $g\left( {x,y} \right) = x + 2y - 7.$ The problem asks us to solve for the minimum value of $f,$ subject to the constraint (see the following graph).

2. We then must calculate the gradients of both *f* and *g*:

$$\begin{array}{l}

{\nabla f\left( {x,y} \right) = \left( {2x - 2} \right)\mathbf{i} + \left( {8y + 8} \right)\mathbf{j}} \\

{\nabla g\left( {x,y} \right) = \mathbf{i} + 2\mathbf{j}.}

\end{array}$$

The equation $\nabla f\left( {x_{0},y_{0}} \right) = \lambda\nabla g\left( {x_{0},y_{0}} \right)$ becomes

$$\left( {2x_{0} - 2} \right)\mathbf{i} + \left( {8y_{0} + 8} \right)\mathbf{j} = \lambda\left( {\mathbf{i} + 2\mathbf{j}} \right),$$

which can be rewritten as

$$\left( {2x_{0} - 2} \right)\mathbf{i} + \left( {8y_{0} + 8} \right)\mathbf{j} = \lambda\mathbf{i} + {2\lambda}\mathbf{j}.$$

Next, we set the coefficients of $\mathbf{i}\ \text{and}\ \mathbf{j}$ equal to each other:

$$\begin{array}{l}

{2x_{0} - 2 = \lambda} \\

{8y_{0} + 8 = 2\lambda.}

\end{array}$$

The equation $g\left( {x_{0},y_{0}} \right) = 0$ becomes $x_{0} + 2y_{0} - 7 = 0.$ Therefore, the system of equations that needs to be solved is

$$\begin{array}{rll}

{2x_{0} - 2} & = & \lambda \\

{8y_{0} + 8} & = & {2\lambda} \\

{x_{0} + 2y_{0} - 7} & = & 0.

\end{array}$$

3. This is a linear system of three equations in three variables. We start by solving the second equation for $\lambda$ and substituting it into the first equation. This gives $\lambda = 4y_{0} + 4,$ so substituting this into the first equation gives

$$2x_{0} - 2 = 4y_{0} + 4.$$

Solving this equation for $x_{0}$ gives $x_{0} = 2y_{0} + 3.$ We then substitute this into the third equation: $$\begin{array}{rll}

{\left( {2y_{0} + 3} \right) + 2y_{0} - 7} & = & 0 \\

{4y_{0} - 4} & = & 0 \\

y_{0} & = & 1.

\end{array}$$

Since $x_{0} = 2y_{0} + 3,$ this gives $x_{0} = 5.$

4. Next, we substitute $\left( {5,1} \right)$ into $f\left( {x,y} \right) = x^{2} + 4y^{2} - 2x + 8y,$ gives $f\left( {5,1} \right) = 5^{2} + 4(1)^{2} - 2(5) + 8(1) = 27.$ To ensure this corresponds to a minimum value on the constraint function, let’s try some other values, such as the intercepts of $g\left( {x,y} \right) = 0,$ Which are $\left( {7,0} \right)$ and $\left( {0,3.5} \right).$ We get $f\left( {7,0} \right) = 35$ and $f\left( {0,3.5} \right) = 77,$ so it appears $f$ has a minimum at $\left( {5,1} \right).$

Use the method of Lagrange multipliers to find the maximum value of $f\left( {x,y} \right) = 9x^{2} + 36xy - 4y^{2} - 18x - 8y$ subject to the constraint $3x + 4y = 32.$

Let’s now return to the problem posed at the beginning of the section.

Golf Balls and Lagrange Multipliers

The golf ball manufacturer, Pro-T, has developed a profit model that depends on the number $x$ of golf balls sold per month (measured in thousands), and the number of hours per month of advertising *y*, according to the function

$$z = f\left( {x,y} \right) = 48x + 96y - x^{2} - 2xy - 9y^{2},$$

where $z$ is measured in thousands of dollars. The budgetary constraint function relating the cost of the production of thousands golf balls and advertising units is given by $20x + 4y = 216.$ Find the values of $x$ and $y$ that maximize profit, and find the maximum profit.

Solution

Again, we follow the problem-solving strategy:

1. The optimization function is $f\left( {x,y} \right) = 48x + 96y - x^{2} - 2xy - 9y^{2}.$ To determine the constraint function, we first subtract 216 from both sides of the constraint, then divide both sides by $4,$ which gives $5x + y - 54 = 0.$ The constraint function is equal to the left-hand side, so $g\left( {x,y} \right) = 5x + y - 54.$ The problem asks us to solve for the maximum value of $f,$ subject to this constraint.

2. So, we calculate the gradients of both $f\ \text{and}$ $g\text{:}$

$$\begin{array}{l}

{\nabla f\left( {x,y} \right) = \left( {48 - 2x - 2y} \right)\mathbf{i} + \left( {96 - 2x - 18y} \right)\mathbf{j}} \\

{\nabla g\left( {x,y} \right) = 5\mathbf{i} + \mathbf{j}.}

\end{array}$$

The equation $\nabla f\left( {x_{0},y_{0}} \right) = \lambda\nabla g\left( {x_{0},y_{0}} \right)$ becomes

$$\left( {48 - 2x_{0} - 2y_{0}} \right)\mathbf{i} + \left( {96 - 2x_{0} - 18y_{0}} \right)\mathbf{j} = \lambda\left( {5\mathbf{i} + \mathbf{j}} \right),$$

which can be rewritten as

$$\left( {48 - 2x_{0} - 2y_{0}} \right)\mathbf{i} + \left( {96 - 2x_{0} - 18y_{0}} \right)\mathbf{j} = \lambda 5\mathbf{i} + \lambda\mathbf{j}.$$

We then set the coefficients of $\mathbf{i}\ \text{and}\ \mathbf{j}$ equal to each other:

$$\begin{array}{rll}

{48 - 2x_{0} - 2y_{0}} & = & {5\lambda} \\

{96 - 2x_{0} - 18y_{0}} & = & {\lambda.}

\end{array}$$

The equation $g\left( {x_{0},y_{0}} \right) = 0$ becomes $5x_{0} + y_{0} - 54 = 0.$ Therefore, the system of equations that needs to be solved is

$$\begin{array}{rll}

{48 - 2x_{0} - 2y_{0}} & = & {5\lambda} \\

{96 - 2x_{0} - 18y_{0}} & = & \lambda \\

{5x_{0} + y_{0} - 54} & = & 0.

\end{array}$$

3. We use the left-hand side of the second equation to replace $\lambda$ in the first equation:

$$\begin{array}{rll}

{48 - 2x_{0} - 2y_{0}} & = & {5\left( {96 - 2x_{0} - 18y_{0}} \right)} \\

{48 - 2x_{0} - 2y_{0}} & = & {480 - 10x_{0} - 90y_{0}} \\

{8x_{0}} & = & {432 - 88y_{0}} \\

x_{0} & = & {54 - 11y_{0}.}

\end{array}$$

Then we substitute this into the third equation:

$$\begin{array}{rll}

{5\left( {54 - 11y_{0}} \right) + y_{0} - 54} & = & 0 \\

{270 - 55y_{0} + y_{0}} & = & 0 \\

{216 - 54y_{0}} & = & 0 \\

y_{0} & = & 4.

\end{array}$$

Since $x_{0} = 54 - 11y_{0},$ this gives $x_{0} = 10.$

4. We then substitute $\left( {10,4} \right)$ into $f\left( {x,y} \right) = 48x + 96y - x^{2} - 2xy - 9y^{2},$ which gives

$$\begin{array}{cl}

{f\left( {10,4} \right)} & {= 48(10) + 96(4) - (10)^{2} - 2(10)(4) - 9(4)^{2}} \\

& {= 480 + 384 - 100 - 80 - 144 = 540.}

\end{array}$$

Therefore the maximum profit that can be attained, subject to budgetary constraints, is $\text{\$}540,000$ with a production level of $10,000$ golf balls and $4$ hours of advertising bought per month. Let’s check to make sure this truly is a maximum. The endpoints of the line that defines the constraint are $\left( {10.8,0} \right)$ and $\left( {0,54} \right)$ Let’s evaluate $f$ at both of these points:

$$\begin{array}{rll}

{f\left( {10.8,0} \right)} & = & {48(10.8) + 96(0) - 10.8^{2} - 2(10.8)(0) - 9\left( 0^{2} \right) = 401.76} \\

{f\left( {0,54} \right)} & = & {48(0) + 96(54) - 0^{2} - 2(0)(54) - 9\left( 54^{2} \right) = -21,060.}

\end{array}$$

The second value represents a loss, since no golf balls are produced. Neither of these values exceed $540,$ so it seems that our extremum is a maximum value of $f.$

A company has determined that its production level is given by the Cobb-Douglas function $f\left( {x,y} \right) = 2.5x^{0.45}y^{0.55}$ where *x* represents the total number of labor hours in $1$ year and *y* represents the total capital input for the company. Suppose $1$ unit of labor costs $\text{\$}40$ and $1$ unit of capital costs $\text{\$}50.$ Use the method of Lagrange multipliers to find the maximum value of $f\left( {x,y} \right) = 2.5x^{0.45}y^{0.55}$ subject to a budgetary constraint of $\text{\$}500,000$ per year.

In the case of an optimization function with three variables and a single constraint function, it is possible to use the method of Lagrange multipliers to solve an optimization problem as well. An example of an optimization function with three variables could be the Cobb-Douglas function in the previous example: $f\left( {x,y,z} \right) = x^{0.2}y^{0.4}z^{0.4},$ where $x$ represents the cost of labor, $y$ represents capital input, and $z$ represents the cost of advertising. The method is the same as for the method with a function of two variables; the equations to be solved are

$$\begin{array}{rll}

{\nabla f\left( {x,y,z} \right)} & = & {\lambda\nabla g\left( {x,y,z} \right)} \\

{g\left( {x,y,z} \right)} & = & 0.

\end{array}$$

Lagrange Multipliers with a Three-Variable Optimization Function

Find the minimum of the function $f\left( {x,y,z} \right) = x^{2} + y^{2} + z^{2}$ subject to the constraint $x + y + z = 1.$

Solution

1. The optimization function is $f\left( {x,y,z} \right) = x^{2} + y^{2} + z^{2}.$ To determine the constraint function, we subtract $1$ from each side of the constraint: $x + y + z - 1 = 0$ which gives the constraint function as $g\left( {x,y,z} \right) = x + y + z - 1.$

2. Next, we calculate $\nabla f\left( {x,y,z} \right)$ and $\nabla g\left( {x,y,z} \right)\text{:}$

$$\begin{array}{l}

{\nabla f\left( {x,y,z} \right) = \left\langle 2x,2y,2z \right\rangle} \\

{\nabla g\left( {x,y,z} \right) = \left\langle 1,1,1 \right\rangle\text{.}}

\end{array}$$

This leads to the equations

$$\begin{array}{rll}

\left\langle 2x_{0},2y_{0},2z_{0} \right\rangle & = & {\lambda\left\langle 1,1,1 \right\rangle} \\

{x_{0} + y_{0} + z_{0} - 1} & = & 0

\end{array}$$

which can be rewritten in the following form:

$$\begin{array}{rll}

{2x_{0}} & = & \lambda \\

{2y_{0}} & = & \lambda \\

{2z_{0}} & = & \lambda \\

{x_{0} + y_{0} + z_{0} - 1} & = & 0.

\end{array}$$

3. Since each of the first three equations has $\lambda$ on the right-hand side, we know that $2x_{0} = 2y_{0} = 2z_{0}$ and all three variables are equal to each other. Substituting $y_{0} = x_{0}$ and $z_{0} = x_{0}$ into the last equation yields $3x_{0} - 1 = 0,$ so $x_{0} = \frac{1}{3}$ and $y_{0} = \frac{1}{3}$ and $z_{0} = \frac{1}{3}$ which corresponds to a critical point on the constraint curve.

4. Then, we evaluate *f* at the point $\left( {\frac{1}{3},\frac{1}{3},\frac{1}{3}} \right)\text{:}$

$$f\left( {\frac{1}{3},\frac{1}{3},\frac{1}{3}} \right) = \left( \frac{1}{3} \right)^{2} + \left( \frac{1}{3} \right)^{2} + \left( \frac{1}{3} \right)^{2} = \frac{3}{9} = \frac{1}{3}.$$

Therefore, an extremum of the function is $\frac{1}{3}.$ To verify it is a minimum, choose other points that satisfy the constraint and calculate $f$ at that point. For example,

$$\begin{array}{rll}

{f\left( {1,0,0} \right)} & = & {1^{2} + 0^{2} + 0^{2} = 1} \\

{f\left( {0,-2,3} \right)} & = & {0^{2} + (-2)^{2} + 3^{2} = 13.}

\end{array}$$

Both of these values are greater than $\frac{1}{3},$ leading us to believe the extremum is a minimum.

Use the method of Lagrange multipliers to find the minimum value of the function

$$f\left( {x,y,z} \right) = x + y + z$$

subject to the constraint $x^{2} + y^{2} + z^{2} = 1.$

Problems with Two Constraints

The method of Lagrange multipliers can be applied to problems with more than one constraint. In this case the optimization function, $w$ is a function of three variables:

$$w = f\left( {x,y,z} \right)$$

and it is subject to two constraints:

$$g\left( {x,y,z} \right) = 0\ \text{and}\ h\left( {x,y,z} \right) = 0.$$

There are two Lagrange multipliers, $\lambda_{1}$ and $\lambda_{2},$ and the system of equations becomes

$$\begin{array}{rll}

{\nabla f\left( {x_{0},y_{0},z_{0}} \right)} & = & {\lambda_{1}\nabla g\left( {x_{0},y_{0},z_{0}} \right) + \lambda_{2}\nabla h\left( {x_{0},y_{0},z_{0}} \right)} \\

{g\left( {x_{0},y_{0},z_{0}} \right)} & = & 0 \\

{h\left( {x_{0},y_{0},z_{0}} \right)} & = & 0.

\end{array}$$

Lagrange Multipliers with Two Constraints

Find the local extreme values of the function

$$f\left( {x,y,z} \right) = x^{2} + y^{2} + z^{2}$$

subject to the constraints $z^{2} = x^{2} + y^{2}$ and $x + y - z + 1 = 0.$

Solution

Let’s follow the problem-solving strategy:

1. The optimization function is $f\left( {x,y,z} \right) = x^{2} + y^{2} + z^{2}.$ To determine the constraint functions, we first subtract $z^{2}$ from both sides of the first constraint, which gives $x^{2} + y^{2} - z^{2} = 0,$ so $g\left( {x,y,z} \right) = x^{2} + y^{2} - z^{2}.$ The second constraint function is $h\left( {x,y,z} \right) = x + y - z + 1.$

2. We then calculate the gradients of $f,g,\ \text{and}\ h\text{:}$

$$\begin{array}{l}

{\nabla f\left( {x,y,z} \right) = 2x\mathbf{i} + 2y\mathbf{j} + 2z\mathbf{k}} \\

{\nabla g\left( {x,y,z} \right) = 2x\mathbf{i} + 2y\mathbf{j} - 2z\mathbf{k}} \\

{\nabla h\left( {x,y,z} \right) = \mathbf{i} + \mathbf{j} - \mathbf{k}.}

\end{array}$$

The equation $\nabla f\left( {x_{0},y_{0},z_{0}} \right) = \lambda_{1}\nabla g\left( {x_{0},y_{0},z_{0}} \right) + \lambda_{2}\nabla h\left( {x_{0},y_{0},z_{0}} \right)$ becomes

$$2x_{0}\mathbf{i} + 2y_{0}\mathbf{j} + 2z_{0}\mathbf{k} = \lambda_{1}\left( {2x_{0}\mathbf{i} + 2y_{0}\mathbf{j} - 2z_{0}\mathbf{k}} \right) + \lambda_{2}\left( {\mathbf{i} + \mathbf{j} - \mathbf{k}} \right),$$

which can be rewritten as

$$2x_{0}\mathbf{i} + 2y_{0}\mathbf{j} + 2z_{0}\mathbf{k} = \left( {2\lambda_{1}x_{0} + \lambda_{2}} \right)\mathbf{i} + \left( {2\lambda_{1}y_{0} + \lambda_{2}} \right)\mathbf{j} - \left( {2\lambda_{1}z_{0} + \lambda_{2}} \right)\mathbf{k}.$$

Next, we set the coefficients of $\mathbf{i}\ \text{,}\ \mathbf{j}\text{, and}\ \mathbf{k}$ equal to each other:

$$\begin{array}{l}

{2x_{0} = 2\lambda_{1}x_{0} + \lambda_{2}} \\

{2y_{0} = 2\lambda_{1}y_{0} + \lambda_{2}} \\

{2z_{0} = -2\lambda_{1}z_{0} - \lambda_{2}.}

\end{array}$$

The two equations that arise from the constraints are $z_{0}{}^{2} = x_{0}{}^{2} + y_{0}{}^{2}$ and $x_{0} + y_{0} - z_{0} + 1 = 0.$ Combining these equations with the previous three equations gives

$$\begin{array}{rll}

{2x_{0}} & = & {2\lambda_{1}x_{0} + \lambda_{2}} \\

{2y_{0}} & = & {2\lambda_{1}y_{0} + \lambda_{2}} \\

{2z_{0}} & = & {-2\lambda_{1}z_{0} - \lambda_{2}} \\

{z_{0}{}^{2}} & = & {x_{0}{}^{2} + y_{0}{}^{2}} \\

{x_{0} + y_{0} - z_{0} + 1} & = & 0.

\end{array}$$

3. The first three equations contain the variable $\lambda_{2}.$ Solving the third equation for $\lambda_{2}$ and replacing into the first and second equations reduces the number of equations to four:

$$\begin{array}{rll}

{2x_{0}} & = & {2\lambda_{1}x_{0} - 2\lambda_{1}z_{0} - 2z_{0}} \\

{2y_{0}} & = & {2\lambda_{1}y_{0} - 2\lambda_{1}z_{0} - 2z_{0}} \\

{z_{0}{}^{2}} & = & {x_{0}{}^{2} + y_{0}{}^{2}} \\

{x_{0} + y_{0} - z_{0} + 1} & = & 0.

\end{array}$$

Next, we solve the first and second equation for $\lambda_{1}.$ The first equation gives $\lambda_{1} = \frac{x_{0} + z_{0}}{x_{0} - z_{0}},$ the second equation gives $\lambda_{1} = \frac{y_{0} + z_{0}}{y_{0} - z_{0}}.$ We set the right-hand side of each equation equal to each other and cross-multiply:

$$\begin{array}{rll}

\frac{x_{0} + z_{0}}{x_{0} - z_{0}} & = & \frac{y_{0} + z_{0}}{y_{0} - z_{0}} \\

{\left( {x_{0} + z_{0}} \right)\left( {y_{0} - z_{0}} \right)} & = & {\left( {x_{0} - z_{0}} \right)\left( {y_{0} + z_{0}} \right)} \\

{x_{0}y_{0} - x_{0}z_{0} + y_{0}z_{0} - z_{0}{}^{2}} & = & {x_{0}y_{0} + x_{0}z_{0} - y_{0}z_{0} - z_{0}2} \\

{2y_{0}z_{0} - 2x_{0}z_{0}} & = & 0 \\

{2z_{0}\left( {y_{0} - x_{0}} \right)} & = & 0.

\end{array}.$$

Therefore, either $z_{0} = 0$ or $y_{0} = x_{0}.$ If $z_{0} = 0,$ then the first constraint becomes $0 = x_{0}{}^{2} + y_{0}{}^{2}.$ The only real solution to this equation is $x_{0} = 0$ and $y_{0} = 0,$ which gives the ordered triple $\left( {0,0,0} \right).$ This point does not satisfy the second constraint, so it is not a solution.

Next, we consider $y_{0} = x_{0},$ which reduces the number of equations to three:

$$\begin{array}{rll}

y_{0} & = & x_{0} \\

{z_{0}{}^{2}} & = & {x_{0}{}^{2} + y_{0}{}^{2}} \\

{x_{0} + y_{0} - z_{0} + 1} & = & 0.

\end{array}$$

We substitute the first equation into the second and third equations:

$$\begin{array}{rll}

{z_{0}{}^{2}} & = & {x_{0}{}^{2} + x_{0}{}^{2}} \\

{x_{0} + x_{0} - z_{0} + 1} & = & 0.

\end{array}$$

Then, we solve the second equation for $z_{0},$ which gives $z_{0} = 2x_{0} + 1.$ We then substitute this into the first equation,

$$\begin{array}{rll}

{z_{0}{}^{2}} & = & {2x_{0}{}^{2}} \\

\left( {2x_{0} + 1} \right)^{2} & = & {2x_{0}{}^{2}} \\

{4x_{0}{}^{2} + 4x_{0} + 1} & = & {2x_{0}{}^{2}} \\

{2x_{0}{}^{2} + 4x_{0} + 1} & = & {0,}

\end{array}$$

and use the quadratic formula to solve for $x_{0}\text{:}$

$$x_{0} = \frac{-4 \pm \sqrt{4^{2} - 4(2)(1)}}{2(2)} = \frac{-4 \pm \sqrt{8}}{4} = \frac{-4 \pm 2\sqrt{2}}{4} = -1 \pm \frac{\sqrt{2}}{2}.$$

Recall $y_{0} = x_{0},$ so this solves for $y_{0}$ as well. Then, $z_{0} = 2x_{0} + 1,$ so

$$z_{0} = 2x_{0} + 1 = 2\left( {-1 \pm \frac{\sqrt{2}}{2}} \right) + 1 = -2 + 1 \pm \sqrt{2} = -1 \pm \sqrt{2}.$$

Therefore, there are two ordered triplet solutions:

$$\left( {-1 + \frac{\sqrt{2}}{2},-1 + \frac{\sqrt{2}}{2},-1 + \sqrt{2}} \right)\ \text{and}\ \left( {-1 - \frac{\sqrt{2}}{2},-1 - \frac{\sqrt{2}}{2},-1 - \sqrt{2}} \right).$$

4. We substitute $\left( {-1 + \frac{\sqrt{2}}{2},-1 + \frac{\sqrt{2}}{2},-1 + \sqrt{2}} \right)$ into $f\left( {x,y,z} \right) = x^{2} + y^{2} + z^{2},$ which gives

$$\begin{array}{cl}

{f\left( {-1 + \frac{\sqrt{2}}{2},-1 + \frac{\sqrt{2}}{2},-1 + \sqrt{2}} \right)} & {= \left( {-1 + \frac{\sqrt{2}}{2}} \right)^{2} + \left( {-1 + \frac{\sqrt{2}}{2}} \right)^{2} + \left( {-1 + \sqrt{2}} \right)^{2}} \\

& {= \left( {1 - \sqrt{2} + \frac{1}{2}} \right) + \left( {1 - \sqrt{2} + \frac{1}{2}} \right) + \left( {1 - 2\sqrt{2} + 2} \right)} \\

& {= 6 - 4\sqrt{2}.}

\end{array}$$

Then, we substitute $\left( {-1 - \frac{\sqrt{2}}{2},-1 - \frac{\sqrt{2}}{2},-1 - \sqrt{2}} \right)$ into $f\left( {x,y,z} \right) = x^{2} + y^{2} + z^{2},$ which gives

$$\begin{array}{cl}

{f\left( {-1 - \frac{\sqrt{2}}{2},-1 - \frac{\sqrt{2}}{2},-1 - \sqrt{2}} \right)} & {= \left( {-1 - \frac{\sqrt{2}}{2}} \right)^{2} + \left( {-1 - \frac{\sqrt{2}}{2}} \right)^{2} + \left( {-1 - \sqrt{2}} \right)^{2}} \\

& {= \left( {1 + \sqrt{2} + \frac{1}{2}} \right) + \left( {1 + \sqrt{2} + \frac{1}{2}} \right) + \left( {1 + 2\sqrt{2} + 2} \right)} \\

& {= 6 + 4\sqrt{2}.}

\end{array}$$

$6 + 4\sqrt{2}$ and $6 - 4\sqrt{2}$ are the local extreme values of $f\left( {x,y,z} \right),$ subject to the given constraints.

Use the method of Lagrange multipliers to find the minimum value of the function

$$f\left( {x,y,z} \right) = x^{2} + y^{2} + z^{2}$$

subject to the constraints $2x + y + 2z = 9$ and $5x + 5y + 7z = 29.$

Section 4.8 Exercises

For the following exercises, use the method of Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraints.

358\.

$f(x,y) = x^{2}y;x^{2} + 2y^{2} = 6$

359.

$f\left( {x,y,z} \right) = xyz,x^{2} + 2y^{2} + 3z^{2} = 6$

360\.

$f(x,y) = xy;4x^{2} + 8y^{2} = 16$

361.

$f(x,y) = 4x^{3} + y^{2};2x^{2} + y^{2} = 1$

362\.

$f\left( {x,y,z} \right) = x^{2} + y^{2} + z^{2},x^{4} + y^{4} + z^{4} = 1$

363.

$f\left( {x,y,z} \right) = yz + xy,xy = 1,y^{2} + z^{2} = 1$

364\.

$f\left( {x,y} \right) = x^{2} + y^{2},\left( {x - 1} \right)^{2} + 4y^{2} = 4$

365.

$f\left( {x,y} \right) = 4xy,\frac{x^{2}}{9} + \frac{y^{2}}{16} = 1$

366\.

$f\left( {x,y,z} \right) = x + y + z,\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1$

367.

$f\left( {x,y,z} \right) = x + 3y - z,x^{2} + y^{2} + z^{2} = 4$

368\.

$f\left( {x,y,z} \right) = x^{2} + y^{2} + z^{2},xyz = 4$

369.

Minimize $f\left( {x,y} \right) = x^{2} + y^{2}$ on the hyperbola $xy = 1.$

370\.

Minimize $f\left( {x,y} \right) = xy$ on the ellipse $b^{2}x^{2} + a^{2}y^{2} = a^{2}b^{2}.$

371.

Maximize $f\left( {x,y,z} \right) = 2x + 3y + 5z$ on the sphere $x^{2} + y^{2} + z^{2} = 19.$

372\.

Maximize $\begin{array}{l}

{f\left( {x,y} \right) = x^{2} - y^{2};x > 0,y > 0;} \\

{g\left( {x,y} \right) = y - x^{2} = 0}

\end{array}$

373.

The curve $x^{3} - y^{3} = 1$ is asymptotic to the line $y = x.$ Find the point(s) on the curve $x^{3} - y^{3} = 1$ farthest from the line $y = x.$

374\.

Maximize $U\left( {x,y} \right) = 8x^{4\text{/}5}y^{1\text{/}5};4x + 2y = 12$

375.

Minimize $f\left( {x,y} \right) = x^{2} + y^{2},x + 2y - 5 = 0.$

376\.

Maximize $f\left( {x,y} \right) = \sqrt{6 - x^{2} - y^{2}},x + y - 2 = 0.$

377.

Minimize $f\left( {x,y,z} \right) = x^{2} + y^{2} + z^{2},x + y + z = 1.$

378\.

Minimize $f\left( {x,y} \right) = x^{2} - y^{2}$ subject to the constraint $x - 2y + 6 = 0.$

379.

Minimize $f\left( {x,y,z} \right) = x^{2} + y^{2} + z^{2}$ when $x + y + z = 9$ and $x + 2y + 3z = 20.$

For the next group of exercises, use the method of Lagrange multipliers to solve the following applied problems.

380\.

A pentagon is formed by placing an isosceles triangle on a rectangle, as shown in the diagram. If the perimeter of the pentagon is $10$ in., find the lengths of the sides of the pentagon that will maximize the area of the pentagon.

381.

A rectangular box without a top (a topless box) is to be made from $12$ ft2 of cardboard. Find the maximum volume of such a box.

382\.

Find the minimum and maximum distances between the ellipse $x^{2} + xy + 2y^{2} = 1$ and the origin.

383.

Find the point on the surface $x^{2} - 2xy + y^{2} - x + y = 0$ closest to the point $\left( {1,2,-3} \right).$

384\.

Show that, of all the triangles inscribed in a circle of radius $R$ (see diagram), the equilateral triangle has the largest perimeter.

385.

Find the minimum distance from point $\left( {0,1} \right)$ to the parabola $x^{2} = 4y.$

386\.

Find the minimum distance from the parabola $y = x^{2}$ to point $\left( {0,3} \right).$

387.

Find the minimum distance from the plane $x + y + z = 1$ to point $\left( {2,1,1} \right).$

388\.

A large container in the shape of a rectangular solid must have a volume of $480$ m3. The bottom of the container costs \$5/m2 to construct whereas the top and sides cost \$3/m2 to construct. Use Lagrange multipliers to find the dimensions of the container of this size that has the minimum cost.

389.

Find the point on the line $y = 2x + 3$ that is closest to point $\left( {4,2} \right).$

390\.

Find the point on the plane $4x + 3y + z = 2$ that is closest to the point $\left( {1,-1,1} \right).$

391.

Find the maximum value of $f\left( {x,y} \right) = \text{sin}\ x\ \text{sin}\ y,$ where $x\ \text{and}\ y$ denote the acute angles of a right triangle. Draw the contours of the function using a CAS.

392\.

A rectangular solid is contained within a tetrahedron with vertices at

$\left( {1,0,0} \right),\left( {0,1,0} \right),\left( {0,0,1} \right),$ and the origin. The base of the box has dimensions $x,y,$ and the height of the box is $z.$ If the sum of $x,y,\ \text{and}\ z$ is 1.0, find the dimensions that maximizes the volume of the rectangular solid.

393.

\[T\] By investing *x* units of labor and *y* units of capital, a watch manufacturer can produce $P\left( {x,y} \right) = 50x^{0.4}y^{0.6}$ watches. Find the maximum number of watches that can be produced on a budget of $\text{\$}20,000$ if labor costs \$100/unit and capital costs \$200/unit. Use a CAS to sketch a contour plot of the function.

---

Chapter Review

> 来源: OpenStax《Calculus Volume 3》| 原页: https://openstax.org/books/calculus-volume-3/pages/4-chapter-review

（该页为章节总览/导航页，无独立正文；本章内容请见其下各小节。）

---

Key Terms

> 来源: OpenStax《Calculus Volume 3》| 原页: https://openstax.org/books/calculus-volume-3/pages/4-key-terms

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Key Terms

Calculus Volume 3Key Terms

------------------------------------------------------------------------

Key Terms

boundary point

a point $P_{0}$ of $R$ is a boundary point if every $\delta$ disk centered around $P_{0}$ contains points both inside and outside $R$

closed set

a set $S$ that contains all its boundary points

connected set

an open set $S$ that cannot be represented as the union of two or more disjoint, nonempty open subsets

constraint

an inequality or equation involving one or more variables that is used in an optimization problem; the constraint enforces a limit on the possible solutions for the problem

contour map

a plot of the various level curves of a given function $f\left( {x,y} \right)$

critical point of a function of two variables

the point $\left( {x_{0},y_{0}} \right)$ is called a critical point of $f\left( {x,y} \right)$ if one of the two following conditions holds:

1. $f_{x}\left( {x_{0},y_{0}} \right) = f_{y}\left( {x_{0},y_{0}} \right) = 0$

2. At least one of $f_{x}\left( {x_{0},y_{0}} \right)$ and $f_{y}\left( {x_{0},y_{0}} \right)$ do not exist

differentiable

a function $f\left( {x,y,z} \right)$ is differentiable at $\left( {x_{0},y_{0}} \right)$ if $f\left( {x,y} \right)$ can be expressed in the form $f(x,y) = f(x_{0},y_{0}) + f_{x}(x_{0},y_{0})\left( {x - x_{0}} \right) + f_{y}(x_{0},y_{0})\left( {y - y_{0}} \right) + E\left( {x,y} \right),$

where the error term $E\left( {x,y} \right)$ satisfies $\underset{{({x,y})}\rightarrow{({x_{0},y_{0}})}}{\text{lim}}\frac{E\left( {x,y} \right)}{\sqrt{\left( {x - x_{0}} \right)^{2} + \left( {y - y_{0}} \right)^{2}}} = 0$

directional derivative

the derivative of a function in the direction of a given unit vector

discriminant

the discriminant of the function $f\left( {x,y} \right)$ is given by the formula $D = f_{xx}(x_{0},y_{0})f_{yy}(x_{0},y_{0}) - \left( {f_{xy}(x_{0},y_{0})} \right)^{2}$

function of two variables

a function $z = f\left( {x,y} \right)$ that maps each ordered pair $\left( {x,y} \right)$ in a subset $D$ of $\mathbb{R}^{2}$ to a unique real number $z$

generalized chain rule

the chain rule extended to functions of more than one independent variable, in which each independent variable may depend on one or more other variables

gradient

the gradient of the function $f\left( {x,y} \right)$ is defined to be $\nabla f\left( {x,y} \right) = \left( {{\partial f}\text{/}{\partial x}} \right)\mathbf{i} + \left( {{\partial f}\text{/}{\partial y}} \right)\mathbf{j},$ which can be generalized to a function of any number of independent variables

graph of a function of two variables

a set of ordered triples $\left( {x,y,z} \right)$ that satisfies the equation $z = f\left( {x,y} \right)$ plotted in three-dimensional Cartesian space

higher-order partial derivatives

second-order or higher partial derivatives, regardless of whether they are mixed partial derivatives

interior point

a point $P_{0}$ of $R$ is a boundary point if there is a $\delta$ disk centered around $P_{0}$ contained completely in $R$

intermediate variable

given a composition of functions (e.g., $f\left( {x(t),y(t)} \right)),$ the intermediate variables are the variables that are independent in the outer function but dependent on other variables as well; in the function $f\left( {x(t),y(t)} \right),$ the variables $x\ \text{and}\ y$ are examples of intermediate variables

Lagrange multiplier

the constant (or constants) used in the method of Lagrange multipliers; in the case of one constant, it is represented by the variable $\lambda$

level curve of a function of two variables

the set of points satisfying the equation $f\left( {x,y} \right) = c$ for some real number $c$ in the range of $f$

level surface of a function of three variables

the set of points satisfying the equation $f\left( {x,y,z} \right) = c$ for some real number $c$ in the range of $f$

linear approximation

given a function $f\left( {x,y} \right)$ and a tangent plane to the function at a point $\left( {x_{0},y_{0}} \right),$ we can approximate $f\left( {x,y} \right)$ for points near $\left( {x_{0},y_{0}} \right)$ using the tangent plane formula

method of Lagrange multipliers

a method of solving an optimization problem subject to one or more constraints

mixed partial derivatives

second-order or higher partial derivatives, in which at least two of the differentiations are with respect to different variables

objective function

the function that is to be maximized or minimized in an optimization problem

open set

a set $S$ that contains none of its boundary points

optimization problem

calculation of a maximum or minimum value of a function of several variables, often using Lagrange multipliers

partial derivative

a derivative of a function of more than one independent variable in which all the variables but one are held constant

partial differential equation

an equation that involves an unknown function of more than one independent variable and one or more of its partial derivatives

region

an open, connected, nonempty subset of $\mathbb{R}^{2}$

saddle point

given the function $z = f(x,y),$ the point $\left( {x_{0},y_{0},f\left( {x_{0},y_{0}} \right)} \right)$ is a saddle point if both $f_{x}\left( {x_{0},y_{0}} \right) = 0$ and $f_{y}\left( {x_{0},y_{0}} \right) = 0,$ but $f$ does not have a local extremum at $\left( {x_{0},y_{0}} \right)$

surface

the graph of a function of two variables, $z = f\left( {x,y} \right)$

tangent plane

given a function $f\left( {x,y} \right)$ that is differentiable at a point $\left( {x_{0},y_{0}} \right),$ the equation of the tangent plane to the surface $z = f\left( {x,y} \right)$ is given by $z = f\left( {x_{0},y_{0}} \right) + f_{x}\left( {x_{0},y_{0}} \right)\left( {x - x_{0}} \right) + f_{y}\left( {x_{0},y_{0}} \right)\left( {y - y_{0}} \right)$

total differential

the total differential of the function $f\left( {x,y} \right)$ at $\left( {x_{0},y_{0}} \right)$ is given by the formula $dz = f_{x}\left( {x_{0},y_{0}} \right)dx + f_{y}\left( {x_{0},y_{0}} \right)dy$

tree diagram

illustrates and derives formulas for the generalized chain rule, in which each independent variable is accounted for

vertical trace

the set of ordered triples $\left( {c,y,z} \right)$ that solves the equation $f\left( {c,y} \right) = z$ for a given constant $x = c$ or the set of ordered triples $\left( {x,d,z} \right)$ that solves the equation $f\left( {x,d} \right) = z$ for a given constant $y = d$

$\delta$ ball

all points in $\mathbb{R}^{3}$ lying at a distance of less than $\delta$ from $\left( {x_{0},y_{0},z_{0}} \right)$

$\delta$ disk

an open disk of radius $\delta$ centered at point $\left( {a,b} \right)$

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Key Equations

> 来源: OpenStax《Calculus Volume 3》| 原页: https://openstax.org/books/calculus-volume-3/pages/4-key-equations

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Key Equations

| | |

|----------------------------------------------------|------------------------------------------------------------------------------------|

| Vertical trace | $f\left( {a,y} \right) = z$ for $x = a$ or $f\left( {x,b} \right) = z$ for $y = b$ |

| Level surface of a function of three variables | $f(x,y,z) = c$ |

| | |

|---------------------------------------------------|------------------------------------------------------------------------------------------------------------------------------------|

| Partial derivative of $f$ with respect to $x$ | $\frac{\partial f}{\partial x} = \underset{h\rightarrow 0}{\text{lim}}\frac{f\left( {x + h,y} \right) - f\left( {x,y} \right)}{h}$ |

| Partial derivative of $f$ with respect to $y$ | $\frac{\partial f}{\partial y} = \underset{k\rightarrow 0}{\text{lim}}\frac{f\left( {x,y + k} \right) - f\left( {x,y} \right)}{k}$ |

| | |

|------------------|--------------------------------------------------------------------------------------------------------------------------------------|

| Discriminant | $D = f_{xx}\left( {x_{0},y_{0}} \right)f_{yy}\left( {x_{0},y_{0}} \right) - \left( {f_{xy}\left( {x_{0},y_{0}} \right)} \right)^{2}$ |

| | |

|-----------------------------------------------------|-----------------------------------------------------------------------------------------------------------------------------------------------------------------------|

| Method of Lagrange multipliers, one constraint | $\begin{array}{rll}

{\nabla f\left( {x_{0},y_{0}} \right)} & = & {\lambda\nabla g\left( {x_{0},y_{0}} \right)} \\

{g\left( {x_{0},y_{0}} \right)} & = & 0

\end{array}$ |

| Method of Lagrange multipliers, two constraints | $\begin{array}{rll}

{\nabla f\left( {x_{0},y_{0},z_{0}} \right)} & = & {\lambda_{1}\nabla g\left( {x_{0},y_{0},z_{0}} \right) + \lambda_{2}\nabla h\left( {x_{0},y_{0},z_{0}} \right)} \\

{g\left( {x_{0},y_{0},z_{0}} \right)} & = & 0 \\

{h\left( {x_{0},y_{0},z_{0}} \right)} & = & 0

\end{array}$ |

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Key Concepts

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Key Concepts

4.1 Functions of Several Variables

4.2 Limits and Continuity

4.3 Partial Derivatives

4.4 Tangent Planes and Linear Approximations

4.5 The Chain Rule

4.6 Directional Derivatives and the Gradient

4.7 Maxima/Minima Problems

4.8 Lagrange Multipliers

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Review Exercises

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Review Exercises

For the following exercises, determine whether the statement is *true or false*. Justify your answer with a proof or a counterexample.

394\.

The domain of $f\left( {x,y} \right) = x^{3}\text{sin}^{-1}(y)$ is $x =$ all real numbers, and $\text{−}\pi \leq y \leq \pi.$

395.

If the function $f\left( {x,y} \right)$ and all its second derivatives are continuous everywhere, then $f_{xy} = f_{yx}.$

396\.

The linear approximation to the function of $f\left( {x,y} \right) = 5x^{2} + x\ \text{tan}(y)$ at $\left( {2,\pi} \right)$ is given by $L\left( {x,y} \right) = 22 + 21\left( {x - 2} \right) + \left( {y - \pi} \right).$

397.

$\left( {\frac{3}{4},\frac{9}{16}} \right)$ is a critical point of $g\left( {x,y} \right) = 4x^{3} - 2x^{2}y + y^{2} - 2.$

For the following exercises, sketch the function in one graph and, in a second, sketch several level curves.

398\.

$f\left( {x,y} \right) = e^{\text{−}{({x^{2} + 2y^{2}})}}.$

399.

$f\left( {x,y} \right) = x + 4y^{2}.$

For the following exercises, evaluate the following limits, if they exist. If they do not exist, prove it.

400\.

$\underset{{({x,y})}\rightarrow{({1,1})}}{\text{lim}}\frac{4xy}{x - 2y^{2}}$

401.

$\underset{{({x,y})}\rightarrow{({0,0})}}{\text{lim}}\frac{4xy}{x - 2y^{2}}$

For the following exercises, find the largest region of continuity for the function.

402\.

$f\left( {x,y} \right) = x^{3}\text{sin}^{-1}(y)$

403.

$g\left( {x,y} \right) = \text{ln}\left( {4 - x^{2} - y^{2}} \right)$

For the following exercises, find all first derivatives, full or partial, as appropriate.

404\.

$f\left( {x,y} \right) = \sqrt{x^{2} - y^{2}}$

405.

$u\left( {x,y} \right) = x^{4} - 3xy + 1,x = 2t,y = t^{3}$

For the following exercises, find all second partial derivatives.

406\.

$g\left( {t,x} \right) = 3t^{2} - \text{sin}\left( {x + t} \right)$

407.

$h\left( {x,y,z} \right) = \frac{x^{3}e^{2y}}{z}$

For the following exercises, find an equation of the tangent plane to the specified surface at the given point.

408\.

$z = x^{3} - 2y^{2} + y - 1$ at point $\left( {1,1,-1} \right)$

409.

$z = e^{x} + \frac{2}{y}$ at point $\left( {0,1,3} \right)$

410\.

Approximate $f\left( {x,y} \right) = e^{x^{2}} + \sqrt{y}$ at $\left( {0.1,9.1} \right).$ Write down your linear approximation function $L\left( {x,y} \right).$ How accurate is the approximation to the exact answer, rounded to four digits?

411.

Find the differential $dz$ of $h\left( {x,y} \right) = 4x^{2} + 2xy - 3y$ and approximate $\text{Δ}z$ at the point $\left( {1,-2} \right).$ Let $\text{Δ}x = 0.1$ and $\text{Δ}y = 0.01.$

412\.

Find the directional derivative of $f\left( {x,y} \right) = x^{2} + 6xy - y^{2}$ in the direction $\text{v} = \mathbf{i} + 4\mathbf{j}.$

413.

Find the maximal directional derivative magnitude and direction for the function $f\left( {x,y} \right) = x^{3} + 2xy - \text{cos}\left( {\pi y} \right)$ at point $\left( {3,0} \right).$

For the following exercises, find the gradient.

414\.

$c\left( {x,t} \right) = e\left( {t - x} \right)^{2} + 3\ \text{cos}(t)$

415.

$f\left( {x,y} \right) = \frac{\sqrt{x} + y^{2}}{xy}$

For the following exercises, find and classify the critical points.

416\.

$z = x^{3} - xy + y^{2} - 1$

For the following exercises, use Lagrange multipliers to find the maximum and minimum values for the functions with the given constraints.

417.

$f\left( {x,y} \right) = x^{2}y,x^{2} + y^{2} = 4$

418\.

$f\left( {x,y} \right) = x^{2} - y^{2},x + 6y = 4$

419.

A machinist is constructing a right circular cone out of a block of aluminum. The machine gives an error of $5\text{\%}$ in height and $2\text{\%}$ in radius. Find the maximum error in the volume of the cone if the machinist creates a cone of height $6$ cm and radius $2$ cm.

420\.

A trash compactor is in the shape of a cuboid. Assume the trash compactor is filled with incompressible liquid. The length and width are decreasing at rates of $2$ ft/sec and $3$ ft/sec, respectively. Find the rate at which the liquid level is rising when the length is $14$ ft, the width is $10$ ft, and the height is $4$ ft.

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