4 Differentiation of Functions of Several Variables 多元函数的微分法

Chapter Outline 本章概要

• 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

• 4.1 多元函数
• 4.2 极限与连续性
• 4.3 偏导数
• 4.4 切平面与线性逼近
• 4.5 链式法则
• 4.6 方向导数与梯度
• 4.7 极大值 / 极小值问题
• 4.8 拉格朗日乘子法

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4.1 Functions of Several Variables 4.1 多元函数

• 4.1.1 Recognize a function of two variables and identify its domain and range.
• 4.1.2 Sketch a graph of a function of two variables.
• 4.1.3 Sketch several traces or level curves of a function of two variables.
• 4.1.4 Recognize a function of three or more variables and identify its level surfaces.

• 4.1.1 识别二元函数，并确定其定义域与值域。
• 4.1.2 作出二元函数的图像。
• 4.1.3 作出二元函数的若干截痕或等高线。
• 4.1.4 识别三元及以上的多元函数，并确定其等值面。

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 二元函数

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 二元函数的定义域与值域

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 解

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 二元函数的作图

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 二元函数的作图

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 解

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 螺母与螺栓

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 解

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 等高线

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 绘制等高线图

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 解

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 求竖直截痕

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 解

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 多于两个变量的函数

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 三元函数的定义域

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 解

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 求等值面

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

Solution 解

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

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

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

Section 4.1 Exercises 4.1 节习题

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 4.2 极限与连续性

• 4.2.1 Calculate the limit of a function of two variables.
• 4.2.2 Learn how a function of two variables can approach different values at a boundary point, depending on the path of approach.
• 4.2.3 State the conditions for continuity of a function of two variables.
• 4.2.4 Verify the continuity of a function of two variables at a point.
• 4.2.5 Calculate the limit of a function of three or more variables and verify the continuity of the function at a point.

• 4.2.1 计算二元函数的极限。
• 4.2.2 了解二元函数在边界点处可随逼近路径不同而趋于不同的值。
• 4.2.3 陈述二元函数连续的条件。
• 4.2.4 验证二元函数在某点处的连续性。
• 4.2.5 计算三元或更多元函数的极限，并验证函数在某点处的连续性。

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 二元函数的极限

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 多元函数的极限法则

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 求二元函数的极限

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 解

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 不存在的极限

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 解

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 内点与边界点

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 边界点处函数的极限

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

Solution 解

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 二元函数的连续性

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 二元函数连续性的验证

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

Solution 解

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 连续函数的和是连续的

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 连续函数的积是连续的

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 连续函数的复合是连续的

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 二元函数连续性的更多例子

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 解

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 三元及更多元变量的函数

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 求三元函数的极限

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

Solution 解

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 4.2 节习题

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 $\underset{h\rightarrow 0}{\text{lim}}\frac{f(1 + h,y) - f(1,y)}{h}.$

4.3 Partial Derivatives 4.3 偏导数

• 4.3.1 Calculate the partial derivatives of a function of two variables.
• 4.3.2 Calculate the partial derivatives of a function of more than two variables.
• 4.3.3 Determine the higher-order derivatives of a function of two variables.
• 4.3.4 Explain the meaning of a partial differential equation and give an example.

• 4.3.1 计算二元函数的偏导数。
• 4.3.2 计算多于两个变量的函数的偏导数。
• 4.3.3 确定二元函数的高阶偏导数。
• 4.3.4 解释偏微分方程的含义并给出例子。

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 二元函数的导数

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 由定义计算偏导数

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 解

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 计算偏导数

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 解

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 $x\text{–}z~\text{plane.}$ 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 由等高线图求偏导数

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 解

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.

4.4 Tangent Planes and Linear Approximations 4.4 切平面与线性逼近

• 4.4.1 Determine the equation of a plane tangent to a given surface at a point.
• 4.4.2 Use the tangent plane to approximate a function of two variables at a point.
• 4.4.3 Explain when a function of two variables is differentiable.
• 4.4.4 Use the total differential to approximate the change in a function of two variables.

• 4.4.1 求给定曲面在一点处的切平面方程。
• 4.4.2 利用切平面逼近二元函数在一点处的值。
• 4.4.3 说明二元函数何时可微。
• 4.4.4 利用全微分逼近二元函数的改变量。

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 切平面

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 求切平面

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 解

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 求另一个切平面

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 解

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 线性逼近

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 利用切平面逼近

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 解

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 可微性

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 证明可微性

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

Solution 解

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 $\left( {-1,2} \right).$

$$f\left( {x,y} \right) = \begin{cases}\frac{xy}{\sqrt{x^{2} + y^{2}}} & {\left( {x,y} \right) \neq \left( {0,0} \right)} \\0 & {\left( {x,y} \right) = \left( {0,0} \right)}\end{cases}$$ 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 可微性蕴含连续性

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 一阶偏导数的连续性蕴含可微性

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 微分

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 用微分作逼近

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 解

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 三元函数的可微性

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 4.4 节习题

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 4.5 链式法则

• 4.5.1 State the chain rules for one or two independent variables.
• 4.5.2 Use tree diagrams as an aid to understanding the chain rule for several independent and intermediate variables.
• 4.5.3 Perform implicit differentiation of a function of two or more variables.

• 4.5.1 陈述关于一个或两个自变量的链式法则。
• 4.5.2 使用树状图来帮助理解含多个自变量与中间变量的链式法则。
• 4.5.3 对二元或多元函数进行隐函数求导。

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.

The Generalized Chain Rule 广义链式法则

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 广义链式法则

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 使用广义链式法则

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 解

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 绘制树状图

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 解

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 隐函数求导

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 多元函数的隐函数求导

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 用偏导数进行隐函数求导

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 解

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 4.5 节习题

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 $\frac{\partial u}{\partial r}.$

4.6 Directional Derivatives and the Gradient 4.6 方向导数与梯度

• 4.6.1 Determine the directional derivative in a given direction for a function of two variables.
• 4.6.2 Determine the gradient vector of a given real-valued function.
• 4.6.3 Explain the significance of the gradient vector with regard to direction of change along a surface.
• 4.6.4 Use the gradient to find the tangent to a level curve of a given function.
• 4.6.5 Calculate directional derivatives and gradients in three dimensions.

• 4.6.1 对二元函数，在给定方向上求方向导数。
• 4.6.2 求给定实值函数的梯度向量。
• 4.6.3 说明梯度向量关于沿曲面变化方向的意义。
• 4.6.4 利用梯度求给定函数等高线的切线。
• 4.6.5 在三维空间中计算方向导数与梯度。

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 方向导数

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 由定义求方向导数

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 解

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 二元函数的方向导数

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 证明

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 求方向导数的另一种方法

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 解

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 梯度

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 求梯度

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 解

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 梯度的性质

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 求最大方向导数

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 解

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.

4.7 Maxima/Minima Problems 4.7 最大值/最小值问题

• 4.7.1 Use partial derivatives to locate critical points for a function of two variables.
• 4.7.2 Apply a second derivative test to identify a critical point as a local maximum, local minimum, or saddle point for a function of two variables.
• 4.7.3 Examine critical points and boundary points to find absolute maximum and minimum values for a function of two variables.

• 4.7.1 利用偏导数确定二元函数的临界点。
• 4.7.2 用二阶导数判别法判定二元函数的某个临界点是局部极大值点、局部极小值点还是鞍点。
• 4.7.3 考察临界点与边界点，求二元函数的绝对最大值与绝对最小值。

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.

Absolute Maxima and Minima 绝对极值

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 极值定理

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 求二元函数的极值

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 求绝对最大值与最小值

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 求绝对极值

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 解

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 章首页：盈利的高尔夫球

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 解

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 4.7 节习题

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.

4.8 Lagrange Multipliers 4.8 拉格朗日乘子

• 4.8.1 Use the method of Lagrange multipliers to solve optimization problems with one constraint.
• 4.8.2 Use the method of Lagrange multipliers to solve optimization problems with two constraints.

• 4.8.1 用拉格朗日乘子法求解含一个约束的最优化问题。
• 4.8.2 用拉格朗日乘子法求解含两个约束的最优化问题。

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 拉格朗日乘子

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 拉格朗日乘子法：单约束

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 证明

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 使用拉格朗日乘子法的步骤

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 应用拉格朗日乘子法

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 解

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 高尔夫球与拉格朗日乘子

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 解

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 $f\ \text{and}$ $g\text{:}$

$$\begin{array}{l} {\nabla f\left( {x,y} \right) = \left( {48 - 2x - 2y} \right)\mathbf{i} + \left( {96 - 2x - 18y} \right)\mathbf{j}} \\ {\nabla g\left( {x,y} \right) = 5\mathbf{i} + \mathbf{j}.} \end{array}$$