3 Vector-Valued Functions 向量值函数

Chapter Outline 本章概要

• 3.1 Vector-Valued Functions and Space Curves
• 3.2 Calculus of Vector-Valued Functions
• 3.3 Arc Length and Curvature
• 3.4 Motion in Space

• 3.1 向量值函数与空间曲线
• 3.2 向量值函数的微积分
• 3.3 弧长与曲率
• 3.4 空间中的运动

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3.1 Vector-Valued Functions and Space Curves 3.1 向量值函数与空间曲线

• 3.1.1 Write the general equation of a vector-valued function in component form and unit-vector form.
• 3.1.2 Recognize parametric equations for a space curve.
• 3.1.3 Describe the shape of a helix and write its equation.
• 3.1.4 Define the limit of a vector-valued function.

• 3.1.1 用分量形式和单位向量形式写出向量值函数的一般方程。
• 3.1.2 识别空间曲线的参数方程。
• 3.1.3 描述螺旋线的形状并写出其方程。
• 3.1.4 定义向量值函数的极限。

Our study of vector-valued functions combines ideas from our earlier examination of single-variable calculus with our description of vectors in three dimensions from the preceding chapter. In this section we extend concepts from earlier chapters and also examine new ideas concerning curves in three-dimensional space. These definitions and theorems support the presentation of material in the rest of this chapter and also in the remaining chapters of the text.

Definition of a Vector-Valued Function 向量值函数的定义

Our first step in studying the calculus of vector-valued functions is to define what exactly a vector-valued function is. We can then look at graphs of vector-valued functions and see how they define curves in both two and three dimensions.

A vector-valued function is a function of the form

$$\mathbf{\text{r}}(t) = f(t)\mspace{2mu}\mathbf{\text{i}} + g(t)\mspace{2mu}\mathbf{\text{j}}\quad\text{or}\quad\mathbf{\text{r}}(t) = f(t)\mspace{2mu}\mathbf{\text{i}} + g(t)\mspace{2mu}\mathbf{\text{j}} + h(t)\mspace{2mu}\mathbf{\text{k}},$$ (3.1)

where the component functions *f, g,* and *h*, are real-valued functions of the parameter *t.* Vector-valued functions are also written in the form

$$\mathbf{\text{r}}(t) = \left\langle {f(t),g(t)} \right\rangle\quad\text{or}\quad\mathbf{\text{r}}(t) = \left\langle {f(t),g(t),h(t)} \right\rangle.$$ (3.2)

In both cases, the first form of the function defines a two-dimensional vector-valued function; the second form describes a three-dimensional vector-valued function.

The parameter *t* can lie between two real numbers: $a \leq t \leq b.$ Another possibility is that the value of *t* might take on all real numbers. Last, the component functions themselves may have domain restrictions that enforce restrictions on the value of *t.* We often use *t* as a parameter because *t* can represent time.

Evaluating Vector-Valued Functions and Determining Domains 向量值函数的求值与定义域的确定

For each of the following vector-valued functions, evaluate $\mathbf{\text{r}}(0),\mspace{2mu}\mathbf{\text{r}}\left( \frac{\pi}{2} \right),\text{and}\ \mathbf{\text{r}}\left( \frac{2\pi}{3} \right).$ Do any of these functions have domain restrictions?

1. $\mathbf{\text{r}}(t) = 4\mspace{2mu}\text{cos}\mspace{2mu} t\mspace{2mu}\mathbf{\text{i}} + 3\mspace{2mu}\text{sin}\mspace{2mu} t\mspace{2mu}\mathbf{\text{j}}$

2. $\mathbf{\text{r}}(t) = 3\mspace{2mu}\text{tan}\mspace{2mu} t\mspace{2mu}\mathbf{\text{i}} + 4\mspace{2mu}\text{sec}\mspace{2mu} t\mspace{2mu}\mathbf{\text{j}} + 5t\mspace{2mu}\mathbf{\text{k}}$

Solution 解

1. To calculate each of the function values, substitute the appropriate value of *t* into the function:

$$\begin{array}{cll} {\mathbf{\text{r}}(0)} & = & {4\mspace{2mu}\text{cos}(0)\mspace{2mu}\mathbf{\text{i}} + 3\mspace{2mu}\text{sin}(0)\mspace{2mu}\mathbf{\text{j}}} \\ & = & {4\mspace{2mu}\mathbf{\text{i}} + 0\mspace{2mu}\mathbf{\text{j}} = 4\mspace{2mu}\mathbf{\text{i}}} \\ {\mathbf{\text{r}}\left( \frac{\pi}{2} \right)} & = & {4\mspace{2mu}\text{cos}\left( \frac{\pi}{2} \right)\mspace{2mu}\mathbf{\text{i}} + 3\mspace{2mu}\text{sin}\left( \frac{\pi}{2} \right)\mspace{2mu}\mathbf{\text{j}}} \\ & = & {0\mspace{2mu}\mathbf{\text{i}} + 3\mspace{2mu}\mathbf{\text{j}} = 3\mspace{2mu}\mathbf{\text{j}}} \\ {\mathbf{\text{r}}\left( \frac{2\pi}{3} \right)} & = & {4\mspace{2mu}\text{cos}\left( \frac{2\pi}{3} \right)\mspace{2mu}\mathbf{\text{i}} + 3\mspace{2mu}\text{sin}\left( \frac{2\pi}{3} \right)\mspace{2mu}\mathbf{\text{j}}} \\ & = & {4\left( {- \frac{1}{2}} \right)\mspace{2mu}\mathbf{\text{i}} + 3\left( \frac{\sqrt{3}}{2} \right)\mspace{2mu}\mathbf{\text{j}} = -2\mspace{2mu}\mathbf{\text{i}} + \frac{3\sqrt{3}}{2}\mspace{2mu}\mathbf{\text{j}}.} \end{array}$$

To determine whether this function has any domain restrictions, consider the component functions separately. The first component function is $f(t) = 4\mspace{2mu}\text{cos}\mspace{2mu} t$ and the second component function is $g(t) = 3\mspace{2mu}\text{sin}\mspace{2mu} t.$ Neither of these functions has a domain restriction, so the domain of $\mathbf{\text{r}}(t) = 4\mspace{2mu}\text{cos}\mspace{2mu} t\mspace{2mu}\mathbf{\text{i}} + 3\mspace{2mu}\text{sin}\mspace{2mu} t\mspace{2mu}\mathbf{\text{j}}$ is all real numbers.

2. To calculate each of the function values, substitute the appropriate value of *t* into the function:

$$\begin{array}{cll} {\mspace{2mu}\mathbf{\text{r}}(0)} & = & {3\mspace{2mu}\text{tan}(0)\mspace{2mu}\mathbf{\text{i}} + 4\mspace{2mu}\text{sec}(0)\mspace{2mu}\mathbf{\text{j}} + 5(0)\mspace{2mu}\mathbf{\text{k}}} \\ & = & {0\mspace{2mu}\mathbf{\text{i}} + 4\mspace{2mu}\mathbf{\text{j}} + 0\mspace{2mu}\mathbf{\text{k}} = 4\mspace{2mu}\mathbf{\text{j}}} \\ {\mspace{2mu}\mathbf{\text{r}}\left( \frac{\pi}{2} \right)} & = & {3\mspace{2mu}\text{tan}\left( \frac{\pi}{2} \right)\mspace{2mu}\mathbf{\text{i}} + 4\mspace{2mu}\text{sec}\left( \frac{\pi}{2} \right)\mspace{2mu}\mathbf{\text{j}} + 5\left( \frac{\pi}{2} \right)\mspace{2mu}\mathbf{\text{k}},\text{which does not exist}} \\ {\mspace{2mu}\mathbf{\text{r}}\left( \frac{2\pi}{3} \right)} & = & {3\mspace{2mu}\text{tan}\left( \frac{2\pi}{3} \right)\mspace{2mu}\mathbf{\text{i}} + 4\mspace{2mu}\text{sec}\left( \frac{2\pi}{3} \right)\mspace{2mu}\mathbf{\text{j}} + 5\left( \frac{2\pi}{3} \right)\mspace{2mu}\mathbf{\text{k}}} \\ & = & {3\left( {- \sqrt{3}} \right)\mspace{2mu}\mathbf{\text{i}} + 4(-2)\mspace{2mu}\mathbf{\text{j}} + \frac{10\pi}{3}\mspace{2mu}\mathbf{\text{k}}} \\ & = & {-3\sqrt{3}\mspace{2mu}\mathbf{\text{i}} - 8\mspace{2mu}\mathbf{\text{j}} + \frac{10\pi}{3}\mspace{2mu}\mathbf{\text{k}}.} \end{array}$$

To determine whether this function has any domain restrictions, consider the component functions separately. The first component function is $f(t) = 3\mspace{2mu}\text{tan}\mspace{2mu} t,$ the second component function is $g(t) = 4\mspace{2mu}\text{sec}\mspace{2mu} t,$ and the third component function is $h(t) = 5t.$ The first two functions are not defined for odd multiples of $\pi\text{/}2,$ so the function is not defined for odd multiples of $\pi\text{/}2.$ Therefore, $\text{dom}\left( {\mspace{2mu}\mathbf{\text{r}}(t)} \right) = \left\{ {t\mspace{2mu}\left| {t \neq \frac{\left( {2n + 1} \right)\pi}{2}} \right.} \right\},$ where *n* is any integer.

For the vector-valued function $\mathbf{\text{r}}(t) = \left( {t^{2} - 3t} \right)\mspace{2mu}\mathbf{\text{i}} + \left( {4t + 1} \right)\mspace{2mu}\mathbf{\text{j}},$ evaluate $\mathbf{\text{r}}(0),\mspace{2mu}\mathbf{\text{r}}(1),\text{and}\ \mathbf{\text{r}}(-4).$ Does this function have any domain restrictions?

Example 3.1 illustrates an important concept. The domain of a vector-valued function consists of real numbers. The domain can be all real numbers or a subset of the real numbers. The range of a vector-valued function consists of vectors. Each real number in the domain of a vector-valued function is mapped to either a two- or a three-dimensional vector.

Graphing Vector-Valued Functions 向量值函数的作图

Recall that a plane vector consists of two quantities: direction and magnitude. Given any point in the plane (the *initial point*), if we move in a specific direction for a specific distance, we arrive at a second point. This represents the *terminal point* of the vector. We calculate the components of the vector by subtracting the coordinates of the initial point from the coordinates of the terminal point.

A vector is considered to be in *standard position* if the initial point is located at the origin. When graphing a vector-valued function, we typically graph the vectors in the domain of the function in standard position, because doing so guarantees the uniqueness of the graph. This convention applies to the graphs of three-dimensional vector-valued functions as well. The graph of a vector-valued function of the form $\mathbf{\text{r}}(t) = f(t)\mspace{2mu}\mathbf{\text{i}} + g(t)\mspace{2mu}\mathbf{\text{j}}$ consists of the set of all $\left( {t,\mspace{2mu}\mathbf{\text{r}}(t)} \right),$ and the path it traces is called a plane curve. The graph of a vector-valued function of the form $\mathbf{\text{r}}(t) = f(t)\mspace{2mu}\mathbf{\text{i}} + g(t)\mspace{2mu}\mathbf{\text{j}} + h(t)\mspace{2mu}\mathbf{\text{k}}$ consists of the set of all $\left( {t,\mspace{2mu}\mathbf{\text{r}}(t)} \right),$ and the path it traces is called a space curve. Any representation of a plane curve or space curve using a vector-valued function is called a vector parameterization of the curve.

Graphing a Vector-Valued Function 向量值函数的作图

Create a graph of each of the following vector-valued functions:

1. The plane curve represented by $\mathbf{\text{r}}(t) = 4\mspace{2mu}\text{cos}\mspace{2mu} t\mspace{2mu}\mathbf{\text{i}} + 3\mspace{2mu}\text{sin}\mspace{2mu} t\mspace{2mu}\mathbf{\text{j}},$ $0 \leq t \leq 2\pi$

2. The plane curve represented by $\mathbf{\text{r}}(t) = 4\mspace{2mu}\text{cos}\mspace{2mu} 2t\textbf{i} + 3\mspace{2mu}\text{sin}\mspace{2mu} 2\mspace{2mu} t\textbf{j},$ $0 \leq t \leq \pi$

3. The space curve represented by $\textbf{r}(t) = 4\text{cos}\mspace{2mu} t\mspace{2mu}\textbf{i} + 4\text{sin}\mspace{2mu} t\mspace{2mu}\textbf{j} + t\mspace{2mu}\textbf{k},$ $0 \leq t \leq 4\pi$

Solution 解

1. As with any graph, we start with a table of values. We then graph each of the vectors in the second column of the table in standard position and connect the terminal points of each vector to form a curve (Figure 3.2). This curve turns out to be an ellipse centered at the origin.

| | | | |

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

| *t* | $\mathbf{\text{r}}(t)$ | *t* | $\mathbf{\text{r}}(t)$ |

| 0 | $4\mspace{2mu}\mathbf{\text{i}}$ | $\pi$ | $-4\mspace{2mu}\mathbf{\text{i}}$ |

| $\frac{\pi}{4}$ | $2\sqrt{2}\mspace{2mu}\mathbf{\text{i}} + \frac{3\sqrt{2}}{2}\mspace{2mu}\mathbf{\text{j}}$ | $\frac{5\pi}{4}$ | $-2\sqrt{2}\mspace{2mu}\mathbf{\text{i}} - \frac{3\sqrt{2}}{2}\mspace{2mu}\mathbf{\text{j}}$ |

| $\frac{\pi}{2}$ | $3\mspace{2mu}\mathbf{\text{j}}$ | $\frac{3\pi}{2}$ | $-3\mspace{2mu}\mathbf{\text{j}}$ |

| $\frac{3\pi}{4}$ | $-2\sqrt{2}\mspace{2mu}\mathbf{\text{i}} + \frac{3\sqrt{2}}{2}\mspace{2mu}\mathbf{\text{j}}$ | $\frac{7\pi}{4}$ | $2\sqrt{2}\mspace{2mu}\mathbf{\text{i}} - \frac{3\sqrt{2}}{2}\mspace{2mu}\mathbf{\text{j}}$ |

| $2\pi$ | $4\mspace{2mu}\mathbf{\text{i}}$ | | |

Table 3.1 Table of Values for $\mathbf{\text{r}}(t) = 4\mspace{2mu}\text{cos}\mspace{2mu} t\mspace{2mu}\mathbf{\text{i}} + 3\mspace{2mu}\text{sin}\mspace{2mu} t\mspace{2mu}\mathbf{\text{j}},$ $0 \leq t \leq 2\pi$

2. The table of values for $\textbf{r}(t) = 4\text{cos}\mspace{2mu} 2t\textbf{i} + 3\mspace{2mu}\text{sin}\mspace{2mu} 2\mspace{2mu} t\textbf{j},$ $0 \leq t \leq \pi$ is as follows:

| | | | |

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

| *t* | $\mathbf{\text{r}}(t)$ | *t* | $\mathbf{\text{r}}(t)$ |

| 0 | $4\mspace{2mu}\mathbf{\text{i}}$ | $\frac{\pi}{2}$ | $-4\mspace{2mu}\mathbf{\text{i}}$ |

| $\frac{\pi}{8}$ | $2\sqrt{2}\mspace{2mu}\mathbf{\text{i}} + \frac{3\sqrt{2}}{2}\mspace{2mu}\mathbf{\text{j}}$ | $\frac{5\pi}{8}$ | $-2\sqrt{2}\mspace{2mu}\mathbf{\text{i}} - \frac{3\sqrt{2}}{2}\mspace{2mu}\mathbf{\text{j}}$ |

| $\frac{\pi}{4}$ | $3\mspace{2mu}\mathbf{\text{j}}$ | $\frac{3\pi}{4}$ | $-3\mspace{2mu}\mathbf{\text{j}}$ |

| $\frac{3\pi}{8}$ | $-2\sqrt{2}\mspace{2mu}\mathbf{\text{i}} + \frac{3\sqrt{2}}{2}\mspace{2mu}\mathbf{\text{j}}$ | $\frac{7\pi}{8}$ | $2\sqrt{2}\mspace{2mu}\mathbf{\text{i}} - \frac{3\sqrt{2}}{2}\mspace{2mu}\mathbf{\text{j}}$ |

| | | $\pi$ | $4\mspace{2mu}\mathbf{\text{i}}$ |

Table 3.2 Table of Values for $\textbf{r}(t) = 4\mspace{2mu}\text{cos}\mspace{2mu}\left( {\text{2}t} \right)\mspace{2mu}\textbf{i} + 3\mspace{2mu}\text{sin}\left( {2t} \right)\mspace{2mu}\textbf{j},$ $0 \leq t \leq \pi$

The graph of this curve is also an ellipse centered at the origin.

3. We go through the same procedure for a three-dimensional vector function.

| | | | |

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

| *t* | $\mathbf{\text{r}}(t)$ | *t* | $\mathbf{\text{r}}(t)$ |

| 0 | $4\mspace{2mu}\mathbf{\text{i}}$ | $\pi$ | $-4\mspace{2mu}\mathbf{\text{j}} + \pi\mspace{2mu}\mathbf{\text{k}}$ |

| $\frac{\pi}{4}$ | $2\sqrt{2}\mspace{2mu}\mathbf{\text{i}} + 2\sqrt{2}\mspace{2mu}\mathbf{\text{j}} + \frac{\pi}{4}\mspace{2mu}\mathbf{\text{k}}$ | $\frac{5\pi}{4}$ | $-2\sqrt{2}\mspace{2mu}\mathbf{\text{i}} - 2\sqrt{2}\mspace{2mu}\mathbf{\text{j}} + \frac{5\pi}{4}\mspace{2mu}\mathbf{\text{k}}$ |

| $\frac{\pi}{2}$ | $4\mspace{2mu}\mathbf{\text{j}} + \frac{\pi}{2}\mspace{2mu}\mathbf{\text{k}}$ | $\frac{3\pi}{2}$ | $-4\mspace{2mu}\mathbf{\text{j}} + \frac{3\pi}{2}\mspace{2mu}\mathbf{\text{k}}$ |

| $\frac{3\pi}{4}$ | $-2\sqrt{2}\mspace{2mu}\mathbf{\text{i}} + 2\sqrt{2}\mspace{2mu}\mathbf{\text{j}} + \frac{3\pi}{4}\mspace{2mu}\mathbf{\text{k}}$ | $\frac{7\pi}{4}$ | $2\sqrt{2}\mspace{2mu}\mathbf{\text{i}} - 2\sqrt{2}\mspace{2mu}\mathbf{\text{j}} + \frac{7\pi}{4}\mspace{2mu}\mathbf{\text{k}}$ |

| $2\pi$ | $4\mspace{2mu}\mathbf{\text{i}} + 2\pi\mspace{2mu}\mathbf{\text{k}}$ | | |

Table 3.3 Table of Values for $\textbf{r}(t) = 4\text{cos}\mspace{2mu} t\mspace{2mu}\textbf{i} + 4\text{sin}\mspace{2mu} t\mspace{2mu}\textbf{j} + t\mspace{2mu}\textbf{k},$ $0 \leq t \leq 4\pi$

The values then repeat themselves, except for the fact that the coefficient of k is always increasing (Figure 3.4). This curve is called a helix. Notice that if the k component is eliminated, then the function becomes $\mathbf{\text{r}}(t) = \text{cos}\mspace{2mu} t\mspace{2mu}\mathbf{\text{i}} + \text{sin}\mspace{2mu} t\mspace{2mu}\mathbf{\text{j}},$ which is a unit circle centered at the origin.

You may notice that the graphs in parts a. and b. are identical. This happens because the function describing curve b is a so-called reparameterization of the function describing curve a. In fact, any curve has an infinite number of reparameterizations; for example, we can replace *t* with $2t$ in any of the three previous curves without changing the shape of the curve. The interval over which *t* is defined may change, but that is all. We return to this idea later in this chapter when we study arc-length parameterization.

As mentioned, the name of the shape of the curve of the graph in Example 3.2c. is a helix (Figure 3.4). The curve resembles a spring, with a circular cross-section looking down along the *z*-axis. It is possible for a helix to be elliptical in cross-section as well. For example, the vector-valued function $\mathbf{\text{r}}(t) = 4\mspace{2mu}\text{cos}\mspace{2mu} t\mspace{2mu}\mathbf{\text{i}} + 3\mspace{2mu}\text{sin}\mspace{2mu} t\mspace{2mu}\mathbf{\text{j}} + t\mspace{2mu}\mathbf{\text{k}}$ describes an elliptical helix. The projection of this helix into the $x,y\text{-plane}$ is an ellipse. Last, the arrows in the graph of this helix indicate the orientation of the curve as *t* progresses from 0 to $4\pi.$

Create a graph of the vector-valued function $\mathbf{\text{r}}(t) = \left( {t^{2} - 1} \right)\mspace{2mu}\mathbf{\text{i}} + \left( {2t - 3} \right)\mspace{2mu}\mathbf{\text{j}},$ $0 \leq t \leq 3.$

At this point, you may notice a similarity between vector-valued functions and parameterized curves. Indeed, given a vector-valued function $\mathbf{\text{r}}(t) = f(t)\mspace{2mu}\mathbf{\text{i}} + g(t)\mspace{2mu}\mathbf{\text{j}},$ we can define $x = f(t)$ and $y = g(t).$ If a restriction exists on the values of *t* (for example, *t* is restricted to the interval $\left\lbrack {a,b} \right\rbrack$ for some constants $a < b),$ then this restriction is enforced on the parameter. The graph of the parameterized function would then agree with the graph of the vector-valued function, except that the vector-valued graph would represent vectors rather than points. Since we can parameterize a curve defined by a function $y = f(x),$ it is also possible to represent an arbitrary plane curve by a vector-valued function.

Limits and Continuity of a Vector-Valued Function 向量值函数的极限与连续性

We now take a look at the limit of a vector-valued function. This is important to understand to study the calculus of vector-valued functions.

A vector-valued function r approaches the limit L as *t* approaches *a,* written

$$\underset{t\rightarrow a}{\text{lim}}\mspace{2mu}\mathbf{\text{r}}(t) = \mathbf{\text{L}},$$

provided

$$\underset{t\rightarrow a}{\text{lim}}\left| \middle| {\mspace{2mu}\mathbf{\text{r}}(t) - \mathbf{\text{L}}} \middle| \right| = 0.$$

This is a rigorous definition of the limit of a vector-valued function. In practice, we use the following theorem:

Limit of a Vector-Valued Function 向量值函数的极限

Let *f, g,* and *h* be functions of *t.* Then the limit of the vector-valued function $\mathbf{\text{r}}(t) = f(t)\mspace{2mu}\mathbf{\text{i}} + g(t)\mspace{2mu}\mathbf{\text{j}}$ as *t* approaches *a* is given by

$$\underset{t\rightarrow a}{\text{lim}}\mspace{2mu}\mathbf{\text{r}}(t) = \left\lbrack {\underset{t\rightarrow a}{\text{lim}}f(t)} \right\rbrack\mspace{2mu}\mathbf{\text{i}} + \left\lbrack {\underset{t\rightarrow a}{\text{lim}}g(t)} \right\rbrack\mspace{2mu}\mathbf{\text{j}},$$ (3.3)

provided the limits $\underset{t\rightarrow a}{\text{lim}}f(t)\ \text{and}\ \underset{t\rightarrow a}{\text{lim}}g(t)$ exist. Similarly, the limit of the vector-valued function $\mathbf{\text{r}}(t) = f(t)\mspace{2mu}\mathbf{\text{i}} + g(t)\mspace{2mu}\mathbf{\text{j}} + h(t)\mspace{2mu}\mathbf{\text{k}}$ as *t* approaches *a* is given by

$$\underset{t\rightarrow a}{\text{lim}}\mspace{2mu}\mathbf{\text{r}}(t) = \left\lbrack {\underset{t\rightarrow a}{\text{lim}}f(t)} \right\rbrack\mspace{2mu}\mathbf{\text{i}} + \left\lbrack {\underset{t\rightarrow a}{\text{lim}}g(t)} \right\rbrack\mspace{2mu}\mathbf{\text{j}} + \left\lbrack {\underset{t\rightarrow a}{\text{lim}}h(t)} \right\rbrack\mspace{2mu}\mathbf{\text{k}},$$ (3.4)

provided the limits $\underset{t\rightarrow a}{\text{lim}}f(t),\ \underset{t\rightarrow a}{\text{lim}}g(t)\text{and}\ \underset{t\rightarrow a}{\text{lim}}h(t)$ exist.

In the following example, we show how to calculate the limit of a vector-valued function.

Evaluating the Limit of a Vector-Valued Function 计算向量值函数的极限

For each of the following vector-valued functions, calculate $\underset{t\rightarrow 3}{\text{lim}}\mspace{2mu}\mathbf{\text{r}}(t)$ for

1. $\mathbf{\text{r}}(t) = \left( {t^{2} - 3t + 4} \right)\mspace{2mu}\mathbf{\text{i}} + \left( {4t + 3} \right)\mspace{2mu}\mathbf{\text{j}}$

2. $\mathbf{\text{r}}(t) = \frac{2t - 4}{t + 1}\mspace{2mu}\mathbf{\text{i}} + \frac{t}{t^{2} + 1}\mspace{2mu}\mathbf{\text{j}} + \left( {4t - 3} \right)\mspace{2mu}\mathbf{\text{k}}$

Solution 解

1. Use Equation 3.3 and substitute the value $t = 3$ into the two component expressions:

$$\begin{array}{cl}{\underset{t\rightarrow 3}{\text{lim}}\mspace{2mu}\mathbf{\text{r}}(t)} & {= \underset{t\rightarrow 3}{\text{lim}}\left\lbrack {\left( {t^{2} - 3t + 4} \right)\mspace{2mu}\mathbf{\text{i}} + \left( {4t + 3} \right)\mspace{2mu}\mathbf{\text{j}}} \right\rbrack} \\ & {= \left\lbrack {\underset{t\rightarrow 3}{\text{lim}}\left( {t^{2} - 3t + 4} \right)} \right\rbrack\mspace{2mu}\mathbf{\text{i}} + \left\lbrack {\underset{t\rightarrow 3}{\text{lim}}\left( {4t + 3} \right)} \right\rbrack\mspace{2mu}\mathbf{\text{j}}} \\ & {= 4\mspace{2mu}\mathbf{\text{i}} + 15\mspace{2mu}\mathbf{\text{j}}.}\end{array}$$

2. Use Equation 3.4 and substitute the value $t = 3$ into the three component expressions:

$$\begin{array}{cl}{\underset{t\rightarrow 3}{\text{lim}}\mspace{2mu}\mathbf{\text{r}}(t)} & {= \underset{t\rightarrow 3}{\text{lim}}\left( {\frac{2t - 4}{t + 1}\mspace{2mu}\mathbf{\text{i}} + \frac{t}{t^{2} + 1}\mspace{2mu}\mathbf{\text{j}} + \left( {4t - 3} \right)\mspace{2mu}\mathbf{\text{k}}} \right)} \\ & {= \left\lbrack {\underset{t\rightarrow 3}{\text{lim}}\left( \frac{2t - 4}{t + 1} \right)} \right\rbrack\mspace{2mu}\mathbf{\text{i}} + \left\lbrack {\underset{t\rightarrow 3}{\text{lim}}\left( \frac{t}{t^{2} + 1} \right)} \right\rbrack\mspace{2mu}\mathbf{\text{j}} + \left\lbrack {\underset{t\rightarrow 3}{\text{lim}}\left( {4t - 3} \right)} \right\rbrack\mspace{2mu}\mathbf{\text{k}}} \\ & {= \frac{1}{2}\mspace{2mu}\mathbf{\text{i}} + \frac{3}{10}\mspace{2mu}\mathbf{\text{j}} + 9\mspace{2mu}\mathbf{\text{k}}.}\end{array}$$

Calculate $\underset{t\rightarrow-2}{\text{lim}}\mspace{2mu}\mathbf{\text{r}}(t)$ for the function $\mathbf{\text{r}}(t) = \sqrt{t^{2} - 3t - 1}\mspace{2mu}\mathbf{\text{i}} + \left( {4t + 3} \right)\mspace{2mu}\mathbf{\text{j}} + \text{sin}\ \frac{\left( {t + 1} \right)\pi}{2}\mspace{2mu}\mathbf{\text{k}}.$

Now that we know how to calculate the limit of a vector-valued function, we can define continuity at a point for such a function.

Let *f, g,* and *h* be functions of *t.* Then, the vector-valued function $\mathbf{\text{r}}(t) = f(t)\mspace{2mu}\mathbf{\text{i}} + g(t)\mspace{2mu}\mathbf{\text{j}}$ is continuous at point $t = a$ if the following three conditions hold:

1. $\mathbf{\text{r}}(a)$ exists

2. $\underset{t\rightarrow a}{\text{lim}}\mspace{2mu}\mathbf{\text{r}}(t)$ exists

3. $\underset{t\rightarrow a}{\text{lim}}\mspace{2mu}\mathbf{\text{r}}(t) = \mathbf{\text{r}}(a)$

Similarly, the vector-valued function $\mathbf{\text{r}}(t) = f(t)\mspace{2mu}\mathbf{\text{i}} + g(t)\mspace{2mu}\mathbf{\text{j}} + h(t)\mspace{2mu}\mathbf{\text{k}}$ is continuous at point $t = a$ if the following three conditions hold:

1. $\mathbf{\text{r}}(a)$ exists

2. $\underset{t\rightarrow a}{\text{lim}}\mspace{2mu}\mathbf{\text{r}}(t)$ exists

3. $\underset{t\rightarrow a}{\text{lim}}\mspace{2mu}\mathbf{\text{r}}(t) = \mathbf{\text{r}}(a)$

Section 3.1 Exercises 3.1 节习题

1.

Give the component functions $x = f(t)$ and $y = g(t)$ for the vector-valued function $\mathbf{\text{r}}(t) = 3\mspace{2mu}\text{sec}\mspace{2mu} t\mspace{2mu}\mathbf{\text{i}} + 2\mspace{2mu}\text{tan}\mspace{2mu} t\mspace{2mu}\mathbf{\text{j}}.$

2\.

Given $\mathbf{\text{r}}(t) = 3\mspace{2mu}\text{sec}\mspace{2mu} t\mspace{2mu}\mathbf{\text{i}} + 2\mspace{2mu}\text{tan}\mspace{2mu} t\mspace{2mu}\mathbf{\text{j}},$ find the following values (if possible).

1. $\mathbf{\text{r}}\left( \frac{\pi}{4} \right)$

2. $\mathbf{\text{r}}(\pi)$

3. $\mathbf{\text{r}}\left( \frac{\pi}{2} \right)$

3.

Sketch the curve of the vector-valued function $\mathbf{\text{r}}(t) = 3\mspace{2mu}\text{sec}\mspace{2mu} t\mspace{2mu}\mathbf{\text{i}} + 2\mspace{2mu}\text{tan}\mspace{2mu} t\mspace{2mu}\mathbf{\text{j}}$ and give the orientation of the curve. Sketch asymptotes as a guide to the graph.

4\.

Evaluate $\underset{t\rightarrow 0}{\text{lim}}\left\langle {e^{t}\mspace{2mu},\frac{\text{sin}\mspace{2mu} t}{t}\mspace{2mu},e^{\text{−}t}} \right\rangle.$

5.

Given the vector-valued function $\mathbf{\text{r}}(t) = \left\langle {\text{cos}\mspace{2mu} t,\text{sin}\mspace{2mu} t} \right\rangle,$ find the following values:

1. $\underset{t\rightarrow\frac{\pi}{3}}{\text{lim}}\mspace{2mu}\mathbf{\text{r}}(t)$

2. $\mathbf{\text{r}}\left( \frac{\pi}{3} \right)$

3. Is $\mathbf{\text{r}}(t)$ continuous at $t = \frac{\pi}{3}?$

4. Graph $\mathbf{\text{r}}(t).$

6\.

Given the vector-valued function $\mathbf{\text{r}}(t) = \left\langle {t,t^{2} + 1} \right\rangle,$ find the following values:

1. $\underset{t\rightarrow-3}{\text{lim}}\mspace{2mu}\mathbf{\text{r}}(t)$

2. $\mathbf{\text{r}}(-3)$

3. Is $\mathbf{\text{r}}(t)$ continuous at $t = -3?$

4. $\mathbf{\text{r}}(t + 2) - \mathbf{\text{r}}(t)$

7.

Let $\mathbf{\text{r}}(t) = e^{t}\mspace{2mu}\mathbf{\text{i}} + \text{sin}\mspace{2mu} t\mspace{2mu}\mathbf{\text{j}} + \text{ln}\mspace{2mu} t\mspace{2mu}\mathbf{\text{k}}.$ Find the following values:

1. $\mathbf{\text{r}}\left( \frac{\pi}{4} \right)$

2. $\underset{t\rightarrow\pi\text{/}4}{\text{lim}}\mspace{2mu}\mathbf{\text{r}}(t)$

3. Is $\mathbf{\text{r}}(t)$ continuous at $t = \frac{\pi}{4}?$

Find the limit of the following vector-valued functions at the indicated value of *t*.

8\.

$\underset{t\rightarrow 4}{\text{lim}}\left\langle {\sqrt{t - 3},\frac{\sqrt{t} - 2}{t - 4},\text{tan}\left( \frac{\pi}{t} \right)} \right\rangle$

9.

$\underset{t\rightarrow\pi\text{/}2}{\text{lim}}\mspace{2mu}\mathbf{\text{r}}(t)$ for $\mathbf{\text{r}}(t) = e^{t}\mspace{2mu}\mathbf{\text{i}} + \text{sin}\mspace{2mu} t\mspace{2mu}\mathbf{\text{j}} + \text{ln}\mspace{2mu} t\mspace{2mu}\mathbf{\text{k}}$

10\.

$\underset{t\rightarrow\infty}{\text{lim}}\left\langle {e^{-2t},\frac{2t + 3}{3t - 1},\text{arctan}(2t)} \right\rangle$

11.

$\underset{t\rightarrow e^{2}}{\text{lim}}\left\langle {t\mspace{2mu}\text{ln}(t),\frac{\text{ln}\mspace{2mu} t}{t^{2}},\sqrt{\text{ln}\mspace{2mu}\left( t^{2} \right)}} \right\rangle$

12\.

$\underset{t\rightarrow\pi\text{/}6}{\text{lim}}\left\langle {\text{cos}^{2}t,\text{sin}^{2}t,1} \right\rangle$

13.

$\underset{t\rightarrow\infty}{\text{lim}}\mspace{2mu}\mathbf{\text{r}}(t)$ for $\mathbf{\text{r}}(t) = 2e^{\text{−}t}\mspace{2mu}\mathbf{\text{i}} + e^{\text{−}t}\mspace{2mu}\mathbf{\text{j}} + \text{ln}(t - 1)\mspace{2mu}\mathbf{\text{k}}$

14\.

Describe the curve defined by the vector-valued function $\mathbf{\text{r}}(t) = (1 + t)\mspace{2mu}\mathbf{\text{i}} + (2 + 5t)\mspace{2mu}\mathbf{\text{j}} + (-1 + 6t)\mspace{2mu}\mathbf{\text{k}}.$

Find the domain of the vector-valued functions.

15.

Domain: $\mathbf{\text{r}}(t) = \left\langle {t^{2},\text{tan}\mspace{2mu} t,\text{ln}\mspace{2mu} t} \right\rangle$

16\.

Domain: $\mathbf{\text{r}}(t) = \left\langle {t^{2},\sqrt{t - 3},\frac{3}{2t + 1}} \right\rangle$

17.

Domain: $\mathbf{\text{r}}(t) = \left\langle {\text{csc}(t),\frac{1}{\sqrt{t - 3}},\text{ln}(t - 2)} \right\rangle$

Let $\mathbf{\text{r}}(t) = \left\langle {\text{cos}\mspace{2mu} t,t,\text{sin}\mspace{2mu} t} \right\rangle$ and use it to answer the following questions.

18\.

For what values of *t* is $\mathbf{\text{r}}(t)$ continuous?

19.

Sketch the graph of $\mathbf{\text{r}}(t).$

20\.

Find the domain of $\mathbf{\text{r}}(t) = 2e^{\text{−}t}\mspace{2mu}\mathbf{\text{i}} + e^{\text{−}t}\mspace{2mu}\mathbf{\text{j}} + \text{ln}(t - 1)\mspace{2mu}\mathbf{\text{k}}.$

21.

For what values of *t* is $\mathbf{\text{r}}(t) = 2e^{\text{−}t}\mspace{2mu}\mathbf{\text{i}} + e^{\text{−}t}\mspace{2mu}\mathbf{\text{j}} + \text{ln}(t - 1)\mspace{2mu}\mathbf{\text{k}}$ continuous?

Eliminate the parameter *t*, write the equation in Cartesian coordinates, then sketch the graphs of the vector-valued functions.

22\.

$\mathbf{\text{r}}(t) = 2t\mspace{2mu}\mathbf{\text{i}} + t^{2}\mspace{2mu}\mathbf{\text{j}}$ (*Hint:* Let $x = 2t$ and $y = t^{2}.$ Solve the first equation for *x* in terms of *t* and substitute this result into the second equation.)

23.

$\mathbf{\text{r}}(t) = t^{3}\mspace{2mu}\mathbf{\text{i}} + 2t\mspace{2mu}\mathbf{\text{j}}$

24\.

$\mathbf{\text{r}}(t) = 2\left( {\text{sinh}\mspace{2mu} t} \right)\mspace{2mu}\mathbf{\text{i}} + 2\left( {\text{cosh}\mspace{2mu} t} \right)\mspace{2mu}\mathbf{\text{j}},t > 0$

25.

$\mathbf{\text{r}}(t) = 3\left( {\text{cos}\mspace{2mu} t} \right)\mspace{2mu}\mathbf{\text{i}} + 3\left( {\text{sin}\mspace{2mu} t} \right)\mspace{2mu}\mathbf{\text{j}}$

26\.

$\mathbf{\text{r}}(t) = \left\langle {3\mspace{2mu}\text{sin}\mspace{2mu} t,3\mspace{2mu}\text{cos}\mspace{2mu} t} \right\rangle$

Use a graphing utility to sketch each of the following vector-valued functions:

27.

\[T\] $\textbf{r}(t) = \left( {2\mspace{2mu}\text{cos}^{2}\mspace{2mu} t} \right)\mspace{2mu}\textbf{i} + \left( 2 - \sqrt{t} \right)\mspace{2mu}\textbf{j}$

28\.

\[T\] $\mathbf{\text{r}}(t) = \left\langle {e^{\text{cos}(3t)},e^{\text{−}\text{sin}(t)}} \right\rangle$

29.

\[T\] $\mathbf{\text{r}}(t) = \left\langle {2 - \text{sin}(2t),3 + 2\mspace{2mu}\text{cos}\mspace{2mu} t} \right\rangle$

30\.

$4x^{2} + 9y^{2} = 36;$ clockwise and counterclockwise

31.

$\mathbf{\text{r}}(t) = \left\langle {t,t^{2}} \right\rangle;$ from left to right

32\.

The line through *P* and *Q* where *P* is $\left( {1,4,-2} \right)$ and *Q* is $\left( {3,9,6} \right)$

Consider the curve described by the vector-valued function $\mathbf{\text{r}}(t) = \left( {50e^{\text{−}t}\text{cos}\mspace{2mu} t} \right)\mspace{2mu}\mathbf{\text{i}} + \left( {50e^{\text{−}t}\text{sin}\mspace{2mu} t} \right)\mspace{2mu}\mathbf{\text{j}} + (5 - 5e^{\text{−}t})\mspace{2mu}\mathbf{\text{k}}.$

33.

What is the initial point of the path corresponding to $\mathbf{\text{r}}(0)?$

34\.

What is $\underset{t\rightarrow\infty}{\text{lim}}\mspace{2mu}\mathbf{\text{r}}(t)?$

35.

\[T\] Use technology to sketch the curve.

36\.

Eliminate the parameter *t* to show that $z = 5 - \frac{r}{10}$ where $r^{2} = x^{2} + y^{2}.$

37.

\[T\] Let $r(t) = \text{cos}\mspace{2mu} t\mspace{2mu}\mathbf{\text{i}} + \text{sin}\mspace{2mu} t\mspace{2mu}\mathbf{\text{j}} + 0.3\mspace{2mu}\text{sin}(2t)\mspace{2mu}\mathbf{\text{k}}.$ Use technology to graph the curve (called the *roller-coaster curve*) over the interval $\left\lbrack {0,2\pi} \right).$ Choose at least two views to determine the peaks and valleys.

38\.

\[T\] Use the result of the preceding problem to construct an equation of a roller coaster with a steep drop from the peak and steep incline from the "valley." Then, use technology to graph the equation.

39.

Use the results of the preceding two problems to construct an equation of a path of a roller coaster with more than two turning points (peaks and valleys).

40\.

1. Graph the curve $\mathbf{\text{r}}(t) = \left( {4 + \text{cos}(18t)} \right)\text{cos}(t)\mspace{2mu}\mathbf{\text{i}} + \left( {4 + \text{cos}(18t)\text{sin}(t)} \right)\mspace{2mu}\mathbf{\text{j}} + 0.3\mspace{2mu}\text{sin}(18t)\mspace{2mu}\mathbf{\text{k}}$ using two viewing angles of your choice to see the overall shape of the curve.

2. Does the curve resemble a "slinky"?

3. What changes to the equation should be made to increase the number of coils of the slinky?

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3.2 Calculus of Vector-Valued Functions 3.2 向量值函数的微积分

• 3.2.1 Write an expression for the derivative of a vector-valued function.
• 3.2.2 Find the tangent vector at a point for a given position vector.
• 3.2.3 Find the unit tangent vector at a point for a given position vector and explain its significance.
• 3.2.4 Calculate the definite integral of a vector-valued function.

• 3.2.1 写出向量值函数导数的表达式。
• 3.2.2 对给定位置向量，求其在某点处的切向量。
• 3.2.3 对给定位置向量，求其在某点处的单位切向量，并说明其意义。
• 3.2.4 计算向量值函数的定积分。

To study the calculus of vector-valued functions, we follow a similar path to the one we took in studying real-valued functions. First, we define the derivative, then we examine applications of the derivative, then we move on to defining integrals. However, we will find some interesting new ideas along the way as a result of the vector nature of these functions and the properties of space curves.

3.3 Arc Length and Curvature 3.3 弧长与曲率

• 3.3.1 Determine the length of a particle’s path in space by using the arc-length function.
• 3.3.2 Explain the meaning of the curvature of a curve in space and state its formula.
• 3.3.3 Describe the meaning of the normal and binormal vectors of a curve in space.

• 3.3.1 利用弧长函数确定空间中粒子运动路径的长度。
• 3.3.2 解释空间中曲线曲率的含义并写出其公式。
• 3.3.3 描述空间中曲线的法向量与副法向量的含义。

In this section, we study formulas related to curves in both two and three dimensions, and see how they are related to various properties of the same curve. For example, suppose a vector-valued function describes the motion of a particle in space. We would like to determine how far the particle has traveled over a given time interval, which can be described by the arc length of the path it follows. Or, suppose that the vector-valued function describes a road we are building and we want to determine how sharply the road curves at a given point. This is described by the curvature of the function at that point. We explore each of these concepts in this section.

Arc Length for Vector Functions 向量值函数的弧长

We have seen how a vector-valued function describes a curve in either two or three dimensions. Recall Arc Length of a Parametric Curve, which states that the formula for the arc length of a curve defined by the parametric functions $x = x(t),y = y(t),t_{1} \leq t \leq t_{2}$ is given by

$$s = {\int_{t_{1}}^{t_{2}}{\sqrt{\left( {x^{\prime}(t)} \right)^{2} + \left( {y^{\prime}(t)} \right)^{2}}dt.}}$$

In a similar fashion, if we define a smooth curve using a vector-valued function $\mathbf{\text{r}}(t) = f(t)\ \mathbf{\text{i}} + g(t)\ \mathbf{\text{j}},$ where $a \leq t \leq b,$ the arc length is given by the formula

$$s = {\int_{a}^{b}\sqrt{\left( {f^{\prime}(t)} \right)^{2} + \left( {g^{\prime}(t)} \right)^{2}}}dt.$$

In three dimensions, if the vector-valued function is described by $\mathbf{\text{r}}(t) = f(t)\ \mathbf{\text{i}} + g(t)\ \mathbf{\text{j}} + h(t)\mspace{2mu}\mathbf{\text{k}}$ over the same interval $a \leq t \leq b,$ the arc length is given by

$$s = {\int_{a}^{b}\sqrt{\left( {f^{\prime}(t)} \right)^{2} + \left( {g^{\prime}(t)} \right)^{2} + \left( {h^{\prime}(t)} \right)^{2}}}dt.$$

Arc-Length Formulas 弧长公式

1. *Plane curve*: Given a smooth curve *C* defined by the function $\mathbf{\text{r}}(t) = f(t)\ \mathbf{\text{i}} + g(t)\ \mathbf{\text{j}},$ where *t* lies within the interval $\left\lbrack {a,b} \right\rbrack,$ the arc length of *C* over the interval is

$$s = {\int_{a}^{b}\sqrt{\ \left\lbrack {f^{\prime}(t)} \right\rbrack^{2} + \left\lbrack {g^{\prime}(t)} \right\rbrack^{2}}}dt = {\int_{a}^{b}\left\| {\mathbf{r^{\prime}}(t)} \right\|}\ dt.$$ (3.11)

2. *Space curve*: Given a smooth curve *C* defined by the function $\mathbf{\text{r}}(t) = f(t)\ \mathbf{\text{i}} + g(t)\ \mathbf{\text{j}} + h(t)\ \mathbf{\text{k}},$ where *t* lies within the interval $\left\lbrack {a,b} \right\rbrack,$ the arc length of *C* over the interval is

$$s = {\int_{a}^{b}{\sqrt{\ \left\lbrack {f^{\prime}(t)} \right\rbrack^{2} + \left\lbrack {g^{\prime}(t)} \right\rbrack^{2} + \left\lbrack {h^{\prime}(t)} \right\rbrack^{2}}dt}} = {\int_{a}^{b}\left\| {\mathbf{r^{\prime}}(t)} \right\|}\ dt.$$ (3.12)

The two formulas are very similar; they differ only in the fact that a space curve has three component functions instead of two. Note that the formulas are defined for smooth curves: curves where the vector-valued function $\mathbf{\text{r}}(t)$ is continuously differentiable with a non-zero derivative. The smoothness condition guarantees that the curve has no cusps (or corners) that could make the formula problematic.

Finding the Arc Length 计算弧长

Calculate the arc length for each of the following vector-valued functions:

1. $\mathbf{\text{r}}(t) = \left( {3t - 2} \right)\mspace{2mu}\mathbf{\text{i}} + \left( {4t + 5} \right)\mspace{2mu}\mathbf{\text{j}},1 \leq t \leq 5$

2. $\mathbf{\text{r}}(t) = \left\langle {t\mspace{2mu}\text{cos}\mspace{2mu} t,t\mspace{2mu}\text{sin}\mspace{2mu} t,2t} \right\rangle,0 \leq t \leq 2\pi$

Solution 解

1. Using Equation 3.11, $\mathbf{r^{\prime}}(t) = 3\mathbf{\text{i}} + 4\mathbf{\text{j}},$ so

$$\begin{array}{cl} s & {= {\int_{a}^{b}\left\| {\mathbf{r^{\prime}}(t)} \right\|}\ dt} \\ & {= {\int_{a}^{5}\sqrt{3^{2} + 4^{2}}}dt} \\ & {= {\int_{1}^{5}{5\ dt}} = \left. {5t} \right|_{1}^{5} = 20.} \end{array}$$

2. Using Equation 3.12, $\mathbf{r^{\prime}}(t) = \left\langle {\text{cos}\mspace{2mu} t - t\mspace{2mu}\text{sin}\mspace{2mu} t,\text{sin}\mspace{2mu} t + t\mspace{2mu}\text{cos}\mspace{2mu} t,2} \right\rangle,$ so

$$\begin{array}{cl} s & {= {\int_{a}^{b}\left\| {\mathbf{r^{\prime}}(t)} \right\|}\ dt} \\ & {= {\int_{0}^{2\pi}\sqrt{\left( {\text{cos}\mspace{2mu} t - t\mspace{2mu}\text{sin}\mspace{2mu} t} \right)^{2} + \left( {\text{sin}\mspace{2mu} t + t\mspace{2mu}\text{cos}\mspace{2mu} t} \right)^{2} + 2^{2}}}dt} \\ & {= {\int_{0}^{2\pi}{\sqrt{\left( {\text{cos}^{2}t - 2t\mspace{2mu}\text{sin}\mspace{2mu} t\mspace{2mu}\text{cos}\mspace{2mu} t + t^{2}\text{sin}^{2}t} \right) + \left( {\text{sin}^{2}t + 2t\mspace{2mu}\text{sin}\mspace{2mu} t\mspace{2mu}\text{cos}\mspace{2mu} t + t^{2}\text{cos}^{2}t} \right) + 4}\ dt}}} \\ & {= {\int_{0}^{2\pi}{\sqrt{\text{cos}^{2}t + \text{sin}^{2}t + t^{2}\left( {\text{cos}^{2}t + \text{sin}^{2}t} \right) + 4}\ dt}}} \\ & {= {\int_{0}^{2\pi}{\sqrt{t^{2} + 5}\ dt}}.} \end{array}$$

Here we can use a table integration formula

$${\int{\sqrt{u^{2} + a^{2}}\ du}} = \frac{u}{2}\sqrt{u^{2} + a^{2}} + \frac{a^{2}}{2}\text{ln}\mspace{2mu}\left| {u + \sqrt{u^{2} + a^{2}}} \right| + C,$$

so we obtain

$$\begin{array}{cl} {\int_{0}^{2\pi}{\sqrt{t^{2} + 5}\ dt}} & {= \frac{1}{2}\left( {t\sqrt{t^{2} + 5} + 5\mspace{2mu}\text{ln}\mspace{2mu}\left| {t + \sqrt{t^{2} + 5}} \right|} \right)_{0}^{2\pi}} \\ & {= \frac{1}{2}\left( {2\pi\sqrt{4\pi^{2} + 5} + 5\mspace{2mu}\text{ln}\mspace{2mu}\left( {2\pi + \sqrt{4\pi^{2} + 5}} \right)} \right) - \frac{5}{2}\mspace{2mu}\text{ln}\mspace{2mu}\sqrt{5}} \\ & {\approx 25.343.} \end{array}$$

Calculate the arc length of the parameterized curve

$$\mathbf{\text{r}}(t) = \left\langle {2t^{2} + 1,2t^{2} - 1,t^{3}} \right\rangle,0 \leq t \leq 3.$$

We now return to the helix introduced earlier in this chapter. A vector-valued function that describes a helix can be written in the form

$$\mathbf{\text{r}}(t) = R\mspace{2mu}\text{cos}\left( \frac{2\pi Nt}{h} \right)\mspace{2mu}\mathbf{\text{i}} + R\mspace{2mu}\text{sin}\left( \frac{2\pi Nt}{h} \right)\mspace{2mu}\mathbf{\text{j}} + t\ \mathbf{\text{k}},\ 0 \leq t \leq h,$$

where *R* represents the radius of the helix, *h* represents the height (distance between two consecutive turns), and the helix completes *N* turns. Let’s derive a formula for the arc length of this helix using Equation 3.12. First of all,

$$\mathbf{r^{\prime}}(t) = - \frac{2\pi NR}{h}\text{sin}\left( \frac{2\pi Nt}{h} \right)\mspace{2mu}\mathbf{\text{i}} + \frac{2\pi NR}{h}\text{cos}\left( \frac{2\pi Nt}{h} \right)\mspace{2mu}\mathbf{\text{j}} + \mathbf{\text{k}}.$$

Therefore,

$$\begin{array}{cl} s & {= {\int_{a}^{b}\left\| {\mathbf{r^{\prime}}(t)} \right\|}\ dt} \\ & {= {\int_{0}^{h}{\sqrt{\left( {- \frac{2\pi NR}{h}\text{sin}\left( \frac{2\pi Nt}{h} \right)} \right)^{2} + \left( {\frac{2\pi NR}{h}\text{cos}\left( \frac{2\pi Nt}{h} \right)} \right)^{2} + 1^{2}}dt}}} \\ & {= {\int_{0}^{h}\sqrt{\frac{4\pi^{2}N^{2}R^{2}}{h^{2}}\left( {\text{sin}^{2}\left( \frac{2\pi Nt}{h} \right) + \text{cos}^{2}\left( \frac{2\pi Nt}{h} \right)} \right) + 1}}dt} \\ & {= {\int_{0}^{h}\sqrt{\frac{4\pi^{2}N^{2}R^{2}}{h^{2}} + 1}}dt} \\ & {= \left\lbrack {t\sqrt{\frac{4\pi^{2}N^{2}R^{2}}{h^{2}} + 1}} \right\rbrack_{0}^{h}} \\ & {= h\sqrt{\frac{4\pi^{2}N^{2}R^{2} + h^{2}}{h^{2}}}} \\ & {= \sqrt{4\pi^{2}N^{2}R^{2} + h^{2}}.} \end{array}$$

This gives a formula for the length of a wire needed to form a helix with *N* turns that has radius *R* and height *h.*

Arc-Length Parameterization 弧长参数化

We now have a formula for the arc length of a curve defined by a vector-valued function. Let’s take this one step further and examine what an arc-length function is.

If a vector-valued function represents the position of a particle in space as a function of time, then the arc-length function measures how far that particle travels as a function of time. The formula for the arc-length function follows directly from the formula for arc length:

$$s(t) = {\int_{a}^{t}\sqrt{\left( {f^{\prime}(u)} \right)^{2} + \left( {g^{\prime}(u)} \right)^{2} + \left( {h^{\prime}(u)} \right)^{2}}}du.$$ (3.13)

If the curve is in two dimensions, then only two terms appear under the square root inside the integral. The reason for using the independent variable *u* is to distinguish between time and the variable of integration. Since $s(t)$ measures distance traveled as a function of time, $s^{\prime}(t)$ measures the speed of the particle at any given time. Since we have a formula for $s(t)$ in Equation 3.13, we can differentiate both sides of the equation:

$$\begin{array}{cl} {s^{\prime}(t)} & {= \frac{d}{dt}\left\lbrack {{\int_{a}^{t}\sqrt{\left( {f^{\prime}(u)} \right)^{2} + \left( {g^{\prime}(u)} \right)^{2} + \left( {h^{\prime}(u)} \right)^{2}}}du} \right\rbrack} \\ & {= \frac{d}{dt}\left\lbrack {\int_{a}^{t}{\left\| {\mathbf{r^{\prime}}(u)} \right\|\ du}} \right\rbrack} \\ & {= \left\| {\mathbf{r^{\prime}}(t)} \right\|.} \end{array}$$

If we assume that $\mathbf{\text{r}}(t)$ defines a smooth curve, then the arc length is always increasing, so $s^{\prime}(t) > 0$ for $t > a.$ Last, if $\mathbf{\text{r}}(t)$ is a curve on which $\left\| {\mathbf{r^{\prime}}(t)} \right\| = 1$ for all *t*, then

$$s(t) = {\int_{a}^{t}{\left\| {\mathbf{r^{\prime}}(u)} \right\|\ du}} = {\int_{a}^{t}{1\ du}} = t - a,$$

which means that *t* represents the arc length as long as $a = 0.$

Arc-Length Function 弧长函数

Let $\mathbf{\text{r}}(t)$ describe a smooth curve for $t \geq a.$ Then the arc-length function is given by

$$s(t) = {\int_{a}^{t}{\left\| {\mathbf{r^{\prime}}(u)} \right\|\ du}}.$$ (3.14)

Furthermore, $\frac{ds}{dt} = \left\| {\mathbf{r^{\prime}}(t)} \right\| > 0.$ If $\left\| {\mathbf{r^{\prime}}(t)} \right\| = 1$ for all $t \geq a,$ then the parameter *t* represents the arc length from the starting point at $t = a.$

A useful application of this theorem is to find an alternative parameterization of a given curve, called an arc-length parameterization. Recall that any vector-valued function can be reparameterized via a change of variables. For example, if we have a function $\mathbf{\text{r}}(t) = \left\langle {3\mspace{2mu}\text{cos}\mspace{2mu} t,3\mspace{2mu}\text{sin}\mspace{2mu} t} \right\rangle,0 \leq t \leq 2\pi$ that parameterizes a circle of radius 3, we can change the parameter from *t* to $4t,$ obtaining a new parameterization $\mathbf{\text{r}}(t) = \left\langle {3\mspace{2mu}\text{cos}\mspace{2mu} 4t,3\mspace{2mu}\text{sin}\mspace{2mu} 4t} \right\rangle.$ The new parameterization still defines a circle of radius 3, but now we need only use the values $0 \leq t \leq {\pi\text{/}2}$ to traverse the circle once.

Suppose that we find the arc-length function $s(t)$ and are able to solve this function for *t* as a function of *s.* We can then reparameterize the original function $\mathbf{\text{r}}(t)$ by substituting the expression for *t* back into $\mathbf{\text{r}}(t).$ The vector-valued function is now written in terms of the parameter *s.* Since the variable *s* represents the arc length, we call this an *arc-length parameterization* of the original function $\mathbf{\text{r}}(t).$ One advantage of finding the arc-length parameterization is that the distance traveled along the curve starting from $s = 0$ is now equal to the parameter *s.* The arc-length parameterization also appears in the context of curvature (which we examine later in this section) and line integrals, which we study in the Introduction to Vector Calculus.

Finding an Arc-Length Parameterization 求弧长参数化

Find the arc-length parameterization for each of the following curves:

1. $\mathbf{\text{r}}(t) = 4\mspace{2mu}\text{cos}\mspace{2mu} t\ \mathbf{\text{i}} + 4\mspace{2mu}\text{sin}\mspace{2mu} t\ \mathbf{\text{j}},t \geq 0$

2. $\mathbf{\text{r}}(t) = \left\langle {t + 3,\ 2t - 4,2t} \right\rangle,t \geq 3$

Solution 解

1. First we find the arc-length function using Equation 3.14:

$$\begin{array}{cl} {s(t)} & {= {\int_{a}^{t}{\left\| {\mathbf{r^{\prime}}(u)} \right\|\ du}}} \\ & {= {\int_{0}^{t}{\left\| \left\langle {-4\mspace{2mu}\text{sin}\mspace{2mu} u,4\mspace{2mu}\text{cos}\mspace{2mu} u} \right\rangle \right\|\ du}}} \\ & {= {\int_{0}^{t}{\sqrt{\left( {-4\mspace{2mu}\text{sin}\mspace{2mu} u} \right)^{2} + \left( {4\mspace{2mu}\text{cos}\mspace{2mu} u} \right)^{2}}\ du}}} \\ & {= {\int_{0}^{t}{\sqrt{16\mspace{2mu}\text{sin}^{2}u + 16\mspace{2mu}\text{cos}^{2}u}\ du}}} \\ & {= {\int_{0}^{t}{4\ du}} = 4t,} \end{array}$$

which gives the relationship between the arc length *s* and the parameter *t* as $s = 4t;$ so, $t = {s\text{/}4.}$ Next we replace the variable *t* in the original function $\mathbf{\text{r}}(t) = 4\mspace{2mu}\text{cos}\mspace{2mu} t\ \mathbf{\text{i}} + 4\mspace{2mu}\text{sin}\mspace{2mu} t\ \mathbf{\text{j}}$ with the expression $s\text{/}4$ to obtain

$$\mathbf{\text{r}}(t) = 4\mspace{2mu}\text{cos}\left( \frac{s}{4} \right)\mspace{2mu}\mathbf{\text{i}} + 4\mspace{2mu}\text{sin}\left( \frac{s}{4} \right)\mspace{2mu}\mathbf{\text{j}}\text{.}$$

This is the arc-length parameterization of $\mathbf{\text{r}}(t).$ Since the original restriction on *t* was given by $t \geq 0,$ the restriction on *s* becomes ${s\text{/}4} \geq 0,$ or $s \geq 0.$

2. The arc-length function is given by Equation 3.14:

$$\begin{array}{cl} {s(t)} & {= {\int_{a}^{t}{\left\| {\mathbf{r^{\prime}}(u)} \right\|\ du}}} \\ & {= {\int_{3}^{t}{\left\| \left\langle {1,2,2} \right\rangle \right\|\ du}}} \\ & {= {\int_{3}^{t}{\sqrt{1^{2} + 2^{2} + 2^{2}}\ du}}} \\ & {= {\int_{3}^{t}{3\ du}}} \\ & {= 3t - 9.} \end{array}$$

Therefore, the relationship between the arc length *s* and the parameter *t* is $s = 3t - 9,$ so $t = \frac{s}{3} + 3.$ Substituting this into the original function $\mathbf{\text{r}}(t) = \left\langle {t + 3,\ 2t - 4,2t} \right\rangle$ yields

$$\mathbf{\text{r}}(s) = \left\langle {\left( {\frac{s}{3} + 3} \right) + 3,\ 2\left( {\frac{s}{3} + 3} \right) - 4,2\left( {\frac{s}{3} + 3} \right)} \right\rangle = \left\langle {\frac{s}{3} + 6,\ \frac{2s}{3} + 2,\frac{2s}{3} + 6} \right\rangle.$$

This is an arc-length parameterization of $\mathbf{\text{r}}(t).$ The original restriction on the parameter $t$ was $t \geq 3,$ so the restriction on *s* is $\left( {s\text{/}3} \right) + 3 \geq 3,$ or $s \geq 0.$

Find the arc-length function for the helix

$$\mathbf{\text{r}}(t) = \left\langle {3\mspace{2mu}\text{cos}\mspace{2mu} t,3\mspace{2mu}\text{sin}\mspace{2mu} t,4t} \right\rangle,t \geq 0.$$

Then, use the relationship between the arc length and the parameter *t* to find an arc-length parameterization of $\mathbf{\text{r}}(t).$

Curvature 曲率

An important topic related to arc length is curvature. The concept of curvature provides a way to measure how sharply a smooth curve turns. A circle has constant curvature. The smaller the radius of the circle, the greater the curvature.

Think of driving down a road. Suppose the road lies on an arc of a large circle. In this case you would barely have to turn the wheel to stay on the road. Now suppose the radius is smaller. In this case you would need to turn more sharply to stay on the road. In the case of a curve other than a circle, it is often useful first to inscribe a circle to the curve at a given point so that it is tangent to the curve at that point and “hugs” the curve as closely as possible in a neighborhood of the point (Figure 3.6). The curvature of the graph at that point is then defined to be the same as the curvature of the inscribed circle.

Let *C* be a smooth curve in the plane or in space given by $\mathbf{\text{r}}(s),$ where $s$ is the arc-length parameter. The curvature $\kappa$ at *s* is

$$\kappa = \left\| \frac{d\ \mathbf{\text{T}}}{ds} \right\| = \left\| {\mathbf{T^{\prime}}(s)} \right\|.$$

Visit this website for more information about the curvature of a space curve.

The formula in the definition of curvature is not very useful in terms of calculation. In particular, recall that $\mathbf{\text{T}}(t)$ represents the unit tangent vector to a given vector-valued function $\mathbf{\text{r}}(t),$ and the formula for $\mathbf{\text{T}}(t)$ is $\mathbf{\text{T}}(t) = \frac{\mathbf{r^{\prime}}(t)}{\left\| {\mathbf{r^{\prime}}(t)} \right\|}.$ To use the formula for curvature, it is first necessary to express $\mathbf{\text{r}}(t)$ in terms of the arc-length parameter *s*, then find the unit tangent vector $\mathbf{\text{T}}(s)$ for the function $\mathbf{\text{r}}(s),$ then take the derivative of $\mathbf{\text{T}}(s)$ with respect to *s.* This is a tedious process. Fortunately, there are equivalent formulas for curvature.

Alternative Formulas for Curvature 曲率的另算公式

If *C* is a smooth curve given by $\mathbf{\text{r}}(t),$ then the curvature $\kappa$ of *C* at *t* is given by

$$\kappa = \frac{\left\| {\mathbf{T^{\prime}}(t)} \right\|}{\left\| {\mathbf{r^{\prime}}(t)} \right\|}.$$ (3.15)

If *C* is a three-dimensional curve, then the curvature can be given by the formula

$$\kappa = \frac{\left\| {\mathbf{r^{\prime}}(t)\ \times \ \mathbf{r^{''}}(t)} \right\|}{\left\| {\mathbf{r^{\prime}}(t)} \right\|^{3}}.$$ (3.16)

If *C* is the graph of a function $y = f(x)$ and both $y^{\prime}$ and $y^{''}$ exist, then the curvature $\kappa$ at point $\left( {x,y} \right)$ is given by

$$\kappa = \frac{\left| y^{''} \right|}{\left\lbrack {1 + \left( y^{\prime} \right)^{2}} \right\rbrack^{3\text{/}2}}.$$ (3.17)

Proof 证明

The first formula follows directly from the chain rule:

$$\frac{d\mathbf{\text{T}}}{dt} = \frac{d\mathbf{\text{T}}}{ds}\ \frac{ds}{dt},$$

where *s* is the arc length along the curve *C.* Dividing both sides by ${ds}\text{/}{dt,}$ and taking the magnitude of both sides gives

$$\left\| \frac{d\mathbf{\text{T}}}{ds} \right\| = \left\| \frac{\mathbf{T^{\prime}}(t)}{\frac{ds}{dt}} \right\|.$$

Since ${{ds}\text{/}{dt}} = \left\| {\mathbf{r^{\prime}}(t)} \right\|,$ this gives the formula for the curvature $\kappa$ of a curve *C* in terms of any parameterization of *C*:

$$\kappa = \frac{\left\| {\mathbf{T^{\prime}}(t)} \right\|}{\left\| {\mathbf{r^{\prime}}(t)} \right\|}.$$

In the case of a three-dimensional curve, we start with the formulas $\mathbf{\text{T}}(t) = {\left( {\mathbf{r^{\prime}}(t)} \right)\text{/}\left\| {\mathbf{r^{\prime}}(t)} \right\|}$ and ${{ds}\text{/}{dt}} = \left\| {\mathbf{r^{\prime}}(t)} \right\|.$ Therefore, $\mathbf{r^{\prime}}(t) = \left( {{ds}\text{/}{dt}} \right)\mspace{2mu}\mathbf{\text{T}}(t).$ We can take the derivative of this function using the scalar product formula:

$$\mathbf{r^{''}}(t) = \frac{d^{2}s}{dt^{2}}\mathbf{\text{T}}(t) + \frac{ds}{dt}\mathbf{T^{\prime}}(t).$$

Using these last two equations we get

$$\begin{array}{cl}

{\mathbf{r^{\prime}}(t)\ \times \ \mathbf{r^{''}}(t)} & {= \frac{ds}{dt}\mathbf{\text{T}}(t)\ \times \ \left( {\frac{d^{2}s}{dt^{2}}\mathbf{\text{T}}(t) + \frac{ds}{dt}\mathbf{T^{\prime}}(t)} \right)} \\

& {= \frac{ds}{dt}\ \frac{d^{2}s}{dt^{2}}\mathbf{\text{T}}(t)\ \times \ \mathbf{\text{T}}(t) + \left( \frac{ds}{dt} \right)^{2}\mathbf{\text{T}}(t)\ \times \ \mathbf{T^{\prime}}(t).}

\end{array}$$

Since $\mathbf{\text{T}}(t)\ \times \ \mathbf{\text{T}}(t) = \mathbf{0},$ this reduces to

$$\mathbf{r^{\prime}}(t)\ \times \ \mathbf{r^{''}}(t) = \left( \frac{ds}{dt} \right)^{2}\mathbf{\text{T}}(t)\ \times \ \mathbf{T^{\prime}}(t).$$

Because $\mathbf{T}(t)$ is a unit vector, $\left\| \mathbf{T}(t) \right\| = 1$ for all values of $t$. So, $\mathbf{T}(t) \cdot \mathbf{T}(t) =$ $\left\| \mathbf{T}(t) \right\|^{2} = 1$.

By Alternative Formulas for Curvature, we have

$$0 = \frac{d}{dt}\left( \mathbf{T}(t) \cdot \mathbf{T}(t) \right) = {\mathbf{T^{\prime}}(t) \cdot \mathbf{T}(t) + \mathbf{T}(t) \cdot \mathbf{T^{\prime}}(t)} = 2\mathbf{T^{\prime}}(t) \cdot \mathbf{T}(t).$$

Therefore, $\mathbf{T^{\prime}} \cdot \mathbf{T} = 0$ showing that $\mathbf{T}$ and $\mathbf{T^{\prime}}$ are perpendicular. This means that

$$\left. \left\| \mathbf{T} \times \mathbf{T^{\prime}} \right. \right\|\left. = \right\|\left. \mathbf{T} \right\|\left. .~ \right\|\left. \mathbf{T^{\prime}} \right\|\left. \text{sin}\frac{\pi}{2}\text{=} \right\|\left. \mathbf{T}^{'} \right\|\text{,~so}$$ $$\left\| \mathbf{r}^{'}(t)\ \times \ \mathbf{r}^{''}(t) \right\| = \left( \frac{ds}{dt} \right)^{2}\left\| \mathbf{T}^{'}(t) \right\|.$$

Now we solve this equation for $\left\| {\mathbf{T^{\prime}}(t)} \right\|$ and use the fact that ${{ds}\text{/}{dt}} = \left\| {\mathbf{r^{\prime}}(t)} \right\|\text{:}$

$$\left\| {\mathbf{T^{\prime}}(t)} \right\| = \frac{\left\| {\mathbf{r^{\prime}}(t)\ \times \ \mathbf{r^{''}}(t)} \right\|}{\left\| {\mathbf{r^{\prime}}(t)} \right\|^{2}}.$$

Then, we divide both sides by $\left\| {\mathbf{r^{\prime}}(t)} \right\|.$ This gives

$$\kappa = \frac{\left\| {\mathbf{T^{\prime}}(t)} \right\|}{\left\| {\mathbf{r^{\prime}}(t)} \right\|} = \frac{\left\| {\mathbf{r^{\prime}}(t)\ \times \ \mathbf{r^{''}}(t)} \right\|}{\left\| {\mathbf{r^{\prime}}(t)} \right\|^{3}}.$$

This proves Equation 3.16. To prove Equation 3.17, we start with the assumption that curve *C* is defined by the function $y = f(x).$ Then, we can define $\mathbf{\text{r}}(t) = x\ \mathbf{\text{i}} + f(x)\ \mathbf{\text{j}} + 0\ \mathbf{\text{k}}.$ Using the previous formula for curvature:

$$\begin{aligned}

{\mathbf{r^{\prime}}(t)} & {= \mathbf{\text{i}} + f^{\prime}(x)\ \mathbf{\text{j}}} \\

{\mathbf{r^{''}}(t)} & {= f^{''}(x)\ \mathbf{\text{j}}} \\

{\mathbf{r^{\prime}}(t)\ \times \ \mathbf{r^{''}}(t)} & {= \left| \begin{matrix}

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

1 & {f^{\prime}(x)} & 0 \\

0 & {f^{''}(x)} & 0

\end{matrix} \right| = f^{''}(x)\ \mathbf{\text{k}}.}

\end{aligned}$$

Therefore,

$$\kappa = \frac{\left\| {\mathbf{r^{\prime}}(t)\ \times \ \mathbf{r^{''}}(t)} \right\|}{\left\| {\mathbf{r^{\prime}}(t)} \right\|^{3}} = \frac{\left| {f^{''}(x)} \right|}{\left( {1 + \left\lbrack \left. \left( f^{\prime}(x) \right)^{2} \right\rbrack \right.} \right)^{3\text{/}2}}.$$

□

Finding Curvature 求曲率

Find the curvature for each of the following curves at the given point:

1. $\mathbf{\text{r}}(t) = 4\mspace{2mu}\text{cos}\mspace{2mu} t\ \mathbf{\text{i}} + 4\mspace{2mu}\text{sin}\mspace{2mu} t\ \mathbf{\text{j}} + 3t\ \mathbf{\text{k}},t = \frac{4\pi}{3}$

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

Solution 解

1. This function describes a helix.

The curvature of the helix at $t = {\left( {4\pi} \right)\text{/}3}$ can be found by using Equation 3.15. First, calculate $\mathbf{\text{T}}(t)\text{:}$

$$\begin{array}{cl}

{\mathbf{\text{T}}(t)} & {= \frac{\mathbf{r^{\prime}}(t)}{\left\| {\mathbf{r^{\prime}}(t)} \right\|}} \\

& {= \frac{\left\langle {-4\mspace{2mu}\text{sin}\mspace{2mu} t,4\mspace{2mu}\text{cos}\mspace{2mu} t,3} \right\rangle}{\sqrt{\left( {-4\mspace{2mu}\text{sin}\mspace{2mu} t} \right)^{2} + \left( {4\mspace{2mu}\text{cos}\mspace{2mu} t} \right)^{2} + 3^{2}}}} \\

& {= \left\langle {- \frac{4}{5}\text{sin}\mspace{2mu} t,\frac{4}{5}\text{cos}\mspace{2mu} t,\frac{3}{5}} \right\rangle.}

\end{array}$$

Next, calculate $\mathbf{T^{\prime}}(t)\text{:}$

$$\mathbf{T^{\prime}}(t) = \left\langle {- \frac{4}{5}\text{cos}\mspace{2mu} t, - \frac{4}{5}\text{sin}\mspace{2mu} t,0} \right\rangle.$$

Last, apply Equation 3.15:

$$\begin{array}{cl}

\kappa & {= \frac{\left\| {\mathbf{T^{\prime}}(t)} \right\|}{\left\| {\mathbf{r^{\prime}}(t)} \right\|} = \frac{\left\| \left\langle {- \frac{4}{5}\text{cos}\mspace{2mu} t, - \frac{4}{5}\text{sin}\mspace{2mu} t,0} \right\rangle \right\|}{\left\| \left\langle {-4\mspace{2mu}\text{sin}\mspace{2mu} t,4\mspace{2mu}\text{cos}\mspace{2mu} t,3} \right\rangle \right\|}} \\

& {= \frac{\sqrt{\left( {- \frac{4}{5}\text{cos}\mspace{2mu} t} \right)^{2} + \left( {- \frac{4}{5}\text{sin}\mspace{2mu} t} \right)^{2} + 0^{2}}}{\sqrt{\left( {-4\mspace{2mu}\text{sin}\mspace{2mu} t} \right)^{2} + \left( {4\mspace{2mu}\text{cos}\mspace{2mu} t} \right)^{2} + 3^{2}}}} \\

& {= \frac{4\text{/}5}{5} = \frac{4}{25}.}

\end{array}$$

The curvature of this helix is constant at all points on the helix.

2. This function describes a semicircle.

To find the curvature of this graph, we must use Equation 3.17. First, we calculate $y^{\prime}$ and $y^{''}\text{:}$

$$\begin{array}{cl}

y & {= \sqrt{4x - x^{2}} = \left( {4x - x^{2}} \right)^{1\text{/}2}} \\

y^{\prime} & {= \frac{1}{2}\left( {4x - x^{2}} \right)^{- {1\text{/}2}}\left( {4 - 2x} \right) = \left( {2 - x} \right)\left( {4x - x^{2}} \right)^{- {1\text{/}2}} \\

y^{''} & {= - \left( {4x - x^{2}} \right)^{- {1\text{/}2}} + \left( {2 - x} \right)\left( {- \frac{1}{2}} \right)\left( {4x - x^{2}} \right)^{- {3\text{/}2}}\left( {4 - 2x} \right)} \\

& {= - \frac{4x - x^{2}}{\left( {4x - x^{2}} \right)^{3\text{/}2}} - \frac{\left( {2 - x} \right)^{2}}{\left( {4x - x^{2}} \right)^{3\text{/}2}} \\

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

& {= - \frac{4}{\left( {4x - x^{2}} \right)^{3\text{/}2}}.}

\end{array}$$

Then, we apply Equation 3.17:

$$\begin{array}{cl}

\kappa & {= \frac{\left| y^{''} \right|}{\left\lbrack {1 + \left( y^{\prime} \right)^{2}} \right\rbrack^{3\text{/}2}}} \\

& {= \frac{\left| {- \frac{4}{\left( {4x - x^{2}} \right)^{3\text{/}2}}} \right|}{\left\lbrack {1 + \left( {\left( {2 - x} \right)\left( {4x - x^{2}} \right)^{- {1\text{/}2}} \right)^{2}} \right\rbrack^{3\text{/}2}} = \frac{\left| \frac{4}{\left( {4x - x^{2}} \right)^{3\text{/}2}} \right|}{\left\lbrack {1 + \frac{\left( {2 - x} \right)^{2}}{4x - x^{2}}} \right\rbrack^{3\text{/}2}}} \\

& {= \frac{\left| \frac{4}{\left( {4x - x^{2}} \right)^{3\text{/}2}} \right|}{\left\lbrack \frac{4x - x^{2} + x^{2} - 4x + 4}{4x - x^{2}} \right\rbrack^{3\text{/}2}} = \left| \frac{4}{\left( {4x - x^{2}} \right)^{3\text{/}2}} \right| \cdot \frac{\left( {4x - x^{2}} \right)^{3\text{/}2}}{8}} \\

& {= \frac{1}{2}.}

\end{array}$$

The curvature of this circle is equal to the reciprocal of its radius.

Find the curvature of the curve defined by the function

$$y = 3x^{2} - 2x + 4$$

at the point $x = 2.$

The Normal and Binormal Vectors 法向量与副法向量

We have seen that the derivative $\mathbf{r^{\prime}}(t)$ of a vector-valued function is a tangent vector to the curve defined by $\mathbf{\text{r}}(t),$ and the unit tangent vector $\mathbf{\text{T}}(t)$ can be calculated by dividing $\mathbf{r^{\prime}}(t)$ by its magnitude. When studying motion in three dimensions, two other vectors are useful in describing the motion of a particle along a path in space: the principal unit normal vector and the binormal vector.

Let *C* be a three-dimensional smooth curve represented by r over an open interval *I.* If $\mathbf{T^{\prime}}(t) \neq \mathbf{0},$ then the principal unit normal vector at *t* is defined to be

$$\mathbf{\text{N}}(t) = \frac{\mathbf{T^{\prime}}(t)}{\left\| {\mathbf{T^{\prime}}(t)} \right\|}.$$ (3.18)

The binormal vector at *t* is defined as

$$\mathbf{\text{B}}(t) = \mathbf{\text{T}}(t)\ \times \ \mathbf{\text{N}}(t),$$ (3.19)

where $\mathbf{\text{T}}(t)$ is the unit tangent vector.

Note that, by definition, the binormal vector is orthogonal to both the unit tangent vector and the normal vector. Furthermore, $\mathbf{\text{B}}(t)$ is always a unit vector. This can be shown using the formula for the magnitude of a cross product

$$\left\| {\mathbf{\text{B}}(t)} \right\| = \left\| {\mathbf{\text{T}}(t)\ \times \ \mathbf{\text{N}}(t)} \right\| = \left\| {\mathbf{\text{T}}(t)} \right\|\left\| {\mathbf{\text{N}}(t)} \right\|\text{sin}\mspace{2mu}\theta,$$

where $\theta$ is the angle between $\mathbf{\text{T}}(t)$ and $\mathbf{\text{N}}(t).$ Since $\mathbf{\text{N}}(t)$ is the derivative of a unit vector, property (vii) of the derivative of a vector-valued function tells us that $\mathbf{\text{T}}(t)$ and $\mathbf{\text{N}}(t)$ are orthogonal to each other, so $\theta = {\pi\text{/}2}.$ Furthermore, they are both unit vectors, so their magnitude is 1. Therefore, $\left\| {\mathbf{\text{T}}(t)} \right\|\left\| {\mathbf{\text{N}}(t)} \right\|\text{sin}\mspace{2mu}\theta = (1)(1)\text{sin}\left( {\pi\text{/}2} \right) = 1$ and $\mathbf{\text{B}}(t)$ is a unit vector.

The principal unit normal vector can be challenging to calculate because the unit tangent vector involves a quotient, and this quotient often has a square root in the denominator. In the three-dimensional case, finding the cross product of the unit tangent vector and the unit normal vector can be even more cumbersome. Fortunately, we have alternative formulas for finding these two vectors, and they are presented in Motion in Space.

Finding the Principal Unit Normal Vector and Binormal Vector 求主单位法向量与副法向量

For each of the following vector-valued functions, find the principal unit normal vector. Then, if possible, find the binormal vector.

1. $\mathbf{\text{r}}(t) = 4\mspace{2mu}\text{cos}\mspace{2mu} t\ \mathbf{\text{i}} - 4\mspace{2mu}\text{sin}\mspace{2mu} t\ \mathbf{\text{j}}$

2. $\mathbf{\text{r}}(t) = \left( {6t + 2} \right)\ \mathbf{\text{i}} + 5t^{2}\ \mathbf{\text{j}} - 8t\ \mathbf{\text{k}}$

Solution 解

1. This function describes a circle.

To find the principal unit normal vector, we first must find the unit tangent vector $\mathbf{\text{T}}(t)\text{:}$

$$\begin{array}{cl}

{\mathbf{\text{T}}(t)} & {= \frac{\mathbf{r^{\prime}}(t)}{\left\| {\mathbf{r^{\prime}}(t)} \right\|}} \\

& {= \frac{-4\mspace{2mu}\text{sin}\mspace{2mu} t\ \mathbf{\text{i}} - 4\mspace{2mu}\text{cos}\mspace{2mu} t\ \mathbf{\text{j}}}{\sqrt{\left( {-4\mspace{2mu}\text{sin}\mspace{2mu} t} \right)^{2} + \left( {-4\mspace{2mu}\text{cos}\mspace{2mu} t} \right)^{2}}}} \\

& {= \frac{-4\mspace{2mu}\text{sin}\mspace{2mu} t\ \mathbf{\text{i}} - 4\mspace{2mu}\text{cos}\mspace{2mu} t\ \mathbf{\text{j}}}{\sqrt{16\mspace{2mu}\text{sin}^{2}t + 16\mspace{2mu}\text{cos}^{2}t}}} \\

& {= \frac{-4\mspace{2mu}\text{sin}\mspace{2mu} t\ \mathbf{\text{i}} - 4\mspace{2mu}\text{cos}\mspace{2mu} t\ \mathbf{\text{j}}}{\sqrt{16\left( {\text{sin}^{2}t + \text{cos}^{2}t} \right)}}} \\

& {= \frac{-4\mspace{2mu}\text{sin}\mspace{2mu} t\ \mathbf{\text{i}} - 4\mspace{2mu}\text{cos}\mspace{2mu} t\ \mathbf{\text{j}}}{4}} \\

& {= - \text{sin}\mspace{2mu} t\ \mathbf{\text{i}} - \text{cos}\mspace{2mu} t\ \mathbf{\text{j}}.}

\end{array}$$

Next, we use Equation 3.18:

$$\begin{array}{cl}

{\mathbf{\text{N}}(t)} & {= \frac{\mathbf{T^{\prime}}(t)}{\left\| {\mathbf{T^{\prime}}(t)} \right\|}} \\

& {= \frac{\text{−}\text{cos}\mspace{2mu} t\ \mathbf{\text{i}} + \text{sin}\mspace{2mu} t\ \mathbf{\text{j}}}{\sqrt{\left( {\text{−}\text{cos}\mspace{2mu} t} \right)^{2} + \left( {\text{sin}\mspace{2mu} t} \right)^{2}}}} \\

& {= \frac{\text{−}\text{cos}\mspace{2mu} t\ \mathbf{\text{i}} + \text{sin}\mspace{2mu} t\ \mathbf{\text{j}}}{\sqrt{\text{cos}^{2}t + \text{sin}^{2}t}}} \\

& {= - \text{cos}\mspace{2mu} t\ \mathbf{\text{i}} + \text{sin}\mspace{2mu} t\ \mathbf{\text{j}}.}

\end{array}$$

Notice that the unit tangent vector and the principal unit normal vector are orthogonal to each other for all values of *t*:

$$\begin{array}{cl}

{\mathbf{\text{T}}(t) \cdot \mathbf{\text{N}}(t)} & {= \left\langle {\text{−}\text{sin}\mspace{2mu} t, - \text{cos}\mspace{2mu} t} \right\rangle \cdot \left\langle {\text{−}\text{cos}\mspace{2mu} t,\text{sin}\mspace{2mu} t} \right\rangle} \\

& {= \text{sin}\mspace{2mu} t\mspace{2mu}\text{cos}\mspace{2mu} t - \text{cos}\mspace{2mu} t\mspace{2mu}\text{sin}\mspace{2mu} t} \\

& {= 0.}

\end{array}$$

Furthermore, the principal unit normal vector points toward the center of the circle from every point on the circle. Since $\mathbf{\text{r}}(t)$ defines a curve in two dimensions, we cannot calculate the binormal vector.

2. This function looks like this:

To find the principal unit normal vector, we first find the unit tangent vector $\mathbf{\text{T}}(t)\text{:}$

$$\begin{array}{cl}

{\mathbf{\text{T}}(t)} & {= \frac{\mathbf{r^{\prime}}(t)}{\left\| {\mathbf{r^{\prime}}(t)} \right\|}} \\

& {= \frac{6\ \mathbf{\text{i}} + 10t\ \mathbf{\text{j}} - 8\ \mathbf{\text{k}}}{\sqrt{6^{2} + \left( {10t} \right)^{2} + (-8)^{2}}}} \\

& {= \frac{6\ \mathbf{\text{i}} + 10t\ \mathbf{\text{j}} - 8\ \mathbf{\text{k}}}{\sqrt{36 + 100t^{2} + 64}}} \\

& {= \frac{6\ \mathbf{\text{i}} + 10t\ \mathbf{\text{j}} - 8\mathbf{\text{k}}}{\sqrt{100\left( {t^{2} + 1} \right)}}} \\

& {= \frac{3\ \mathbf{\text{i}} + 5t\ \mathbf{\text{j}} - 4\mathbf{\text{k}}}{5\sqrt{t^{2} + 1}}} \\

& {= \frac{3}{5}\left( {t^{2} + 1} \right)^{- {1\text{/}2}}\mathbf{\text{i}} + t\left( {t^{2} + 1} \right)^{- {1\text{/}2}}\mathbf{\text{j}} - \frac{4}{5}\left( {t^{2} + 1} \right)^{- {1\text{/}2}}\mathbf{\text{k}}.}

\end{array}$$

Next, we calculate $\mathbf{T^{\prime}}(t)$ and $\left\| {\mathbf{T^{\prime}}(t)} \right\|\text{:}$

$$\begin{array}{cl}

{\mathbf{T^{\prime}}(t)} & {= \frac{3}{5}\left( {- \frac{1}{2}} \right)\left( {t^{2} + 1} \right)^{- {3\text{/}2}}\left( {2t} \right)\mspace{2mu}\mathbf{\text{i}} + \left( {\left( {t^{2} + 1} \right)^{- {1\text{/}2}} - t\left( \frac{1}{2} \right)\left( {t^{2} + 1} \right)^{- {3\text{/}2}}\left( {2t} \right)} \right)\mspace{2mu}\mathbf{\text{j}}} \\

& {\mspace{54mu} - \frac{4}{5}\left( {- \frac{1}{2}} \right)\left( {t^{2} + 1} \right)^{- {3\text{/}2}}\left( {2t} \right)\mspace{2mu}\mathbf{\text{k}}} \\

& {= - \frac{3t}{5\left( {t^{2} + 1} \right)^{3\text{/}2}}\mathbf{\text{i}} + \frac{1}{\left( {t^{2} + 1} \right)^{3\text{/}2}}\mathbf{\text{j}} + \frac{4t}{5\left( {t^{2} + 1} \right)^{3\text{/}2}}\mathbf{\text{k}}} \\

\left\| {\mathbf{T^{\prime}}(t)} \right\| & {= \sqrt{\left( {- \frac{3t}{5\left( {t^{2} + 1} \right)^{3\text{/}2}}} \right)^{2} + \left( {- \frac{1}{\left( {t^{2} + 1} \right)^{3\text{/}2}}} \right)^{2} + \left( \frac{4t}{5\left( {t^{2} + 1} \right)^{3\text{/}2}} \right)^{2}}} \\

& {= \sqrt{\frac{9t^{2}}{25\left( {t^{2} + 1} \right)^{3}} + \frac{1}{\left( {t^{2} + 1} \right)^{3}} + \frac{16t^{2}}{25\left( {t^{2} + 1} \right)^{3}}}} \\

& {= \sqrt{\frac{25t^{2} + 25}{25\left( {t^{2} + 1} \right)^{3}}}} \\

& {= \sqrt{\frac{1}{\left( {t^{2} + 1} \right)^{2}}}} \\

& {= \frac{1}{t^{2} + 1}.}

\end{array}$$

Therefore, according to Equation 3.18:

$$\begin{array}{cl}

{\mathbf{\text{N}}(t)} & {= \frac{\mathbf{T^{\prime}}(t)}{\left\| {\mathbf{T^{\prime}}(t)} \right\|}} \\

& {= \left( {- \frac{3t}{5\left( {t^{2} + 1} \right)^{3\text{/}2}}\mathbf{\text{i}} + \frac{1}{\left( {t^{2} + 1} \right)^{3\text{/}2}}\mathbf{\text{j}} + \frac{4t}{5\left( {t^{2} + 1} \right)^{3\text{/}2}}\mathbf{\text{k}}} \right)\left( {t^{2} + 1} \right)} \\

& {= - \frac{3t}{5\left( {t^{2} + 1} \right)^{1\text{/}2}}\mathbf{\text{i}} + \frac{5}{5\left( {t^{2} + 1} \right)^{1\text{/}2}}\mathbf{\text{j}} + \frac{4t}{5\left( {t^{2} + 1} \right)^{1\text{/}2}}\mathbf{\text{k}}} \\

& {= - \frac{3t\ \mathbf{\text{i}} - 5\mathbf{\text{j}} - 4t\ \mathbf{\text{k}}}{5\sqrt{t^{2} + 1}}.}

\end{array}$$

Once again, the unit tangent vector and the principal unit normal vector are orthogonal to each other for all values of *t*:

$$\begin{array}{cl}

{\mathbf{\text{T}}(t) \cdot \mathbf{\text{N}}(t)} & {= \left( \frac{3\ \mathbf{\text{i}} + 5t\ \mathbf{\text{j}} - 4\mathbf{\text{k}}}{5\sqrt{t^{2} + 1}} \right) \cdot \left( {- \frac{3t\ \mathbf{\text{i}} - 5\mathbf{\text{j}} - 4t\ \mathbf{\text{k}}}{5\sqrt{t^{2} + 1}}} \right)} \\

& {= \frac{3\left( {-3t} \right) - 5t(-5) - 4\left( {4t} \right)}{5\sqrt{t^{2} + 1}}} \\

& {= \frac{-9t + 25t - 16t}{5\sqrt{t^{2} + 1}}} \\

& {= 0.}

\end{array}$$

Last, since $\mathbf{\text{r}}(t)$ represents a three-dimensional curve, we can calculate the binormal vector using Equation 3.17:

$$\begin{matrix}

{\textbf{B}(t)} & {= \textbf{T}(t)\ \times \ \textbf{N}(t)} \\

& {= \left| \begin{matrix}

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

\frac{3}{5\sqrt{t^{2} + 1}} & {+ \frac{5t}{5\sqrt{t^{2} + 1}}} & {- \frac{4}{5\sqrt{t^{2} + 1}}} \\

{- \frac{3t}{5\sqrt{t^{2} + 1}}} & {+ \frac{5}{5\sqrt{t^{2} + 1}}} & \frac{4t}{5\sqrt{t^{2} + 1}}

\end{matrix} \right|} \\

& {= \left( \left( + \frac{5t}{5\sqrt{t^{2} + 1}} \right)\left( \frac{4t}{5\sqrt{t^{2} + 1}} \right) - \left( - \frac{4}{5\sqrt{t^{2} + 1}} \right)\left( + \frac{5}{5\sqrt{t^{2} + 1}} \right) \right)\mspace{2mu}\textbf{i}} \\

& {\quad + \left( \left( \frac{3}{5\sqrt{t^{2} + 1}} \right)\left( \frac{4t}{5\sqrt{t^{2} + 1}} \right) - \left( - \frac{4}{5\sqrt{t^{2} + 1}} \right)\left( - \frac{3t}{5\sqrt{t^{2} + 1}} \right) \right)\mspace{2mu}\textbf{j}} \\

& {\quad + \left( \left( \frac{3}{5\sqrt{t^{2} + 1}} \right)\left( + \frac{5}{5\sqrt{t^{2} + 1}} \right) - \left( + \frac{5t}{5\sqrt{t^{2} + 1}} \right)\left( - \frac{3t}{5\sqrt{t^{2} + 1}} \right) \right)\mspace{2mu}\textbf{k}} \\

& {= \left( \frac{20t^{2} + 20}{25\left( t^{2} + 1 \right)} \right)\mspace{2mu}\textbf{i} + \left( \frac{-15 - 15t^{2}}{25\left( t^{2} + 1 \right)} \right)\mspace{2mu}\textbf{k}} \\

& {= 20\left( \frac{t^{2} + 1}{25\left( t^{2} + 1 \right)} \right)\mspace{2mu}\textbf{i+}15\left( \frac{t^{2} + 1}{25\left( t^{2} + 1 \right)} \right)\mspace{2mu}\textbf{k}} \\

& {= \frac{4}{5}\textbf{i+}\frac{3}{5}\textbf{k}.}

\end{matrix}$$

Find the unit normal vector for the vector-valued function $\mathbf{\text{r}}(t) = \left( {t^{2} - 3t} \right)\ \mathbf{\text{i}} + \left( {4t + 1} \right)\ \mathbf{\text{j}}$ and evaluate it at $t = 2.$

For any smooth curve in three dimensions that is defined by a vector-valued function, we now have formulas for the unit tangent vector T, the unit normal vector N, and the binormal vector B. The unit normal vector and the binormal vector form a plane that is perpendicular to the curve at any point on the curve, called the normal plane. In addition, these three vectors form a frame of reference in three-dimensional space called the Frenet frame of reference (also called the TNB frame) (Figure 3.7). Last, the plane determined by the vectors T and N forms the osculating plane of *C* at any point *P* on the curve.

Suppose we form a circle in the osculating plane of *C* at point *P* on the curve. Assume that the circle has the same curvature as the curve does at point *P* and let the circle have radius *r.* Then, the curvature of the circle is given by ${1\text{/}r}.$ We call *r* the radius of curvature of the curve, and it is equal to the reciprocal of the curvature. If this circle lies on the concave side of the curve and is tangent to the curve at point *P,* then this circle is called the osculating circle of *C* at *P*, as shown in the following figure.

For more information on osculating circles, see this demonstration on curvature and torsion, this article on osculating circles, and this discussion of Serret formulas.

To find the equation of an osculating circle in two dimensions, we need find only the center and radius of the circle.

Finding the Equation of an Osculating Circle 求密切圆的方程

Find an equation of the osculating circle of the curve defined by the function $y = x^{3} - 3x + 1$ at $x = 1.$

Solution 解

Figure 3.9 shows the graph of $y = x^{3} - 3x + 1.$

First, let's calculate the curvature at $x = 1\text{:}$

$$\kappa = \frac{\left| {f^{''}(x)} \right|}{\left( {1 + \left\lbrack {f^{\prime}(x)} \right\rbrack^{2}} \right)^{3\text{/}2}} = \frac{\left| {6x} \right|}{\left( {1 + \left\lbrack {3x^{2} - 3} \right\rbrack^{2}} \right)^{3\text{/}2}}.$$

This gives $\kappa = 6.$ Therefore, the radius of the osculating circle is given by $R = \frac{1}{\kappa} = \frac{1}{6}.$ Next, we then calculate the coordinates of the center of the circle. When $x = 1,$ the slope of the tangent line is zero. Therefore, the center of the osculating circle is directly above the point on the graph with coordinates $\left( {1,-1} \right).$ The center is located at $\left( {1, - \frac{5}{6}} \right).$ The formula for a circle with radius *r* and center $\left( {h,k} \right)$ is given by $\left( {x - h} \right)^{2} + \left( {y - k} \right)^{2} = r^{2}.$ Therefore, the equation of the osculating circle is $\left( {x - 1} \right)^{2} + \left( {y + \frac{5}{6}} \right)^{2} = \frac{1}{36}.$ The graph and its osculating circle appears in the following graph.

Find an equation of the osculating circle of the curve defined by the vector-valued function $y = 2x^{2} - 4x + 5$ at $x = 1.$

Section 3.3 Exercises 3.3 节习题

Find the arc length of the curve on the given interval.

102\.

$\mathbf{\text{r}}(t) = t^{2}\mathbf{\text{i}} + 14t\mathbf{\text{j}},\ 0 \leq t \leq 7.$ This portion of the graph is shown here:

103.

$\mathbf{\text{r}}(t) = t^{2}\mathbf{\text{i}} + (2t^{2} + 1)\mathbf{\text{j}},\ 1 \leq t \leq 3$

104\.

$\mathbf{\text{r}}(t) = \left\langle {2\mspace{2mu}\text{sin}\mspace{2mu} t,5t,2\mspace{2mu}\text{cos}\mspace{2mu} t} \right\rangle,0 \leq t \leq \pi.$ This portion of the graph is shown here:

105.

$\mathbf{\text{r}}(t) = \left\langle {t^{2} + 1,4t^{3} + 3} \right\rangle,\ - 1 \leq t \leq 0$

106\.

$\mathbf{\text{r}}(t) = \left\langle {e^{\text{−}t}\text{cos}\mspace{2mu} t,e^{\text{−}t}\text{sin}\mspace{2mu} t} \right\rangle$ over the interval $\left\lbrack {0,\frac{\pi}{2}} \right\rbrack.$ Here is the portion of the graph on the indicated interval:

107.

Find the length of one turn of the helix given by $\mathbf{\text{r}}(t) = \frac{1}{2}\text{cos}\mspace{2mu} t\mathbf{\text{i}} + \frac{1}{2}\text{sin}\mspace{2mu} t\mathbf{\text{j}} + \sqrt{\frac{3}{4}}\ t\ \mathbf{\text{k}}.$

108\.

Find the arc length of the vector-valued function $\mathbf{\text{r}}(t) = - t\mathbf{\text{i}} + 4t\mathbf{\text{j}} + 3t\mathbf{\text{k}}$ over $\lbrack 0,1\rbrack.$

109.

A particle travels once around a circle with the equation of motion $\mathbf{\text{r}}(t) = 3\mspace{2mu}\text{cos}\mspace{2mu} t\mathbf{\text{i}} + 3\mspace{2mu}\text{sin}\mspace{2mu} t\mathbf{\text{j}} + 0\mathbf{\text{k}}.$ Find the distance traveled around the circle by the particle.

110\.

Set up an integral to find the circumference of the ellipse with the equation $\mathbf{\text{r}}(t) = \text{cos}\mspace{2mu} t\mathbf{\text{i}} + 2\mspace{2mu}\text{sin}\mspace{2mu} t\mathbf{\text{j}} + 0\mathbf{\text{k}}.$

111.

Find the length of the curve $\mathbf{\text{r}}(t) = \left\langle {\sqrt{2}t,e^{t},e^{\text{−}t}} \right\rangle$ over the interval $0 \leq t \leq 1.$ The graph is shown here:

112\.

Find the length of the curve $\mathbf{\text{r}}(t) = \left\langle {2\mspace{2mu}\text{sin}\mspace{2mu} t,5t,2\mspace{2mu}\text{cos}\mspace{2mu} t} \right\rangle$ for $t \in \left\lbrack {-10,10} \right\rbrack.$

113.

The position function for a particle is $\mathbf{\text{r}}(t) = a\mspace{2mu}\text{cos}(\omega t)\mathbf{\text{i}} + b\mspace{2mu}\text{sin}(\omega t)\mathbf{\text{j}}.$ Find the unit tangent vector and the unit normal vector at $t = 0.$

114\.

Given $\mathbf{\text{r}}(t) = a\mspace{2mu}\text{cos}(\omega t)\mathbf{\text{i}} + b\mspace{2mu}\text{sin}(\omega t)\mathbf{\text{j}},$ find the binormal vector $\mathbf{\text{B}}(0).$

115.

Given $\mathbf{\text{r}}(t) = \left\langle {2e^{t},e^{t}\text{cos}\mspace{2mu} t,e^{t}\text{sin}\mspace{2mu} t} \right\rangle,$ determine the tangent vector $\mathbf{\text{T}}(t).$

116\.

Given $\mathbf{\text{r}}(t) = \left\langle {2e^{t},e^{t}\text{cos}\mspace{2mu} t,e^{t}\text{sin}\mspace{2mu} t} \right\rangle,$ determine the unit tangent vector $\mathbf{\text{T}}(t)$ evaluated at $t = 0.$

117.

Given $\mathbf{\text{r}}(t) = \left\langle {2e^{t},e^{t}\text{cos}\mspace{2mu} t,e^{t}\text{sin}\mspace{2mu} t} \right\rangle,$ find the unit normal vector $\mathbf{\text{N}}(t)$ evaluated at $t = 0,$ $\mathbf{\text{N}}(0).$

118\.

Given $\mathbf{\text{r}}(t) = \left\langle {2e^{t},e^{t}\text{cos}\mspace{2mu} t,e^{t}\text{sin}\mspace{2mu} t} \right\rangle,$ find the unit binormal vector evaluated at $t = 0.$

119.

Given $\mathbf{\text{r}}(t) = t\mathbf{\text{i}} + t^{2}\mathbf{\text{j}} + t\mathbf{\text{k}},$ find the unit tangent vector $\mathbf{\text{T}}(t).$ The graph is shown here:

120\.

Find the unit tangent vector $\mathbf{\text{T}}(t)$ and unit normal vector $\mathbf{\text{N}}(t)$ at $t = 0$ for the plane curve $\mathbf{\text{r}}(t) = \left\langle {t^{3} - 4t,5t^{2} - 2} \right\rangle.$ The graph is shown here:

121.

Find the unit tangent vector $\mathbf{\text{T}}(t)$ for $\mathbf{\text{r}}(t) = 3t\mathbf{\text{i}} + 5t^{2}\mathbf{\text{j}} + 2t\mathbf{\text{k}}$

122\.

Find the principal normal vector to the curve $\mathbf{\text{r}}(t) = \left\langle {6\mspace{2mu}\text{cos}\mspace{2mu} t,6\mspace{2mu}\text{sin}\mspace{2mu} t} \right\rangle$ at the point determined by $t = {\pi\text{/}3}.$

123.

Find $\mathbf{\text{T}}(t)$ for the curve $\mathbf{\text{r}}(t) = \left( {t^{3} - 4t} \right)\mspace{2mu}\mathbf{\text{i}} + \left( {5t^{2} - 2} \right)\mspace{2mu}\mathbf{\text{j}}.$

124\.

Find $\mathbf{\text{N}}(t)$ for the curve $\mathbf{\text{r}}(t) = \left( {t^{3} - 4t} \right)\mspace{2mu}\mathbf{\text{i}} + \left( {5t^{2} - 2} \right)\mspace{2mu}\mathbf{\text{j}}.$

125.

Find the unit normal vector $\mathbf{\text{N}}(t)$ for $\mathbf{\text{r}}(t) = \left\langle {2\mspace{2mu}\text{sin}\mspace{2mu} t,5t,2\mspace{2mu}\text{cos}\mspace{2mu} t} \right\rangle.$

126\.

Find the unit tangent vector $\mathbf{\text{T}}(t)$ for $\mathbf{\text{r}}(t) = \left\langle {2\mspace{2mu}\text{sin}\mspace{2mu} t,5t,2\mspace{2mu}\text{cos}\mspace{2mu} t} \right\rangle.$

127.

Find the arc-length function $s(t)$ for the line segment given by $\mathbf{\text{r}}(t) = \left\langle {3 - 3t,4t} \right\rangle.$ Write *r* as a parameter of *s.*

128\.

Parameterize the helix $\mathbf{\text{r}}(t) = \text{cos}\mspace{2mu} t\mathbf{\text{i}} + \text{sin}\mspace{2mu} t\mathbf{\text{j}} + t\mathbf{\text{k}}$ using the arc-length parameter *s*, from $t = 0.$

129.

Parameterize the curve using the arc-length parameter *s*, at the point at which $t = 0$ for $\mathbf{\text{r}}(t) = e^{t}\text{sin}\mspace{2mu} t\mathbf{\text{i}} + e^{t}\text{cos}\mspace{2mu} t\mathbf{\text{j}}.$

130\.

Find the curvature of the curve $\mathbf{\text{r}}(t) = 5\mspace{2mu}\text{cos}\mspace{2mu} t\mathbf{\text{i}} + 4\mspace{2mu}\text{sin}\mspace{2mu} t\mathbf{\text{j}}$ at $t = {\pi\text{/}3}.$ (*Note:* The graph is an ellipse.)

131.

Find the *x*-coordinate at which the curvature of the curve $y = {1\text{/}x}$ is a maximum value.

132\.

Find the curvature of the curve $\mathbf{\text{r}}(t) = 5\mspace{2mu}\text{cos}\mspace{2mu} t\mathbf{\text{i}} + 5\mspace{2mu}\text{sin}\mspace{2mu} t\mathbf{\text{j}}.$ Does the curvature depend upon the parameter *t*?

133.

Find the curvature $\kappa$ for the curve $y = x - \frac{1}{4}x^{2}$ at the point $x = 2.$

134\.

Find the curvature $\kappa$ for the curve $y = \frac{1}{3}x^{3}$ at the point $x = 1.$

135.

Find the curvature $\kappa$ of the curve $\mathbf{\text{r}}(t) = t\mathbf{\text{i}} + 6t^{2}\mathbf{\text{j}} + 4t\mspace{2mu}\mathbf{\text{k}}.$ The graph is shown here:

136\.

Find the curvature of $\mathbf{\text{r}}(t) = \left\langle {2\mspace{2mu}\text{sin}\mspace{2mu} t,5t,2\mspace{2mu}\text{cos}\mspace{2mu} t} \right\rangle.$

137.

Find the curvature of $\mathbf{\text{r}}(t) = \sqrt{2}t\mathbf{\text{i}} + e^{t}\mathbf{\text{j}} + e^{\text{−}t}\mathbf{\text{k}}$ at point $P\left( {0,1,1} \right).$

138\.

At what point does the curve $y = e^{x}$ have maximum curvature?

139.

What happens to the curvature as $x\rightarrow\infty$ for the curve $y = e^{x}?$

140\.

Find the point of maximum curvature on the curve $y = \text{ln}\mspace{2mu} x.$

141.

Find the equations of the normal plane and the osculating plane of the curve $\mathbf{\text{r}}(t) = \left\langle {2\mspace{2mu}\text{sin}(3t),t,2\mspace{2mu}\text{cos}(3t)} \right\rangle$ at point $\left( {0,\pi,-2} \right).$

142\.

Find equations of the osculating circles of the ellipse $4y^{2} + 9x^{2} = 36$ at the points $(2,0)$ and $(0,3).$

143.

Find the equation for the osculating plane at point $t = {\pi\text{/}4}$ on the curve $\mathbf{\text{r}}(t) = \text{cos}(2t)\mathbf{\text{i}} + \text{sin}(2t)\mathbf{\text{j}} + t\mathbf{\text{k}}.$

144\.

Find the radius of curvature of $6y = x^{3}$ at the point $\left( {2,\frac{4}{3}} \right).$

145.

Find the curvature at each point $\left( {x,y} \right)$ on the hyperbola $\mathbf{\text{r}}(t) = \left\langle {a\mspace{2mu}\text{cosh}(t),b\mspace{2mu}\text{sinh}(t)} \right\rangle.$

146\.

Calculate the curvature of the circular helix $\mathbf{\text{r}}(t) = r\mspace{2mu}\text{sin}(t)\mathbf{\text{i}} + r\mspace{2mu}\text{cos}(t)\mathbf{\text{j}} + t\mspace{2mu}\mathbf{\text{k}}.$

147.

Find the radius of curvature of $y = \text{ln}(x + 1)$ at point $\left( {2,\text{ln}\mspace{2mu} 3} \right).$

148\.

Find the radius of curvature of the hyperbola $xy = 1$ at point $(1,1).$

A particle moves along the plane curve C described by $\mathbf{\text{r}}(t) = t\mathbf{\text{i}} + t^{2}\mathbf{\text{j}}.$ Solve the following problems.

149.

Find the length of the curve over the interval $\left\lbrack {0,2} \right\rbrack.$

150\.

Find the curvature of the plane curve at $t = 0,1,2.$

151.

Describe the curvature as *t* increases from $t = 0$ to $t = 2.$

The surface of a large cup is formed by revolving the graph of the function $y = 0.25x^{1.6}$ from $x = 0$ to $x = 5$ about the *y*-axis (measured in centimeters).

152\.

\[T\] Use technology to graph the surface.

153.

Find the curvature $\kappa$ of the generating curve as a function of *x.*

154\.

\[T\] Use technology to graph the curvature function.

---

Components of the Acceleration Vector 加速度向量的分量

We can combine some of the concepts discussed in Arc Length and Curvature with the acceleration vector to gain a deeper understanding of how this vector relates to motion in the plane and in space. Recall that the unit tangent vector T and the unit normal vector N form an osculating plane at any point *P* on the curve defined by a vector-valued function $\mathbf{\text{r}}(t).$ The following theorem shows that the acceleration vector $\mathbf{\text{a}}(t)$ lies in the osculating plane and can be written as a linear combination of the unit tangent and the unit normal vectors.

The Plane of the Acceleration Vector 加速度向量所在的平面

The acceleration vector $\mathbf{\text{a}}(t)$ of an object moving along a curve traced out by a twice-differentiable function $\mathbf{\text{r}}(t)$ lies in the plane formed by the unit tangent vector $\mathbf{\text{T}}(t)$ and the principal unit normal vector $\mathbf{\text{N}}(t)$ to *C.* Furthermore,

$$\mathbf{\text{a}}(t) = v^{\prime}(t)\mspace{2mu}\mathbf{\text{T}}(t) + \left\lbrack {v(t)} \right\rbrack^{2}\kappa\mspace{2mu}\mathbf{\text{N}}(t).$$

Here, $v(t)$ is the speed of the object and $\kappa$ is the curvature of *C* traced out by $\mathbf{\text{r}}(t).$

Proof 证明

Because $\mathbf{\text{v}}(t) = \mathbf{r^{\prime}}(t)$ and $\mathbf{\text{T}}(t) = \frac{\mathbf{r^{\prime}}(t)}{\left\| {\mathbf{r^{\prime}}(t)} \right\|},$ we have $\mathbf{\text{v}}(t) = \left\| {\mathbf{r^{\prime}}(t)} \right\|\mathbf{\text{T}}(t) = v(t)\mspace{2mu}\mathbf{\text{T}}(t).$ Now we differentiate this equation:

$$\mathbf{\text{a}}(t) = \mathbf{v^{\prime}}(t) = \frac{d}{dt}\left( {v(t)\mspace{2mu}\mathbf{\text{T}}(t)} \right) = v^{\prime}(t)\mspace{2mu}\mathbf{\text{T}}(t) + v(t)\mspace{2mu}\mathbf{T^{\prime}}(t).$$

Since $\mathbf{\text{N}}(t) = \frac{\mathbf{T^{\prime}}(t)}{\left\| {\mathbf{T^{\prime}}(t)} \right\|},$ we know $\mathbf{T^{\prime}}(t) = \left\| {\mathbf{T^{\prime}}(t)} \right\|\mathbf{\text{N}}(t),$ so

$$\mathbf{\text{a}}(t) = v^{\prime}(t)\mspace{2mu}\mathbf{\text{T}}(t) + v(t)\left\| {\mathbf{T^{\prime}}(t)} \right\|\mathbf{\text{N}}(t).$$

A formula for curvature is $\kappa = \frac{\left\| {\mathbf{T^{\prime}}(t)} \right\|}{\left\| {\mathbf{r^{\prime}}(t)} \right\|},$ so $\left\| {\mathbf{T^{\prime}}(t)} \right\| = \kappa\left\| {\mathbf{r^{\prime}}(t)} \right\| = \kappa v(t).$ This gives $\mathbf{\text{a}}(t) = v^{\prime}(t)\mspace{2mu}\mathbf{\text{T}}(t) + \kappa\left( {v(t)} \right)^{2}\mathbf{\text{N}}(t).$

□

The coefficients of $\mathbf{\text{T}}(t)$ and $\mathbf{\text{N}}(t)$ are referred to as the tangential component of acceleration and the normal component of acceleration, respectively. We write $a_{\mathbf{\text{T}}}$ to denote the tangential component and $a_{\mathbf{\text{N}}}$ to denote the normal component.

Tangential and Normal Components of Acceleration 加速度的切向分量与法向分量

Let $\mathbf{\text{r}}(t)$ be a vector-valued function that denotes the position of an object as a function of time. Then $\mathbf{\text{a}}(t) = \mathbf{\text{r''}}(t)$ is the acceleration vector. The tangential and normal components of acceleration $a_{\mathbf{\text{T}}}$ and $a_{\mathbf{\text{N}}}$ are given by the formulas

$$a_{\mathbf{\text{T}}} = \mathbf{\text{a}} \cdot \mathbf{\text{T}} = \frac{\mathbf{\text{v}} \cdot \mathbf{\text{a}}}{\left\| \mathbf{\text{v}} \right\|}$$ (3.23)

and

$$a_{\mathbf{\text{N}}} = \mathbf{\text{a}} \cdot \mathbf{\text{N}} = \frac{\left\| {\mathbf{\text{v}}\ \times \ \mathbf{\text{a}}} \right\|}{\left\| \mathbf{\text{v}} \right\|} = \sqrt{\left\| \mathbf{\text{a}} \right\|^{2} - a_{\mathbf{\text{T}}}^{2}}.$$ (3.24)

These components are related by the formula

$$\mathbf{\text{a}}(t) = a_{\mathbf{\text{T}}}\mathbf{\text{T}}(t) + a_{\mathbf{\text{N}}}\mathbf{\text{N}}(t).$$ (3.25)

Here $\mathbf{\text{T}}(t)$ is the unit tangent vector to the curve defined by $\mathbf{\text{r}}(t),$ and $\mathbf{\text{N}}(t)$ is the unit normal vector to the curve defined by $\mathbf{\text{r}}(t).$

The normal component of acceleration is also called the *centripetal component of acceleration* or sometimes the *radial component of acceleration*. To understand centripetal acceleration, suppose you are traveling in a car on a circular track at a constant speed. Then, as we saw earlier, the acceleration vector points toward the center of the track at all times. As a rider in the car, you feel a pull toward the *outside* of the track because you are constantly turning. This sensation acts in the opposite direction of centripetal acceleration. The same holds true for noncircular paths. The reason is that your body tends to travel in a straight line and resists the force resulting from acceleration that push it toward the side. Note that at point *B* in Figure 3.14 the acceleration vector is pointing backward. This is because the car is decelerating as it goes into the curve.

The tangential and normal unit vectors at any given point on the curve provide a frame of reference at that point. The tangential and normal components of acceleration are the projections of the acceleration vector onto T and N, respectively.

Finding Components of Acceleration 求加速度的分量

A particle moves in a path defined by the vector-valued function $\mathbf{\text{r}}(t) = t^{2}\mathbf{\text{i}} + \left( {2t - 3} \right)\mspace{2mu}\mathbf{\text{j}} + \left( {3t^{2} - 3t} \right)\mspace{2mu}\mathbf{\text{k}},$ where *t* measures time in seconds and distance is measured in feet.

1. Find $a_{\mathbf{\text{T}}}$ and $a_{\mathbf{\text{N}}}$ as functions of *t*.

2. Find $a_{\mathbf{\text{T}}}$ and $a_{\mathbf{\text{N}}}$ at time $t = 2.$

Solution 解

1. Let’s start with Equation 3.23:

$$\begin{array}{cll} {\mathbf{\text{v}}(t)} & = & {\mathbf{r^{\prime}}(t) = 2t\mspace{2mu}\mathbf{\text{i}} + 2\mathbf{\text{j}} + \left( {6t - 3} \right)\mspace{2mu}\mathbf{\text{k}}} \\ {\mathbf{\text{a}}(t)} & = & {\mathbf{v^{\prime}}(t) = 2\mathbf{\text{i}} + 6\mathbf{\text{k}}} \\ a_{\mathbf{\text{T}}} & = & \frac{\mathbf{\text{v}} \cdot \mathbf{\text{a}}}{\left\| \mathbf{\text{v}} \right\|} \\ & = & \frac{\left( {2t\mspace{2mu}\mathbf{\text{i}} + 2\mathbf{\text{j}} + \left( {6t - 3} \right)\mspace{2mu}\mathbf{\text{k}}} \right) \cdot \left( {2\mathbf{\text{i}} + 6\mathbf{\text{k}}} \right)}{\left\| {2t\mspace{2mu}\mathbf{\text{i}} + 2\mathbf{\text{j}} + \left( {6t - 3} \right)\mspace{2mu}\mathbf{\text{k}}} \right\|} \\ & = & \frac{4t + 6\left( {6t - 3} \right)}{\sqrt{\left( {2t} \right)^{2} + 2^{2} + \left( {6t - 3} \right)^{2}}} \\ & = & {\frac{40t - 18}{\sqrt{40t^{2} - 36t + 13}}.} \end{array}$$

Then we apply Equation 3.24:

$$\begin{array}{cl} a_{\mathbf{\text{N}}} & {= \sqrt{\left\| \mathbf{\text{a}} \right\|^{2} - {a\mspace{2mu}\mathbf{\text{T}}}^{2}}} \\ & {= \sqrt{\left\| {2\mathbf{\text{i}} + 6\mathbf{\text{k}}} \right\|^{2} - \left( \frac{40t - 18}{\sqrt{40t^{2} - 36t + 13}} \right)^{2}}} \\ & {= \sqrt{4 + 36 - \frac{\left( {40t - 18} \right)^{2}}{40t^{2} - 36t + 13}}} \\ & {= \sqrt{\frac{40\left( {40t^{2} - 36t + 13} \right) - \left( {1600t^{2} - 1440t + 324} \right)}{40t^{2} - 36t + 13}}} \\ & {= \sqrt{\frac{196}{40t^{2} - 36t + 13}}} \\ & {= \frac{14}{\sqrt{40t^{2} - 36t + 13}}.} \end{array}$$

2. We must evaluate each of the answers from part a. at $t = 2\text{:}$

$$\begin{array}{cll} {a_{\mathbf{\text{T}}}(2)} & = & \frac{40(2) - 18}{\sqrt{40(2)^{2} - 36(2) + 13}} \\ & = & {\frac{80 - 18}{\sqrt{160 - 72 + 13}} = \frac{62}{\sqrt{101}}} \\ {a_{\mathbf{\text{N}}}(2)} & = & \frac{14}{\sqrt{40(2)^{2} - 36(2) + 13}} \\ & = & {\frac{14}{\sqrt{160 - 72 + 13}} = \frac{14}{\sqrt{101}}.} \end{array}$$

The units of acceleration are feet per second squared, as are the units of the normal and tangential components of acceleration.

An object moves in a path defined by the vector-valued function $\mathbf{\text{r}}(t) = 4t\mspace{2mu}\mathbf{\text{i}} + t^{2}\mathbf{\text{j}},$ where *t* measures time in seconds.

1. Find $a_{\mathbf{\text{T}}}$ and $a_{\mathbf{\text{N}}}$ as functions of *t*.

2. Find $a_{\mathbf{\text{T}}}$ and $a_{\mathbf{\text{N}}}$ at time $t = -3.$

Projectile Motion 抛体运动

Now let’s look at an application of vector functions. In particular, let’s consider the effect of gravity on the motion of an object as it travels through the air, and how it determines the resulting trajectory of that object. In the following, we ignore the effect of air resistance. This situation, with an object moving with an initial velocity but with no forces acting on it other than gravity, is known as projectile motion. It describes the motion of objects from golf balls to baseballs, and from arrows to cannonballs.

First we need to choose a coordinate system. If we are standing at the origin of this coordinate system, then we choose the positive *y*-axis to be up, the negative *y-*axis to be down, and the positive *x-*axis to be forward (i.e., away from the thrower of the object). The effect of gravity is in a downward direction, so Newton’s second law tells us that the force on the object resulting from gravity is equal to the mass of the object times the acceleration resulting from to gravity, or $F_{g} = mg,$ where $F_{g}$ represents the force from gravity and *g* represents the acceleration resulting from gravity at Earth’s surface. The value of *g* in the English system of measurement is approximately 32 ft/sec² and it is approximately 9.8 m/sec² in the metric system. This is the only force acting on the object. Since gravity acts in a downward direction, we can write the force resulting from gravity in the form $F_{g} = \text{−}mg\mspace{2mu}\mathbf{\text{j}},$ as shown in the following figure.

Visit this website for a video showing projectile motion.

Newton’s second law also tells us that $F = m\mspace{2mu}\mathbf{\text{a}},$ where a represents the acceleration vector of the object. This force must be equal to the force of gravity at all times, so we therefore know that

$$\begin{array}{rll} F & = & F_{g} \\ {m\mspace{2mu}\mathbf{\text{a}}} & = & {\text{−}mg\mspace{2mu}\mathbf{\text{j}}} \\ \mathbf{\text{a}} & = & {\text{−}g\mspace{2mu}\mathbf{\text{j}}.} \end{array}$$

Now we use the fact that the acceleration vector is the first derivative of the velocity vector. Therefore, we can rewrite the last equation in the form

$$\mathbf{v^{\prime}}(t) = \text{−}g\mspace{2mu}\mathbf{\text{j}}.$$

By taking the antiderivative of each side of this equation we obtain

$$\begin{array}{cl} {\mathbf{\text{v}}(t)} & {= {\int{\text{−}g\mspace{2mu}\mathbf{\text{j}}dt}}} \\ & {= \text{−}gt\mspace{2mu}\mathbf{\text{j}} + \mathbf{\text{C}}_{1}} \end{array}$$

for some constant vector $\mathbf{\text{C}}_{1}.$ To determine the value of this vector, we can use the velocity of the object at a fixed time, say at time $t = 0.$ We call this velocity the *initial velocity*: $\mathbf{\text{v}}(0) = \mathbf{\text{v}}_{0}.$ Therefore, $\mathbf{\text{v}}(0) = \text{−}g(0)\mspace{2mu}\mathbf{\text{j}} + \mathbf{\text{C}}_{1} = \mathbf{\text{v}}_{0}$ and $\mathbf{\text{C}}_{1} = \mathbf{\text{v}}_{0}.$ This gives the velocity vector as $\mathbf{\text{v}}(t) = \text{−}gt\mspace{2mu}\mathbf{\text{j}} + \mathbf{\text{v}}_{0}.$

Next we use the fact that velocity $\mathbf{\text{v}}(t)$ is the derivative of position $\mathbf{\text{s}}(t).$ This gives the equation

$$\mathbf{s^{\prime}}(t) = \text{−}gt\mspace{2mu}\mathbf{\text{j}} + \mathbf{\text{v}}_{0}.$$

Taking the antiderivative of both sides of this equation leads to

$$\begin{array}{cl} {\mathbf{\text{s}}(t)} & {= {\int{\text{−}gt\mspace{2mu}\mathbf{\text{j}} + \mathbf{\text{v}}_{0}dt}}} \\ & {= - \frac{1}{2}gt^{2}\mathbf{\text{j}} + \mathbf{\text{v}}_{0}t + \mathbf{\text{C}}_{2},} \end{array}$$

with another unknown constant vector $\mathbf{\text{C}}_{2}.$ To determine the value of $\mathbf{\text{C}}_{2},$ we can use the position of the object at a given time, say at time $t = 0.$ We call this position the *initial position*: $\mathbf{\text{s}}(0) = \mathbf{\text{s}}_{0}.$ Therefore, $\mathbf{\text{s}}(0) = \text{−}\left( {1\text{/}2} \right)g(0)^{2}\mathbf{\text{j}} + \mathbf{\text{v}}_{0}(0) + \mathbf{\text{C}}_{2} = \mathbf{\text{s}}_{0}$ and $\mathbf{\text{C}}_{2} = \mathbf{\text{s}}_{0}.$ This gives the position of the object at any time as

$$\mathbf{\text{s}}(t) = - \frac{1}{2}gt^{2}\mathbf{\text{j}} + \mathbf{\text{v}}_{0}t + \mathbf{\text{s}}_{0}.$$

Let’s take a closer look at the initial velocity and initial position. In particular, suppose the object is thrown upward from the origin at an angle $\theta$ to the horizontal, with initial speed $v_{0}.$ How can we modify the previous result to reflect this scenario? First, we can assume it is thrown from the origin. If not, then we can move the origin to the point from where it is thrown. Therefore, $\mathbf{\text{s}}_{0} = \mathbf{0},$ as shown in the following figure.

We can rewrite the initial velocity vector in the form $\mathbf{\text{v}}_{0} = v_{0}\text{cos}\mspace{2mu}\theta\mspace{2mu}\mathbf{\text{i}} + v_{0}\text{sin}\mspace{2mu}\theta\mspace{2mu}\mathbf{\text{j}}.$ Then the equation for the position function $\mathbf{\text{s}}(t)$ becomes

$$\begin{array}{cl} {\mathbf{\text{s}}(t)} & {= - \frac{1}{2}gt^{2}\mspace{2mu}\mathbf{\text{j}} + v_{0}t\mspace{2mu}\text{cos}\mspace{2mu}\theta\mspace{2mu}\mathbf{\text{i}} + v_{0}t\mspace{2mu}\text{sin}\mspace{2mu}\theta\mspace{2mu}\mathbf{\text{j}}} \\ & {= v_{0}t\mspace{2mu}\text{cos}\mspace{2mu}\theta\mspace{2mu}\mathbf{\text{i}} + v_{0}t\mspace{2mu}\text{sin}\mspace{2mu}\theta\mspace{2mu}\mathbf{\text{j}} - \frac{1}{2}gt^{2}\mspace{2mu}\mathbf{\text{j}}} \\ & {= v_{0}t\mspace{2mu}\text{cos}\mspace{2mu}\theta\mspace{2mu}\mathbf{\text{i}} + \left( {v_{0}t\mspace{2mu}\text{sin}\mspace{2mu}\theta\mspace{2mu} - \frac{1}{2}gt^{2}} \right)\mspace{2mu}\mathbf{\text{j}}.} \end{array}$$

The coefficient of i represents the horizontal component of $\mathbf{\text{s}}(t)$ and is the horizontal distance of the object from the origin at time *t.* The maximum value of the horizontal distance (measured at the same initial and final altitude) is called the range *R*. The coefficient of j represents the vertical component of $\mathbf{\text{s}}(t)$ and is the altitude of the object at time *t.* The maximum value of the vertical distance is the height *H*.

Motion of a Cannonball 炮弹的运动

During an Independence Day celebration, a cannonball is fired from a cannon on a cliff toward the water. The cannon is aimed at an angle of 30° above horizontal and the initial speed of the cannonball is $600\ \text{ft/sec}\text{.}$ The cliff is 100 ft above the water (Figure 3.17).

1. Find the maximum height of the cannonball.

2. How long will it take for the cannonball to splash into the sea?

3. How far out to sea will the cannonball hit the water?

Solution 解

We use the equation

$$\mathbf{\text{s}}(t) = v_{0}t\mspace{2mu}\text{cos}\mspace{2mu}\theta\mspace{2mu}\mathbf{\text{i}} + \left( {v_{0}t\mspace{2mu}\text{sin}\mspace{2mu}\theta - \frac{1}{2}gt^{2}} \right)\mspace{2mu}\mathbf{\text{j}}$$

with $\theta = 30\text{°},$ $g = 32{\ \text{ft/sec}}^{2},$ and $v_{0} = 600$ ft/sec. Then the position equation becomes

$$\begin{array}{cl} {\mathbf{\text{s}}(t)} & {= 600t\left( {\text{cos}\mspace{2mu} 30} \right)\mspace{2mu}\mathbf{\text{i}} + \left( {600t\mspace{2mu}\text{sin}\mspace{2mu} 30 - \frac{1}{2}(32)t^{2}} \right)\mspace{2mu}\mathbf{\text{j}}} \\ & {= 300t\sqrt{3}\mathbf{\text{i}} + \left( {300t - 16t^{2}} \right)\mspace{2mu}\mathbf{\text{j}}.} \end{array}$$

1. The cannonball reaches its maximum height when the vertical component of its velocity is zero, because the cannonball is neither rising nor falling at that point. The velocity vector is

$$\begin{array}{cl} {\mathbf{\text{v}}(t)} & {= \mathbf{s^{\prime}}(t)} \\ & {= 300\sqrt{3}\mathbf{\text{i}} + \left( {300 - 32t} \right)\mspace{2mu}\mathbf{\text{j}}.} \end{array}$$

Therefore, the vertical component of velocity is given by the expression $300 - 32t.$ Setting this expression equal to zero and solving for *t* gives $t = 9.375$ sec. The height of the cannonball at this time is given by the vertical component of the position vector, evaluated at $t = 9.375.$

$$\begin{array}{cl} {\mathbf{\text{s}}(9.375)} & {= 300(9.375)\sqrt{3}\mathbf{\text{i}} + \left( {300(9.375) - 16(9.375)^{2}} \right)\mspace{2mu}\mathbf{\text{j}}} \\ & {= 4871.39\mathbf{\text{i}} + 1406.25\mathbf{\text{j}}} \end{array}$$

Therefore, the maximum height of the cannonball is 1406.39 ft above the cannon, or 1506.39 ft above sea level.

2. When the cannonball lands in the water, it is 100 ft below the cannon. Therefore, the vertical component of the position vector is equal to $-100.$ Setting the vertical component of $\mathbf{\text{s}}(t)$ equal to $-100$ and solving, we obtain

$$\begin{array}{cll} & & \\ {300t - 16t^{2}} & = & -100 \\ {16t^{2} - 300t - 100} & = & 0 \\ {4t^{2} - 75t - 25} & = & 0 \\ t & = & \frac{75 \pm \sqrt{(-75)^{2} - 4(4)(-25)}}{2(4)} \\ & = & \frac{75 \pm \sqrt{6025}}{8} \\ & = & {\frac{75 \pm 5\sqrt{241}}{8}.} \end{array}$$

The positive value of *t* that solves this equation is approximately 19.08. Therefore, the cannonball hits the water after approximately 19.08 sec.

3. To find the distance out to sea, we simply substitute the answer from part (b) into $\mathbf{\text{s}}(t)\text{:}$

$$\begin{array}{cl} {\mathbf{\text{s}}(19.08)} & {= 300(19.08)\sqrt{3}\mathbf{\text{i}} + \left( {300(19.08) - 16(19.08)^{2}} \right)\mspace{2mu}\mathbf{\text{j}}} \\ & {= 9914.26\mspace{2mu}\mathbf{\text{i}} - 100.7424\mspace{2mu}\mathbf{\text{j}}.} \end{array}$$

Therefore, the ball hits the water about 9914.26 ft away from the base of the cliff. Notice that the vertical component of the position vector is very close to $-100,$ which tells us that the ball just hit the water. Note that 9914.26 feet is not the true range of the cannon since the cannonball lands in the ocean at a location below the cannon. The range of the cannon would be determined by finding how far out the cannonball is when its height is 100 ft above the water (the same as the altitude of the cannon).

An archer fires an arrow at an angle of 40° above the horizontal with an initial speed of 98 m/sec. The height of the archer is 171.5 cm. Find the horizontal distance the arrow travels before it hits the ground.

One final question remains: In general, what is the maximum distance a projectile can travel, given its initial speed? To determine this distance, we assume the projectile is fired from ground level and we wish it to return to ground level. In other words, we want to determine an equation for the range. In this case, the equation of projectile motion is

$$\mathbf{\text{s}}(t) = v_{0}t\mspace{2mu}\text{cos}\mspace{2mu}\theta\mspace{2mu}\mathbf{\text{i}} + \left( {v_{0}t\mspace{2mu}\text{sin}\mspace{2mu}\theta - \frac{1}{2}gt^{2}} \right)\mspace{2mu}\mathbf{\text{j}}.$$

Setting the second component equal to zero and solving for *t* yields

$$\begin{array}{rll} & & \\ {v_{0}t\mspace{2mu}\text{sin}\mspace{2mu}\theta - \frac{1}{2}gt^{2}} & = & 0 \\ {t\left( {v_{0}\text{sin}\mspace{2mu}\theta - \frac{1}{2}gt} \right)} & = & 0. \end{array}$$

Therefore, either $t = 0$ or $t = \frac{2v_{0}\text{sin}\mspace{2mu}\theta}{g}.$ We are interested in the second value of *t*, so we substitute this into $\mathbf{\text{s}}(t),$ which gives

$$\begin{array}{cl} {\mathbf{\text{s}}\left( \frac{2v_{0}\text{sin}\mspace{2mu}\theta}{g} \right)} & {= v_{0}\left( \frac{2v_{0}\text{sin}\mspace{2mu}\theta}{g} \right)\text{cos}\mspace{2mu}\theta\mspace{2mu}\mathbf{\text{i}} + \left( {v_{0}\left( \frac{2v_{0}\text{sin}\mspace{2mu}\theta}{g} \right)\text{sin}\mspace{2mu}\theta - \frac{1}{2}g\left( \frac{2v_{0}\text{sin}\mspace{2mu}\theta}{g} \right)^{2}} \right)\mspace{2mu}\mathbf{\text{j}}} \\ & {= \left( \frac{2v_{0}^{2}\text{sin}\mspace{2mu}\theta\mspace{2mu}\text{cos}\mspace{2mu}\theta}{g} \right)\mspace{2mu}\mathbf{\text{i}}} \\ & {= \frac{v_{0}^{2}\text{sin}\mspace{2mu} 2\theta}{g}\mathbf{\text{i}}.} \end{array}$$

Thus, the expression for the range of a projectile fired at an angle $\theta$ is

$$R = \frac{v_{0}^{2}\text{sin}\mspace{2mu} 2\theta}{g}\mathbf{\text{i}}.$$

The only variable in this expression is $\theta.$ To maximize the distance traveled, take the derivative of the coefficient of i with respect to $\theta$ and set it equal to zero:

$$\begin{array}{rll} & & \\ & & \\ {\frac{d}{d\theta}\left( \frac{v_{0}^{2}\text{sin}\mspace{2mu} 2\theta}{g} \right)} & = & 0 \\ \frac{2v_{0}^{2}\text{cos}\mspace{2mu} 2\theta}{g} & = & 0 \\ \theta & = & {45\text{°}.} \end{array}$$

This value of $\theta$ is the smallest positive value that makes the derivative equal to zero. Therefore, in the absence of air resistance, the best angle to fire a projectile (to maximize the range) is at a $45\text{°}$ angle. The distance it travels is given by

$$\mathbf{\text{s}}\left( \frac{2v_{0}\text{sin}\mspace{2mu} 45}{g} \right) = \frac{v_{0}^{2}\text{sin}\mspace{2mu} 90}{g}\mathbf{\text{i}} = \frac{v_{0}^{2}}{g}\mathbf{\text{j}}.$$

Therefore, the range for an angle of $45\text{°}$ is $v_{0}^{2}\text{/}{g.}$

Kepler’s Laws 开普勒定律

During the early 1600s, Johannes Kepler was able to use the amazingly accurate data from his mentor Tycho Brahe to formulate his three laws of planetary motion, now known as Kepler’s laws of planetary motion. These laws also apply to other objects in the solar system in orbit around the Sun, such as comets (e.g., Halley’s comet) and asteroids. Variations of these laws apply to satellites in orbit around Earth.

Kepler’s Laws of Planetary Motion 开普勒行星运动定律

1. The path of any planet about the Sun is elliptical in shape, with the center of the Sun located at one focus of the ellipse (the law of ellipses).

2. A line drawn from the center of the Sun to the center of a planet sweeps out equal areas in equal time intervals (the law of equal areas) (Figure 3.18).

3. The ratio of the squares of the periods of any two planets is equal to the ratio of the cubes of the lengths of their semimajor orbital axes (the law of harmonies).

Kepler’s third law is especially useful when using appropriate units. In particular, *1 astronomical unit* is defined to be the average distance from Earth to the Sun, and is now recognized to be 149,597,870,700 m or, approximately 93,000,000 mi. We therefore write 1 A.U. = 93,000,000 mi. Since the time it takes for Earth to orbit the Sun is 1 year, we use Earth years for units of time. Then, substituting 1 year for the period of Earth and 1 A.U. for the average distance to the Sun, Kepler’s third law can be written as

$$T_{p}^{2} = D_{p}^{3}$$

for any planet in the solar system, where $T_{P}$ is the period of that planet measured in Earth years and $D_{P}$ is the average distance from that planet to the Sun measured in astronomical units. Therefore, if we know the average distance from a planet to the Sun (in astronomical units), we can then calculate the length of its year (in Earth years), and vice versa.

Kepler’s laws were formulated based on observations from Brahe; however, they were not proved formally until Sir Isaac Newton was able to apply calculus. Furthermore, Newton was able to generalize Kepler’s third law to other orbital systems, such as a moon orbiting around a planet. Kepler’s original third law only applies to objects orbiting the Sun.

Proof 证明

Let’s now prove Kepler’s first law using the calculus of vector-valued functions. First we need a coordinate system. Let’s place the Sun at the origin of the coordinate system and let the vector-valued function $\mathbf{\text{r}}(t)$ represent the location of a planet as a function of time. Newton proved Kepler’s law using his second law of motion and his law of universal gravitation. Newton’s second law of motion can be written as $\mathbf{\text{F}} = m\mspace{2mu}\mathbf{\text{a}},$ where F represents the net force acting on the planet. His law of universal gravitation can be written in the form $\mathbf{\text{F}} = - \frac{GmM}{\left\| \mathbf{\text{r}} \right\|^{2}} \cdot \frac{\mathbf{\text{r}}}{\left\| \mathbf{\text{r}} \right\|},$ which indicates that the force resulting from the gravitational attraction of the Sun points back toward the Sun, and has magnitude $\frac{GmM}{\left\| \mathbf{\text{r}} \right\|^{2}}$ (Figure 3.19).

Setting these two forces equal to each other, and using the fact that $\mathbf{\text{a}}(t) = \mathbf{v^{\prime}}(t),$ we obtain

$$m\mspace{2mu}\mathbf{v^{\prime}}(t) = - \frac{GmM}{\left\| \mathbf{\text{r}} \right\|^{2}} \cdot \frac{\mathbf{\text{r}}}{\left\| \mathbf{\text{r}} \right\|},$$

which can be rewritten as

$$\frac{d\mspace{2mu}\mathbf{\text{v}}}{dt} = - \frac{GM}{\left\| \mathbf{\text{r}} \right\|^{3}}\mathbf{\text{r}}.$$

This equation shows that the vectors ${d\mspace{2mu}\mathbf{\text{v}}}\text{/}{dt}$ and r are parallel to each other, so ${{d\mspace{2mu}\mathbf{\text{v}}}\text{/}{dt}}\ \times \ \mathbf{\text{r}} = \mathbf{0}.$ Next, let’s differentiate $\mathbf{\text{r}}\ \times \ \mathbf{\text{v}}$ with respect to time:

$$\frac{d}{dt}\left( {\mathbf{\text{r}}\ \times \ \mathbf{\text{v}}} \right) = \frac{d\mspace{2mu}\mathbf{\text{r}}}{dt}\ \times \ \mathbf{\text{v}} + \mathbf{\text{r}}\ \times \ \frac{d\mspace{2mu}\mathbf{\text{v}}}{dt} = \mathbf{\text{v}}\ \times \ \mathbf{\text{v}} + \mathbf{0} = \mathbf{0}.$$

This proves that $\mathbf{\text{r}}\ \times \ \mathbf{\text{v}}$ is a constant vector, which we call C. Since $\mathbf{\text{r}}$ and v are both perpendicular to C for all values of *t*, they must lie in a plane perpendicular to C. Therefore, the motion of the planet lies in a plane.

Next we calculate the expression ${{d\mspace{2mu}\mathbf{\text{v}}}\text{/}{dt}}\ \times \ \mathbf{\text{C}}\text{:}$

$$\frac{d\mspace{2mu}\mathbf{\text{v}}}{dt}\ \times \ \mathbf{\text{C}} = - \frac{GM}{\left\| \mathbf{\text{r}} \right\|^{3}}\mathbf{\text{r}}\ \times \ \left( {\mathbf{\text{r}}\ \times \ \mathbf{\text{v}}} \right) = - \frac{GM}{\left\| \mathbf{\text{r}} \right\|^{3}}{\left\lbrack {\left( {\mathbf{\text{r}} \cdot \mathbf{\text{v}}} \right)\mspace{2mu}\mathbf{\text{r}} - \left( {\mathbf{\text{r}} \cdot \mathbf{\text{r}}} \right)\mspace{2mu}\mathbf{\text{v}}} \right\rbrack.}$$ (3.26)

The last equality in Equation 3.26 is from the triple cross product formula.

Visit this website for an explanation of the triple cross product formula.

We need an expression for $\mathbf{\text{r}} \cdot \mathbf{\text{v}}.$ To calculate this, we differentiate $\mathbf{\text{r}} \cdot \mathbf{\text{r}}$ with respect to time:

$$\frac{d}{dt}\left( {\mathbf{\text{r}} \cdot \mathbf{\text{r}}} \right) = \frac{d\mspace{2mu}\mathbf{\text{r}}}{dt} \cdot \mathbf{\text{r}} + \mathbf{\text{r}} \cdot \frac{d\mspace{2mu}\mathbf{\text{r}}}{dt} = 2\mathbf{\text{r}} \cdot \frac{d\mspace{2mu}\mathbf{\text{r}}}{dt} = 2\mathbf{\text{r}} \cdot \mathbf{\text{v}}.$$ (3.27)

Since $\mathbf{\text{r}} \cdot \mathbf{\text{r}} = \left\| \mathbf{\text{r}} \right\|^{2},$ we also have

$$\frac{d}{dt}\left( {\mathbf{\text{r}} \cdot \mathbf{\text{r}}} \right) = \frac{d}{dt}\left\| \mathbf{\text{r}} \right\|^{2} = 2\left\| \mathbf{\text{r}} \right\|\frac{d}{dt}\left\| \mathbf{\text{r}} \right\|.$$ (3.28)

Combining Equation 3.27 and Equation 3.28, we get

$$\begin{array}{rll}

{2\mathbf{\text{r}} \cdot \mathbf{\text{v}}} & = & {2\left\| \mathbf{\text{r}} \right\|\frac{d}{dt}\left\| \mathbf{\text{r}} \right\|} \\

{\mathbf{\text{r}} \cdot \mathbf{\text{v}}} & = & {\left\| \mathbf{\text{r}} \right\|\frac{d}{dt}\left\| \mathbf{\text{r}} \right\|.}

\end{array}$$

Substituting this into Equation 3.26 gives us

$$\begin{array}{cl}

{\frac{d\mspace{2mu}\mathbf{\text{v}}}{dt}\ \times \ \mathbf{\text{C}}} & {= - \frac{GM}{\left\| \mathbf{\text{r}} \right\|^{3}}\left\lbrack {\left( {\mathbf{\text{r}} \cdot \mathbf{\text{v}}} \right)\mspace{2mu}\mathbf{\text{r}} - \left( {\mathbf{\text{r}} \cdot \mathbf{\text{r}}} \right)\mspace{2mu}\mathbf{\text{v}}} \right\rbrack} \\

& {= - \frac{GM}{\left\| \mathbf{\text{r}} \right\|^{3}}\left\lbrack {\left\| \mathbf{\text{r}} \right\|\left( {\frac{d}{dt}\left\| \mathbf{\text{r}} \right\|} \right)\mspace{2mu}\mathbf{\text{r}} - \left\| \mathbf{\text{r}} \right\|^{2}\mathbf{\text{v}}} \right\rbrack} \\

& {= \text{−}GM\left\lbrack {\frac{1}{\left\| \mathbf{\text{r}} \right\|^{2}}\left( {\frac{d}{dt}\left\| \mathbf{\text{r}} \right\|} \right)\mspace{2mu}\mathbf{\text{r}} - \frac{1}{\left\| \mathbf{\text{r}} \right\|}\mathbf{\text{v}}} \right\rbrack} \\

& {= GM\left\lbrack {\frac{\mathbf{\text{v}}}{\left\| \mathbf{\text{r}} \right\|} - \frac{\mathbf{\text{r}}}{\left\| \mathbf{\text{r}} \right\|^{2}}\left( {\frac{d}{dt}\left\| \mathbf{\text{r}} \right\|} \right)} \right\rbrack.}

\end{array}$$ (3.29)

However,

$$\begin{array}{cl}

{\frac{d}{dt}\ \frac{\mathbf{\text{r}}}{\left\| \mathbf{\text{r}} \right\|}} & {= \frac{\frac{d}{dt}\left( \mathbf{\text{r}} \right)\left\| \mathbf{\text{r}} \right\| - \mathbf{\text{r}}\frac{d}{dt}\left\| \mathbf{\text{r}} \right\|}{\left\| \mathbf{\text{r}} \right\|^{2}}} \\

& {= \frac{\frac{d\mspace{2mu}\mathbf{\text{r}}}{dt}}{\left\| \mathbf{\text{r}} \right\|} - \frac{\mathbf{\text{r}}}{\left\| \mathbf{\text{r}} \right\|^{2}}\ \frac{d}{dt}\left\| \mathbf{\text{r}} \right\|} \\

& {= \frac{\mathbf{\text{v}}}{\left\| \mathbf{\text{r}} \right\|} - \frac{\mathbf{\text{r}}}{\left\| \mathbf{\text{r}} \right\|^{2}}\ \frac{d}{dt}\left\| \mathbf{\text{r}} \right\|.}

\end{array}$$

Therefore, Equation 3.29 becomes

$$\frac{d\mspace{2mu}\mathbf{\text{v}}}{dt}\ \times \ \mathbf{\text{C}} = GM\left( {\frac{d}{dt}\ \frac{\mathbf{\text{r}}}{\left\| \mathbf{\text{r}} \right\|}} \right).$$

Since C is a constant vector, we can integrate both sides and obtain

$$\mathbf{\text{v}}\ \times \ \mathbf{\text{C}} = GM\frac{\mathbf{\text{r}}}{\left\| \mathbf{\text{r}} \right\|} + \mathbf{\text{D}},$$

where D is a constant vector. Our goal is to solve for $\left\| \mathbf{\text{r}} \right\|.$ Let’s start by calculating $\mathbf{\text{r}} \cdot \left( {\mathbf{\text{v}}\ \times \ \mathbf{\text{C}}} \right)\text{:}$

$$\mathbf{\text{r}} \cdot \left( {\mathbf{\text{v}}\ \times \ \mathbf{\text{C}}} \right) = \mathbf{\text{r}} \cdot \left( {GM\frac{\mathbf{\text{r}}}{\left\| \mathbf{\text{r}} \right\|} + \mathbf{\text{D}}} \right) = GM\frac{\left\| \mathbf{\text{r}} \right\|^{2}}{\left\| \mathbf{\text{r}} \right\|} + \mathbf{\text{r}} \cdot \mathbf{\text{D}} = GM\left\| \mathbf{\text{r}} \right\| + \mathbf{\text{r}} \cdot \mathbf{\text{D}}.$$

However, $\mathbf{\text{r}} \cdot \left( {\mathbf{\text{v}}\ \times \ \mathbf{\text{C}}} \right) = \left( {\mathbf{\text{r}}\ \times \ \mathbf{\text{v}}} \right) \cdot \mathbf{\text{C}},$ so

$$\left( {\mathbf{\text{r}}\ \times \ \mathbf{\text{v}}} \right) \cdot \mathbf{\text{C}} = GM\left\| \mathbf{\text{r}} \right\| + \mathbf{\text{r}} \cdot \mathbf{\text{D}}.$$

Since $\mathbf{\text{r}}\ \times \ \mathbf{\text{v}} = \mathbf{\text{C}},$ we have

$$\left\| \mathbf{\text{C}} \right\|^{2} = GM\left\| \mathbf{\text{r}} \right\| + \mathbf{\text{r}} \cdot \mathbf{\text{D}}.$$

Note that $\mathbf{\text{r}} \cdot \mathbf{\text{D}} = \left\| \mathbf{\text{r}} \right\|\left\| \mathbf{\text{D}} \right\|\text{cos}\mspace{2mu}\theta,$ where $\theta$ is the angle between r and D. Therefore,

$$\left\| \mathbf{\text{C}} \right\|^{2} = GM\left\| \mathbf{\text{r}} \right\| + \left\| \mathbf{\text{r}} \right\|\left\| \mathbf{\text{D}} \right\|\text{cos}\mspace{2mu}\theta.$$

Solving for $\left\| \mathbf{\text{r}} \right\|,$

$$\left\| \mathbf{\text{r}} \right\| = \frac{\left\| \mathbf{\text{C}} \right\|^{2}}{GM + \left\| \mathbf{\text{D}} \right\|\text{cos}\mspace{2mu}\theta} = \frac{\left\| \mathbf{\text{C}} \right\|^{2}}{GM}\left( \frac{1}{1 + e\mspace{2mu}\text{cos}\mspace{2mu}\theta} \right),$$

where $e = {\left\| \mathbf{\text{D}} \right\|\text{/}{GM}}.$ This is the polar equation of a conic with a focus at the origin, which we set up to be the Sun. It is a hyperbola if $e > 1,$ a parabola if $e = 1,$ or an ellipse if $e < 1.$ Since planets have closed orbits, the only possibility is an ellipse. However, at this point it should be mentioned that hyperbolic comets do exist. These are objects that are merely passing through the solar system at speeds too great to be trapped into orbit around the Sun. As they pass close enough to the Sun, the gravitational field of the Sun deflects the trajectory enough so the path becomes hyperbolic.

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Using Kepler’s Third Law for Nonheliocentric Orbits 将开普勒第三定律用于非日心轨道

Kepler’s third law of planetary motion can be modified to the case of one object in orbit around an object other than the Sun, such as the Moon around the Earth. In this case, Kepler’s third law becomes

$$P^{2} = \frac{4\pi^{2}a^{3}}{G\left( {m + M} \right)},$$ (3.30)

where *m* is the mass of the Moon and *M* is the mass of Earth, *a* represents the length of the major axis of the elliptical orbit, and *P* represents the period.

Given that the mass of the Moon is $7.35\ \times \ 10^{22}\ \text{kg,}$ the mass of Earth is $5.97\ \times \ 10^{24}\ \text{kg,}$ $G = 6.67\ \times \ 10^{-11}{\text{m}^{3}\text{/}{\text{kg} \cdot \text{sec}^{2}}},$ and the period of the moon is 27.3 days, let’s find the length of the major axis of the orbit of the Moon around Earth.

Solution 解

It is important to be consistent with units. Since the universal gravitational constant contains seconds in the units, we need to use seconds for the period of the Moon as well:

$$27.3\ \text{days}\ \times \ \frac{24\ \text{hr}}{1\ \text{day}}\ \times \ \frac{3600\mspace{2mu}\text{sec}}{1\mspace{2mu}\text{hour}} = 2,358,720\mspace{2mu}\text{sec}\text{.}$$

Substitute all the data into Equation 3.30 and solve for *a:*

$$\begin{array}{cll}

\left( {2,358,720\mspace{2mu}\text{sec}} \right)^{2} & = & \frac{4\pi^{2}a^{3}}{\left( {6.67\ \times \ 10^{-11}\mspace{2mu}\frac{\text{m}^{3}}{\text{kg} \cdot \mspace{2mu}\text{sec}^{2}}} \right)\left( {7.35\ x\ 10^{22}\text{kg} + 5.97\ x\ 10^{24}\text{kg}} \right)} \\

{5.563\ \times \ 10^{12}} & = & \frac{4\pi^{2}a^{3}}{\left( {6.67\ \times \ 10^{-11}\mspace{2mu}\text{m}^{3}} \right)\left( {6.04\ x\ 10^{24}} \right)} \\

{\left( {5.563\ \times \ 10^{12}} \right)\left( {6.67\ \times \ 10^{-11}\mspace{2mu}\text{m}^{3}} \right)\left( {6.04\ \times \ 10^{24}} \right)} & = & {4\pi^{2}a^{3}} \\

a^{3} & = & {\frac{2.241\ \times \ 10^{27}}{4\pi^{2}}\text{m}^{3}} \\

a & = & {3.84\ \times \ 10^{8}\text{m}} \\

& \approx & {384,000\ \text{km.}}

\end{array}$$

Analysis 分析

According to solarsystem.nasa.gov, the actual average distance from the Moon to Earth is 384,400 km. This is calculated using reflectors left on the Moon by Apollo astronauts back in the 1960s.

Titan is the largest moon of Saturn. The mass of Titan is approximately $1.35\ \times \ 10^{23}$ kg. The mass of Saturn is approximately $5.68\ \times \ 10^{26}$ kg. Titan takes approximately 16 days to orbit Saturn. Use this information, along with the universal gravitation constant $G = 6.67\ \times \ 10^{-11}{\text{m}^{3}\text{/}{\text{kg} \cdot \text{sec}^{2}}}$ to estimate the distance from Titan to Saturn.

Chapter Opener: Halley’s Comet 章首引例：哈雷彗星

We now return to the chapter opener, which discusses the motion of Halley’s comet around the Sun. Kepler’s first law states that Halley’s comet follows an elliptical path around the Sun, with the Sun as one focus of the ellipse. The period of Halley’s comet is approximately 76.1 years, depending on how closely it passes by Jupiter and Saturn as it passes through the outer solar system. Let’s use $T = 76.1$ years. What is the average distance of Halley’s comet from the Sun?

Solution 解

Using the equation $T^{2} = D^{3}$ with $T = 76.1,$ we obtain $D^{3} = 5791.21,$ so $D \approx 17.96$ A.U. This comes out to approximately $1.67\ \times \ 10^{9}$ mi.

A natural question to ask is: What are the maximum (aphelion) and minimum (perihelion) distances from Halley’s Comet to the Sun? The eccentricity of the orbit of Halley’s Comet is 0.967 (Source: nssdc.gsfc.nasa.gov/planetary/factsheet/cometfact.html). Recall that the formula for the eccentricity of an ellipse is $e = {c\text{/}a},$ where *a* is the length of the semimajor axis and *c* is the distance from the center to either focus. Therefore, $0.967 = {c\text{/}17.96}$ and $c \approx 17.37$ A.U. Subtracting this from *a* gives the perihelion distance $p = a - c = 17.96 - 17.37 = 0.59$ A.U. According to the National Space Science Data Center (Source: nssdc.gsfc.nasa.gov/planetary/factsheet/cometfact.html), the perihelion distance for Halley’s comet is 0.587 A.U. To calculate the aphelion distance, we add

$$P = a + c = 17.96 + 17.37 = 35.33\ \text{A}\text{.U}\text{.}$$

This is approximately $3.3\ \times \ 10^{9}$ mi. The average distance from Pluto to the Sun is 39.5 A.U. (Source: www.oarval.org/furthest.htm), so it would appear that Halley’s Comet stays just within the orbit of Pluto.

Navigating a Banked Turn 通过倾斜弯道

How fast can a racecar travel through a circular turn without skidding and hitting the wall? The answer could depend on several factors:

• The weight of the car;
• The friction between the tires and the road;
• The radius of the circle;
• The “steepness” of the turn.

• 汽车的重量；
• 轮胎与路面之间的摩擦；
• 圆弧的半径；
• 弯道的“陡度”。

In this project we investigate this question for NASCAR racecars at the Bristol Motor Speedway in Tennessee. Before considering this track in particular, we use vector functions to develop the mathematics and physics necessary for answering questions such as this.

A car of mass *m* moves with constant angular speed $\omega$ around a circular curve of radius *R* (Figure 3.20). The curve is banked at an angle $\theta.$ If the height of the car off the ground is *h*, then the position of the car at time *t* is given by the function $r(t) = \left\langle {R\mspace{2mu}\text{cos}\left( {\omega t} \right),R\mspace{2mu}\text{sin}\left( {\omega t} \right),h} \right\rangle.$

1. Find the velocity function $\mathbf{\text{v}}(t)$ of the car. Show that v is tangent to the circular curve. This means that, without a force to keep the car on the curve, the car will shoot off of it.

2. Show that the speed of the car is $\omega R.$ Use this to show that ${\left( {2\pi r} \right)\text{/}{\left| \mathbf{\text{v}} \right| =}}{\left( {2\pi} \right)\text{/}\omega}.$

3. Find the acceleration a. Show that this vector points toward the center of the circle and that $\left| \mathbf{\text{a}} \right| = R\omega^{2}.$

4. The force required to produce this circular motion is called the *centripetal force*, and it is denoted F_cent. This force points toward the center of the circle (not toward the ground). Show that $\left| \mathbf{\text{F}}_{\text{cent}} \right| = {\left( {m\left| \mathbf{\text{v}} \right|^{2}} \right)\text{/}R}.$

As the car moves around the curve, three forces act on it: gravity, the force exerted by the road (this force is perpendicular to the ground), and the friction force (Figure 3.21). Because describing the frictional force generated by the tires and the road is complex, we use a standard approximation for the frictional force. Assume that $\left| \mathbf{\text{f}} \middle| = \mu \middle| \mathbf{\text{N}} \right|$ for some positive constant $\mu.$ The constant $\mu$ is called the *coefficient of friction*.

Let $v_{\text{max}}$ denote the maximum speed the car can attain through the curve without skidding. In other words, $v_{\text{max}}$ is the fastest speed at which the car can navigate the turn. When the car is traveling at this speed, the magnitude of the centripetal force is

$$\left| \mathbf{\text{F}}_{\text{cent}} \right| = \frac{mv_{\text{max}}^{2}}{R}.$$

The next three questions deal with developing a formula that relates the speed $v_{\text{max}}$ to the banking angle $\theta.$

5. Show that $\left| \mathbf{\text{N}} \middle| \mspace{2mu}\text{cos}\mspace{2mu}\theta = mg + \middle| \mathbf{\text{f}} \middle| \mspace{2mu}\text{sin}\mspace{2mu}\theta. \right.$ Conclude that $\left| \mathbf{\text{N}} \middle| = {\left( {mg} \right)\text{/}{\left( {\text{cos}\mspace{2mu}\theta - \mu\mspace{2mu}\text{sin}\mspace{2mu}\theta} \right).}} \right.$

6. The centripetal force is the sum of the forces in the horizontal direction, since the centripetal force points toward the center of the circular curve. Show that

$$\left| \mathbf{\text{F}}_{\text{cent}|} = \middle| \mathbf{\text{N}} \middle| \mspace{2mu}\text{sin}\mspace{2mu}\theta + \middle| \mathbf{\text{f}} \middle| \mspace{2mu}\text{cos}\mspace{2mu}\theta. \right.$$

Conclude that

$$|\mathbf{\text{F}}_{\text{cent}|} = \frac{\text{sin}\mspace{2mu}\theta + \mu\mspace{2mu}\text{cos}\mspace{2mu}\theta}{\text{cos}\mspace{2mu}\theta - \mu\mspace{2mu}\text{sin}\mspace{2mu}\theta}mg.$$

7. Show that $\text{v}_{\text{max}}^{2} = \left( {\left( {\text{sin}\mspace{2mu}\theta + \mu\mspace{2mu}\text{cos}\mspace{2mu}\theta} \right)\text{/}\left( {\text{cos}\mspace{2mu}\theta - \mu\mspace{2mu}\text{sin}\mspace{2mu}\theta} \right)} \right)gR.$ Conclude that the maximum speed does not actually depend on the mass of the car.

Now that we have a formula relating the maximum speed of the car and the banking angle, we are in a position to answer the questions like the one posed at the beginning of the project.

The Bristol Motor Speedway is a NASCAR short track in Bristol, Tennessee. The track has the approximate shape shown in Figure 3.22. Each end of the track is approximately semicircular, so when cars make turns they are traveling along an approximately circular curve. If a car takes the inside track and speeds along the bottom of turn 1, the car travels along a semicircle of radius approximately 211 ft with a banking angle of 24°. If the car decides to take the outside track and speeds along the top of turn 1, then the car travels along a semicircle with a banking angle of 28°. (The track has variable angle banking.)

The coefficient of friction for a normal tire in dry conditions is approximately 0.7. Therefore, we assume the coefficient for a NASCAR tire in dry conditions is approximately 0.98.

Before answering the following questions, note that it is easier to do computations in terms of feet and seconds, and then convert the answers to miles per hour as a final step.

8. In dry conditions, how fast can the car travel through the bottom of the turn without skidding?

9. In dry conditions, how fast can the car travel through the top of the turn without skidding?

10. In wet conditions, the coefficient of friction can become as low as 0.1. If this is the case, how fast can the car travel through the bottom of the turn without skidding?

11. Suppose the measured speed of a car going along the outside edge of the turn is 105 mph. Estimate the coefficient of friction for the car’s tires.

Section 3.4 Exercises 3.4 节习题

155.

Given $\mathbf{\text{r}}(t) = (3t^{2} - 2)\mathbf{\text{i}} + (2t - \text{sin}(t))\mathbf{\text{j}},$ find the velocity of a particle moving along this curve.

156\.

Given $\mathbf{\text{r}}(t) = (3t^{2} - 2)\mathbf{\text{i}} + (2t - \text{sin}(t))\mathbf{\text{j}},$ find the acceleration vector of a particle moving along the curve in the preceding exercise.

Given the following position functions, find the velocity, acceleration, and speed in terms of the parameter *t*.

157.

$\mathbf{\text{r}}(t) = \left\langle {3\mspace{2mu}\text{cos}\mspace{2mu} t,3\mspace{2mu}\text{sin}\mspace{2mu} t,t^{2}} \right\rangle$

158\.

$\mathbf{\text{r}}(t) = e^{\text{−}t}\mathbf{\text{i}} + t^{2}\mathbf{\text{j}} + \text{tan}\mspace{2mu} t\mspace{2mu}\mathbf{\text{k}}$

159.

$\mathbf{\text{r}}(t) = 2\mspace{2mu}\text{cos}\mspace{2mu} t\mspace{2mu}\mathbf{\text{j}} + 3\mspace{2mu}\text{sin}\mspace{2mu} t\mspace{2mu}\mathbf{\text{k}}.$

Find the velocity, acceleration, and speed of a particle with the given position function.

160\.

$\mathbf{\text{r}}(t) = \left\langle {t^{2} - 1,t} \right\rangle$

161.

$\mathbf{\text{r}}(t) = \left\langle {e^{t},e^{\text{−}t}} \right\rangle$. The graph is shown here:

162\.

$\mathbf{\text{r}}(t) = \left\langle {\text{sin}\mspace{2mu} t,t,\text{cos}\mspace{2mu} t} \right\rangle.$

163.

The position function of an object is given by $\mathbf{\text{r}}(t) = \left\langle {t^{2},5t,t^{2} - 16t} \right\rangle.$ At what time is the speed a minimum?

164\.

Let $\mathbf{\text{r}}(t) = r\mspace{2mu}\text{cosh}(\omega t)\mathbf{\text{i}} + r\mspace{2mu}\text{sinh}(\omega t)\mathbf{\text{j}}.$ Find the velocity and acceleration vectors and show that the acceleration is proportional to $\mathbf{\text{r}}(t).$

Consider the motion of a point on the circumference of a rolling circle. As the circle rolls, it generates the cycloid $\mathbf{\text{r}}(t) = \left( {\omega t - \text{sin}(\omega t)} \right)\mspace{2mu}\mathbf{\text{i}} + \left( {1 - \text{cos}(\omega t)} \right)\mspace{2mu}\mathbf{\text{j}},$ where $\omega$ is the angular velocity of the circle:

165.

Find the equations for the velocity, acceleration, and speed of the particle at any time.

A person on a hang glider is spiraling upward as a result of the rapidly rising air on a path having position vector $\mathbf{\text{r}}(t) = (3\mspace{2mu}\text{cos}\mspace{2mu} t)\mathbf{\text{i}} + (3\mspace{2mu}\text{sin}\mspace{2mu} t)\mathbf{\text{j}} + t^{2}\mathbf{\text{k}}.$ The path is similar to that of a helix, although it is not a helix. The graph is shown here:

Find the following quantities:

166\.

The velocity and acceleration vectors

167.

The glider’s speed at any time

168\.

The times, if any, at which the glider’s acceleration is orthogonal to its velocity

Given that $\mathbf{\text{r}}(t) = \left\langle {e^{-5t}\text{sin}\mspace{2mu} t,e^{-5t}\text{cos}\mspace{2mu} t,4e^{-5t}} \right\rangle$ is the position vector of a moving particle, find the following quantities:

169.

The velocity of the particle

170\.

The speed of the particle

171.

The acceleration of the particle

172\.

Find the maximum speed of a point on the circumference of an automobile tire of radius 1 ft when the automobile is traveling at 55 mph.

A projectile is shot in the air from ground level with an initial velocity of 500 m/sec at an angle of 60° with the horizontal. The graph is shown here:

173.

At what time does the projectile reach maximum height?

174\.

What is the approximate maximum height of the projectile?

175.

At what time is the maximum range of the projectile attained?

176\.

What is the maximum range?

177.

What is the total flight time of the projectile?

A projectile is fired at a height of 1.5 m above the ground with an initial velocity of 100 m/sec and at an angle of 30° above the horizontal. Use this information to answer the following questions:

178\.

Determine the maximum height of the projectile.

179.

Determine the range of the projectile.

180\.

A golf ball is hit in a horizontal direction off the top edge of a building that is 100 ft tall. How fast must the ball be launched to land 450 ft away?

181.

A projectile is fired from ground level at an angle of 8° with the horizontal. The projectile is to have a range of 50 m. Find the minimum velocity necessary to achieve this range.

182\.

Prove that an object moving in a straight line at a constant speed has an acceleration of zero.

183.

The acceleration of an object is given by $\mathbf{\text{a}}(t) = t\mspace{2mu}\mathbf{\text{j}} + t\mspace{2mu}\mathbf{\text{k}}.$ The velocity at $t = 1$ sec is $\mathbf{\text{v}}(1) = 5\mathbf{\text{j}}$ and the position of the object at $t = 1$ sec is $\mathbf{\text{r}}(1) = 0\mathbf{\text{i}} + 0\mathbf{\text{j}} + 0\mathbf{\text{k}}.$ Find the object’s position at any time.

184\.

Find $\mathbf{\text{r}}(t)$ given that $\mathbf{\text{a}}(t) = -32\mathbf{\text{j}},$ $\mathbf{\text{v}}(0) = 600\sqrt{3}\mathbf{\text{i}} + 600\mathbf{\text{j}},$ and $\mathbf{\text{r}}(0) = \mathbf{0}.$

185.

Find the tangential and normal components of acceleration for $\mathbf{\text{r}}(t) = a\mspace{2mu}\text{cos}(\omega t)\mathbf{\text{i}} + a\mspace{2mu}\text{sin}(\omega t)\mathbf{\text{j}}$ at $t = 0.$

186\.

Given $\mathbf{\text{r}}(t) = t^{2}\mathbf{\text{i}} + 2t\mspace{2mu}\mathbf{\text{j}}$ and $t = 1,$ find the tangential and normal components of acceleration.

For each of the following problems, find the tangential and normal components of acceleration.

187.

$\mathbf{\text{r}}(t) = \left\langle {e^{t}\text{cos}\mspace{2mu} t,e^{t}\text{sin}\mspace{2mu} t,e^{t}} \right\rangle.$ The graph is shown here:

188\.

$\mathbf{\text{r}}(t) = \left\langle {\text{cos}(2t),\text{sin}(2t),1} \right\rangle$

189.

$\mathbf{\text{r}}(t) = \left\langle {2t,t^{2},\frac{t^{3}}{3}} \right\rangle$

190\.

$\mathbf{\text{r}}(t) = \left\langle {\frac{2}{3}\left( {1 + t} \right)^{3\text{/}2},\frac{2}{3}\left( {1 - t} \right)^{3\text{/}2},\sqrt{2}t} \right\rangle$

191.

$\mathbf{\text{r}}(t) = \left\langle {6t,3t^{2},2t^{3}} \right\rangle$

192\.

$\mathbf{\text{r}}(t) = t^{2}\mathbf{\text{i}} + t^{2}\mathbf{\text{j}} + t^{3}\mathbf{\text{k}}$

193.

$\mathbf{\text{r}}(t) = 3\mspace{2mu}\text{cos}\left( {2\pi t} \right)\mspace{2mu}\mathbf{\text{i}} + 3\mspace{2mu}\text{sin}\left( {2\pi t} \right)\mspace{2mu}\mathbf{\text{j}}$

194\.

Find the position vector-valued function $\mathbf{\text{r}}(t),$ given that $\mathbf{\text{a}}(t) = \mathbf{\text{i}} + e^{t}\mathbf{\text{j}},$ $\mathbf{\text{v}}(0) = 2\mathbf{\text{j}},$ and $\mathbf{\text{r}}(0) = 2\mathbf{\text{i}}.$

195.

The force on a particle is given by $\mathbf{\text{f}}(t) = \left( {\text{cos}\mspace{2mu} t} \right)\mspace{2mu}\mathbf{\text{i}} + \left( {\text{sin}\mspace{2mu} t} \right)\mspace{2mu}\mathbf{\text{j}}.$ The particle is located at point $(c,0)$ at $t = 0.$ The initial velocity of the particle is given by $\mathbf{\text{v}}(0) = v_{0}\mathbf{\text{j}}.$ Find the path of the particle of mass m. (Recall, $\mathbf{\text{F}} = m \cdot \mathbf{\text{a}}.)$

196\.

An automobile that weighs 2700 lb makes a turn on a flat road while traveling at 56 ft/sec. If the radius of the turn is 70 ft, what is the required frictional force to keep the car from skidding?

197.

Using Kepler’s laws, it can be shown that $v_{0} = \sqrt{\frac{2GM}{r_{0}}}$ is the minimum speed needed when $\theta = 0$ so that an object will escape from the pull of a central force resulting from mass *M*. Use this result to find the minimum speed when $\theta = 0$ for a space capsule to escape from the gravitational pull of Earth if the probe is at an altitude of 300 km above Earth’s surface.

198\.

Find the time in years it takes the dwarf planet Pluto to make one orbit about the Sun given that $a = 39.5$ A.U.

Suppose that the position function for an object in three dimensions is given by the equation $\mathbf{\text{r}}(t) = t\mspace{2mu}\text{cos}(t)\mathbf{\text{i}} + t\mspace{2mu}\text{sin}(t)\mathbf{\text{j}} + 3t\mspace{2mu}\mathbf{\text{k}}.$

199.

Show that the particle moves on a circular cone.

200\.

Find the angle between the velocity and acceleration vectors when $t = 1.5.$

201.

Find the tangential and normal components of acceleration when $t = 1.5.$

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Key Terms 关键术语

acceleration vector

the second derivative of the position vector

arc-length function

a function $s(t)$ that describes the arc length of curve *C* as a function of *t*

arc-length parameterization

a reparameterization of a vector-valued function in which the parameter is equal to the arc length

binormal vector

a unit vector orthogonal to the unit tangent vector and the unit normal vector

component functions

the component functions of the vector-valued function $\mathbf{\text{r}}(t) = f(t)\mspace{2mu}\mathbf{\text{i}} + g(t)\mspace{2mu}\mathbf{\text{j}}$ are $f(t)$ and $g(t),$ and the component functions of the vector-valued function $\mathbf{\text{r}}(t) = f(t)\mspace{2mu}\mathbf{\text{i}} + g(t)\mspace{2mu}\mathbf{\text{j}} + h(t)\mspace{2mu}\mathbf{\text{k}}$ are $f(t),$ $g(t)$ and $h(t)$

curvature

the derivative of the unit tangent vector with respect to the arc-length parameter

definite integral of a vector-valued function

the vector obtained by calculating the definite integral of each of the component functions of a given vector-valued function, then using the results as the components of the resulting function

derivative of a vector-valued function

the derivative of a vector-valued function $\mathbf{\text{r}}(t)$ is $\mathbf{r^{\prime}}(t) = \underset{\text{Δ}t\rightarrow 0}{\text{lim}}\frac{\mathbf{\text{r}}\left( {t + \text{Δ}t} \right) - \mathbf{\text{r}}(t)}{\text{Δ}t},$ provided the limit exists

Frenet frame of reference

(TNB frame) a frame of reference in three-dimensional space formed by the unit tangent vector, the unit normal vector, and the binormal vector

helix

a three-dimensional curve in the shape of a spiral

indefinite integral of a vector-valued function

a vector-valued function with a derivative that is equal to a given vector-valued function

Kepler’s laws of planetary motion

three laws governing the motion of planets, asteroids, and comets in orbit around the Sun

limit of a vector-valued function

a vector-valued function $\mathbf{\text{r}}(t)$ has a limit L as *t* approaches *a* if $\underset{t\rightarrow a}{\text{lim}}\left| {\mspace{2mu}\mathbf{\text{r}}(t) - \mathbf{\text{L}}} \right| = 0$

normal component of acceleration

the coefficient of the unit normal vector N when the acceleration vector is written as a linear combination of $\mathbf{\text{T}}$ and $\mathbf{\text{N}}$

normal plane

a plane that is perpendicular to a curve at any point on the curve

osculating circle

a circle that is tangent to a curve *C* at a point *P* and that shares the same curvature

osculating plane

the plane determined by the unit tangent and the unit normal vector

plane curve

the set of ordered pairs $\left( {f(t),g(t)} \right)$ together with their defining parametric equations $x = f(t)$ and $y = g(t)$

principal unit normal vector

a vector orthogonal to the unit tangent vector, given by the formula $\frac{\mathbf{T^{\prime}}(t)}{\left\| {\mathbf{T^{\prime}}(t)} \right\|}$

principal unit tangent vector

a unit vector tangent to a curve *C*

projectile motion

motion of an object with an initial velocity but no force acting on it other than gravity

radius of curvature

the reciprocal of the curvature

reparameterization

an alternative parameterization of a given vector-valued function

smooth

curves where the vector-valued function $\mathbf{\text{r}}(t)$ is differentiable with a non-zero derivative

space curve

the set of ordered triples $\left( {f(t),g(t),h(t)} \right)$ together with their defining parametric equations $x = f(t),$ $y = g(t)$ and $z = h(t)$

tangent vector

to $\mathbf{\text{r}}(t)$ at $t = t_{0}$ any vector v such that, when the tail of the vector is placed at point $\mathbf{\text{r}}\left( t_{0} \right)$ on the graph, vector v is tangent to curve *C*

tangential component of acceleration

the coefficient of the unit tangent vector T when the acceleration vector is written as a linear combination of $\mathbf{\text{T}}$ and $\mathbf{\text{N}}$

vector parameterization

any representation of a plane or space curve using a vector-valued function

vector-valued function

a function of the form $\mathbf{\text{r}}(t) = f(t)\mspace{2mu}\mathbf{\text{i}} + g(t)\mspace{2mu}\mathbf{\text{j}}$ or $\mathbf{\text{r}}(t) = f(t)\mspace{2mu}\mathbf{\text{i}} + g(t)\mspace{2mu}\mathbf{\text{j}} + h(t)\mspace{2mu}\mathbf{\text{k}},$ where the component functions *f, g,* and *h* are real-valued functions of the parameter *t*

velocity vector

the derivative of the position vector

Key Equations 关键公式

Key Concepts 关键概念

3.1 Vector-Valued Functions and Space Curves 3.1 向量值函数与空间曲线

• A vector-valued function is a function of the form $\mathbf{\text{r}}(t) = f(t)\mspace{2mu}\mathbf{\text{i}} + g(t)\mspace{2mu}\mathbf{\text{j}}$ or $\mathbf{\text{r}}(t) = f(t)\mspace{2mu}\mathbf{\text{i}} + g(t)\mspace{2mu}\mathbf{\text{j}} + h(t)\mspace{2mu}\mathbf{\text{k}},$ where the component functions *f, g,* and *h* are real-valued functions of the parameter *t*.
• The graph of a vector-valued function of the form $\mathbf{\text{r}}(t) = f(t)\mspace{2mu}\mathbf{\text{i}} + g(t)\mspace{2mu}\mathbf{\text{j}}$ is called a *plane curve*. The graph of a vector-valued function of the form $\mathbf{\text{r}}(t) = f(t)\mspace{2mu}\mathbf{\text{i}} + g(t)\mspace{2mu}\mathbf{\text{j}} + h(t)\mspace{2mu}\mathbf{\text{k}}$ is called a *space curve*.
• It is possible to represent an arbitrary plane curve by a vector-valued function.
• To calculate the limit of a vector-valued function, calculate the limits of the component functions separately.

• 向量值函数是一种形如 $\mathbf{\text{r}}(t) = f(t)\mspace{2mu}\mathbf{\text{i}} + g(t)\mspace{2mu}\mathbf{\text{j}}$ 或 $\mathbf{\text{r}}(t) = f(t)\mspace{2mu}\mathbf{\text{i}} + g(t)\mspace{2mu}\mathbf{\text{j}} + h(t)\mspace{2mu}\mathbf{\text{k}},$ 的函数，其中分量函数 *f、g* 和 *h* 是参数 *t* 的实值函数。
• 形如 $\mathbf{\text{r}}(t) = f(t)\mspace{2mu}\mathbf{\text{i}} + g(t)\mspace{2mu}\mathbf{\text{j}}$ 的向量值函数的图像称为*平面曲线*；形如 $\mathbf{\text{r}}(t) = f(t)\mspace{2mu}\mathbf{\text{i}} + g(t)\mspace{2mu}\mathbf{\text{j}} + h(t)\mspace{2mu}\mathbf{\text{k}}$ 的向量值函数的图像称为*空间曲线*。
• 任意平面曲线都可以用向量值函数表示。
• 要计算向量值函数的极限，需分别求出各分量函数的极限。

3.2 Calculus of Vector-Valued Functions 3.2 向量值函数的微积分

• To calculate the derivative of a vector-valued function, calculate the derivatives of the component functions, then put them back into a new vector-valued function.
• Many of the properties of differentiation from the Introduction to Derivatives also apply to vector-valued functions.
• The derivative of a vector-valued function $\mathbf{\text{r}}(t)$ is also a tangent vector to the curve. The unit tangent vector $\mathbf{\text{T}}(t)$ is calculated by dividing the derivative of a vector-valued function by its magnitude.
• The antiderivative of a vector-valued function is found by finding the antiderivatives of the component functions, then putting them back together in a vector-valued function.
• The definite integral of a vector-valued function is found by finding the definite integrals of the component functions, then putting them back together in a vector-valued function.

• 要计算向量值函数的导数，先求各分量函数的导数，再将它们组合成新的向量值函数。
• 导数导论中介绍的许多求导性质同样适用于向量值函数。
• 向量值函数 $\mathbf{\text{r}}(t)$ 的导数也是曲线的一条切向量。单位切向量 $\mathbf{\text{T}}(t)$ 由向量值函数的导数除以其模得到。
• 向量值函数的原函数，通过先求各分量函数的原函数，再将它们组合成向量值函数得到。
• 向量值函数的定积分，通过先求各分量函数的定积分，再将它们组合成向量值函数得到。

3.3 Arc Length and Curvature 3.3 弧长与曲率

• The arc-length function for a vector-valued function is calculated using the integral formula $s(t) = {\int_{a}^{t}{\left\| {\mathbf{r^{\prime}}(u)} \right\|\ du}}.$ This formula is valid in both two and three dimensions.
• The curvature of a curve at a point in either two or three dimensions is defined to be the curvature of the inscribed circle at that point. The arc-length parameterization is used in the definition of curvature.
• There are several different formulas for curvature. The curvature of a circle is equal to the reciprocal of its radius.
• The principal unit normal vector at *t* is defined to be

• 向量值函数的弧长函数由积分公式 $s(t) = {\int_{a}^{t}{\left\| {\mathbf{r^{\prime}}(u)} \right\|\ du}}$ 算出。该公式在二维与三维中都成立。
• 曲线在二维或三维中某点的曲率，定义为该点处内切圆的曲率。曲率的定义用到了弧长参数化。
• 曲率有若干不同的公式。圆的曲率等于其半径的倒数。
• 在 *t* 处的主单位法向量定义为

$$\mathbf{\text{N}}(t) = \frac{\mathbf{T^{\prime}}(t)}{\left\| {\mathbf{T^{\prime}}(t)} \right\|}.$$

• The binormal vector at *t* is defined as $\mathbf{\text{B}}(t) = \mathbf{\text{T}}(t)\ \times \ \mathbf{\text{N}}(t),$ where $\mathbf{\text{T}}(t)$ is the unit tangent vector.
• The Frenet frame of reference is formed by the unit tangent vector, the principal unit normal vector, and the binormal vector.
• The osculating circle is tangent to a curve at a point and has the same curvature as the tangent curve at that point.

• 在 *t* 处的副法向量定义为 $\mathbf{\text{B}}(t) = \mathbf{\text{T}}(t)\ \times \ \mathbf{\text{N}}(t)$，其中 $\mathbf{\text{T}}(t)$ 是单位切向量。
• 弗勒内（Frenet）标架由单位切向量、主单位法向量与副法向量构成。
• 曲率圆在点处与曲线相切，且在该点处与切线曲线具有相同的曲率。

3.4 Motion in Space 3.4 空间中的运动

• If $\mathbf{\text{r}}(t)$ represents the position of an object at time *t*, then $\mathbf{\text{r}}\prime(t)$ represents the velocity and $\mathbf{\text{r''}}(t)$ represents the acceleration of the object at time *t.* The magnitude of the velocity vector is speed.
• The acceleration vector always points toward the concave side of the curve defined by $\mathbf{\text{r}}(t).$ The tangential and normal components of acceleration $a_{\mathbf{\text{T}}}$ and $a_{\mathbf{\text{N}}}$ are the projections of the acceleration vector onto the unit tangent and unit normal vectors to the curve.
• Kepler’s three laws of planetary motion describe the motion of objects in orbit around the Sun. His third law can be modified to describe motion of objects in orbit around other celestial objects as well.
• Newton was able to use his law of universal gravitation in conjunction with his second law of motion and calculus to prove Kepler’s three laws.

• 若 $\mathbf{\text{r}}(t)$ 表示物体在时刻 *t* 的位置，则 $\mathbf{\text{r}}\prime(t)$ 表示速度，$\mathbf{\text{r''}}(t)$ 表示物体在时刻 *t* 的加速度。速度向量的模为速率。
• 加速度向量总是指向由 $\mathbf{\text{r}}(t)$ 所定义曲线的凹侧。加速度的切向分量与法向分量 $a_{\mathbf{\text{T}}}$、$a_{\mathbf{\text{N}}}$ 是加速度向量到曲线的单位切向量与单位法向量的投影。
• 开普勒行星运动三定律描述绕太阳运行轨道上的物体运动。他的第三定律也可经修改用于描述绕其他天体运行的轨道物体运动。
• 牛顿得以将万有引力定律与他的第二运动定律及微积分结合，证明开普勒三定律。

Review Exercises 复习题

*True or False*? Justify your answer with a proof or a counterexample.

202\.

A parametric equation that passes through points P and Q can be given by $\mathbf{\text{r}}(t) = \left\langle {t^{2},3t + 1,t - 2} \right\rangle,$ where $P(1,4,-1)$ and $Q(16,11,2).$

203.

$\frac{d}{dt}\left\lbrack {\mathbf{\text{u}}(t)\ \times \ \mathbf{\text{u}}(t)} \right\rbrack = 2\mathbf{\text{u}^{\prime}}(t)\ \times \ \mathbf{\text{u}}(t)$

204\.

The curvature of a circle of radius $r$ is constant everywhere. Furthermore, the curvature is equal to ${1\text{/}r}.$

205.

The speed of a particle with a position function $\mathbf{\text{r}}(t)$ is ${\left( {\mathbf{\text{r}^{\prime}}(t)} \right)\text{/}\left( \left| {\mathbf{\text{r}^{\prime}}(t)} \right| \right)}.$

Find the domains of the vector-valued functions.

206\.

$\mathbf{\text{r}}(t) = \left\langle {\text{sin}(t),\text{ln}(t),\sqrt{t}} \right\rangle$

207.

$\mathbf{\text{r}}(t) = \left\langle {e^{t},\frac{1}{\sqrt{4 - t}},\text{sec}(t)} \right\rangle$

Sketch the curves for the following vector equations. Use a calculator if needed.

208\.

\[T\] $\mathbf{\text{r}}(t) = \left\langle {t^{2},t^{3}} \right\rangle$

209.

\[T\] $\mathbf{\text{r}}(t) = \left\langle {\text{sin}\left( {20t} \right)e^{\text{−}t},\text{cos}\left( {20t} \right)e^{\text{−}t},e^{\text{−}t}} \right\rangle$

Find a vector function that describes the following curves.

210\.

Intersection of the cylinder $x^{2} + y^{2} = 4$ with the plane $x + z = 6$

211.

Intersection of the cone $z = \sqrt{x^{2} + y^{2}}$ and plane $z = y - 4$

Find the derivatives of $\mathbf{\text{u}}(t),$ $\mathbf{\text{u}^{\prime}}(t),$ $\mathbf{\text{u}^{\prime}}(t)\ \times \ \mathbf{\text{u}}(t),$ $\mathbf{\text{u}}(t)\ \times \ \mathbf{\text{u}^{\prime}}(t),$ and $\mathbf{\text{u}}(t) \cdot \mathbf{\text{u}^{\prime}}(t).$ Find the unit tangent vector.

212\.

$\mathbf{\text{u}}(t) = \left\langle {e^{t},e^{\text{−}t}} \right\rangle$

213.

$\mathbf{\text{u}}(t) = \left\langle {t^{2},2t + 6,4t^{5} - 12} \right\rangle$

Evaluate the following integrals.

214\.

$\int{\left( {\text{tan}(t)\text{sec}(t)\mathbf{\text{i}} - te^{3t}\mathbf{\text{j}}} \right)dt}$

215.

${\int\limits_{1}^{4}{\mathbf{\text{u}}(t)dt}},$ with $\mathbf{\text{u}}(t) = \left\langle {\frac{\text{ln}(t)}{t},\frac{1}{\sqrt{t}},\text{sin}\left( \frac{t\pi}{4} \right)} \right\rangle$

Find the length for the following curves.

216\.

$\mathbf{\text{r}}(t) = \left\langle {3t,4\mspace{2mu}\text{cos}(t),4\mspace{2mu}\text{sin}(t)} \right\rangle$ for $1 \leq t \leq 4$

217.

$\mathbf{\text{r}}(t) = 2\mathbf{\text{i}} + \text{t}\ \mathbf{\text{j}} + 3t^{2}\mathbf{\text{k}}$ for $0 \leq t \leq 1$

Reparameterize the following functions with respect to their arc length measured from $t = 0$ in direction of increasing $t.$

218\.

$\mathbf{\text{r}}(t) = 2t\mspace{2mu}\mathbf{\text{i}} + (4t - 5)\mathbf{\text{j}} + (1 - 3t)\mathbf{\text{k}}$

219.

$\mathbf{\text{r}}(t) = \text{cos}(2t)\mathbf{\text{i}} + 8t\mspace{2mu}\mathbf{\text{j}} - \text{sin}(2t)\mathbf{\text{k}}$

Find the curvature for the following vector functions.

220\.

$\mathbf{\text{r}}(t) = (2\mspace{2mu}\text{sin}\mspace{2mu} t)\mathbf{\text{i}} - 4t\mspace{2mu}\mathbf{\text{j}} + (2\mspace{2mu}\text{cos}\mspace{2mu} t)\mathbf{\text{k}}$

221.

$\mathbf{\text{r}}(t) = \sqrt{2}e^{t}\mathbf{\text{i}} + \sqrt{2}e^{\text{−}t}\mathbf{\text{j}} + 2t\mspace{2mu}\mathbf{\text{k}}$

222\.

Find the unit tangent vector, the unit normal vector, and the binormal vector for $\mathbf{\text{r}}(t) = 2\mspace{2mu}\text{cos}\mspace{2mu} t\mspace{2mu}\mathbf{\text{i}} + 3t\mspace{2mu}\mathbf{\text{j}} + 2\mspace{2mu}\text{sin}\mspace{2mu} t\mspace{2mu}\mathbf{\text{k}}.$

223.

Find the tangential and normal acceleration components with the position vector $\mathbf{\text{r}}(t) = \left\langle {\text{cos}\mspace{2mu} t,\text{sin}\mspace{2mu} t,e^{t}} \right\rangle.$

224\.

A Ferris wheel car is moving at a constant speed $v$ and has a constant radius $r.$ Find the tangential and normal acceleration of the Ferris wheel car.

225.

The position of a particle is given by $\mathbf{\text{r}}(t) = \left\langle {t^{2},\text{ln}(t),\text{sin}\left( {\pi t} \right)} \right\rangle,$ where $t$ is measured in seconds and $\mathbf{\text{r}}$ is measured in meters. Find the velocity, acceleration, and speed functions. What are the position, velocity, speed, and acceleration of the particle at 1 sec?

The following problems consider launching a cannonball out of a cannon. The cannonball is shot out of the cannon with an angle $\theta$ and initial velocity $\textbf{v}_{0}.$ The only force acting on the cannonball is gravity, so we begin with a constant acceleration $\mathbf{\text{a}}(t) = \text{−}g\mspace{2mu}\mathbf{\text{j}}.$

226\.

Find the velocity vector function $\mathbf{\text{v}}(t).$

227.

Find the position vector $\mathbf{\text{r}}(t)$ and the parametric representation for the position.

228\.

At what angle do you need to fire the cannonball for the horizontal distance to be greatest? What is the total distance it would travel?