Chapter 12: Vector-Valued Functions and Motion in Space

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OVERVIEW In this chapter we introduce the calculus of vector-valued functions. The domains of these functions are sets of real numbers, as before, but their ranges consist of vectors instead of scalars. When a vector-valued function changes, the change can occur in both magnitude and direction, so the derivative is itself a vector. The integral of a vectorvalued function is also a vector. We use the calculus of these functions to describe the paths and motions of objects moving in a plane or in space, so their velocities and accelerations are given by vectors.

12.1 Curves in Space and Their Tangents

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FIGURE 12.1 The position vector $𝐫 = ⟶ 𝑂 𝑃$ of a particle moving through space is a function of time.

When a particle moves through space during a time interval I, we think of the particle’s coordinates as functions defined on I:

𝑥 = 𝑓 (𝑡), 𝑦 = 𝑔 (𝑡), 𝑧 = ℎ (𝑡), 𝑡 \in 𝐼 . (1)

The points $(𝑥, 𝑦, 𝑧) = (𝑓 (𝑡), 𝑔 (𝑡), ℎ (𝑡)), 𝑡 \in 𝐼,$ make up the curve in space that we call the particle’s path. The equations and interval in Equation (1) parametrize the curve.

A curve in space can also be represented in vector form. The vector

𝐫 (𝑡) = ⟶ 𝑂 𝑃 = 𝑓 (𝑡) 𝐢 + 𝑔 (𝑡) 𝐣 + ℎ (𝑡) 𝐤 (2)

from the origin to the particle’s position $𝑃 (𝑓 (𝑡), 𝑔 (𝑡), ℎ (𝑡))$ at time t is the particle’s position vector (Figure 12.1). The functions f, g, and h are the component functions (or components) of the position vector. We think of the particle’s path as the curve traced by r during the time interval I. Figure 12.2 displays several space curves generated by a computer graphing program.

Equation (2) defines r as a vector function of the real variable t on the interval I. More generally, a vector-valued function or vector function on a domain set D is a rule that assigns a vector in space to each element in D. For now, the domains will be intervals of real numbers, and the graph of the function represents a curve in space. Vector functions on a domain in the plane or in space give rise to “vector fields,” which are important to the study of fluid flows, gravitational fields, and electromagnetic phenomena. We investigate vector fields and their applications in Chapter 15.

Real-valued functions are often called scalar functions to distinguish them from vector functions. The components of r in Equation (2) are scalar functions of t. The domain of a vector-valued function is the common domain of its components.

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FIGURE 12.2 Space curves are defined by the position vectors r( ). t

EXAMPLE 1 Graph the vector function

𝐫 (𝑡) = (c o s 𝑡) 𝐢 + (s i n 𝑡) 𝐣 + 𝑡 𝐤 .

FIGURE 12.3 The helix r i j k ( ) cos sint t t t = + + ( ) ( ) (Example 1).

Solution This vector function r( ) is defined for all real values of t t. The curve traced by r winds around the circular cylinder $𝑥 2 + 𝑦 2 = 1$ (Figure 12.3). The curve lies on the cylinder because the i- and j-components of r, being the x- and y-coordinates of the tip of r, satisfy the cylinder’s equation:

𝑥 2 + 𝑦 2 = (c o s 𝑡) 2 + (s i n 𝑡) 2 = 1 .

The curve rises as the k-component $𝑧 = 𝑡$ increases. Each time t increases by 2 , theπ curve completes one turn around the cylinder. The curve is called a helix (from an ancient Greek word for “spiral”). The equations

𝑥 = c o s 𝑡, 𝑦 = s i n 𝑡, 𝑧 = 𝑡

parametrize the helix. The domain is the largest set of points t for which all three equations are defined, $o r - \infty < 𝑡 < \infty$ for this example. Figure 12.4 shows more helices. ■

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FIGURE 12.4 Helices spiral upward around a cylinder, like coiled springs.

Limits and Continuity

The way we define limits of vector-valued functions is similar to the way we define limits of real-valued functions.

DEFINITION Let $𝐫 (𝑡) = 𝑓 (𝑡) 𝐢 + 𝑔 (𝑡) 𝐣 + ℎ (𝑡) 𝐤$ be a vector function with domain $𝐷,$ and let L be a vector. We say that r has limit L as t approaches $𝑡 0$ and write
$l i m 𝑡 \to 𝑡 0 𝐫 (𝑡) = 𝐋$
if, for every number $𝜀 > 0$ , there exists a corresponding number $𝛿 > 0$ such that, for all $𝑡 \in 𝐷,$
$| 𝐫 (𝑡) - 𝐋 | < 𝜀 w h e n e v e r 0 < | 𝑡 - 𝑡 0 | < 𝛿 .$

I $˙ 𝐋 = 𝐿 1 𝐢 + 𝐿 2 𝐣 + 𝐿 3 𝐤$ , then it can be shown that $l i m 𝑡 \to 𝑡 0 𝐫 (𝑡) = 𝐋$ precisely when

l i m 𝑡 \to 𝑡 0 𝑓 (𝑡) = 𝐿 1, l i m 𝑡 \to 𝑡 0 𝑔 (𝑡) = 𝐿 2, a n d l i m 𝑡 \to 𝑡 0 ℎ (𝑡) = 𝐿 3 .

We omit the proof. The equation

To calculate the limit of a vector function, we find the limit of each component scalar function.

l i m 𝑡 \to 𝑡 0 𝐫 (𝑡) = (l i m 𝑡 \to 𝑡 0 𝑓 (𝑡)) 𝐢 + (l i m 𝑡 \to 𝑡 0 𝑔 (𝑡)) 𝐣 + (l i m 𝑡 \to 𝑡 0 ℎ (𝑡)) 𝐤 (3)

provides a practical way to calculate limits of vector functions.

EXAMPLE 2

I f 𝐫 (𝑡) = (c o s 𝑡) 𝐢 + (s i n 𝑡) 𝐣 + 𝑡 𝐤, t h e n

l i m 𝑡 \to 𝜋 / 4 𝐫 (𝑡) = (l i m 𝑡 \to 𝜋 / 4 c o s 𝑡) 𝐢 + (l i m 𝑡 \to 𝜋 / 4 s i n 𝑡) 𝐣 + (l i m 𝑡 \to 𝜋 / 4 𝑡) 𝐤 = \sqrt 2 2 𝐢 + \sqrt 2 2 𝐣 + 𝜋 4 𝐤 .

We define continuity for vector functions the same way we define continuity for scalar functions defined over an interval.

DEFINITION A vector function $𝐫 (𝑡)$ is continuous at a point $𝑡 = 𝑡 0$ in its domain if lim $𝐫 (𝑡) = 𝐫 (𝑡 0)$ . The function is continuous if it is continuous at →t t ₀ every point in its domain.

From Equation (3), we see that $𝐫 (𝑡)$ is continuous at $𝑡 = 𝑡 0$ if and only if each component function is continuous there (Exercise 45).

EXAMPLE 3

(a) All the space curves shown in Figures 12.2 and 12.4 are continuous because their component functions are continuous at every value of t in $(- \infty, \infty)$

(b) The function

𝐠 (𝑡) = (c o s 𝑡) 𝐢 + (s i n 𝑡) 𝐣 + ⌊ 𝑡 ⌋ 𝐤

is discontinuous at every integer, because the greatest integer function $⌊ 𝑡 ⌋$ is discontinuous at every integer. ■

Derivatives and Motion

Suppose that $𝐫 (𝑡) = 𝑓 (𝑡) 𝐢 + 𝑔 (𝑡) 𝐣 + ℎ (𝑡) 𝐤$ is the position vector of a particle moving along a curve in space and that $𝑓, 𝑔 .$ , and h are differentiable functions of t. Then the differ ence between the particle’s positions at time t and time $𝑡 + Δ 𝑡$ is the vector

Δ 𝐫 = 𝐫 (𝑡 + Δ 𝑡) - 𝐫 (𝑡)

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FIGURE 12.5 $A s Δ 𝑡 \to 0,$ the point Q approaches the point P along the curve $𝐶 .$ In the limit, the vector $⟶ 𝑃 𝑄 / Δ 𝑡$ becomes the tangent vector $𝐫' (𝑡)$

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FIGURE 12.6 A piecewise smooth curve made up of five smooth curves connected end to end in a continuous fashion. The curve here is not smooth at the points joining the five smooth curves.

(Figure 12.5a). In terms of components,

Δ 𝐫 = 𝐫 (𝑡 + Δ 𝑡) - 𝐫 (𝑡) = [𝑓 (𝑡 + Δ 𝑡) 𝐢 + 𝑔 (𝑡 + Δ 𝑡) 𝐣 + ℎ (𝑡 + Δ 𝑡) 𝐤] - [𝑓 (𝑡) 𝐢 + 𝑔 (𝑡) 𝐣 + ℎ (𝑡) 𝐤] = [𝑓 (𝑡 + Δ 𝑡) - 𝑓 (𝑡)] 𝐢 + [𝑔 (𝑡 + Δ 𝑡) - 𝑔 (𝑡)] 𝐣 + [ℎ (𝑡 + Δ 𝑡) - ℎ (𝑡)] 𝐤 .

As $Δ 𝑡$ approaches zero, three things seem to happen simultaneously. First, Q approaches P along the curve. Second, the secant line $𝑃 𝑄$ seems to approach a limiting position tangent to the curve at P. Third, the quotient $Δ 𝐫 / Δ 𝑡$ (Figure 12.5b) approaches the limit

l i m Δ 𝑡 \to 0 Δ 𝐫 Δ 𝑡 = [l i m Δ 𝑡 \to 0 𝑓 (𝑡 + Δ 𝑡) - 𝑓 (𝑡) Δ 𝑡] 𝐢 + [l i m Δ 𝑡 \to 0 𝑔 (𝑡 + Δ 𝑡) - 𝑔 (𝑡) Δ 𝑡] 𝐣 + [l i m Δ 𝑡 \to 0 ℎ (𝑡 + Δ 𝑡) - ℎ (𝑡) Δ 𝑡] 𝐤 = [𝑑 𝑓 𝑑 𝑡] 𝐢 + [𝑑 𝑔 𝑑 𝑡] 𝐣 + [𝑑 ℎ 𝑑 𝑡] 𝐤 .

These observations lead us to the following definition.

DEFINITION The vector function $𝐫 (𝑡) = 𝑓 (𝑡) 𝐢 + 𝑔 (𝑡) 𝐣 + ℎ (𝑡) 𝐤$ has a derivative (is differentiable) at t if $𝑓, 𝑔,$ and h have derivatives at t. The derivative is the vector function
$𝐫' (𝑡) = 𝑑 𝐫 𝑑 𝑡 = l i m Δ 𝑡 \to 0 𝐫 (𝑡 + Δ 𝑡) - 𝐫 (𝑡) Δ 𝑡 = 𝑑 𝑓 𝑑 𝑡 𝐢 + 𝑑 𝑔 𝑑 𝑡 𝐣 + 𝑑 ℎ 𝑑 𝑡 𝐤 .$

A vector function r is differentiable if it is differentiable at every point of its domain.

The geometric significance of the definition of derivative is shown in Figure 12.5. The points P and Q have position vectors $𝐫 (𝑡)$ and $𝐫 (𝑡 + Δ 𝑡)$ , and the vector $⟶ 𝑃 𝑄$ is represented by $𝐫 (𝑡 + Δ 𝑡) - 𝐫 (𝑡)$ . For $Δ 𝑡 > 0$ , the scalar multiple $(1 / Δ 𝑡) (𝐫 (𝑡 + Δ 𝑡) - 𝐫 (𝑡))$ points in the same direction as the vector $⟶ 𝑃 𝑄 . A s Δ 𝑡 0$ , this vector approaches the vector $𝐫' (𝑡)$ which is a vector tangent to the curve at $𝑃,$ as long as it is different from the zero vector 0 (Figure 12.5b).

The curve traced by r is smooth if $𝑑 𝐫 / 𝑑 𝑡$ is continuous and never 0, that is, $i f 𝑓, 𝑔,$ and h have continuous first derivatives that are not simultaneously 0. We require $𝑑 𝐫 / 𝑑 𝑡 \neq 0$ for a smooth curve to make sure the curve has a continuously turning tangent at each point. On a smooth curve, there are no sharp corners or cusps.

A curve that is made up of a finite number of smooth curves pieced together in a continuous fashion is called piecewise smooth (Figure 12.6).

Look once again at Figure 12.5. We drew the figure for $Δ 𝑡$ positive, so $Δ 𝐫$ points forward, in the direction of the motion. The vector $Δ 𝐫 / Δ 𝑡$ , having the same direction as $Δ 𝐫 .$ points forward too. Had $Δ 𝑡$ been negative, $Δ 𝐫$ would have pointed backward, against the direction of motion. The quotient $Δ 𝐫 / Δ 𝑡 .$ , however, being a negative scalar multiple of $Δ 𝐫$ would once again have pointed forward. No matter how $Δ 𝐫$ points, $Δ 𝐫 / Δ 𝑡$ points forward, and we expect the vector $𝑑 𝐫 / 𝑑 𝑡 = l i m Δ 𝑡 \to 0 Δ 𝐫 / Δ 𝑡$ , when different from $𝟎,$ to do the same. This means that the derivative $𝑑 𝐫 / 𝑑 𝑡,$ , which is the rate of change of position with respect to time, always points in the direction of motion. For a smooth curve, $𝑑 𝐫 / 𝑑 𝑡$ is never zero; the particle does not stop or reverse direction.

DEFINITIONS If r is the position vector of a particle moving along a smooth curve in space, then
$𝐯 (𝑡) = 𝑑 𝐫 𝑑 𝑡$
is the particle’s velocity vector. If v is a nonzero vector, then it is tangent to the curve, and its direction is the direction of motion. The magnitude of v is the particle’s speed, and the derivative $𝐚 = 𝑑 𝐯 / 𝑑 𝑡$ , when it exists, is the particle’s acceleration vector. In summary,

Velocity is the derivative of position: $𝐯 = 𝑑 𝐫 𝑑 𝑡 .$

Speed is the magnitude of velocity: Speed .= v

Acceleration is the derivative of velocity:

$𝐚 = 𝑑 𝐯 𝑑 𝑡 = 𝑑 2 𝐫 𝑑 𝑡 2 .$

The unit vector v v is the direction of motion at time t.

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FIGURE 12.7 The curve and the velocity vector when $𝑡 = 7 𝜋 / 4$ for the motion given in Example 4.

EXAMPLE 4 Find the velocity, speed, and acceleration of a particle whose motion in space is given by the position vector $𝐫 (𝑡) = 2 c o s 𝑡 𝐢 + 2 s i n 𝑡 𝐣 + 5 c o s 2$ k.t Sketch the velocity vector $𝐯 (7 𝜋 / 4)$ .

Solution The velocity and acceleration vectors at time t are

𝐯 (𝑡) = 𝐫' (𝑡) = - 2 s i n 𝑡 𝐢 + 2 c o s 𝑡 𝐣 - 10 c o s 𝑡 s i n 𝑡 𝐤 = - 2 s i n 𝑡 𝐢 + 2 c o s 𝑡 𝐣 - 5 s i n 2 𝑡 𝐤, 𝐚 (𝑡) = 𝐫'' (𝑡) = - 2 c o s 𝑡 𝐢 - 2 s i n 𝑡 𝐣 - 10 c o s 2 𝑡 𝐤,

and the speed is

| 𝐯 (𝑡) | = \sqrt (- 2 s i n 𝑡) 2 + (2 c o s 𝑡) 2 + (- 5 s i n 2 𝑡) 2 = \sqrt 4 + 25 s i n 2 2 𝑡 .

When $𝑡 = 7 𝜋 / 4$ , we have

𝐯 (7 𝜋 4) = \sqrt 2 𝐢 + \sqrt 2 𝐣 + 5 𝐤, 𝐚 (7 𝜋 4) = - \sqrt 2 𝐢 + \sqrt 2 𝐣, ∣ 𝐯 (7 𝜋 4) ∣ = \sqrt 29 .

A sketch of the curve of motion, and the velocity vector when $𝑡 = 7 𝜋 / 4$ , can be seen inFigure 12.7. 一

We can express the velocity of a moving particle as the product of its speed and direction:

V e l o c i t y = | 𝐯 | (𝐯 | 𝐯 |) = (s p e e d) (d i r e c t i o n) .

Differentiation Rules

Because the derivatives of vector functions may be computed component by component, the rules for differentiating vector functions have the same form as the rules for differentiating scalar functions.

Differentiation Rules for Vector Functions

Let u and v be differentiable vector functions of $𝑡, 𝐂$ a constant vector, c any scalar, and f any differentiable scalar function.

Constant Function Rule:

𝑑 𝑑 𝑡 𝐂 = 𝟎

Scalar Multiple Rules:

𝑑 𝑑 𝑡 [𝑐 𝐮 (𝑡)] = 𝑐 𝐮' (𝑡)

𝑑 𝑑 𝑡 [𝑓 (𝑡) 𝐮 (𝑡)] = 𝑓' (𝑡) 𝐮 (𝑡) + 𝑓 (𝑡) 𝐮' (𝑡)

Sum Rule:

𝑑 𝑑 𝑡 [𝐮 (𝑡) + 𝐯 (𝑡)] = 𝐮' (𝑡) + 𝐯' (𝑡)

Difference Rule:

𝑑 𝑑 𝑡 [𝐮 (𝑡) - 𝐯 (𝑡)] = 𝐮' (𝑡) - 𝐯' (𝑡)

When you use the Cross Product Rule, remember to preserve the order of the factors. If u comes first on the left side of the equation, it must also come first on the right, or the signs will be wrong.

Dot Product Rule:

𝑑 𝑑 𝑡 [𝐮 (𝑡) \cdot 𝐯 (𝑡)] = 𝐮' (𝑡) \cdot 𝐯 (𝑡) + 𝐮 (𝑡) \cdot 𝐯' (𝑡)

Cross Product Rule:
Chain Rule:

𝑑 𝑑 𝑡 [𝐮 (𝑡) \times 𝐯 (𝑡)] = 𝐮' (𝑡) \times 𝐯 (𝑡) + 𝐮 (𝑡) \times 𝐯' (𝑡)

𝑑 𝑑 𝑡 [𝐮 (𝑓 (𝑡))] = 𝑓' (𝑡) 𝐮' (𝑓 (𝑡))

We will prove the product rules and the Chain Rule but will leave the rules for constants, scalar multiples, sums, and differences as exercises.

Proof of the Dot Product Rule Suppose that

𝐮 = 𝑢 1 (𝑡) 𝐢 + 𝑢 2 (𝑡) 𝐣 + 𝑢 3 (𝑡) 𝐤

and

𝐯 = 𝑣 1 (𝑡) 𝐢 + 𝑣 2 (𝑡) 𝐣 + 𝑣 3 (𝑡) 𝐤 .

Then

𝑑 𝑑 𝑡 (𝐮 \cdot 𝐯) = 𝑑 𝑑 𝑡 (𝑢 1 𝑣 1 + 𝑢 2 𝑣 2 + 𝑢 3 𝑣 3) = 𝑢' 1 𝑣 1 + 𝑢' 2 𝑣 2 + 𝑢' 3 𝑣 3 ⏟____⏟____⏟ 𝐮' \cdot 𝐯 + 𝑢 1 𝑣' 1 + 𝑢 2 𝑣' 2 + 𝑢 3 𝑣' 3 ⏟____⏟____⏟ 𝐮 \cdot 𝐯' .

Proof of the Cross Product Rule We model the proof after the proof of the Product Rule for scalar functions. According to the definition of derivative,

𝑑 𝑑 𝑡 (𝐮 \times 𝐯) = l i m ℎ \to 0 𝐮 (𝑡 + ℎ) \times 𝐯 (𝑡 + ℎ) - 𝐮 (𝑡) \times 𝐯 (𝑡) ℎ .

To change this fraction into an equivalent one that contains the difference quotients for the derivatives of u and v, we subtract and add $𝐮 (𝑡) \times 𝐯 (𝑡 + ℎ)$ in the numerator. Then

𝑑 𝑑 𝑡 (𝐮 \times 𝐯) = l i m ℎ \to 0 𝐮 (𝑡 + ℎ) \times 𝐯 (𝑡 + ℎ) - 𝐮 (𝑡) \times 𝐯 (𝑡 + ℎ) + 𝐮 (𝑡) \times 𝐯 (𝑡 + ℎ) - 𝐮 (𝑡) \times 𝐯 (𝑡) ℎ = l i m ℎ \to 0 [𝐮 (𝑡 + ℎ) - 𝐮 (𝑡) ℎ \times 𝐯 (𝑡 + ℎ) + 𝐮 (𝑡) \times 𝐯 (𝑡 + ℎ) - 𝐯 (𝑡) ℎ] = l i m ℎ \to 0 𝐮 (𝑡 + ℎ) - 𝐮 (𝑡) ℎ \times l i m ℎ \to 0 𝐯 (𝑡 + ℎ) + l i m ℎ \to 0 𝐮 (𝑡) \times l i m ℎ \to 0 𝐯 (𝑡 + ℎ) - 𝐯 (𝑡) ℎ .

As an algebraic convenience, we sometimes write the product of a scalar c and a vector v as vc instead of cv. This permits us, for instance, to write the Chain Rule in a familiar form:

𝑑 𝐮 𝑑 𝑡 = 𝑑 𝐮 𝑑 𝑠 𝑑 𝑠 𝑑 𝑡,

where $𝑠 = 𝑓 (𝑡)$

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FIGURE 12.8 If a particle moves on a sphere in such a way that its position r is a differentiable function of time, then $𝐫 \cdot (𝑑 𝐫 / 𝑑 𝑡) = 0$

The last of these equalities holds because the limit of the cross product of two vector functions is the cross product of their limits if the latter exist (Exercise 46). As h approaches zero, $𝐯 (𝑡 + ℎ)$ approaches $𝐯 (𝑡)$ because v, being differentiable at t, is continuous at t (Exercise 47). The two fractions approach the values of $𝑑 𝐮 / 𝑑 𝑡$ and $𝑑 𝐯 / 𝑑 𝑡$ at t. In short,

𝑑 𝑑 𝑡 (𝐮 \times 𝐯) = (𝑑 𝐮 𝑑 𝑡 \times 𝐯) + (𝐮 \times 𝑑 𝐯 𝑑 𝑡) .

Proof of the Chain Rule Suppose that $𝐮 (𝑠) = 𝑎 (𝑠) 𝐢 + 𝑏 (𝑠) 𝐣 + 𝑐 (𝑠) 𝐤$ is a differentiable vector function of s and that $𝑠 = 𝑓 (𝑡)$ is a differentiable scalar function of t. Then a, $𝑏,$ and c are differentiable functions of t, and the Chain Rule for differentiable real-valued functions gives

𝑑 𝑑 𝑡 [𝐮 (𝑠)] = 𝑑 𝑎 𝑑 𝑡 𝐢 + 𝑑 𝑏 𝑑 𝑡 𝐣 + 𝑑 𝑐 𝑑 𝑡 𝐤 = 𝑑 𝑎 𝑑 𝑠 𝑑 𝑠 𝑑 𝑡 𝐢 + 𝑑 𝑏 𝑑 𝑠 𝑑 𝑠 𝑑 𝑡 𝐣 + 𝑑 𝑐 𝑑 𝑠 𝑑 𝑠 𝑑 𝑡 𝐤 = 𝑑 𝑠 𝑑 𝑡 (𝑑 𝑎 𝑑 𝑠 𝐢 + 𝑑 𝑏 𝑑 𝑠 𝐣 + 𝑑 𝑐 𝑑 𝑠 𝐤) = 𝑑 𝑠 𝑑 𝑡 𝑑 𝐮 𝑑 𝑠 = 𝑓' (𝑡) 𝐮' (𝑓 (𝑡)) . 𝑠 = 𝑓 (𝑡)

Vector Functions of Constant Length

When we track a particle moving on a sphere centered at the origin (Figure 12.8), the position vector has a constant length equal to the radius of the sphere. The velocity vector $𝑑 𝐫 / 𝑑 𝑡$ tangent to the path of motion, is tangent to the sphere and hence perpendicular to r. This is always the case for a differentiable vector function of constant length: The vector and its first derivative are orthogonal. By direct calculation,

𝐫 (𝑡) \cdot 𝐫 (𝑡) = | 𝐫 (𝑡) | 2 = 𝑐 2 | 𝐫 (𝑡) | = 𝑐 i s c o n s t a n t . 𝑑 𝑑 𝑡 [𝐫 (𝑡) \cdot 𝐫 (𝑡)] = 0 D i f f e r e n t i a t e b o t h s i d e s . 𝐫' (𝑡) \cdot 𝐫 (𝑡) + 𝐫 (𝑡) \cdot 𝐫' (𝑡) = 0 R u l e 5 w i t h 𝐫 (𝑡) = 𝐮 (𝑡) = 𝐯 (𝑡) 2 𝐫' (𝑡) \cdot 𝐫 (𝑡) = 0 .

Thus the vectors $𝐫' (𝑡)$ and $𝐫 (𝑡)$ are orthogonal because their dot product is 0. In summary, the following holds.

If r is a differentiable vector function of t and the length of $𝐫 (𝑡)$ is constant, then

𝐫 \cdot 𝑑 𝐫 𝑑 𝑡 = 0 . (4)

We will use this observation repeatedly in Section 12.4. The converse is also true (see Exercise 41).

Exercises 12.1

In Exercises 1–4, find the given limits.

lim $[(s i n 𝑡 2) 𝐢 + (c o s 23 𝑡) 𝐣 + (t a n 54 𝑡) 𝐤]$ t →π
$l i m 𝑡 - 1 [𝑡 3 𝐢 + (s i n 𝜋 2 𝑡) 𝐣 + (l n (𝑡 + 2)) 𝐤]$
lim t _i1^{2⎛ −}⎜ ^⎞⎟_{⎟ −} ^t j k t ¹ arctan ( ) ^{− ⎞}⎟⎟_⎟ + t 1→ tln t1 −

l i m 𝑡 \to 0 [(s i n 𝑡 𝑡) 𝐢 + (t a n 2 𝑡 s i n 2 𝑡) 𝐣 - (𝑡 3 - 8 𝑡 + 2) 𝐤]

Motion in the Plane

In Exercises 5–8, r( ) is the position of a particle in the t xy-plane at time t. Find an equation in x and y whose graph is the path of the particle. Then find the particle’s velocity and acceleration vectors at the given value of t.

𝟓 . 𝐫 (𝑡) = (𝑡 + 1) 𝐢 + (𝑡 2 - 1) 𝐣, 𝑡 = 1

𝟔 . 𝐫 (𝑡) = 𝑡 𝑡 + 1 𝐢 + 1 𝑡 𝐣, 𝑡 = - 12

𝟕 . 𝐫 (𝑡) = 𝑒 𝑡 𝐢 + 29 𝑒 2 𝑡 𝐣, 𝑡 = l n 3

𝟖 . 𝐫 (𝑡) = (c o s 2 𝑡) 𝐢 + (3 s i n 2 𝑡) 𝐣, 𝑡 = 0

Exercises 9–12 give the position vectors of particles moving along various curves in the xy-plane. In each case, find the particle’s velocity and acceleration vectors at the stated times and sketch them as vectors on the curve.

Motion on the circle $𝑥 2 + 𝑦 2 = 1$

𝐫 (𝑡) = (s i n 𝑡) 𝐢 + (c o s 𝑡) 𝐣; 𝑡 = 𝜋 / 4 a n d 𝜋 / 2 10 . $ 𝑀 𝑜 𝑡 𝑖 𝑜 𝑛 𝑜 𝑛 𝑡 ℎ 𝑒 𝑐 𝑖 𝑟 𝑐 𝑙 𝑒 $ 𝑥 2 + 𝑦 2 = 16 $ $ 𝐫 (𝑡) = (4 c o s 𝑡 2) 𝐢 + (4 s i n 𝑡 2) 𝐣; 𝑡 = 𝜋 a n d 3 𝜋 / 2 11 . $ 𝑀 𝑜 𝑡 𝑖 𝑜 𝑛 𝑜 𝑛 𝑡 ℎ 𝑒 𝑐 𝑦 𝑐 𝑙 𝑜 𝑖 𝑑 $ 𝑥 = 𝑡 - s i n 𝑡, 𝑦 = 1 - c o s 𝑡 $ $ 𝐫 (𝑡) = (𝑡 - s i n 𝑡) 𝐢 + (1 - c o s 𝑡) 𝐣; 𝑡 = 𝜋 a n d 3 𝜋 / 2 12 . $ 𝑀 𝑜 𝑡 𝑖 𝑜 𝑛 𝑜 𝑛 𝑡 ℎ 𝑒 𝑝 𝑎 𝑟 𝑎 𝑏 𝑜 𝑙 𝑎 $ 𝑦 = 𝑥 2 + 1 $ $ 𝐫 (𝑡) = 𝑡 𝐢 + (𝑡 2 + 1) 𝐣; 𝑡 = - 1, 0, a n d 1

Motion in Space

In Exercises $13 - 18, 𝐫 (𝑡)$ is the position of a particle in space at time t. Find the particle’s velocity and acceleration vectors. Then find the particle’s speed and direction of motion at the given value of t. Write the particle’s velocity at that time as the product of its speed and direction.

$𝐫 (𝑡) = (𝑡 + 1) 𝐢 + (𝑡 2 - 1) 𝐣 + 2 𝑡 𝐤, 𝑡 = 1$

𝟏 𝟒 . 𝐫 (𝑡) = (1 + 𝑡) 𝐢 + 𝑡 2 \sqrt 2 𝐣 + 𝑡 3 3 𝐤, 𝑡 = 1

𝟏 𝟓 . 𝐫 (𝑡) = (2 c o s 𝑡) 𝐢 + (3 s i n 𝑡) 𝐣 + 4 𝑡 𝐤, 𝑡 = 𝜋 / 2

𝐫 (𝑡) = (s e c 𝑡) 𝐢 + (t a n 𝑡) 𝐣 + 43 𝑡 𝐤, 𝑡 = 𝜋 / 6

$𝐫 (𝑡) = (2 l n (𝑡 + 1)) 𝐢 + 𝑡 2 𝐣 + 𝑡 2 2 𝐤, 𝑡 = 1$
$𝐫 (𝑡) = 𝑒 - 𝑡 𝐢 + (2 c o s 3 𝑡) 𝐣 + (2 s i n 3 𝑡) 𝐤, 𝑡 = 0$

In Exercises 19–22, r( ) is the position of a particle in space at time t t. Find the angle between the velocity and acceleration vectors at time $𝑡 = 0$

𝟏 𝟗 . 𝐫 (𝑡) = (3 𝑡 + 1) 𝐢 + \sqrt 3 𝑡 𝐣 + 𝑡 2 𝐤

$𝐫 (𝑡) = (\sqrt 2 2 𝑡) 𝐢 + (\sqrt 2 2 𝑡 - 16 𝑡 2) 𝐣$
$𝐫 (𝑡) = (l n (𝑡 2 + 1)) 𝐢 + (a r c t a n 𝑡) 𝐣 + \sqrt 𝑡 2 + 1 𝐤$
$𝐫 (𝑡) = 49 (1 + 𝑡) 3 / 2 𝐢 + 49 (1 - 𝑡) 3 / 2 𝐣 + 13 𝑡 𝐤$

Tangents to Curves

As mentioned in the text, the tangent line to a smooth curve $𝐫 (𝑡) = 𝑓 (𝑡) 𝐢 + 𝑔 (𝑡) 𝐣 + ℎ (𝑡) 𝐤 a t 𝑡 = 𝑡 0$ is the line that passes through the point $(𝑓 (𝑡 0), 𝑔 (𝑡 0), ℎ (𝑡 0))$ parallel to $𝐯 (𝑡 0)$ , the curve’s velocity vector a $𝑡 0 .$ In Exercises 23–26, find parametric equations for the line that is tangent to the given curve at the given parameter value $𝑡 = 𝑡 0$

$𝐫 (𝑡) = (s i n 𝑡) 𝐢 + (𝑡 2 - c o s 𝑡) 𝐣 + 𝑒 𝑡 𝐤, 𝑡 0 = 0$

𝟐 𝟒 . 𝐫 (𝑡) = 𝑡 2 𝐢 + (2 𝑡 - 1) 𝐣 + 𝑡 3 𝐤, 𝑡 0 = 2

$𝐫 (𝑡) = l n 𝑡 𝐢 + 𝑡 - 1 𝑡 + 2 𝐣 + 𝑡 l n 𝑡 𝐤, 𝑡 0 = 1$

𝟐 𝟔 . 𝐫 (𝑡) = (c o s 𝑡) 𝐢 + (s i n 𝑡) 𝐣 + (s i n 2 𝑡) 𝐤, 𝑡 0 = 𝜋 2

In Exercises 27–30, find the value(s) of t so that the tangent line to the given curve contains the given point.

$𝐫 (𝑡) = 𝑡 2 𝐢 + (1 + 𝑡) 𝐣 + (2 𝑡 - 3) 𝐤; (- 8, 2, - 1)$

𝟐 𝟖 . 𝐫 (𝑡) = 𝑡 𝐢 + 3 𝐣 + (23 𝑡 3 / 2) 𝐤; (0, 3, - 8 / 3)

𝟐 𝟗 . 𝐫 (𝑡) = 2 𝑡 𝐢 + 𝑡 2 𝐣 - 𝑡 2 𝐤; (0, - 4, 4)

𝟑 𝟎 . 𝐫 (𝑡) = - 𝑡 𝐢 + 𝑡 2 𝐣 + (l n 𝑡) 𝐤; (2, - 5, - 3)

In Exercises 31– $𝛿 - 36, 𝐫 (𝑡)$ is the position of a particle in space at time t. Match each position function with one of the graphs A–F.

$𝐫 (𝑡) = (𝑡 c o s 𝑡) 𝐢 + (𝑡 s i n 𝑡) 𝐣 + 𝑡 𝐤$
$𝐫 (𝑡) = (c o s 𝑡) 𝐢 + (s i n 𝑡) 𝐣 + (s i n 2 𝑡) 𝐤$
$𝐫 (𝑡) = 𝑡 2 𝐢 + (𝑡 2 + 1) 𝐣 + 𝑡 4 𝐤$
$𝐫 (𝑡) = 𝑡 𝐢 + (l n 𝑡) 𝐣 + (s i n 𝑡) 𝐤$

r i j k ( ) cos sin t t t t = + + ( ) ( )

t 36. = + + ( ) ( ) t t t t t r i j( ) sin cos )k +t 12

教材插图

Theory and Examples

Motion along a circle Each of the following equations in parts (a)–(e) describes the motion of a particle having the same path, namely the unit circle $𝑥 2 + 𝑦 2 = 1$ . Although the path of each particle in parts $(a) - (e)$ is the same, the behavior, or “dynamics,” of each particle is different. For each particle, answer the following questions.

i) Does the particle have constant speed? If so, what is its constant speed?

ii) Is the particle’s acceleration vector always orthogonal to its velocity vector?

iii) Does the particle move clockwise or counterclockwise around the circle?

iv) Is the particle initially located at the point ( ) 1, 0 ?

𝐫 (𝑡) = (c o s 𝑡) 𝐢 + (s i n 𝑡) 𝐣, 𝑡 \geq 0

𝐛 . 𝐫 (𝑡) = c o s (2 𝑡) 𝐢 + s i n (2 𝑡) 𝐣, 𝑡 \geq 0

𝐜 . 𝐫 (𝑡) = c o s (𝑡 - 𝜋 / 2) 𝐢 + s i n (𝑡 - 𝜋 / 2) 𝐣, 𝑡 \geq 0

𝐝 . 𝐫 (𝑡) = (c o s 𝑡) 𝐢 - (s i n 𝑡) 𝐣, 𝑡 \geq 0

𝐞 . 𝐫 (𝑡) = c o s (𝑡 2) 𝐢 + s i n (𝑡 2) 𝐣, 𝑡 \geq 0 38 . $ 𝑀 𝑜 𝑡 𝑖 𝑜 𝑛 𝑎 𝑙 𝑜 𝑛 𝑔 𝑎 𝑐 𝑖 𝑟 𝑐 𝑙 𝑒 𝑆 ℎ 𝑜 𝑤 𝑡 ℎ 𝑎 𝑡 𝑡 ℎ 𝑒 𝑣 𝑒 𝑐 𝑡 𝑜 𝑟 - 𝑣 𝑎 𝑙 𝑢 𝑒 𝑑 𝑓 𝑢 𝑛 𝑐 𝑡 𝑖 𝑜 𝑛 $ 𝐫 (𝑡) = (2 𝐢 + 2 𝐣 + 𝐤) + c o s 𝑡 (1 \sqrt 2 𝐢 - 1 \sqrt 2 𝐣) + s i n 𝑡 (1 \sqrt 3 𝐢 + 1 \sqrt 3 𝐣 + 1 \sqrt 3 𝐤)

describes the motion of a particle moving in the circle of radius 1 centered at the point (2, 2,1 and lying in the plane) $𝑥 + 𝑦 - 2 𝑧 = 2$

Motion along a parabola A particle moves along the top of the parabola $𝑦 2 = 2 𝑥$ from left to right at a constant speed of 5 units per second. Find the velocity of the particle as it moves through the point ( 2, 2 .)
Motion along a cycloid A particle moves in the xy-plane in such a way that its position at time t is

𝐫 (𝑡) = (𝑡 - s i n 𝑡) 𝐢 + (1 - c o s 𝑡) 𝐣 .

a. Graph r( ). The resulting curve is a cycloid. tT

b. Find the maximum and minimum values of v and a . (Hint: Find the extreme values of $| 𝐯 | 2$ and $| 𝐚 | 2$ first, and take square roots later.)

Let r be a differentiable vector function of t. Show that if $𝐫 \cdot (𝑑 𝐫 / 𝑑 𝑡) = 0$ for all t, then r is constant.
Derivatives of triple scalar products

a. Show that if u, v, and w are differentiable vector functions of t, then

𝑑 𝑑 𝑡 (𝐮 \cdot 𝐯 \times 𝐰) = 𝑑 𝐮 𝑑 𝑡 \cdot 𝐯 \times 𝐰 + 𝐮 \cdot 𝑑 𝐯 𝑑 𝑡 \times 𝐰 + 𝐮 \cdot 𝐯 \times 𝑑 𝐰 𝑑 𝑡 .

b. Show that

𝑑 𝑑 𝑡 (𝐫 \cdot 𝑑 𝐫 𝑑 𝑡 \times 𝑑 2 𝐫 𝑑 𝑡 2) = 𝐫 \cdot (𝑑 𝐫 𝑑 𝑡 \times 𝑑 3 𝐫 𝑑 𝑡 3) .

(Hint: Differentiate on the left and look for vectors whose products are zero.)

Prove the two Scalar Multiple Rules for vector functions.
Prove the Sum and Difference Rules for vector functions.
Component test for continuity at a point Show that the vector function r defined by $𝐫 (𝑡) = 𝑓 (𝑡) 𝐢 + 𝑔 (𝑡) 𝐣 + ℎ (𝑡) 𝐤$ is continuous at $𝑡 = 𝑡 0$ if and only $i f 𝑓, 𝑔,$ and h are continuous at $𝑡 0 .$
Limits of cross products of vector functions Suppose that $𝐫 1 (𝑡) = 𝑓 1 (𝑡) 𝐢 + 𝑓 2 (𝑡) 𝐣 + 𝑓 3 (𝑡) 𝐤, 𝐫 2 (𝑡) = 𝑔 1 (𝑡) 𝐢 + 𝑔 2 (𝑡) 𝐣 + 𝑔 3 (𝑡) 𝐤$ lim $𝐫 1 (𝑡) = 𝐀,$ , and lim $𝐫 2 (𝑡) = 𝐁 .$ . Use the determinant formula →t t ₀ →t t ₀ for cross products and the Limit Product Rule for scalar functions to show that

l i m 𝑡 \to 𝑡 0 (𝐫 1 (𝑡) \times 𝐫 2 (𝑡)) = 𝐀 \times 𝐁 .

Differentiable vector functions are continuous Show that if $𝐫 (𝑡) = 𝑓 (𝑡) 𝐢 + 𝑔 (𝑡) 𝐣 + ℎ (𝑡) 𝐤$ is differentiable at $𝑡 = 𝑡 0$ , then it is continuous at $𝑡 0$ as well.
Constant Function Rule Prove that if u is the vector function with the constant value C, then du $/ 𝑑 𝑡 = 𝟎$

COMPUTER EXPLORATIONS

Use a CAS to perform the following steps in Exercises 49–52.

a. Plot the space curve traced out by the position vector r.

b. Find the components of the velocity vector dr dt.

c. Evaluate dr dt at the given point $𝑡 0$ and determine the equation of the tangent line to the curve at $𝐫 (𝑡 0)$

d. Plot the tangent line together with the curve over the given interval.

𝟒 𝟗 . 𝐫 (𝑡) = (s i n 𝑡 - 𝑡 c o s 𝑡) 𝐢 + (c o s 𝑡 + 𝑡 s i n 𝑡) 𝐣 + 𝑡 2 𝐤, 0 \leq 𝑡 \leq 6 𝜋, 𝑡 0 = 3 𝜋 / 2

= + + − ≤ ≤ = − r i j k ( ) 2 , 2 3, 1 t t e e t t t t
$𝐫 (𝑡) = (s i n 2 𝑡) 𝐢 + (l n (1 + 𝑡)) 𝐣 + 𝑡 𝐤, 0 \leq 𝑡 \leq 4 𝜋,$ t ₀ = π 4

𝟓 𝟐 . 𝐫 (𝑡) = (l n (𝑡 2 + 2)) 𝐢 + (a r c t a n 3 𝑡) 𝐣 + \sqrt 𝑡 2 + 1 𝐤, - 3 \leq 𝑡 \leq 5, 𝑡 0 = 3

In Exercises 53 and 54, you will explore graphically the behavior of the helix

𝐫 (𝑡) = (c o s 𝑎 𝑡) 𝐢 + (s i n 𝑎 𝑡) 𝐣 + 𝑏 𝑡 𝐤

as you change the values of the constants a and b. Use a CAS to perform the steps in each exercise.

Set $𝑏 = 1$ . Plot the helix r( ) together with the tangent linet to the curve at $𝑡 = 3 𝜋 / 2$ for $𝑎 = 1, 2, 4,$ , and 6 over the interval $0 \leq 𝑡 \leq 4 𝜋$ . Describe in your own words what happens to the graph of the helix and the position of the tangent line as a increases through these positive values.
Set $𝑎 = 1$ . Plot the helix r( ) together with the tangent line to thet curve at $𝑡 = 3 𝜋 / 2$ for $𝑏 = 1 / 4, 1 / 2, 2$ , and 4 over the interval $0 \leq 𝑡 \leq$ 4 . Describe in your own words what happens to theπ graph of the helix and the position of the tangent line as b increases through these positive values.

12.2 Integrals of Vector Functions; Projectile Motion

In this section we investigate integrals of vector functions and their application to motion along a path in space or in the plane.

Integrals of Vector Functions

A differentiable vector function R( ) is an t antiderivative of a vector function r( ) on ant interval I if $𝑑 𝐑 / 𝑑 𝑡 = 𝐫$ at each point of I. If R is an antiderivative of r on $𝐼,$ it can be shown, working one component at a time, that every antiderivative of r on I has the form $𝐑 + 𝐂$ for some constant vector C (Exercise 45). The set of all antiderivatives of r on I is the indefinite integral of r on I.

DEFINITION The indefinite integral of r with respect to t is the set of all antiderivatives of r, denoted by
$\int 𝐫 (𝑡) 𝑑 𝑡 .$

The usual arithmetic rules for indefinite integrals apply.

EXAMPLE 1 To integrate a vector function, we integrate each of its components.

\int ((c o s 𝑡) 𝐢 + 𝐣 - 2 𝑡 𝐤) 𝑑 𝑡 = (\int c o s 𝑡 𝑑 𝑡) 𝐢 + (\int 𝑑 𝑡) 𝐣 - (\int 2 𝑡 𝑑 𝑡) 𝐤 (1)

= (s i n 𝑡 + 𝐶 1) 𝐢 + (𝑡 + 𝐶 2) 𝐣 - (𝑡 2 + 𝐶 3) 𝐤 (2)

= (s i n 𝑡) 𝐢 + 𝑡 𝐣 - 𝑡 2 𝐤 + 𝐂 𝐶 = 𝐶 1 𝐢 + 𝐶 2 𝐣 - 𝐶 3 𝐤

As in the integration of scalar functions, we recommend that you skip the steps inEqua tions (1) and (2) and go directly to the final form. Find an antiderivative for eachcomponent and add a constant vector at the end. 一

Definite integrals of vector functions are best defined in terms of components. The definition is consistent with how we compute limits and derivatives of vector functions.

DEFINITION If the components of $𝐫 (𝑡) = 𝑓 (𝑡) 𝐢 + 𝑔 (𝑡) 𝐣 + ℎ (𝑡) 𝐤$ are integrable over $[𝑎, 𝑏]$ , then so is r, and the definite integral of r from a to b is
$\int 𝑏 𝑎 𝐫 (𝑡) 𝑑 𝑡 = (\int 𝑏 𝑎 𝑓 (𝑡) 𝑑 𝑡) 𝐢 + (\int 𝑏 𝑎 𝑔 (𝑡) 𝑑 𝑡) 𝐣 + (\int 𝑏 𝑎 ℎ (𝑡) 𝑑 𝑡) 𝐤 .$

EXAMPLE 2 As in Example 1, we integrate each component.

\int 𝜋 0 ((c o s 𝑡) 𝐢 + 𝐣 - 2 𝑡 𝐤) 𝑑 𝑡 = (\int 𝜋 0 c o s 𝑡 𝑑 𝑡) 𝐢 + (\int 𝜋 0 𝑑 𝑡) 𝐣 - (\int 𝜋 0 2 𝑡 𝑑 𝑡) 𝐤 = [s i n 𝑡] 𝜋 0 𝐢 + [𝑡] 𝜋 0 𝐣 - [𝑡 2] 𝜋 0 𝐤 = [0 - 0] 𝐢 + [𝜋 - 0] 𝐣 - [𝜋 2 - 02] 𝐤 = 𝜋 𝐣 - 𝜋 2 𝐤

The Fundamental Theorem of Calculus for continuous vector functions says that

\int 𝑏 𝑎 𝐫 (𝑡) 𝑑 𝑡 = 𝐑 (𝑡)] 𝑏 𝑎 = 𝐑 (𝑏) - 𝐑 (𝑎),

where R is any antiderivative of r, so that $𝐑' (𝑡) = 𝐫 (𝑡)$ (Exercise 46). Notice that an antiderivative of a vector function is also a vector function, whereas a definite integral of a vector function is a single constant vector.

EXAMPLE 3 Suppose we do not know the path of a hang glider, but only its acceleration vector $𝐚 (𝑡) = - (3 c o s 𝑡) 𝐢 - (3 s i n 𝑡) 𝐣 + 2 𝐤$ . We also know that initially (at time $𝑡 = 0)$ the glider departed from the point $(4, 0, 0)$ with velocity $𝐯 (0) = 3 𝐣 .$ . Find the glider’s position as a function of t.

Solution Our goal is to find r( ) knowingt

The differential equation:

𝐚 = 𝑑 2 𝐫 𝑑 𝑡 2 = - (3 c o s 𝑡) 𝐢 - (3 s i n 𝑡) 𝐣 + 2 𝐤

The initial conditions:

𝐯 (0) = 3 𝐣 a n d 𝐫 (0) = 4 𝐢 + 0 𝐣 + 0 𝐤 .

Integrating both sides of the differential equation with respect to t gives

𝐯 (𝑡) = - (3 s i n 𝑡) 𝐢 + (3 c o s 𝑡) 𝐣 + 2 𝑡 𝐤 + 𝐂 1 .

We use $𝐯 (0) = 3 𝐣$ to find $𝐂 1 :$

3 𝐣 = - (3 s i n 0) 𝐢 + (3 c o s 0) 𝐣 + (0) 𝐤 + 𝐂 1 3 𝐣 = 3 𝐣 + 𝐂 1 𝐂 1 = 0 .

The glider’s velocity as a function of time is

𝑑 𝐫 𝑑 𝑡 = 𝐯 (𝑡) = - (3 s i n 𝑡) 𝐢 + (3 c o s 𝑡) 𝐣 + 2 𝑡 𝐤 .

教材插图

FIGURE 12.9 The path of the hang glider in Example 3. Although the path spirals around the z-axis, it is not a helix.

教材插图

(a)

教材插图

FIGURE 12.10 (a) Position, velocity, acceleration, and launch angle at $𝑡 = 0 .$ (b) Position, velocity, and acceleration at a later time t.

Integrating both sides of this last differential equation gives

𝐫 (𝑡) = (3 c o s 𝑡) 𝐢 + (3 s i n 𝑡) 𝐣 + 𝑡 2 𝐤 + 𝐂 2 .

We then use the initial condition r i (0) 4 to find $𝐂 2 :$ :

4 𝐢 = (3 c o s 0) 𝐢 + (3 s i n 0) 𝐣 + (02) 𝐤 + 𝐂 2 4 𝐢 = 3 𝐢 + (0) 𝐣 + (0) 𝐤 + 𝐂 2 𝐂 2 = 𝐢 .

The glider’s position as a function of t is

𝐫 (𝑡) = (1 + 3 c o s 𝑡) 𝐢 + (3 s i n 𝑡) 𝐣 + 𝑡 2 𝐤 .

This is the path of the glider shown in Figure 12.9. Although the path resembles that of a helix due to its spiraling nature around the z-axis, it is not a helix because of the way it is rising. (We say more about this in Section 12.5.)

The Vector and Parametric Equations for Ideal Projectile Motion

A classic example of integrating vector functions is the derivation of the equations for the motion of a projectile. In physics, projectile motion describes how an object fired at some angle from an initial position, and acted upon by only the force of gravity, moves in a vertical coordinate plane. In the classic example, we ignore the effects of any frictional drag on the object, which may vary with its speed and altitude, and also the fact that the force of gravity changes slightly with the projectile’s changing height. In addition, we ignore the long-distance effects of Earth turning beneath the projectile, such as in a rocket launch or the firing of a projectile from a cannon. Ignoring these effects gives us a reasonable approximation of the motion in most cases.

To derive equations for projectile motion, we assume that the projectile behaves like a particle moving in a vertical coordinate plane and that the only force acting on the projectile during its flight is the constant force of gravity, which always points straight down. The magnitude of the gravitational acceleration $𝑔$ is approximately $9.8 m / s 2$ at sea level. We assume that the projectile is launched from the origin at time $𝑡 = 0$ into the first quadrant with an initial velocity $𝐯 0$ (Figure 12.10). If $𝐯 0$ makes an angle B with the horizontal, then

𝐯 0 = (| 𝐯 0 | c o s 𝛼) 𝐢 + (| 𝐯 0 | s i n 𝛼) 𝐣 .

If we use the simpler notation $\boldsymbol𝑣0$ for the initial speed $| 𝐯 0 |$ , then

𝐯 0 = (𝑣 0 c o s 𝛼) 𝐢 + (𝑣 0 s i n 𝛼) 𝐣 . (3)

The projectile’s initial position is

𝐫 0 = 0 𝐢 + 0 𝐣 = 0 . (4)

Newton’s second law of motion says that the force acting on the projectile is equal to the projectile’s mass m times its acceleration, or m $(𝑑 2 𝐫 / 𝑑 𝑡 2)$ , if r is the projectile’s position vector and t is time. If the force is solely the gravitational force mgj, then

𝑚 𝑑 2 𝐫 𝑑 𝑡 2 = - 𝑚 𝑔 𝐣 a n d 𝑑 2 𝐫 𝑑 𝑡 2 = - 𝑔 𝐣,

where $𝑔$ is the acceleration due to gravity. We find r as a function of t by solving the following initial value problem.

Differential equation:

𝑑 2 𝐫 𝑑 𝑡 2 = - 𝑔 𝐣

Initial conditions:

𝐫 = 𝐫 0 a n d 𝑑 𝐫 𝑑 𝑡 = 𝐯 0 w h e n 𝑡 = 0

The first integration gives

𝑑 𝐫 𝑑 𝑡 = - (𝑔 𝑡) 𝐣 + 𝐯 0 .

A second integration gives

𝐫 = - 12 𝑔 𝑡 2 𝐣 + 𝐯 0 𝑡 + 𝐫 0 .

Substituting the values of $𝐯 0$ and $𝐫 0$ from Equations (3) and (4) gives

𝐫 = - 12 𝑔 𝑡 2 𝐣 + (𝑣 0 c o s 𝛼) 𝑡 𝐢 + (𝑣 0 s i n 𝛼) 𝑡 𝐣 ⏟_____⏟_____⏟ 𝐯 0 𝑡 + 𝟎 .

Collecting terms, we obtain the following.

Ideal Projectile Motion Equation

𝐫 = (𝑣 0 c o s 𝛼) 𝑡 𝐢 + ((𝑣 0 s i n 𝛼) 𝑡 - 12 𝑔 𝑡 2) 𝐣 . (5)

Equation (5) is the vector equation of the path for ideal projectile motion. The angle α is the projectile’s launch angle (firing angle, angle of elevation), and $\boldsymbol𝑣0,$ as we said before, is the projectile’s initial speed. The components of r give the parametric equations

𝑥 = (𝑣 0 c o s 𝛼) 𝑡 a n d 𝑦 = (𝑣 0 s i n 𝛼) 𝑡 - 12 𝑔 𝑡 2, (6)

where x is the distance downrange and $𝑦$ is the height of the projectile at time $𝑡 \geq 0$

EXAMPLE 4 A projectile is fired from the origin over horizontal ground at an initial speed of 500 m s and a launch angle of 60 . Where will the projectile be 10 s later?°

Solution We use Equation (5) with $𝑣 0 = 500, 𝛼 = 6 0 \circ, 𝑔 = 9.8$ , and $𝑡 = 10$ to find the projectile’s components 10 s after firing.

𝐫 = (𝑣 0 c o s 𝛼) 𝑡 𝐢 + ((𝑣 0 s i n 𝛼) 𝑡 - 12 𝑔 𝑡 2) 𝐣 = (500) (12) (10) 𝐢 + ((500) (\sqrt 3 2) 10 - (12) (9 . 8) (100)) 𝐣 \approx 2500 𝐢 + 3840 𝐣

Ten seconds after firing, the projectile is about 3840 m above ground and 2500 m downrange from the origin. 1

Ideal projectiles move along parabolas, as we now deduce from Equations (6). If we substitute $𝑡 = 𝑥 / (𝑣 0 c o s 𝛼)$ from the first equation into the second, we obtain the Cartesian coordinate equation

𝑦 = - (𝑔 2 𝑣 0 2 c o s 2 𝛼) 𝑥 2 + (t a n 𝛼) 𝑥 .

This equation has the form $𝑦 = 𝑎 𝑥 2 + 𝑏 𝑥$ , so its graph is a parabola.

教材插图

FIGURE 12.11 The path of a projectile fired from $(𝑥 0, 𝑦 0)$ with an initial velocity $𝐯 0$ at an angle of α degrees with the horizontal.

A projectile reaches its highest point when its vertical velocity component is zero. When fired over horizontal ground, the projectile lands when its vertical component equals zero in Equation (5), and the range R is the distance from the origin to the point of impact. We summarize the results here, which you are asked to verify in Exercise 31.

Height, Flight Time, and Range for Ideal Projectile Motion

For ideal projectile motion when an object is launched from the origin over a horizontal surface with initial speed $𝑣 0$ and launch angle α:

Maximum height:

𝑦 m a x = (𝑣 0 s i n 𝛼) 2 2 𝑔

Flight time:

𝑡 = 2 𝑣 0 s i n 𝛼 𝑔

Range:

𝑅 = 𝑣 20 𝑔 s i n 2 𝛼

If we fire our ideal projectile from the point $(𝑥 0, 𝑦 0)$ instead of the origin (Figure 12.11), the position vector for the path of motion is

𝐫 = (𝑥 0 + (𝑣 0 c o s 𝛼) 𝑡) 𝐢 + (𝑦 0 + (𝑣 0 s i n 𝛼) 𝑡 - 12 𝑔 𝑡 2) 𝐣, (7)

as you are asked to show in Exercise 33.

Projectile Motion with Wind Gusts

The next example shows how to account for another force acting on a projectile due to a gust of wind. We assume that the path of the baseball in Example 5 lies in a vertical plane.

EXAMPLE 5 A baseball is hit when it is 1 m above the ground. It leaves the bat with initial speed of $50 m / s,$ , making an angle of $2 0 \circ$ with the horizontal. At the instant the ball is hit, an instantaneous gust of wind blows in the horizontal direction directly opposite the direction the ball is taking toward the outfield, adding a component of $- 2.5 𝐢 (m / s)$ to the ball’s initial velocity $(2.5 m / s : = : 9 k m / h)$

(a) Find a vector equation (position vector) for the path of the baseball.

(b) How high does the baseball go, and when does it reach maximum height?

Solution

(a) Using Equation (3) and accounting for the gust of wind, the initial velocity of the baseball is

𝐯 0 = (𝑣 0 c o s 𝛼) 𝐢 + (𝑣 0 s i n 𝛼) 𝐣 - 2 . 5 𝐢 = (50 c o s 2 0 \circ) 𝐢 + (50 s i n 2 0 \circ) 𝐣 - (2 . 5) 𝐢 = (50 c o s 2 0 \circ - 2 . 5) 𝐢 + (50 s i n 2 0 \circ) 𝐣 .

The initial position is $𝐫 0 = 0 𝐢 + 1 𝐣$ . Integration of $𝑑 2 𝐫 / 𝑑 𝑡 2 = - 𝑔 𝐣 g i v e s$

𝑑 𝐫 𝑑 𝑡 = - (𝑔 𝑡) 𝐣 + 𝐯 0 .

A second integration gives

𝐫 = - 12 𝑔 𝑡 2 𝐣 + 𝐯 0 𝑡 + 𝐫 0 .

Substituting the values of $𝐯 0$ and $𝐫 0$ into the last equation gives the position vector of the baseball.

𝐫 = - 12 𝑔 𝑡 2 𝐣 + 𝐯 0 𝑡 + 𝐫 0 = - 4 . 9 𝑡 2 𝐣 + (50 c o s 2 0 \circ - 2 . 5) 𝑡 𝐢 + (50 s i n 2 0 \circ) 𝑡 𝐣 + 1 𝐣 = (50 c o s 2 0 \circ - 2 . 5) 𝑡 𝐢 + (1 + (50 s i n 2 0 \circ) 𝑡 - 4 . 9 𝑡 2) 𝐣 .

(b) The baseball reaches its highest point when the vertical component of velocity is zero, or

𝑑 𝑦 𝑑 𝑡 = 50 s i n 2 0 \circ - 9 . 8 𝑡 = 0 .

Solving for t we find

𝑡 = 50 s i n 2 0 \circ 9.8 \approx 1 . 75 s .

Substituting this time into the vertical component for r gives the maximum height

𝑦 m a x = 1 + (50 s i n 2 0 \circ) (1 . 75) - 4 . 9 (1 . 75) 2 \approx 1 5 . 9 m .

That is, the maximum height of the baseball is about 15.9 m, reached about 11.75 s after leaving the bat.

1 + (50 s i n 2 0 \circ) 𝑡 - 4 . 9 𝑡 2 = 0 1 + (1 7 . 1) 𝑡 - 4 . 9 𝑡 2 = 0 .

The solution values are about $𝑡 = 3.55 s$ and $𝑡 = - 0.06 s$ . Substituting the positive time into the horizontal component for r, we find the range

𝑅 = (50 c o s 2 0 \circ - 2 . 5) (3 . 55) \approx 15 7 . 8 m .

Thus, the horizontal range is about 157.8 m, and the flight time is about 3.55 s.

In Exercises 41 and 42, we consider projectile motion when there is air resistance slowing down the flight.

Exercises 12.2

Integrating Vector-Valued Functions

Evaluate the integrals in Exercises 1–10.

$\int 10 [𝑡 3 𝐢 + 7 𝐣 + (𝑡 + 1) 𝐤] 𝑑 𝑡$

\int 21 [(6 - 6 𝑡) 𝐢 + 3 \sqrt 𝑡 𝐣 + (4 𝑡 2) 𝐤] 𝑑 𝑡

\int 𝜋 / 4 - 𝜋 / 4 [(s i n 𝑡) 𝐢 + (1 + c o s 𝑡) 𝐣 + (s e c 2 𝑡) 𝐤] 𝑑 𝑡

π 3 4. [ ] ( ) ( ) ( ) + + sec tan tan 2 sin cos t t t t t dt i j k

$\int 41 [1 𝑡 𝐢 + 1 5 - 𝑡 𝐣 + 1 2 𝑡 𝐤] 𝑑 𝑡$
$\int 10 [2 \sqrt 1 - 𝑡 2 𝐢 + \sqrt 3 1 + 𝑡 2 𝐤] 𝑑 𝑡$
$\int 10 [𝑡 𝑒 𝑡 2 ˙ 𝐢 + 𝑒 - 𝑡 ˙ 𝐣 + 𝐤] 𝑑 𝑡$
$\int l n 3 1 [𝑡 𝑒 𝑡 𝐢 + 𝑒 𝑡 𝐣 + l n 𝑡 𝐤] 𝑑 𝑡$
$\int 𝜋 / 2 0 [c o s 𝑡 𝐢 - s i n 2 𝑡 𝐣 + s i n 2 𝑡 𝐤] 𝑑 𝑡$
$\int 𝜋 / 4 0 [s e c 𝑡 𝐢 + t a n 2 𝑡 𝐣 - 𝑡 s i n 𝑡 𝐤] 𝑑 𝑡$

Initial Value Problems

Solve the initial value problems in Exercises 11–20 for r as a vector function of t.

Differential equation: $𝑑 𝐫 𝑑 𝑡 = - 𝑡 𝐢 - 𝑡 𝐣 - 𝑡 𝐤$

𝐫 (0) = 𝐢 + 2 𝐣 + 3 𝐤

Differential equation: $𝑑 𝐫 𝑑 𝑡 = (180 𝑡) 𝐢 + (180 𝑡 - 16 𝑡 2) 𝐣$ Initial condition: $𝐫 (0) = 100 𝐣$
Differential equation: $𝑑 𝐫 𝑑 𝑡 = 32 (𝑡 + 1) 1 / 2 𝐢 + 𝑒 - 𝑡 𝐣 + 1 𝑡 + 1 𝐤$ Initial condition: $𝐫 (0) = 𝐤$
Differential equation: $𝑑 𝐫 𝑑 𝑡 = (𝑡 3 + 4 𝑡) 𝐢 + 𝑡 𝐣 + 2 𝑡 2 𝐤$ Initial condition: $𝐫 (0) = 𝐢 + 𝐣$
Differential equation:

𝑑 𝐫 𝑑 𝑡 = (t a n 𝑡) 𝐢 + (c o s (12 𝑡)) 𝐣 - (s e c 2 𝑡) 𝐤, - 𝜋 4 < 𝑡 < 𝜋 4

Initial condition: $𝐫 (0) = 3 𝐢 - 2 𝐣 + 𝐤$

Differential equation:

𝑑 𝐫 𝑑 𝑡 = (𝑡 𝑡 2 + 2) 𝐢 - (𝑡 2 + 1 𝑡 - 2) 𝐣 + (𝑡 2 + 4 𝑡 2 + 3) 𝐤, 𝑡 < 2

Initial condition: $𝐫 (0) = 𝐢 - 𝐣 + 𝐤$

Differential equation: $𝑑 2 𝐫 𝑑 𝑡 2 = - 32 𝐤$ Initial conditions: $𝐫 (0) = 100 𝐤 a n d$

𝑑 𝐫 𝑑 𝑡 ∣ 𝑡 = 0 = 8 𝐢 + 8 𝐣

Differential equation: $𝑑 2 𝐫 𝑑 𝑡 2 = - (𝐢 + 𝐣 + 𝐤)$

Initial conditions:

𝐫 (0) = 10 𝐢 + 10 𝐣 + 10 𝐤 a n d

𝑑 𝐫 𝑑 𝑡 ∣ 𝑡 = 0 = 𝟎

𝑑 2 𝐫 𝑑 𝑡 2 = 𝑒 𝑡 𝐢 - 𝑒 - 𝑡 𝐣 + 4 𝑒 2 𝑡 𝐤

Initial conditions:

𝐫 (0) = 3 𝐢 + 𝐣 + 2 𝐤 a n d

𝑑 𝐫 𝑑 𝑡 ∣ 𝑡 = 0 = - 𝐢 + 4 𝐣 20 . $ 𝐷 𝑖 𝑓 𝑓 𝑒 𝑟 𝑒 𝑛 𝑡 𝑖 𝑎 𝑙 𝑒 𝑞 𝑢 𝑎 𝑡 𝑖 𝑜 𝑛 : $ 𝑑 2 𝐫 𝑑 𝑡 2 = (s i n 𝑡) 𝐢 - (c o s 𝑡) 𝐣 + (4 s i n 𝑡 c o s 𝑡) 𝐤

Initial conditions: $𝐫 (0) = 𝐢 - 𝐤$ and

𝑑 𝐫 𝑑 𝑡 ∣ 𝑡 = 0 = 𝐢

Motion Along a Straight Line

At time $𝑡 = 0,$ a particle is located at the point (1, 2, 3 . It travels) in a straight line to the point (4, 1, 4 , has speed 2 at ) (1, 2, 3 , and) has constant acceleration $3 𝐢 - 𝐣 + 𝐤 .$ . Find an equation for the position vector r( ) of the particle at time t t.
A particle traveling in a straight line is located at the point (1, 1, 2 and has speed 2 at time− ) $𝑡 = 0$ . The particle moves toward the point (3, 0, 3 with constant acceleration) $2 𝐢 + 𝐣 + 𝐤 .$ Find its position vector r( ) at time t t.

Projectile Motion

Projectile flights in Exercises 23–40 are to be treated as ideal unless stated otherwise. All launch angles are assumed to be measured from the horizontal. All projectiles are assumed to be launched from the origin over a horizontal surface unless stated otherwise. For some exercises, a calculator may be helpful.

Travel time A projectile is fired at a speed of 840 m s at an angle of 60 . How long will it take to get 21 km downrange?°
Range and height versus speed

a. Show that doubling a projectile’s initial speed at a given launch angle multiplies its range by 4.

b. By about what percentage should you increase the initial speed to double the height and range?

Flight time and height A projectile is fired with an initial speed of 500 m s at an angle of elevation of 45 .°

a. When and how far away will the projectile strike?

b. How high overhead will the projectile be when it is 5 km downrange?

c. What is the greatest height reached by the projectile?

Throwing a baseball A baseball is thrown from the stands 9.8 m above the field at an angle of 30 up from the horizontal.° When and how far away will the ball strike the ground if its initial speed is $9.8 m / s ?$
Firing golf balls A spring gun at ground level fires a golf ball at an angle of 45 . The ball lands 10 m away. °

a. What was the ball’s initial speed?

b. For the same initial speed, find the two firing angles that make the range 6 m.

Beaming electrons An electron in a cathode-ray tube (CRT) is beamed horizontally at a speed o $𝛥 5 \times 106$ m s toward the face of the tube 40 cm away. About how far will the electron drop before it hits?
Equal-range firing angles What two angles of elevation will enable a projectile to reach a target 16 km downrange on the same level as the gun if the projectile’s initial speed is 400 m s?
Finding muzzle speed Find the muzzle speed of a gun whose maximum range is 24.5 km.
Verify the results given in the text (following Example 4) for the maximum height, flight time, and range for ideal projectile motion.
Colliding marbles The accompanying figure shows an experiment with two marbles. Marble A was launched toward marble B with launch angle α and initial speed $\boldsymbol𝑣0.$ At the same instant, marble B was released to fall from rest at R tan units directly aboveα a spot R units downrange from A. The marbles were found to collide regardless of the value of $\boldsymbol𝑣0.$ . Was this mere coincidence, or must this happen? Give reasons for your answer.

教材插图

Firing from $(𝑥 0, 𝑦 0)$ Derive the equations

𝑥 = 𝑥 0 + (𝑣 0 c o s 𝛼) 𝑡, 𝑦 = 𝑦 0 + (𝑣 0 s i n 𝛼) 𝑡 - 12 𝑔 𝑡 2

(see Equation (7) in the text) by solving the following initial value problem for a vector r in the plane.

Differential equation:

Initial conditions:

𝑑2𝐫𝑑𝑡2=−𝑔𝐣𝐫(0)=𝑥0𝐢+𝑦0𝐣𝑑𝐫𝑑𝑡(0)=(𝑣0cos⁡𝛼)𝐢+(𝑣0sin⁡𝛼)𝐣34.$𝑊ℎ𝑒𝑟𝑒𝑡𝑟𝑎𝑗𝑒𝑐𝑡𝑜𝑟𝑖𝑒𝑠𝑐𝑟𝑒𝑠𝑡𝐹𝑜𝑟𝑎𝑝𝑟𝑜𝑗𝑒𝑐𝑡𝑖𝑙𝑒𝑓𝑖𝑟𝑒𝑑𝑓𝑟𝑜𝑚𝑡ℎ𝑒𝑔𝑟𝑜𝑢𝑛𝑑𝑎𝑡𝑙𝑎𝑢𝑛𝑐ℎ𝑎𝑛𝑔𝑙𝑒𝛼𝑤𝑖𝑡ℎ𝑖𝑛𝑖𝑡𝑖𝑎𝑙𝑠𝑝𝑒𝑒𝑑$\boldsymbol𝑣0,$,𝑐𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝛼𝑎𝑠𝑎𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑎𝑛𝑑$𝑣0$𝑎𝑠𝑎𝑓𝑖𝑥𝑒𝑑𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡.𝐹𝑜𝑟𝑒𝑎𝑐ℎ$𝛼,0<𝛼<𝜋/2$,𝑤𝑒𝑜𝑏𝑡𝑎𝑖𝑛𝑎𝑝𝑎𝑟𝑎𝑏𝑜𝑙𝑖𝑐𝑡𝑟𝑎𝑗𝑒𝑐𝑡𝑜𝑟𝑦𝑎𝑠𝑠ℎ𝑜𝑤𝑛𝑖𝑛𝑡ℎ𝑒𝑎𝑐𝑐𝑜𝑚𝑝𝑎𝑛𝑦𝑖𝑛𝑔𝑓𝑖𝑔𝑢𝑟𝑒.𝑆ℎ𝑜𝑤𝑡ℎ𝑎𝑡𝑡ℎ𝑒𝑝𝑜𝑖𝑛𝑡𝑠𝑖𝑛𝑡ℎ𝑒𝑝𝑙𝑎𝑛𝑒𝑡ℎ𝑎𝑡𝑔𝑖𝑣𝑒𝑡ℎ𝑒𝑚𝑎𝑥𝑖𝑚𝑢𝑚ℎ𝑒𝑖𝑔ℎ𝑡𝑠𝑜𝑓𝑡ℎ𝑒𝑠𝑒𝑝𝑎𝑟𝑎𝑏𝑜𝑙𝑖𝑐𝑡𝑟𝑎𝑗𝑒𝑐𝑡𝑜𝑟𝑖𝑒𝑠𝑎𝑙𝑙𝑙𝑖𝑒𝑜𝑛𝑡ℎ𝑒𝑒𝑙𝑙𝑖𝑝𝑠𝑒$𝑥2+4(𝑦−𝑣024𝑔)2=𝑣044𝑔2,

where $𝑥 \geq 0 .$

教材插图

Launching downhill An ideal projectile is launched straight down an inclined plane as shown in the accompanying figure.

a. Show that the greatest downhill range is achieved when the initial velocity vector bisects angle AOR.

b. If the projectile were fired uphill instead of down, what launch angle would maximize its range? Give reasons for your answer.

教材插图

Elevated green A golf ball is hit with an initial speed of 35.5 m s at an angle of elevation of $4 5 \circ$ from the tee to a green that is elevated 14 m above the tee as shown in the diagram. Assuming that the pin, 112 m downrange, does not get in the way, where will the ball land in relation to the pin?

教材插图

Volleyball A volleyball is hit when it is 1.3 m above the ground and 4 m from a 2-m-high net. It leaves the point of impact with an initial velocity of 12 m s at an angle o $2 7 \circ$ and slips by the opposing team untouched.

a. Find a vector equation for the path of the volleyball.

b. How high does the volleyball go, and when does it reach maximum height?

c. Find its range and flight time.

d. When is the volleyball 2.3 m above the ground? How far (ground distance) is the volleyball from where it will land?

e. Suppose that the net is raised to 2.5 m. Does this change things? Explain.

Shot put In Moscow in 1987, Natalya Lisouskaya set a women’s world record by putting a 4kg shot 22.63 m. Assuming that she launched the shot at $a 4 0 \circ$ angle to the horizontal from 2 m above the ground, what was the shot’s initial speed?
A child throws a ball with an initial speed of 18 m s at an angle of elevation of $6 0 \circ$ toward a tall building that is 7 m from the child. If the child’s hand is 1.6 m from the ground, show that the ball hits the building, and find the height above the ground of the point where the ball hits the building.
Hitting a baseball under a wind gust A baseball is hit when it is 0.8 m above the ground. It leaves the bat with an initial velocity of 40 m s at a launch angle of $2 3 \circ .$ . At the instant the ball is hit, an instantaneous gust of wind blows against the ball, adding a component of $- 4 𝐢 (m / s)$ to the ball’s initial velocity. A 5-m-high fence lies 90 m from home plate in the direction of the flight.

a. Find a vector equation for the path of the baseball.

b. How high does the baseball go, and when does it reach maximum height?

c. Find the range and flight time of the baseball, assuming that the ball is not caught.

d. When is the baseball 6 m high? How far (ground distance) is the baseball from home plate at that height?

e. Has the batter hit a home run? Explain.

Projectile Motion with Linear Drag

The main force affecting the motion of a projectile, other than gravity, is air resistance. This slowing down force is drag force, and it acts in a direction opposite to the velocity of the projectile (see accompanying figure). For projectiles moving through the air at relatively low speeds, however, the drag force is (very nearly) proportional to the speed (to the first power) and so is called linear.

教材插图

Linear drag Derive the equations

𝑥 = 𝑣 0 𝑘 (1 - 𝑒 - 𝑘 𝑡) c o s 𝛼

𝑦 = 𝑣 0 𝑘 (1 - 𝑒 - 𝑘 𝑡) (s i n 𝛼) + 𝑔 𝑘 2 (1 - 𝑘 𝑡 - 𝑒 - 𝑘 𝑡)

by solving the following initial value problem for a vector r in the plane.

Differential equation: $𝑑 2 𝐫 𝑑 𝑡 2 = - 𝑔 𝐣 - 𝑘 𝐯 = - 𝑔 𝐣 - 𝑘 𝑑 𝐫 𝑑 𝑡$

Initial conditions: $𝐫 (0) = 𝟎$

𝑑 𝐫 𝑑 𝑡 ∣ 𝑡 = 0 = 𝐯 0 = (𝑣 0 c o s 𝛼) 𝐢 + (𝑣 0 s i n 𝛼) 𝐣

The drag coefficient k is a positive constant representing resistance due to air density, $𝑣 0$ and B are the projectile’s initial speed and launch angle, and g is the acceleration of gravity.

Hitting a baseball with linear drag Consider the baseball problem in Example 5 when there is linear drag (see Exercise 41). Assume a drag coefficient $𝑘 = 0.12$ , but no gust of wind.

a. From Exercise 41, find a vector form for the path of the baseball.

b. How high does the baseball go, and when does it reach maximum height?

c. Find the range and flight time of the baseball.

d. When is the baseball 9 m high? How far (ground distance) is the baseball from home plate at that height?

e. A 3-m-high outfield fence is 115 m from home plate in the direction of the flight of the baseball. The outfielder can jump and catch any ball up to 3.3 m off the ground to stop it from going over the fence. Has the batter hit a home run?

Theory and Examples

Establish the following properties of integrable vector functions.

a. The Constant Scalar Multiple Rule:

\int 𝑏 𝑎 𝑘 𝐫 (𝑡) 𝑑 𝑡 = 𝑘 \int 𝑏 𝑎 𝐫 (𝑡) 𝑑 𝑡 (a n y s c a l a r 𝑘)

The Rule for Negatives,

\int 𝑏 𝑎 (- 𝐫 (𝑡)) 𝑑 𝑡 = - \int 𝑏 𝑎 𝐫 (𝑡) 𝑑 𝑡,

is obtained by taking $𝑘 = - 1$

b. The Sum and Difference Rules:

\int 𝑏 𝑎 (𝐫 1 (𝑡) \pm 𝐫 2 (𝑡)) 𝑑 𝑡 = \int 𝑏 𝑎 𝐫 1 (𝑡) 𝑑 𝑡 \pm \int 𝑏 𝑎 𝐫 2 (𝑡) 𝑑 𝑡

c. The Constant Vector Multiple Rules:

\int 𝑏 𝑎 𝐂 \cdot 𝐫 (𝑡) 𝑑 𝑡 = 𝐂 \cdot \int 𝑏 𝑎 𝐫 (𝑡) 𝑑 𝑡 (a n y c o n s t a n t v e c t o r 𝐂)

and

\int 𝑏 𝑎 𝐂 \times 𝐫 (𝑡) 𝑑 𝑡 = 𝐂 \times \int 𝑏 𝑎 𝐫 (𝑡) 𝑑 𝑡 (a n y c o n s t a n t v e c t o r 𝐂)

Products of scalar and vector functions Suppose that the scalar function u t( ) and the vector function r( ) are both defined fort $𝑎 \leq 𝑡 \leq 𝑏 .$

a. Show that ur is continuous on $[𝑎, 𝑏]$ if u and r are continuous on $[𝑎, 𝑏] .$

b. If u and r are both differentiable on $[𝑎, 𝑏],$ , show that ur is differentiable on $[𝑎, 𝑏]$ and that

𝑑 𝑑 𝑡 (𝑢 𝐫) = 𝑢 𝑑 𝐫 𝑑 𝑡 + 𝑑 𝑢 𝑑 𝑡 𝐫 .

Antiderivatives of vector functions

a. Use Corollary 2 of the Mean Value Theorem for scalar functions to show that if two vector functions $𝐑 1 (𝑡)$ and $𝐑 2 (𝑡)$ have identical derivatives on an interval I, then the functions differ by a constant vector value throughout I.

b. Use the result in part (a) to show that if $𝐑 (𝑡)$ is any antiderivative of r( ) on t I, then any other antiderivative of r on I equals $𝐑 (𝑡) + 𝐂$ for some constant vector C.

The Fundamental Theorem of Calculus The Fundamental Theorem of Calculus for scalar functions of a real variable holds for vector functions of a real variable as well. Prove this by using the theorem for scalar functions to show first that if a vector function r( ) is continuous fort $𝑎 \leq 𝑡 \leq 𝑏 \cdot$ , then

𝑑 𝑑 𝑡 \int 𝑡 𝑎 𝐫 (𝜏) 𝑑 𝜏 = 𝐫 (𝑡)

at every point t of $(𝑎, 𝑏)$ . Then use the conclusion in part (b) of Exercise 45 to show that if R is any antiderivative of r on $[𝑎, 𝑏] .$ then

\int 𝑏 𝑎 𝐫 (𝑡) 𝑑 𝑡 = 𝐑 (𝑏) - 𝐑 (𝑎) .

Hitting a baseball with linear drag under a wind gust Consider again the baseball problem in Example 5. This time, assume a drag coefficient of 0.08 and an instantaneous gust of wind that adds a component o $- 5 𝐢 (𝐦 / 𝐬)$ to the initial velocity at the instant the baseball is hit.

a. Find a vector equation for the path of the baseball.

b. How high does the baseball go, and when does it reach maximum height?

c. Find the range and flight time of the baseball.

d. When is the baseball 10 m high? How far (ground distance) is the baseball from home plate at that height?

e. A 6-m-high outfield fence is 120 m from home plate in the direction of the flight of the baseball. Has the batter hit a home run? $I f * y e s, *$ what change in the horizontal component of the ball’s initial velocity would have kept the ball in the park? If $″ n o, ″$ what change would have allowed it to be a home run?

Height versus time Show that a projectile attains three-quarters of its maximum height in half the time it takes to reach the maximum height.

教材插图

FIGURE 12.12 Smooth curves can be scaled like number lines, the coordinate of each point being its directed distance along the curve from a preselected base point.

In this and the next two sections, we study the mathematical features of a curve’s shape that describe the sharpness of its turning and its twisting.

Arc Length Along a Space Curve

One of the features of smooth space and plane curves is that they have a measurable length. This enables us to locate points along these curves by giving their directed distance s along the curve from some base point, the way we locate points on coordinate axes by giving their directed distance from the origin (Figure 12.12). This is what we did for plane curves in Section 10.2.

FIGURE 12.13 The helix in Example 1, $𝐫 (𝑡) = (c o s 𝑡) 𝐢 + (s i n 𝑡) 𝐣 + 𝑡 𝐤 .$

To measure distance along a smooth curve in space, we add a z-term to the formula we use for curves in the plane.

DEFINITION The length of a smooth curve $𝐫 (𝑡) = 𝑥 (𝑡) 𝐢 + 𝑦 (𝑡) 𝐣 + 𝑧 (𝑡) 𝐤$ $𝑎 \leq 𝑡 \leq 𝑏 :$ , that is traced exactly once as t increases from $𝑡 = 𝑎 t a n 𝑡 = 𝑏$ is

教材插图

$𝐿 = \int 𝑏 𝑎 \sqrt (𝑑 𝑥 𝑑 𝑡) 2 + (𝑑 𝑦 𝑑 𝑡) 2 + (𝑑 𝑧 𝑑 𝑡) 2 𝑑 𝑡 . (1)$

Just as for plane curves, we can calculate the length of a curve in space from any convenient parametrization that meets the stated conditions. We omit the proof.

The square root in Equation (1) is v , the length of a velocity vector $𝑑 𝐫 / 𝑑 𝑡$ . This enables us to write the formula for length a shorter way.

Arc Length Formula

𝐿 = \int 𝑏 𝑎 | 𝐯 | 𝑑 𝑡 (2)

EXAMPLE 1 A glider is soaring upward along the helix

𝐫 (𝑡) = (c o s 𝑡) 𝐢 + (s i n 𝑡) 𝐣 + 𝑡 𝐤 .

How long is the glider’s path from $𝑡 = 0 t o 𝑡 = 2 𝜋 ?$

Solution The path segment during this time corresponds to one full turn of the helix (Figure 12.13). The length of this portion of the curve is

𝐿 = \int 𝑏 𝑎 | 𝐯 | 𝑑 𝑡 = \int 2 𝜋 0 \sqrt (- s i n 𝑡) 2 + (c o s 𝑡) 2 + (1) 2 𝑑 𝑡 = \int 2 𝜋 0 \sqrt 2 𝑑 𝑡 = 2 𝜋 \sqrt 2 u n i t s o f l e n g t h .

This is $\sqrt 2$ times the circumference of the circle in the xy-plane over which the helix stands.

If we choose a base point $𝑃 (𝑡 0)$ on a smooth curve C parametrized by t, each value of t determines a point $𝑃 (𝑡) = (𝑥 (𝑡), 𝑦 (𝑡), 𝑧 (𝑡))$ on C and a “directed distance”

𝑠 (𝑡) = \int 𝑡 𝑡 0 | 𝐯 (𝜏) | 𝑑 𝜏,

教材插图

FIGURE 12.14 The directed distance along the curve from $𝑃 (𝑡 0)$ to any point $𝑃 (𝑡)$ is $𝑠 (𝑡) = \int 𝑡 𝑡 0 | 𝐯 (𝜏) | 𝑑 𝜏$

U is the Greek letter tau (rhymes with “now”)

measured along C from the base point (Figure 12.14). This is the arc length function we defined in Section 10.2 for plane curves that have no z-component. If $˙ 𝜌 𝑡 > 𝑡 0, 𝑠 (𝑡)$ is the distance along the curve from $𝑃 (𝑡 0)$ to $𝑃 (𝑡)$ . If $𝑡 < 𝑡 0, 𝑠 (𝑡)$ is the negative of the distance. Each value of s determines a point on $𝐶,$ , and this parametrizes C with respect to s. We call s an arc length parameter for the curve. The parameter’s value increases in the direction of increasing t. We will see that the arc length parameter is particularly effective for investigating the turning and twisting nature of a space curve.

Arc Length Parameter with Base Point $𝑃 (𝑡 0)$

𝑠 (𝑡) = \int 𝑡 𝑡 0 \sqrt [𝑥' (𝜏)] 2 + [𝑦' (𝜏)] 2 + [𝑧' (𝜏)] 2 𝑑 𝜏 = \int 𝑡 𝑡 0 | 𝐯 (𝜏) | 𝑑 𝜏 (3)

We use the Greek letter $𝜏(6∘tau\dag)$ as the variable of integration in Equation (3) because the letter t is already in use as the upper limit.

If a curve $𝐫 (𝑡)$ is already given in terms of some parameter $𝑡,$ and $𝑠 (𝑡)$ is the arc length function given by Equation (3), then we may be able to solve for t as a function of s: $𝑡 = 𝑡 (𝑠)$ . Then the curve can be reparametrized in terms of s by substituting for t: $𝐫 = 𝐫 (𝑡 (𝑠))$ . The new parametrization identifies a point on the curve with its directed distance along the curve from the base point.

EXAMPLE 2 This is an example for which we can actually find the arc length parametrization of a curve. If $𝑡 0 = 0$ , then the arc length parameter along the helix

𝐫 (𝑡) = (c o s 𝑡) 𝐢 + (s i n 𝑡) 𝐣 + 𝑡 𝐤

from $𝑡 0$ to t is

𝑠 (𝑡) = \int 𝑡 𝑡 0 | 𝐯 (𝜏) | 𝑑 𝜏 E q . (3) = \int 𝑡 0 \sqrt 2 𝑑 𝜏 V a l u e f r o m E x a m p l e 1 = \sqrt 2 𝑡 .

Solving this equation for t gives $𝑡 = 𝑠 / \sqrt 2$ . Substituting into the position vector r gives the following arc length parametrization for the helix:

𝐫 (𝑡 (𝑠)) = (c o s 𝑠 \sqrt 2) 𝐢 + (s i n 𝑠 \sqrt 2) 𝐣 + 𝑠 \sqrt 2 𝐤 .

HISTORICAL BIOGRAPHY Josiah Willard Gibbs (1839–1903)

Gibbs, born in Connecticut, USA, taught at Yale as a professor of mathematics. He made contributions to thermodynamics, electromagnetics, and statistical mechanics. For his foundational work, Gibbs is known as the father of vector analysis.

To know more, visit the companion Website.

Unlike the case that appears in Example 2, the arc length parametrization is generally difficult to find analytically for a curve already given in terms of some other parameter t. Fortunately, however, we rarely need an exact formula for s( ) or its inverse t t s( ).

Speed on a Smooth Curve

Since the derivatives beneath the radical in Equation (3) are continuous (the curve is smooth), the Fundamental Theorem of Calculus tells us that s is a differentiable function of t with derivative

𝑑 𝑠 𝑑 𝑡 = ∣ 𝐯 (𝑡) ∣ . (4)

教材插图

FIGURE 12.15 We find the unit tangent vector T by dividing v by its length v .

教材插图

FIGURE 12.16 Counterclockwise motion around the unit circle.

Although the base point $𝑃 (𝑡 0)$ plays a role in defining s in Equation (3), it plays no role in Equation (4). The rate at which a moving particle covers distance along its path is independent of how far away it is from the base point. Equation (4) says that this rate is the magnitude of v.

Notice that $𝑑 𝑠 / 𝑑 𝑡 > 0$ since, by definition, v is never zero for a smooth curve. We see once again that s is an increasing function of t.

Unit Tangent Vector

On a smooth curve, we already know that the velocity vector $𝐯 = 𝑑 𝐫 / 𝑑 𝑡$ is tangent to the curve $𝐫 (𝑡)$ and that the vector

𝐓 = 𝐯 | 𝐯 |

is therefore a unit vector tangent to the curve, called the unit tangent vector (Figure 12.15). The unit tangent vector T for a smooth curve is a differentiable function of t whenever v is a differentiable function of t. As we will see in Section 12.5, T is one of three unit vectors in a traveling reference frame that is used to describe the motion of objects traveling in three dimensions.

EXAMPLE 3 Find the unit tangent vector of the curve

𝐫 (𝑡) = (1 + 3 c o s 𝑡) 𝐢 + (3 s i n 𝑡) 𝐣 + 𝑡 2 𝐤

representing the path of the glider in Example 3, Section 12.2.

Solution In that example, we found

𝐯 = 𝑑 𝐫 𝑑 𝑡 = - (3 s i n 𝑡) 𝐢 + (3 c o s 𝑡) 𝐣 + 2 𝑡 𝐤

and

| 𝐯 | = \sqrt 9 + 4 𝑡 2 .

Thus,

𝐓 = 𝐯 | 𝐯 | = - 3 s i n 𝑡 \sqrt 9 + 4 𝑡 2 𝐢 + 3 c o s 𝑡 \sqrt 9 + 4 𝑡 2 𝐣 + 2 𝑡 \sqrt 9 + 4 𝑡 2 𝐤 .

For the counterclockwise motion

𝐫 (𝑡) = (c o s 𝑡) 𝐢 + (s i n 𝑡) 𝐣

around the unit circle, we see that

𝐯 = (- s i n 𝑡) 𝐢 + (c o s 𝑡) 𝐣

is already a unit vector, so $𝐓 = 𝐯$ and T is orthogonal to r (Figure 12.16).

The velocity vector is the change in the position vector r with respect to time t, but how does the position vector change with respect to arc length? More precisely, what is the derivative $𝑑 𝐫 / 𝑑 𝑠 ?$ Since $𝑑 𝑠 / 𝑑 𝑡 > 0$ for the curves we are considering, s is one-to-one and has an inverse that gives t as a differentiable function of s (Section 3.8). The derivative of the inverse is

𝑑 𝑡 𝑑 𝑠 = 1 𝑑 𝑠 / 𝑑 𝑡 = 1 | 𝐯 | .

This makes r a differentiable function of s whose derivative can be calculated with the Chain Rule to be

𝑑 𝐫 𝑑 𝑠 = 𝑑 𝐫 𝑑 𝑡 𝑑 𝑡 𝑑 𝑠 = 𝐯 1 | 𝐯 | = 𝐯 | 𝐯 | = 𝐓 . (5)

This equation says that $𝑑 𝐫 / 𝑑 𝑠$ is the unit tangent vector in the direction of the velocity vector v (Figure 12.15).

Exercises 12.3

Finding Tangent Vectors and Lengths

In Exercises 1–8, find the curve’s unit tangent vector. Also, find the length of the indicated portion of the curve.

𝟏 . 𝐫 (𝑡) = (2 c o s 𝑡) 𝐢 + (2 s i n 𝑡) 𝐣 + \sqrt 5 𝑡 𝐤, 0 \leq 𝑡 \leq 𝜋

r i j k ( ) 6 sin 2 6 cos 2 5 , 0 t t t t t = + + ≤ ≤ ( ) ( ) π
$𝐫 (𝑡) = 𝑡 𝐢 + (2 / 3) 𝑡 3 / 2 𝐤, 0 \leq 𝑡 \leq 8$
r i j k ( ) 2 1 , 0 3 t t t t t = + − + + ≤ ≤ ( ) ( )
$𝐫 (𝑡) = (c o s 3 𝑡) 𝐣 + (s i n 3 𝑡) 𝐤, 0 \leq 𝑡 \leq 𝜋 / 2$
r i j k ( ) 6 2 3 , 1 t t t t t = − − ≤ ≤ 2 3 3 3
r i j k ( ) cos sin 2 2 3 , 0 t t t t t t t = + + ≤ ≤ ( ) ( ) ( ) π 3 2
r i j( ) sin cos cos sin , 2 2t t t t t t t t= + + − ≤ ≤( ) ( )
Find the point on the curve

𝐫 (𝑡) = (5 s i n 𝑡) 𝐢 + (5 c o s 𝑡) 𝐣 + 12 𝑡 𝐤

at a distance 26 units along the curve from the point π (0, 5, 0 in) the direction corresponding to increasing t values.

Find the point on the curve

𝐫 (𝑡) = (12 s i n 𝑡) 𝐢 - (12 c o s 𝑡) 𝐣 + 5 𝑡 𝐤

at a distance 13 units along the curve from the point π (0, 12, 0− ) in the direction corresponding to decreasing t values.

Arc Length Parameter

In Exercises 11–14, find the arc length parameter along the curve from the point where $𝑡 = 0$ by evaluating the integral

𝑠 (𝑡) = \int 𝑡 0 | 𝐯 (𝜏) | 𝑑 𝜏

from Equation (3). Then use the formula for s( ) to find the length oft the indicated portion of the curve.

$𝐫 (𝑡) = (4 c o s 𝑡) 𝐢 + (4 s i n 𝑡) 𝐣 + 3 𝑡 𝐤, 0 \leq 𝑡 \leq 𝜋 / 2$
$𝐫 (𝑡) = (c o s 𝑡 + 𝑡 s i n 𝑡) 𝐢 + (s i n 𝑡 - 𝑡 c o s 𝑡) 𝐣, 𝜋 / 2 \leq 𝑡 \leq 𝜋$
r i j k ( ) cos sin , ln 4 0 t e t e t e t = + + − ≤ ≤ ( ) ( ) t t t

𝐫 (𝑡) = (1 + 2 𝑡) 𝐢 + (1 + 3 𝑡) 𝐣 + (6 - 6 𝑡) 𝐤, - 1 \leq 𝑡 \leq 0

Theory and Examples

Arc length Find the length of the curve

𝐫 (𝑡) = (\sqrt 2 𝑡) 𝐢 + (\sqrt 2 𝑡) 𝐣 + (1 - 𝑡 2) 𝐤

from ( ) 0, 0, 1 to ( 2, 2, 0 .)

Length of helix The length $2 𝜋 \sqrt 2$ of the turn of the helix in Example 1 is also the length of the diagonal of a square 2π units on a side. Show how to obtain this square by cutting away and flattening a portion of the cylinder around which the helix winds.
Ellipse

a. Show that the curve $𝐫 (𝑡) = (c o s 𝑡) 𝐢 + (s i n 𝑡) 𝐣 + (1 - c o s 𝑡) 𝐤,$ $0 \leq 𝑡 \leq 2 𝜋$ , is an ellipse by showing that it is the intersection of a right circular cylinder and a plane. Find equations for the cylinder and plane.

b. Sketch the ellipse on the cylinder. Add to your sketch the unit tangent vectors a $𝑡 = 0, 𝜋 / 2,$ , andπ $3 𝜋 / 2$

c. Show that the acceleration vector always lies parallel to the plane (orthogonal to a vector normal to the plane). Thus, if you draw the acceleration as a vector attached to the ellipse, it will lie in the plane of the ellipse. Add the acceleration vectors for $𝑡 = 0, 𝜋 / 2, 𝜋$ , and $3 𝜋 / 2$ to your sketch.

d. Write an integral for the length of the ellipse. Do not try to evaluate the integral; it is nonelementary.

e. Numerical integrator Estimate the length of the ellipse toT two decimal places.

Length is independent of parametrization To illustrate that the length of a smooth space curve does not depend on the parametrization you use to compute it, calculate the length of one turn of the helix in Example 1 with the following parametrizations.

a. r i j k ( ) cos 4 sin 4 4 , 0 2 t t t t t = + + ≤ ≤ ( ) ( ) π

b. r i j k ( ) cos 2 sin 2 2 , 0 4 t t t t t = + + ≤ ≤ [ ] ( ) ( ) [ ] ( ) π

c. r i j k ( ) cos sin , 2 0 t t t t t = − − − ≤ ≤ ( ) ( ) π

The involute of a circle If a string wound around a fixed circle is unwound while held taut in the plane of the circle, its end P traces an involute of the circle. In the accompanying figure, the circle in question is the circle $𝑥 2 + 𝑦 2 = 1$ and the tracing point starts at (1, 0). The unwound portion of the string is tangent to the circle at Q, and t is the radian measure of the angle from the positive x-axis to segment $𝑂 𝑄$ . Derive the parametric equations

𝑥 = c o s 𝑡 + 𝑡 s i n 𝑡, 𝑦 = s i n 𝑡 - 𝑡 c o s 𝑡, 𝑡 > 0

of the point $𝑃 (𝑥, 𝑦)$ for the involute.

教材插图

(Continuation of Exercise 19.) Find the unit tangent vector to the involute of the circle at the point $𝑃 (𝑥, 𝑦)$
Distance along a line Show that if u is a unit vector, then the arc length parameter along the line $𝐫 (𝑡) = 𝑃 0 + 𝑡 𝐮$ from the point $𝑃 0 (𝑥 0, 𝑦 0, 𝑧 0)$ where $𝑡 = 0$ , is t itself.
Use Simpson’s Rule with n = 10 to approximate the length of arc of $𝐫 (𝑡) = 𝑡 𝐢 + 𝑡 2 𝐣 + 𝑡 3 𝐤$ from the origin to the point 2, 4, 8( ).

12.4 Curvature and Normal Vectors of a Curve

教材插图

FIGURE 12.17 As P moves along the curve in the direction of increasing arc length, the unit tangent vector turns. The value of $| 𝑑 𝐓 / 𝑑 𝑠 |$ at $𝑃$ is called the curvature of the curve at $𝑃 .$

κ is the Greek letter kappa.

教材插图

FIGURE 12.18 Along a straight line, T always points in the same direction. The curvature, $| 𝑑 𝐓 / 𝑑 𝑠 |$ , is zero (Example 1).

In this section we study how a curve turns or bends. To gain perspective, we look first at curves in the coordinate plane. Then we consider curves in space.

Curvature of a Plane Curve

As a particle moves along a smooth curve in the plane, $𝐓 = 𝑑 𝐫 / 𝑑 𝑠$ turns as the curve bends. Since T is a unit vector, its length remains constant and only its direction changes as the particle moves along the curve. The rate at which T turns per unit of length along the curve is called the curvature (Figure 12.17). The traditional symbol for the curvature function is the Greek letter κ (“kappa”).

DEFINITION If T is the unit tangent vector of a smooth curve in the plane, then the curvature function of the curve is
$𝜅 = ∣ 𝑑 𝐓 𝑑 𝑠 ∣ .$
If $| 𝑑 𝐓 / 𝑑 𝑠 |$ is large, T turns sharply as the particle passes through P, and the curvature at $𝑃$ is large. $I f ∣ 𝑑 𝐓 / 𝑑 𝑠 ∣$ is close to zero, T turns more slowly, and the curvature at $𝑃$ is smaller.

If a smooth curve r( ) is already given in terms of some parameter t t other than the arc length parameter s, we can calculate the curvature as
$𝜅 = ∣ 𝑑 𝐓 𝑑 𝑠 ∣ = ∣ 𝑑 𝐓 𝑑 𝑡 𝑑 𝑡 𝑑 𝑠 ∣ C h a i n R u l e$ $= 1 | 𝑑 𝑠 / 𝑑 𝑡 | ∣ 𝑑 𝐓 𝑑 𝑡 ∣$ $= 1 | 𝐯 | ∣ 𝑑 𝐓 𝑑 𝑡 ∣ . 𝑑 𝑠 𝑑 𝑡 = | 𝐯 |$

Formula for Calculating Curvature

If $𝐫 (𝑡)$ is a smooth curve in the plane, then the curvature is the scalar function

𝜅 = 1 | 𝐯 | ∣ 𝑑 𝐓 𝑑 𝑡 ∣, (1)

where $𝐓 = 𝐯 / | 𝐯 |$ is the unit tangent vector.

Testing the definition, we see in Examples 1 and 2 below that the curvature is constant for straight lines and circles.

EXAMPLE 1 A straight line is parametrized by $𝐫 (𝑡) = 𝐂 + 𝑡 𝐯$ for constant vectors C and v. Thus, $𝐫' (𝑡) = 𝐯$ , and the unit tangent vector $𝐓 = 𝐯 / | 𝐯 |$ is a constant vector that always points in the same direction and has derivative 0 (Figure 12.18). It follows that, for any value of the parameter t, the curvature of the straight line is zero:

𝜅 = 1 | 𝐯 | ∣ 𝑑 𝐓 𝑑 𝑡 ∣ = 1 | 𝐯 | | 𝟎 | = 0 .

EXAMPLE 2 Here we find the curvature of a circle. We begin with the parametrization

𝐫 (𝑡) = (𝑎 c o s 𝑡) 𝐢 + (𝑎 s i n 𝑡) 𝐣

of a circle of radius a. Then

𝐯 = 𝑑 𝐫 𝑑 𝑡 = - (𝑎 s i n 𝑡) 𝐢 + (𝑎 c o s 𝑡) 𝐣

| 𝐯 | = \sqrt (- 𝑎 s i n 𝑡) 2 + (𝑎 c o s 𝑡) 2 = \sqrt 𝑎 2 = | 𝑎 | = 𝑎 . S i n c e 𝑎 > 0, | 𝑎 | = 𝑎 .

From this we find

𝐓 = 𝐯 | 𝐯 | = - (s i n 𝑡) 𝐢 + (c o s 𝑡) 𝐣

𝑑 𝐓 𝑑 𝑡 = - (c o s 𝑡) 𝐢 - (s i n 𝑡) 𝐣

∣ 𝑑 𝐓 𝑑 𝑡 ∣ = \sqrt c o s 2 𝑡 + s i n 2 𝑡 = 1 .

Hence, for any value of the parameter t, the curvature of the circle is

𝜅 = 1 | 𝐯 | ∣ 𝑑 𝐓 𝑑 𝑡 ∣ = 1 𝑎 (1) = 1 𝑎 = 1 r a d i u s .

Among the vectors orthogonal to the unit tangent vector T, there is one of particular significance because it points in the direction in which the curve is turning. Since T has constant length (because its length is always 1), the derivative $𝑑 𝐓 / 𝑑 𝑠$ is orthogonal to T (Equation 4, Section 12.1). Therefore, if we divide $𝑑 𝐓 / 𝑑 𝑠$ by its length $𝜅,$ we obtain a unit vector N orthogonal to T (Figure 12.19).

DEFINITION At a point where $𝜅 \neq 0$ , the principal unit normal vector for a smooth curve in the plane is

FIGURE 12.19 The vector dT ds, normal to the curve, always points in the direction in which T is turning. The unit normal vector N is the direction of dT ds.

教材插图

$𝐍 = 1 𝜅 𝑑 𝐓 𝑑 𝑠 .$

The vector $𝑑 𝐓 / 𝑑 𝑠$ points in the direction in which T turns as the curve bends. Therefore, if we face in the direction of increasing arc length, the vector $𝑑 𝐓 / 𝑑 𝑠$ points toward the right if T turns clockwise and toward the left if T turns counterclockwise. In other words, the principal normal vector N will point toward the concave side of the curve (Figure 12.19).

If a smooth curve r( ) is already given in terms of some parameter t t other than the arc length parameter s, we can use the Chain Rule to calculate N directly:

𝐍 = 𝑑 𝐓 / 𝑑 𝑠 | 𝑑 𝐓 / 𝑑 𝑠 | = (𝑑 𝐓 / 𝑑 𝑡) (𝑑 𝑡 / 𝑑 𝑠) | 𝑑 𝐓 / 𝑑 𝑡 | | 𝑑 𝑡 / 𝑑 𝑠 | = 𝑑 𝐓 / 𝑑 𝑡 | 𝑑 𝐓 / 𝑑 𝑡 | . 𝑑 𝑡 𝑑 𝑠 = 1 𝑑 𝑠 / 𝑑 𝑡 > 0 c a n c e l s .

This formula enables us to find N without having to find κ and s first.

Formula for Calculating N

If r( ) is a smooth curve in the plane, then the principal unit normal ist

𝐍 = 𝑑 𝐓 / 𝑑 𝑡 | 𝑑 𝐓 / 𝑑 𝑡 |, (2)

where $𝐓 = 𝐯 / | 𝐯 |$ is the unit tangent vector.

EXAMPLE 3 Find T and N for the circular motion

𝐫 (𝑡) = (c o s 2 𝑡) 𝐢 + (s i n 2 𝑡) 𝐣 .

Solution We first find T:

𝐯 = - (2 s i n 2 𝑡) 𝐢 + (2 c o s 2 𝑡) 𝐣

| 𝐯 | = \sqrt 4 s i n 2 2 𝑡 + 4 c o s 2 2 𝑡 = 2

𝐓 = 𝐯 | 𝐯 | = - (s i n 2 𝑡) 𝐢 + (c o s 2 𝑡) 𝐣 .

From this we find

𝑑 𝐓 𝑑 𝑡 = - (2 c o s 2 𝑡) 𝐢 - (2 s i n 2 𝑡) 𝐣

FIGURE 12.20 The center of the osculating circle at $𝑃 (𝑥, 𝑦)$ lies toward the inner side of the curve.

∣ 𝑑 𝐓 𝑑 𝑡 ∣ = \sqrt 4 c o s 2 2 𝑡 + 4 s i n 2 2 𝑡 = 2

and

𝐍 = 𝑑 𝐓 / 𝑑 𝑡 | 𝑑 𝐓 / 𝑑 𝑡 | = - (c o s 2 𝑡) 𝐢 - (s i n 2 𝑡) 𝐣 . (E q . (2))

Notice that $𝐓 \cdot 𝐍 = 0 .$ , verifying that N is orthogonal to T. Notice too, that for the circular motion here, N points from r( ) toward the circle’s center at the origin.t ■

Circle of Curvature for Plane Curves

教材插图

The circle of curvature or osculating circle at a point P on a plane curve where $𝜅 \neq 0$ is the circle in the plane of the curve that

is tangent to the curve at $𝑃$ (has the same tangent line the curve has)
has the same curvature the curve has at P
has center that lies toward the concave or inner side of the curve (as in Figure 12.20).

The radius of curvature of the curve at $𝑃$ is the radius of the circle of curvature, which, according to Example 2, is

R a d i u s o f c u r v a t u r e = 𝜌 = 1 𝜅 .

To find $𝜌,$ we find κ and take the reciprocal. The center of curvature of the curve at $𝑃$ is the center of the circle of curvature.

EXAMPLE 4 Find and graph the osculating circle of the parabola $𝑦 = 𝑥 2$ at the origin.

Solution We parametrize the parabola using the parameter $𝑡 = 𝑥$ (Section 10.1, Example 5):

𝐫 (𝑡) = 𝑡 𝐢 + 𝑡 2 𝐣 .

First we find the curvature of the parabola at the origin, using Equation (1):

𝐯 = 𝑑 𝐫 𝑑 𝑡 = 𝐢 + 2 𝑡 𝐣

教材插图

FIGURE 12.21 The osculating circle for the parabola $𝑦 = 𝑥 2$ at the origin (Example 4).

教材插图

FIGURE 12.22 The helix

| 𝐯 | = \sqrt 1 + 4 𝑡 2

𝐫 (𝑡) = (𝑎 c o s 𝑡) 𝐢 + (𝑎 s i n 𝑡) 𝐣 + 𝑏 𝑡 𝐤,

drawn with a and b positive and $𝑡 \geq 0$ (Example 5).

so that

𝐓 = 𝐯 | 𝐯 | = (1 + 4 𝑡 2) - 1 / 2 𝐢 + 2 𝑡 (1 + 4 𝑡 2) - 1 / 2 𝐣 .

From this we find

𝑑 𝐓 𝑑 𝑡 = - 4 𝑡 (1 + 4 𝑡 2) - 3 / 2 𝐢 + [2 (1 + 4 𝑡 2) - 1 / 2 - 8 𝑡 2 (1 + 4 𝑡 2) - 3 / 2] 𝐣 .

At the origin, $𝑡 = 0$ , so the curvature is

𝜅 (0) = 1 | 𝐯 (0) | ∣ 𝑑 𝐓 𝑑 𝑡 (0) ∣ = 1 \sqrt 1 | 0 𝐢 + 2 𝐣 | = (1) \sqrt 02 + 22 = 2 . (E q . (1))

Therefore, the radius of curvature is $1 / 𝜅 = 1 / 2$ . At the origin we have $𝑡 = 0$ and $𝐓 = 𝐢,$ so $𝐍 = 𝐣 .$ Thus the center of the circle is $(0, 1 / 2)$ . The equation of the osculating circle is

(𝑥 - 0) 2 + (𝑦 - 12) 2 = (12) 2 .

You can see from Figure 12.21 that the osculating circle is a better approximation to the parabola at the origin than is the tangent line approximation $𝑦 = 0$

Curvature and Normal Vectors for Space Curves

If a smooth curve in space is specified by the position vector $𝐫 (𝑡)$ as a function of some parameter t, and if s is the arc length parameter of the curve, then the unit tangent vector T is $𝑑 𝐫 / 𝑑 𝑠 = 𝐯 / | 𝐯 |$ . The curvature in space is then defined to be

𝜅 = ∣ 𝑑 𝐓 𝑑 𝑠 ∣ = 1 | 𝐯 | ∣ 𝑑 𝐓 𝑑 𝑡 ∣ (3)

just as for plane curves. The vector $𝑑 𝐓 / 𝑑 𝑠$ is orthogonal to T, and we define the principal unit normal to be

𝐍 = 1 𝜅 𝑑 𝐓 𝑑 𝑠 = 𝑑 𝐓 / 𝑑 𝑡 | 𝑑 𝐓 / 𝑑 𝑡 | . (4)

EXAMPLE 5 Find the curvature for the helix (Figure 12.22)

𝐫 (𝑡) = (𝑎 c o s 𝑡) 𝐢 + (𝑎 s i n 𝑡) 𝐣 + 𝑏 𝑡 𝐤, 𝑎, 𝑏 \geq 0, 𝑎 2 + 𝑏 2 \neq 0 .

Solution We calculate T from the velocity vector v:

𝐯 = - (𝑎 s i n 𝑡) 𝐢 + (𝑎 c o s 𝑡) 𝐣 + 𝑏 𝐤

| 𝐯 | = \sqrt 𝑎 2 s i n 2 𝑡 + 𝑎 2 c o s 2 𝑡 + 𝑏 2 = \sqrt 𝑎 2 + 𝑏 2

𝐓 = 𝐯 | 𝐯 | = 1 \sqrt 𝑎 2 + 𝑏 2 [- (𝑎 s i n 𝑡) 𝐢 + (𝑎 c o s 𝑡) 𝐣 + 𝑏 𝐤] .

Then we use Equation (3):

𝜅 = 1 | 𝐯 | ∣ 𝑑 𝐓 𝑑 𝑡 ∣ = 1 \sqrt 𝑎 2 + 𝑏 2 ∣ 1 \sqrt 𝑎 2 + 𝑏 2 [- (𝑎 c o s 𝑡) 𝐢 - (𝑎 s i n 𝑡) 𝐣] ∣ = 𝑎 𝑎 2 + 𝑏 2 | - (c o s 𝑡) 𝐢 - (s i n 𝑡) 𝐣 | = 𝑎 𝑎 2 + 𝑏 2 \sqrt (c o s 𝑡) 2 + (s i n 𝑡) 2 = 𝑎 𝑎 2 + 𝑏 2 .

From this equation, we see that increasing b for a fixed a decreases the curvature. Decreasing a for a fixed b eventually decreases the curvature as well.

If $𝑏 = 0,$ , the helix reduces to a circle of radius a, and its curvature reduces to $1 / 𝑎,$ asit should. $I f 𝑎 = 0$ , the helix becomes the $𝑧 - a x i s .$ , and its curvature reduces to 0, again as itshould. 一

EXAMPLE 6 Find N for the helix in Example 5 and describe how the vector is pointing.

Solution We have

𝑑 𝐓 𝑑 𝑡 = - 1 \sqrt 𝑎 2 + 𝑏 2 [(𝑎 c o s 𝑡) 𝐢 + (𝑎 s i n 𝑡) 𝐣]

Example 5

∣ 𝑑 𝐓 𝑑 𝑡 ∣ = 1 \sqrt 𝑎 2 + 𝑏 2 \sqrt 𝑎 2 c o s 2 𝑡 + 𝑎 2 s i n 2 𝑡 = 𝑎 \sqrt 𝑎 2 + 𝑏 2

𝐍 = 𝑑 𝐓 / 𝑑 𝑡 | 𝑑 𝐓 / 𝑑 𝑡 | = - \sqrt 𝑎 2 + 𝑏 2 𝑎 \cdot 1 \sqrt 𝑎 2 + 𝑏 2 [(𝑎 c o s 𝑡) 𝐢 + (𝑎 s i n 𝑡) 𝐣] = - (c o s 𝑡) 𝐢 - (s i n 𝑡) 𝐣 .

Eq. (4)

Thus, N is parallel to the xy-plane and always points toward the z-axis.

EXERCISES 12.4

Plane Curves

Find T, N, and κ for the plane curves in Exercises 1–4.

r i j ( ) ln cos , 2 2 t t t t = + − < < ( ) π π
r i j ( ) ln sec , 2 t t t t = + − < < ( ) π π 2
$𝐫 (𝑡) = (2 𝑡 + 3) 𝐢 + (5 - 𝑡 2) 𝐣$
$𝐫 (𝑡) = (c o s 𝑡 + 𝑡 s i n 𝑡) 𝐢 + (s i n 𝑡 - 𝑡 c o s 𝑡) 𝐣, 𝑡 > 0$
A formula for the curvature of the graph of a function in the xy-plane

a. The graph $𝑦 = 𝑓 (𝑥)$ in the xy-plane automatically has the parametrization $𝑥 = 𝑥, 𝑦 = 𝑓 (𝑥)$ , and the vector formula $𝐫 (𝑥) = 𝑥 𝐢 + 𝑓 (𝑥) 𝐣$ . Use this formula to show that if f is a twice-differentiable function of $𝑥,$ then

𝜅 (𝑥) = | 𝑓'' (𝑥) | [1 + (𝑓' (𝑥)) 2] 3 / 2 .

b. Use the formula for κ in part (a) to find the curvature of $𝑦 = l n (c o s 𝑥), - 𝜋 / 2 < 𝑥 < 𝜋 / 2$ . Compare your answer with the answer in Exercise 1.

c. Show that the curvature is zero at a point of inflection.

A formula for the curvature of a parametrized plane curve

a. Show that the curvature of a smooth curve

$𝐫 (𝑡) = 𝑓 (𝑡) 𝐢 + 𝑔 (𝑡) 𝐣$ defined by twice-differentiable functions $𝑥 = 𝑓 (𝑡)$ and $𝑦 = 𝑔 (𝑡)$ is given by the formula

𝜅 = | 𝑥' 𝑦'' - 𝑦' 𝑥'' | [(𝑥') 2 + (𝑦') 2] 3 / 2 .

Apply this formula to find the curvatures of the following curves.

b. $𝐫 (𝑡) = 𝑡 𝐢 + (l n s i n 𝑡) 𝐣, 0 < 𝑡 < 𝜋$

c. r i j( ) arctan sinh ln cosht t t= +[ ] ( )( )

Normals to plane curves

a. Show that $𝐧 (𝑡) = - 𝑔' (𝑡) 𝐢 + 𝑓' (𝑡) 𝐣$ and $- 𝐧 (𝑡) = 𝑔' (𝑡) 𝐢 - 𝑓' (𝑡) 𝐣$ are both normal to the curve $𝐫 (𝑡) = 𝑓 (𝑡) 𝐢 + 𝑔 (𝑡) 𝐣$ at the point $(𝑓 (𝑡), 𝑔 (𝑡))$

To obtain N for a particular plane curve, we can choose the one of n or −n from part (a) that points toward the concave side of the curve, and make it into a unit vector. (See Figure 12.19.) Apply this method to find N for the following curves.

b. $𝐫 (𝑡) = 𝑡 𝐢 + 𝑒 2 𝑡 𝐣$

𝐫 (𝑡) = \sqrt 4 - 𝑡 2 𝐢 + 𝑡 𝐣, - 2 \leq 𝑡 \leq 2

(Continuation of Exercise 7)

a. Use the method of Exercise 7 to find N for the curve $𝐫 (𝑡) = 𝑡 𝐢 + (1 / 3) 𝑡 3 𝐣$ when $𝑡 < 0;$ when t > 0.

b. Calculate N for $𝑡 \neq 0$ directly from T using Equation (4) for the curve in part (a). Does N exist at t = 0? Graph the curve and explain what is happening to N as t passes from negative to positive values.

Space Curves

Find T, N, and κ for the space curves in Exercises 9–16.

𝐫 (𝑡) = (3 s i n 𝑡) 𝐢 + (3 c o s 𝑡) 𝐣 + 4 𝑡 𝐤

𝟏 𝟎 . 𝐫 (𝑡) = (c o s 𝑡 + 𝑡 s i n 𝑡) 𝐢 + (s i n 𝑡 - 𝑡 c o s 𝑡) 𝐣 + 3 𝐤

$𝐫 (𝑡) = (𝑒 𝑡 c o s 𝑡) 𝐢 + (𝑒 𝑡 s i n 𝑡) 𝐣 + 2 𝐤$
r i j ( ) 6 sin 2 6 cos 2 5 t t t t = + + ( ) ( ) k
$𝐫 (𝑡) = (𝑡 3 / 3) 𝐢 + (𝑡 2 / 2) 𝐣 + 𝐤, 𝑡 > 0$
$𝐫 (𝑡) = (c o s 3 𝑡) 𝐣 + (s i n 3 𝑡) 𝐤, 0 < 𝑡 < 𝜋 / 2$

𝟏 𝟓 . 𝐫 (𝑡) = 𝑡 𝐢 + (𝑎 c o s h (𝑡 / 𝑎)) 𝐤, 𝑎 > 0

$𝐫 (𝑡) = (c o s h 𝑡) 𝐢 - (s i n h 𝑡) 𝐣 + 𝑡 𝐤$

More on Curvature

Show that the parabola $𝑦 = 𝑎 𝑥 2, 𝑎 \neq 0 .$ , has its largest curvature at its vertex and has no minimum curvature. (Note: Since the curvature of a curve remains the same if the curve is translated or rotated, this result is true for any parabola.)
Show that the ellipse $𝑥 = 𝑎 c o s 𝑡, 𝑦 = 𝑏 s i n 𝑡, 𝑎 > 𝑏 > 0$ , has its largest curvature on its major axis and its smallest curvature on its minor axis. (The same is true for any ellipse.)
Maximizing the curvature of a helix In Example 5, we found the curvature of the helix $𝐫 (𝑡) = (𝑎 c o s 𝑡) 𝐢 + (𝑎 s i n 𝑡) 𝐣 + 𝑏 𝑡 𝐤$ $(𝑎, 𝑏 \geq 0)$ to be $𝜅 = 𝑎 / (𝑎 2 + 𝑏 2)$ . What is the largest value κ can have for a given value of b? Give reasons for your answer.
Total curvature We find the total curvature of the portion of a smooth curve that runs from $𝑠 = 𝑠 0 t o 𝑠 = 𝑠 1 > 𝑠 0$ by integrating κ from $𝑠 0$ to s . If the curve has some other parameter, say t, then the total curvature is

𝐾 = \int 𝑠 1 𝑠 0 𝜅 𝑑 𝑠 = \int 𝑡 1 𝑡 0 𝜅 𝑑 𝑠 𝑑 𝑡 𝑑 𝑡 = \int 𝑡 1 𝑡 0 𝜅 | 𝐯 | 𝑑 𝑡,

where $𝑡 0$ and $𝑡 1$ correspond to $𝑠 0$ and $𝑠 1$ . Find the total curvatures of a. The portion of the helix $𝐫 (𝑡) = (3 c o s 𝑡) 𝐢 + (3 s i n 𝑡) 𝐣 + 𝑡 𝐤,$ 0 4 . ≤ ≤t π

b. The parabola $𝑦 = 𝑥 2, - \infty < 𝑥 < \infty .$

Find an equation for the circle of curvature of the curve $𝐫 (𝑡) = 𝑡 𝐢 + (s i n 𝑡) 𝐣$ at the point $(𝜋 / 2, 1)$ . (The curve parametrizes the graph of $𝑦 = s i n 𝑥$ in the xy-plane.)
Find an equation for the circle of curvature of the curve $𝐫 (𝑡) = (2 l n 𝑡) 𝐢 - [𝑡 + (1 / 𝑡)] 𝐣, 𝑒 - 2 \leq 𝑡 \leq 𝑒 2,$ at the point (0, 2 ,− ) where t = 1.

The formulaT

𝜅 (𝑥) = | 𝑓'' (𝑥) | [1 + (𝑓' (𝑥)) 2] 3 / 2,

derived in Exercise 5, expresses the curvature $𝜅 (𝑥)$ of a twicedifferentiable plane curve $𝑦 = 𝑓 (𝑥)$ as a function of x. Find the curvature function of each of the curves in Exercises 23–26. Then graph f ( ) together with x κ( ) x over the given interval. You will find some surprises.

$𝑦 = 𝑥 2, - 2 \leq 𝑥 \leq 22 4 . 𝑦 = 𝑥 4 / 4, - 2 \leq 𝑥 \leq 2$
$𝑦 = s i n 𝑥, 0 \leq 𝑥 \leq 2 𝜋$
$𝑦 = 𝑒 𝑥, - 1 \leq 𝑥 \leq 2$

In Exercises 27 and 28, determine the maximum curvature for the graph of each function.

$𝑓 (𝑥) = l n 𝑥$
$𝑓 (𝑥) = 𝑥 𝑥 + 1 f o r 𝑥 > - 1$
Osculating circle Show that the center of the osculating circle for the parabola $𝑦 = 𝑥 2$ at the point $(𝑎, 𝑎 2)$ is located at $(- 4 𝑎 3, 3 𝑎 2 + 12) .$
Osculating circle Find a parametrization of the osculating circle for the parabola $𝑦 = 𝑥 2 w h e n 𝑥 = 1$

COMPUTER EXPLORATIONS

In Exercises 31–38 you will use $a C A S$ to explore the osculating circle at a point P on a plane curve where $𝜅 \neq 0$ . Use a CAS to perform the following steps:

a. Plot the plane curve given in parametric or function form over the specified interval to see what it looks like.

b. Calculate the curvature κ of the curve at the given value $𝑡 0$ using the appropriate formula from Exercise 5 or 6. Use the parametrization x = t and $𝑦 = 𝑓 (𝑡)$ if the curve is given as a function $𝑦 = 𝑓 (𝑥)$ .

c. Find the unit normal vector N at $𝑡 0 .$ . Notice that the signs of the components of N depend on whether the unit tangent vector T is turning clockwise or counterclockwise at $𝑡 = 𝑡 0$ . (See Exercise 7.)

d. If $𝐂 = 𝑎 𝐢 + 𝑏 𝐣$ is the vector from the origin to the center (a b, ) of the osculating circle, find the center C from the vector equation

𝐂 = 𝐫 (𝑡 0) + 1 𝜅 (𝑡 0) 𝐍 (𝑡 0) .

The point $𝑃 (𝑥 0, 𝑦 0)$ on the curve is given by the position vector $𝐫 (𝑡 0)$

e. Plot implicitly the equation $(𝑥 - 𝑎) 2 + (𝑦 - 𝑏) 2 = 1 / 𝜅 2$ of the osculating circle. Then plot the curve and osculating circle together. You may need to experiment with the size of the viewing window, but be sure the axes are equally scaled.

$𝐫 (𝑡) = (3 c o s 𝑡) 𝐢 + (5 s i n 𝑡) 𝐣, 0 \leq 𝑡 \leq 2 𝜋, 𝑡 0 = 𝜋 / 4$

𝟑 𝟐 . 𝐫 (𝑡) = (c o s 3 𝑡) 𝐢 + (s i n 3 𝑡) 𝐣, 0 \leq 𝑡 \leq 2 𝜋, 𝑡 0 = 𝜋 / 4

$𝐫 (𝑡) = 𝑡 2 𝐢 + (𝑡 3 - 3 𝑡) 𝐣, - 4 \leq 𝑡 \leq 4, 𝑡 0 = 3 / 5$
$𝐫 (𝑡) = (𝑡 3 - 2 𝑡 2 - 𝑡) 𝐢 + 3 𝑡 \sqrt 1 + 𝑡 2 𝐣, - 2 \leq 𝑡 \leq 5, 𝑡 0 = 1$

35 . 𝐫 (𝑡) = (2 𝑡 - s i n 𝑡) 𝐢 + (2 - 2 c o s 𝑡) 𝐣, 0 \leq 𝑡 \leq 3 𝜋, 𝑡 0 = 3 𝜋 / 2

𝟑 𝟔 . 𝐫 (𝑡) = (𝑒 - 𝑡 c o s 𝑡) 𝐢 + (𝑒 - 𝑡 s i n 𝑡) 𝐣, 0 \leq 𝑡 \leq 6 𝜋, 𝑡 0 = 𝜋 / 4 37 . $ $ 𝑦 = 𝑥 2 - 𝑥, - 2 \leq 𝑥 \leq 5, 𝑥 0 = 1 $ $ 𝟑 𝟖 . 𝑦 = 𝑥 (1 - 𝑥) 2 / 5, - 1 \leq 𝑥 \leq 2, 𝑥 0 = 1 / 2

12.5 Tangential and Normal Components of Acceleration

教材插图

FIGURE 12.23 The TNB frame of mutually orthogonal unit vectors traveling along a curve in space.

教材插图

FIGURE 12.24 The vectors T, N, and B (in that order) make a right-handed frame of mutually orthogonal unit vectors in space.

教材插图

FIGURE 12.25 The tangential and normal components of acceleration. The acceleration a always lies in the plane of T and N and is orthogonal to B.

If you are flying in an airplane that is traveling along a curve in space, the Cartesian i, j, and k coordinate system for representing the vectors describing your motion may not be very relevant to you. Vectors that are likely to be more important are those representing your forward direction (the unit tangent vector T) and the direction in which your path is turning (the unit normal vector N), along with a third unit vector perpendicular to the other two. Expressing the acceleration vector along the curve as a linear combination of these three mutually orthogonal unit vectors traveling with the motion (Figure 12.23) can reveal much about the nature of your path and your motion along it.

The TNB Frame

The binormal vector of a curve in space is $𝐁 = 𝐓 \times 𝐍,$ which is a unit vector that is orthogonal to both T and N (Figure 12.24). Together T, N, and B define a moving righthanded vector frame that plays a significant role in analyzing the paths of particles moving through space. It is called the Frenet (“fre-nay”) frame (after Jean-Frédéric Frenet, 1816–1900), or the TNB frame.

Tangential and Normal Components of Acceleration

When an object is accelerated by gravity, brakes, or rocket motors, we often need to know how much of the acceleration acts in the direction of motion, which is the direction of the tangent vector T. We can calculate this using the Chain Rule to rewrite v as

𝐯 = 𝑑 𝐫 𝑑 𝑡 = 𝑑 𝐫 𝑑 𝑠 𝑑 𝑠 𝑑 𝑡 = 𝐓 𝑑 𝑠 𝑑 𝑡 .

Then we differentiate both ends of this string of equalities to get

𝐚 = 𝑑 𝐯 𝑑 𝑡 = 𝑑 𝑑 𝑡 (𝐓 𝑑 𝑠 𝑑 𝑡) = 𝑑 2 𝑠 𝑑 𝑡 2 𝐓 + 𝑑 𝑠 𝑑 𝑡 𝑑 𝐓 𝑑 𝑡 = 𝑑 2 𝑠 𝑑 𝑡 2 𝐓 + 𝑑 𝑠 𝑑 𝑡 (𝑑 𝐓 𝑑 𝑠 𝑑 𝑠 𝑑 𝑡) = 𝑑 2 𝑠 𝑑 𝑡 2 𝐓 + 𝑑 𝑠 𝑑 𝑡 (𝜅 𝐍 𝑑 𝑠 𝑑 𝑡) 𝑑 𝐓 𝑑 𝑠 = 𝜅 𝐍 = 𝑑 2 𝑠 𝑑 𝑡 2 𝐓 + 𝜅 (𝑑 𝑠 𝑑 𝑡) 2 𝐍 .

DEFINITION If the acceleration vector is written as
$𝐚 = 𝑎 T 𝐓 + 𝑎 N 𝐍, (1)$
then
$𝑎 T = 𝑑 2 𝑠 𝑑 𝑡 2 = 𝑑 𝑑 𝑡 | 𝐯 | a n d 𝑎 N = 𝜅 (𝑑 𝑠 𝑑 𝑡) 2 = 𝜅 | 𝐯 | 2 (2)$
are the tangential and the normal scalar components of acceleration.

Notice that the binormal vector B does not appear in Equation (1). No matter how the path of the moving object we are watching may appear to twist and turn in space, the acceleration a always lies in the plane of T and N and therefore is orthogonal to B. The equation also tells us exactly how much of the acceleration takes place tangent to the motion $(𝑑 2 𝑠 / 𝑑 𝑡 2)$ and how much takes place normal to the motion $⌈ 𝜅 (𝑑 𝑠 / 𝑑 𝑡) 2 ⌉$ (Figure 12.25).

What information can we discover from Equations (2)? By definition, acceleration a is the rate of change of velocity v, and in general, both the length and direction of v change as an object moves along its path. The tangential component of acceleration $𝑎 T$ measures the rate of change of the length of v (that is, the change in the speed). The normal component of acceleration $𝑎 N$ is proportional to the rate of change of the direction of v.

教材插图

FIGURE 12.26 The tangential and normal components of the acceleration of an object that is speeding up as it moves counterclockwise around a circle of radius $𝜌 .$

教材插图

FIGURE 12.27 The tangential and normal components of the acceleration of the motion r i( ) cos sint t t t= + +( ) (sin cos , for t t t− )j t > 0. If a string wound around a fixed circle is unwound while held taut in the plane of the circle, its end P traces an involute of the circle (Example 1).

Notice that the normal scalar component of the acceleration is the curvature times the square of the speed. This explains why you have to hold on when your car makes a sharp (large κ), high-speed (large v ) turn. If you double the speed of your car, you will experience four times the normal component of acceleration for the same curvature.

If an object moves in a circle at a constant speed, $𝑑 2 𝑠 / 𝑑 𝑡 2$ is zero and all the acceleration points along N toward the circle’s center. If the object is speeding up or slowing down, a has a positive or negative tangential component (Figure 12.26).

To calculate $\boldsymbol𝑎N,$ , we usually use the formula $𝑎 N = \sqrt | 𝐚 | 2 - 𝑎 T 2$ , which comes from solving the equation $| 𝐚 | 2 = 𝐚 \cdot 𝐚 = 𝑎 T 2 + 𝑎 N 2$ for $𝑎 N$ (which, unlike the tangential component, cannot be negative). With this formula, we can find $𝑎 N$ without having to calculate κ first.

Formula for Calculating the Normal Component of Acceleration

𝑎 N = \sqrt | 𝐚 | 2 - 𝑎 2 T (3)

EXAMPLE 1 Without finding T and N, write the acceleration of the motion

𝐫 (𝑡) = (c o s 𝑡 + 𝑡 s i n 𝑡) 𝐢 + (s i n 𝑡 - 𝑡 c o s 𝑡) 𝐣, 𝑡 > 0

in the form $𝐚 = 𝑎 T 𝐓 + 𝑎 N 𝐍 .$ (The path of the motion is the involute of the circle in Figure 12.27. See also Section 12.3, Exercise 19.)

Solution We use the first of Equations (2) to find $𝑎 T$ :

𝐯 = 𝑑 𝐫 𝑑 𝑡 = (- s i n 𝑡 + s i n 𝑡 + 𝑡 c o s 𝑡) 𝐢 + (c o s 𝑡 - c o s 𝑡 + 𝑡 s i n 𝑡) 𝐣 = (𝑡 c o s 𝑡) 𝐢 + (𝑡 s i n 𝑡) 𝐣 | 𝐯 | = \sqrt 𝑡 2 c o s 2 𝑡 + 𝑡 2 s i n 2 𝑡 = \sqrt 𝑡 2 = | 𝑡 | = 𝑡 𝑎 T = 𝑑 𝑑 𝑡 | 𝐯 | = 𝑑 𝑑 𝑡 (𝑡) = 1 . (t > 0)

Eq. (2)

Knowing $\boldsymbol𝑎T,$ , we use Equation (3) to find $𝑎 N :$

𝐚 = (c o s 𝑡 - 𝑡 s i n 𝑡) 𝐢 + (s i n 𝑡 + 𝑡 c o s 𝑡) 𝐣 | 𝐚 | 2 = 𝑡 2 + 1 𝑎 N = \sqrt | 𝐚 | 2 - 𝑎 2 T = \sqrt (𝑡 2 + 1) - (1) = \sqrt 𝑡 2 = 𝑡 . (A f t e r s o m e a l g e b r a)

We then use Equation (1) to write

𝐚 = 𝑎 T 𝐓 + 𝑎 N 𝐍 = (1) 𝐓 + (𝑡) 𝐍 = 𝐓 + 𝑡 𝐍 .

Torsion

How does $𝑑 𝐁 / 𝑑 𝑠$ behave in relation to T, N, and B? From the rule for differentiating a cross product in Section 12.1, we have

𝑑 𝐁 𝑑 𝑠 = 𝑑 (𝐓 \times 𝐍) 𝑑 𝑠 = 𝑑 𝐓 𝑑 𝑠 \times 𝐍 + 𝐓 \times 𝑑 𝐍 𝑑 𝑠 .

Since N is the direction of $𝑑 𝐓 / 𝑑 𝑠, (𝑑 𝐓 / 𝑑 𝑠) \times 𝐍 = 𝟎$ and

𝑑 𝐁 𝑑 𝑠 = 𝟎 + 𝐓 \times 𝑑 𝐍 𝑑 𝑠 = 𝐓 \times 𝑑 𝐍 𝑑 𝑠 .

From this we see that $𝑑 𝐁 / 𝑑 𝑠$ is orthogonal to $𝐓,$ since a cross product is orthogonal to its factors.

Since $𝑑 𝐁 / 𝑑 𝑠$ is also orthogonal to B (the latter has constant length), it follows that $𝑑 𝐁 / 𝑑 𝑠$ is orthogonal to the plane of B and T. In other words, $𝑑 𝐁 / 𝑑 𝑠$ is parallel to $𝐍,$ so $𝑑 𝐁 / 𝑑 𝑠$ is a scalar multiple of N. In symbols,

𝑑 𝐁 𝑑 𝑠 = - 𝜏 𝐍 .

The negative sign in this equation is traditional. The scalar τ is called the torsion along the curve. Notice that

𝑑 𝐁 𝑑 𝑠 \cdot 𝐍 = - 𝜏 𝐍 \cdot 𝐍 = - 𝜏 (1) = - 𝜏 .

We use this equation for our next definition.

DEFINITION Let $𝐁 = 𝐓 \times 𝐍 .$ . The torsion function of a smooth curve is
$\boldsymbol𝜏=−𝑑𝐁𝑑𝑠⋅𝐍.(4)$

教材插图

FIGURE 12.28 The names of the three planes determined by T, N, and B.

Unlike the curvature $𝜅,$ which is never negative, the torsion τ may be positive, negative, or zero.

The three planes determined by T, N, and B are named and shown in Figure 12.28. The curvature $𝜅 = | 𝑑 𝐓 / 𝑑 𝑠 |$ can be thought of as the rate at which the normal plane turns as the point P moves along its path. Similarly, the torsion $\boldsymbol𝜏 = −(𝑑𝐁/𝑑𝑠)$ ⋅ N is the rate at which the osculating plane turns about T as P moves along the curve. Torsion measures how the curve twists.

Look at Figure 12.29. If P is a train climbing up a curved track, the rate at which the headlight turns from side to side per unit distance is the curvature of the track. The rate at which the engine tends to twist out of the plane formed by T and N is the torsion. It can be shown that a space curve is a helix if and only if it has constant nonzero curvature and constant nonzero torsion.

教材插图

FIGURE 12.29 Every moving body travels with a TNB frame that characterizes the geometry of its path of motion.

Formulas for Computing Curvature and Torsion

We now give easy-to-use formulas for computing the curvature and torsion of a smooth curve. From Equations (1) and (2), we have

𝐯 \times 𝐚 = (𝑑 𝑠 𝑑 𝑡 𝐓) \times [𝑑 2 𝑠 𝑑 𝑡 2 𝐓 + 𝜅 (𝑑 𝑠 𝑑 𝑡) 2 𝐍] 𝐯 = 𝑑 𝑟 / 𝑑 𝑡 = (𝑑 𝑠 / 𝑑 𝑡) 𝐓 = (𝑑 𝑠 𝑑 𝑡 𝑑 2 𝑠 𝑑 𝑡 2) (𝐓 \times 𝐓) + 𝜅 (𝑑 𝑠 𝑑 𝑡) 3 (𝐓 \times 𝐍) = 𝜅 (𝑑 𝑠 𝑑 𝑡) 3 𝐁 . 𝐓 \times 𝐓 = 0 a n d 𝐓 \times 𝐍 = 𝐁

It follows that

| 𝐯 \times 𝐚 | = 𝜅 ∣ 𝑑 𝑠 𝑑 𝑡 ∣ 3 | 𝐁 | = 𝜅 | 𝐯 | 3 . 𝑑 𝑠 𝑑 𝑡 = | 𝐯 | a n d | 𝐁 | = 1

Solving for κ gives the following formula.

Vector Formula for Curvature

𝜅 = | 𝐯 \times 𝐚 | | 𝐯 | 3 (5)

Equation (5) calculates the curvature, a geometric property of the curve, from the velocity and acceleration of any vector representation of the curve in which v is different from zero. From any formula for motion along a curve, no matter how variable the motion may be (as long as v is never zero), we can calculate a geometric property of the curve that seems to have nothing to do with the way the curve is parametrically defined.

The most widely used formula for torsion, derived in more advanced texts, is given in a determinant form.

Formula for Torsion

(6)

Newton’s Dot Notation for Derivatives The dots in Equation (6) denote differentiation with respect to t, one derivative _for _each _dot. _Thus, x _(“x _dot”) _means $𝑑 𝑥 / 𝑑 𝑡, ¨ 𝑥$ (“x double dot”) means $𝑑 2 𝑥 / 𝑑 𝑡 2$ _and x _(“x _triple _dot”) _means $𝑑 3 𝑥 / 𝑑 𝑡 3$ Similarly, $˙ 𝑦 = 𝑑 𝑦 / 𝑑 𝑡$ , and so on.

𝜏 = ∣ ∣ ∣ ∣ ˙ 𝑥 ˙ 𝑦 ˙ 𝑧 ¨ 𝑥 ¨ 𝑦 ¨ 𝑧 ⃛ 𝑥 ⃛ 𝑦 ⃛ 𝑧 ∣ ∣ ∣ ∣ | 𝐯 \times 𝐚 | 2 (i f 𝐯 \times 𝐚 \neq 𝟎)

This formula calculates the torsion directly from the derivatives of the component functions $𝑥 = 𝑓 (𝑡), 𝑦 = 𝑔 (𝑡), 𝑧 = ℎ (𝑡)$ that make up r. The determinant’s first row comes from v, the second row comes from a, and the third row comes from $˙ 𝐚 = 𝑑 𝐚 / 𝑑 𝑡$ . This formula for torsion is traditionally written using Newton’s dot notation for derivatives.

EXAMPLE 2 Use Equations (5) and (6) to find the curvature κ and torsion τ for the helix

𝐫 (𝑡) = (𝑎 c o s 𝑡) 𝐢 + (𝑎 s i n 𝑡) 𝐣 + 𝑏 𝑡 𝐤, 𝑎, 𝑏 \geq 0, 𝑎 2 + 𝑏 2 \neq 0 .

Solution We calculate the curvature with Equation (5):

𝐯 = - (𝑎 s i n 𝑡) 𝐢 + (𝑎 c o s 𝑡) 𝐣 + 𝑏 𝐤 𝐚 = - (𝑎 c o s 𝑡) 𝐢 - (𝑎 s i n 𝑡) 𝐣 𝐯 \times 𝐚 = ∣ ∣ ∣ ∣ 𝐢 𝐣 𝐤 - 𝑎 s i n 𝑡 𝑎 c o s 𝑡 𝑏 - 𝑎 c o s 𝑡 - 𝑎 s i n 𝑡 0 ∣ ∣ ∣ ∣ = (𝑎 𝑏 s i n 𝑡) 𝐢 - (𝑎 𝑏 c o s 𝑡) 𝐣 + 𝑎 2 𝐤 𝜅 = | 𝐯 \times 𝐚 | | 𝐯 | 3 = \sqrt 𝑎 2 𝑏 2 + 𝑎 4 (𝑎 2 + 𝑏 2) 3 / 2 = 𝑎 \sqrt 𝑎 2 + 𝑏 2 (𝑎 2 + 𝑏 2) 3 / 2 = 𝑎 𝑎 2 + 𝑏 2 . (7)

Notice that Equation (7) agrees with the result in Example 5 in Section 12.4, where we calculated the curvature directly from its definition.

To evaluate Equation (6) for the torsion, we find the entries in the determinant by differentiating r with respect to t. We already have v and a, and

˙ 𝐚 = 𝑑 𝐚 𝑑 𝑡 = (𝑎 s i n 𝑡) 𝐢 - (𝑎 c o s 𝑡) 𝐣 .

Hence,

𝜏 = ∣ ∣ ∣ ∣ ∣ ˙ 𝑥 ˙ 𝑦 ˙ 𝑧 ¨ 𝑥 ¨ 𝑦 ¨ 𝑧 ⃛ 𝑥 ⃛ 𝑦 ⃛ 𝑧 ∣ ∣ ∣ ∣ ∣ | 𝐯 \times 𝐚 | 2 = ∣ ∣ ∣ ∣ ∣ - 𝑎 s i n 𝑡 𝑎 c o s 𝑡 𝑏 - 𝑎 c o s 𝑡 - 𝑎 s i n 𝑡 0 𝑎 s i n 𝑡 - 𝑎 c o s 𝑡 0 ∣ ∣ ∣ ∣ ∣ (𝑎 \sqrt 𝑎 2 + 𝑏 2) 2 = 𝑏 (𝑎 2 c o s 2 𝑡 + 𝑎 2 s i n 2 𝑡) 𝑎 2 (𝑎 2 + 𝑏 2) = 𝑏 𝑎 2 + 𝑏 2 . (7)

From this last equation we see that the torsion of a helix about a circular cylinder is constant. In fact, constant curvature and constant torsion characterize the helix among all curves in space.

Computation Formulas for Curves in Space

Unit tangent vector:

𝐓 = 𝐯 | 𝐯 |

Principal unit normal vector:

𝐍 = 𝑑 𝐓 / 𝑑 𝑡 | 𝑑 𝐓 / 𝑑 𝑡 |

Binormal vector:

𝐁 = 𝐓 \times 𝐍

Curvature:

𝜅 = ∣ 𝑑 𝐓 𝑑 𝑠 ∣ = | 𝐯 \times 𝐚 | | 𝐯 | 3

Torsion:

𝜏 = - 𝑑 𝐁 𝑑 𝑠 \cdot 𝐍 = ∣ ∣ ∣ ∣ ˙ 𝑥 ˙ 𝑦 ˙ 𝑧 ¨ 𝑥 ¨ 𝑦 ¨ 𝑧 ⃛ 𝑥 ⃛ 𝑦 ⃛ 𝑧 ∣ ∣ ∣ ∣ | 𝐯 \times 𝐚 | 2

Tangential and normal scalar

𝐚 = 𝑎 T 𝐓 + 𝑎 N 𝐍

components of acceleration:

𝑎 T = 𝑑 𝑑 𝑡 | 𝐯 |

𝑎 N = 𝜅 | 𝐯 | 2 = \sqrt | 𝐚 | 2 - 𝑎 2 T

Exercises 12.5

Finding Tangential and Normal Components

In Exercises 1–4, write a in the form $𝐚 = 𝑎 T 𝐓 + 𝑎 N 𝐍$ without finding T and N.

𝟏 . 𝐫 (𝑡) = (2 𝑡 + 3) 𝐢 + (𝑡 2 - 1) 𝐣

𝟐 . 𝐫 (𝑡) = 3 𝑡 2 𝐢 + 2 𝑡 3 𝐣

𝟑 . 𝐫 (𝑡) = (𝑎 c o s 𝑡) 𝐢 + (𝑎 s i n 𝑡) 𝐣 + 𝑏 𝑡 𝐤

𝐫 (𝑡) = (1 + 3 𝑡) 𝐢 + (𝑡 - 2) 𝐣 - 3 𝑡 𝐤

In Exercises 5–10, write a in the form $𝐚 = 𝑎 T 𝐓 + 𝑎 N 𝐍$ at the given value of t without finding T and N.

𝟓 . 𝐫 (𝑡) = 𝑒 𝑡 𝐢 + 𝑒 - 𝑡 𝐣, 𝑡 = 0

𝟔 . 𝐫 (𝑡) = c o s (𝑡 2) 𝐢 + s i n (𝑡 2) 𝐣, 𝑡 = 12 \sqrt 𝜋

𝟕 . 𝐫 (𝑡) = (𝑡 + 1) 𝐢 + 2 𝑡 𝐣 + 𝑡 2 𝐤, 𝑡 = 1

𝟖 . 𝐫 (𝑡) = (𝑡 c o s 𝑡) 𝐢 + (𝑡 s i n 𝑡) 𝐣 + 𝑡 2 𝐤, 𝑡 = 0

𝐫 (𝑡) = 𝑡 2 𝐢 + (𝑡 + (1 / 3) 𝑡 3) 𝐣 + (𝑡 - (1 / 3) 𝑡 3) 𝐤, 𝑡 = 0

𝟏 𝟎 . 𝐫 (𝑡) = (𝑒 𝑡 c o s 𝑡) 𝐢 + (𝑒 𝑡 s i n 𝑡) 𝐣 + \sqrt 2 𝑒 𝑡 𝐤, 𝑡 = 0

Finding the TNB Frame

In Exercises 11 and 12, find r, T, N, and B at the given value of t. Then find equations for the osculating, normal, and rectifying planes at that value of t.

$𝐫 (𝑡) = (c o s 𝑡) 𝐢 + (s i n 𝑡) 𝐣 - 𝐤, 𝑡 = 𝜋 / 4$

𝟏 𝟐 . 𝐫 (𝑡) = (c o s 𝑡) 𝐢 + (s i n 𝑡) 𝐣 + 𝑡 𝐤, 𝑡 = 0

In Exercises 9–16 of Section 12.4, you found T, N, and κ. Now, in the following Exercises 13–20, find B and τ for these space curves.

$𝐫 (𝑡) = (3 s i n 𝑡) 𝐢 + (3 c o s 𝑡) 𝐣 + 4 𝑡 𝐤$
$𝐫 (𝑡) = (c o s 𝑡 + 𝑡 s i n 𝑡) 𝐢 + (s i n 𝑡 - 𝑡 c o s 𝑡) 𝐣 + 3 𝐤$

𝟏 𝟓 . 𝐫 (𝑡) = (𝑒 𝑡 c o s 𝑡) 𝐢 + (𝑒 𝑡 s i n 𝑡) 𝐣 + 2 𝐤

𝟏 𝟔 . 𝐫 (𝑡) = (6 s i n 2 𝑡) 𝐢 + (6 c o s 2 𝑡) 𝐣 + 5 𝑡 𝐤

𝟏 𝟕 . 𝐫 (𝑡) = (𝑡 3 / 3) 𝐢 + (𝑡 2 / 2) 𝐣 + 𝐤, 𝑡 > 0

$𝐫 (𝑡) = (c o s 3 𝑡) 𝐣 + (s i n 3 𝑡) 𝐤, 0 < 𝑡 < 𝜋 / 2$

𝟏 𝟗 . 𝐫 (𝑡) = 𝑡 𝐢 + (𝑎 c o s h (𝑡 / 𝑎)) 𝐤, 𝑎 > 0

𝟐 𝟎 . 𝐫 (𝑡) = (c o s h 𝑡) 𝐢 - (s i n h 𝑡) 𝐣 + 𝑡 𝐤

Physical Applications

The speedometer on your car reads a steady 35 km/h. Could you be accelerating? Explain.
Can anything be said about the acceleration of a particle that is moving at a constant speed? Give reasons for your answer.
Can anything be said about the speed of a particle whose acceleration is always orthogonal to its velocity? Give reasons for your answer.
An object of mass m travels along the parabola $𝑦 = 𝑥 2$ with a constant speed of 10 units s. What is the force on the object due to its acceleration at $(0, 0) ℧ ˙ a t (2 𝑑 1 / 2, 2) ↕$ Write your answers in terms of i and j. (Remember Newton’s law, F = ma.)

Theory and Examples

Show that κ and τ are both zero for the line

𝐫 (𝑡) = (𝑥 0 + 𝐴 𝑡) 𝐢 + (𝑦 0 + 𝐵 𝑡) 𝐣 + (𝑧 0 + 𝐶 𝑡) 𝐤 .

Show that a moving particle will move in a straight line if the normal component of its acceleration is zero.
A sometime shortcut to curvature If you already know $| 𝑎 N |$ and v , then the formula $𝑎 N = 𝜅 | 𝐯 | 2$ gives a convenient way to find the curvature. Use it to find the curvature and radius of curvature of the curve

𝐫 (𝑡) = (c o s 𝑡 + 𝑡 s i n 𝑡) 𝐢 + (s i n 𝑡 - 𝑡 c o s 𝑡) 𝐣, 𝑡 > 0 .

(Take $𝑎 N$ and v from Example 1.)

What can be said about the torsion of a smooth plane curve $𝐫 (𝑡) = 𝑓 (𝑡) 𝐢 + 𝑔 (𝑡) 𝐣 ?$ Give reasons for your answer.
Differentiable curves with zero torsion lie in planes That a sufficiently differentiable curve with zero torsion lies in a plane is a special case of the fact that a particle whose velocity remains perpendicular to a fixed vector C moves in a plane perpendicular to C. This, in turn, can be viewed as the following result.

Suppose $𝐫 (𝑡) = 𝑓 (𝑡) 𝐢 + 𝑔 (𝑡) 𝐣 + ℎ (𝑡) 𝐥$ k is twice differentiable for all t in an interval $[𝑎, 𝑏],$ that $𝐫 = 0$ when $𝑡 = 𝑎,$ and that $𝐯 \cdot 𝐤 = 0$ for all $𝑡 s i n [𝑎, 𝑏]$ . Show that $ℎ (𝑡) = 0$ for all $𝑡 i n [𝑎, 𝑏] .$ (Hint: Start with $𝐚 = ¯ 𝚪 𝑑 2 𝐫 / 𝑑 𝑡 2$ and apply the initial conditions in reverse order.)

A formula that calculates τ from B and v If we start with the definition $\boldsymbol𝜏 = −(𝑑𝐁/𝑑𝑠)$ ⋅ N and apply the Chain Rule to rewrite dB ds as

𝑑 𝐁 𝑑 𝑠 = 𝑑 𝐁 𝑑 𝑡 𝑑 𝑡 𝑑 𝑠 = 𝑑 𝐁 𝑑 𝑡 1 | 𝐯 |,

we arrive at the formula

𝜏 = - 1 | 𝐯 | (𝑑 𝐁 𝑑 𝑡 \cdot 𝐍) .

Use the formula to find the torsion of the helix in Example 2.

COMPUTER EXPLORATIONS

Rounding the answers to four decimal places, use a CAS to find v, a, speed, T, N, B, κ τ, , and the tangential and normal components of acceleration for the curves in Exercises 31–34 at the given values of t.

$𝐫 (𝑡) = (𝑡 c o s 𝑡) 𝐢 + (𝑡 s i n 𝑡) 𝐣 + 𝑡 𝐤, 𝑡 = \sqrt 3$
$𝐫 (𝑡) = (𝑒 𝑡 c o s 𝑡) 𝐢 + (𝑒 𝑡 s i n 𝑡) 𝐣 + 𝑒 𝑡 𝐤, 𝑡 = l n 2$
r i j k ( ) sin 1 cos , 3 t t t t t t = − + − + − = − ( ) ( ) π
$𝐫 (𝑡) = (3 𝑡 - 𝑡 2) 𝐢 + (3 𝑡 2) 𝐣 + (3 𝑡 + 𝑡 3) 𝐤, 𝑡 = 1$

Velocity and Acceleration in Polar Coordinates

教材插图

FIGURE 12.30 The length of r is the positive polar coordinate r of the point P. Thus $𝐮 𝑟,$ which is r r , is also $𝐫 / 𝑟$ Equations (1) express u and $𝐮 𝜃$ in terms of i and j.

教材插图

FIGURE 12.31 In polar coordinates, the velocity vector is $𝐯 = ˙ 𝑟 𝐮 𝑟 + 𝑟 ˙ 𝜃 𝐮 𝜃$

教材插图

FIGURE 12.32 Position vector and basic unit vectors in cylindrical coordinates. Notice that $| 𝐫 | \neq 𝑟 i f 𝑧 \neq 0$ since $| 𝐫 | = \sqrt 𝑟 2 + 𝑧 2$

In this section we derive equations for velocity and acceleration in polar coordinates. These equations are useful for calculating the paths of planets and satellites in space, and we use them to examine Kepler’s three laws of planetary motion.

Motion in Polar and Cylindrical Coordinates

When a particle at $𝑃 (𝑟, 𝜃)$ moves along a curve in the polar coordinate plane, we express its position, velocity, and acceleration in terms of the moving unit vectors

𝐮 𝑟 = (c o s 𝜃) 𝐢 + (s i n 𝜃) 𝐣, 𝐮 𝜃 = - (s i n 𝜃) 𝐢 + (c o s 𝜃) 𝐣, (1)

shown in Figure 12.30. The vector u points along the position vector $⟶ 𝑂 𝑃$ , so $𝐫 = 𝑟 𝐮 𝑟$ The vector $𝐮 𝜃,$ , orthogonal to u , points in the direction of increasing θ.

We find from Equations (1) that

𝑑 𝐮 𝑟 𝑑 𝜃 = - (s i n 𝜃) 𝐢 + (c o s 𝜃) 𝐣 = 𝐮 𝜃 𝑑 𝐮 𝜃 𝑑 𝜃 = - (c o s 𝜃) 𝐢 - (s i n 𝜃) 𝐣 = - 𝐮 𝑟 .

We next differentiate $𝐮 𝑟$ and $𝐮 𝜃$ with respect to t to find how they change with time. The Chain Rule gives

˙ 𝐮 𝑟 = 𝑑 𝐮 𝑟 𝑑 𝜃 ˙ 𝜃 = ˙ 𝜃 𝐮 𝜃, ˙ 𝐮 𝜃 = 𝑑 𝐮 𝜃 𝑑 𝜃 ˙ 𝜃 = - ˙ 𝜃 𝐮 𝑟 . (2)

Hence, we can express the velocity vector in terms of $𝐮 𝑟$ and $𝐮 𝜃$ as

𝐯 = ˙ 𝐫 = 𝑑 𝑑 𝑡 (𝑟 𝐮 𝑟) = ˙ 𝑟 𝐮 𝑟 + 𝑟 ˙ 𝐮 𝑟 = ˙ 𝑟 𝐮 𝑟 + 𝑟 ˙ 𝜃 𝐮 𝜃 .

See Figure 12.31. As in the previous section, we use Newton’s dot notation for time derivatives to keep the formulas as simple as we can: u means du $𝑟 / 𝑑 𝑡, ˙ 𝜃$ means $𝑑 𝜃 / 𝑑 𝑡$ , and so on.

The acceleration is

𝐚 = ˙ 𝐯 = (¨ 𝑟 𝐮 𝑟 + ˙ 𝑟 ˙ 𝐮 𝑟) + (˙ 𝑟 ˙ 𝜃 𝐮 𝜃 + 𝑟 ¨ 𝜃 𝐮 𝜃 + 𝑟 ˙ 𝜃 ˙ 𝐮 𝜃) .

When Equations (2) are used to evaluate u and $˙ 𝐮 𝜃$ and the components are separated, the equation for acceleration in terms of u and $𝐮 𝜃$ becomes

𝐚 = (¨ 𝑟 - 𝑟 ˙ 𝜃 2) 𝐮 𝑟 + (𝑟 ¨ 𝜃 + 2 ˙ 𝑟 ˙ 𝜃) 𝐮 𝜃 .

To extend these equations of motion to space, we add zk to the right-hand side of the equation $𝐫 = 𝑟 𝐮 𝑟$ . Then, in these cylindrical coordinates, we have

Position:

𝐫 = 𝑟 𝐮 𝑟 + 𝑧 𝐤

Velocity:

𝐯 = ˙ 𝑟 𝐮 𝑟 + 𝑟 ˙ 𝜃 𝐮 𝜃 + ˙ 𝑧 𝐤 (3)

Acceleration:

𝐚 = (¨ 𝑟 - 𝑟 ˙ 𝜃 2) 𝐮 𝑟 + (𝑟 ¨ 𝜃 + 2 ˙ 𝑟 ˙ 𝜃) 𝐮 𝜃 + ¨ 𝑧 𝐤

The vectors $𝐮 𝑟, 𝐮 𝜃,$ and k make a right-handed frame (Figure 12.32) in which

𝐮 𝑟 \times 𝐮 𝜃 = 𝐤, 𝐮 𝜃 \times 𝐤 = 𝐮 𝑟, 𝐤 \times 𝐮 𝑟 = 𝐮 𝜃 .

教材插图

FIGURE 12.33 The force of gravity is directed along the line joining the centers of mass.

教材插图

FIGURE 12.34 A planet that obeys Newton’s laws of gravitation and motion travels in the plane through its sun’s center of mass perpendicular to $𝐂 = 𝐫 \times ˙ 𝐫$

教材插图

FIGURE 12.35 The line joining a planet to its sun sweeps over equal areas in equal times.

Planets Move in Planes

Newton’s law of gravitation says that if r is the radius vector from the center of a sun of mass M to the center of a planet of mass m, then the force F of the gravitational attraction between the planet and sun is

𝐅 = - 𝐺 𝑚 𝑀 | 𝐫 | 2 𝐫 | 𝐫 |

(Figure 12.33). The number G is the universal gravitational constant. If we measure mass in kilograms, force in newtons, and distance in meters, G is about $6.6738 \times 1 0 - 11 N m 2 k g - 2 .$

Combining the gravitation law with Newton’s second law, $𝐅 = 𝑚 ¨ 𝐫 .$ , for the force acting on the planet gives

𝑚 ¨ 𝐫 = - 𝐺 𝑚 𝑀 | 𝐫 | 2 𝐫 | 𝐫 |, ¨ 𝐫 = - 𝐺 𝑀 | 𝐫 | 2 𝐫 | 𝐫 | .

The planet is therefore accelerated toward the sun’s center of mass at all times.

_Since _r _is _a _scalar _multiple _of _r, _we _have

𝐫 \times ¨ 𝐫 = 𝟎 .

From this last equation,

It follows that

𝑑 𝑑 𝑡 (𝐫 \times ˙ 𝐫) = ˙ 𝐫 \times ˙ 𝐫 ⏟ 0 + 𝐫 \times ¨ 𝐫 = 𝐫 \times ¨ 𝐫 = 𝟎 .

𝐫 \times ˙ 𝐫 = 𝐂 (4)

for some constant vector C.

_Equation ₍₄₎ _tells _us _that _r _and _r _always _lie _in _a _plane _{perpendicular} _to _C. _Hence, _the planet moves in a fixed plane through the center of mass of its sun (Figure 12.34). We next see how Kepler’s laws describe the motion in a precise way.

Kepler’s First Law (Ellipse Law)

Kepler’s first law says that a planet’s path is an ellipse with its sun at one focus. The eccentricity of the ellipse is

𝑒 = 𝑟 0 𝑣 20 𝐺 𝑀 - 1, (5)

and the polar equation (see Section 10.7 Equation (5)) is

𝑟 = (1 + 𝑒) 𝑟 0 1 + 𝑒 c o s 𝜃 . (6)

Here $\boldsymbol𝑣0$ is the speed when the planet is positioned at its minimum distance $𝑟 0$ from the sun. We omit the lengthy proof. The sun’s mass M is $1.99 \times 1030 k g .$

Kepler’s Second Law (Equal Area Law)

Kepler’s second law says that the radius vector from the sun to a planet (the vector r in our model) sweeps out equal areas in equal times, as displayed in Figure 12.35. In that figure, we assume the plane of the planet is the xy-plane, so the unit vector in the direction of C is k.

We introduce polar coordinates in the plane, choosing as initial line $𝜃 = 0$ , the direction r when $| 𝐫 | = 𝑟$ is a minimum value. Then at $𝑡 = 0,$ , we have $𝑟 (0) = 𝑟 0$ being a minimum so,

˙ 𝑟 ∣ 𝑡 = 0 = 𝑑 𝑟 𝑑 𝑡 ∣ 𝑡 = 0 = 0 a n d 𝑣 0 = | 𝐯 | 𝑡 = 0 = [𝑟 ˙ 𝜃] 𝑡 = 0 . E q . (3), ˙ 𝑧 = 0

To derive Kepler’s second law, we use Equation (3) to evaluate the cross product $𝐂 = 𝐫 \times ˙ 𝐫$ from Equation (4):

𝐂 = 𝐫 \times ˙ 𝐫 = 𝐫 \times 𝐯 = 𝑟 𝐮 𝑟 \times (˙ 𝑟 𝐮 𝑟 + 𝑟 ˙ 𝜃 𝐮 𝜃) = 𝑟 ˙ 𝑟 (𝐮 𝑟 \times 𝐮 𝑟 ⏟ 0) + 𝑟 (𝑟 ˙ 𝜃) (𝐮 𝑟 \times 𝐮 𝜃 ⏟ k) = 𝑟 (𝑟 ˙ 𝜃) 𝐤 . (7)

Setting t equal to zero shows that

𝐂 = [𝑟 (𝑟 ˙ 𝜃)] 𝑡 = 0 𝐤 = 𝑟 0 𝑣 0 𝐤 .

Substituting this value for C in Equation (7) gives

𝑟 0 𝑣 0 𝐤 = 𝑟 2 ˙ 𝜃 𝐤, o r 𝑟 2 ˙ 𝜃 = 𝑟 0 𝑣 0 .

This is where the area comes in. The area differential in polar coordinates is

𝑑 𝐴 = 12 𝑟 2 𝑑 𝜃

(Section 10.5). Accordingly, dA dt has the constant value

𝑑 𝐴 𝑑 𝑡 = 12 𝑟 2 ˙ 𝜃 = 12 𝑟 0 𝑣 0 . (8)

So dA dt is constant, giving Kepler’s second law.

HISTORICAL BIOGRAPHY Johannes Kepler (1571–1630)

The German astronomer, mathematician, and physicist Johannes Kepler was the first scientist to demand physical explanations of celestial phenomena. His three laws of planetary motion, the results of a lifetime of work, changed astronomy and played a crucial role in the development of Newtonian physics and calculus.

To know more, visit the companion Website.

Kepler’s Third Law (Time–Distance Law)

The time T it takes a planet to go around its sun once is the planet’s orbital period. Kepler’s third law says that T and the orbit’s semimajor axis a are related by the equation

𝑇 2 𝑎 3 = 4 𝜋 2 𝐺 𝑀 .

Since the right-hand side of this equation is constant within a given solar system, the ratio of $𝑇 2$ to $𝑎 3$ is the same for every planet in the system.

Here is a partial derivation of Kepler’s third law. The area enclosed by the planet’s elliptical orbit is calculated as follows:

A r e a = \int 𝑇 0 𝑑 𝐴 = \int 𝑇 0 12 𝑟 0 𝑣 0 𝑑 𝑡 = 12 𝑇 𝑟 0 𝑣 0 . (E q . (8))

If b is the semiminor axis, the area of the ellipse is Qab, so

𝑇 = 2 𝜋 𝑎 𝑏 𝑟 0 𝑣 0 = 2 𝜋 𝑎 2 𝑟 0 𝑣 0 \sqrt 1 - 𝑒 2 . F o r a n y e l l i p s e, 𝑏 = 𝑎 \sqrt 1 - 𝑒 2 . (9)

It remains only to express a and e in terms of $𝑟 0, 𝑣 0, 𝐺$ , and M. Equation (5) does this for e. For a, we observe that setting $𝜃$ equal to Q in Equation (6) gives

𝑟 m a x = 𝑟 0 1 + 𝑒 1 - 𝑒 .

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Hence, from Figure 12.36,

2 𝑎 = 𝑟 0 + 𝑟 m a x = 2 𝑟 0 1 - 𝑒 = 2 𝑟 0 𝐺 𝑀 2 𝐺 𝑀 - 𝑟 0 𝑣 20 . (10)

Squaring both sides of Equation (9) and substituting the results of Equations (5) and (10) produce Kepler’s third law (Exercise 11).

FIGURE 12.36 The length of the major

axis of the ellipse is $2 𝑎 = 𝑟 0 + 𝑟 m a x$

Exercises 12.6

In Exercises 1–7, find the velocity and acceleration vectors in terms of u and $𝐮 𝜃 .$

$𝑟 = 𝜃 a n d 𝑑 𝜃 𝑑 𝑡 = 2$
$𝑟 = 1 𝜃 a n d 𝑑 𝜃 𝑑 𝑡 = 𝑡 2$
$𝑟 = 𝑎 (1 - c o s 𝜃) a n d 𝑑 𝜃 𝑑 𝑡 = 3$
$𝑟 = 𝑎 s i n 2 𝜃 a n d 𝑑 𝜃 𝑑 𝑡 = 2 𝑡$
$𝑟 = 𝑒 𝑎 𝜃 a n d 𝑑 𝜃 𝑑 𝑡 = 2$
$𝑟 = 𝑎 (1 + s i n 𝑡) a n d 𝜃 = 1 - 𝑒 - 𝑡$
r t= 2 cos 4 and θ = 2t
Type of orbit For what values of $𝑣 0$ in Equation (5) is the orbit in Equation (6) a circle? An ellipse? A parabola? A hyperbola?
Circular orbits Show that a planet in a circular orbit moves with a constant speed. (Hint: This is a consequence of one of Kepler’s laws.)
Suppose that r is the position vector of a particle moving along a plane curve and $𝑑 𝐴 / 𝑑 𝑡$ is the rate at which the vector sweeps out area. Without introducing coordinates, and assuming the necessary derivatives exist, give a geometric argument based on increments and limits for the validity of the equation

𝑑 𝐴 𝑑 𝑡 = 12 | 𝐫 \times ˙ 𝐫 | .

Kepler’s third law Complete the derivation of Kepler’s third law (the part following Equation (10)).
Do the data in the accompanying table support Kepler’s third law? Give reasons for your answer.

Planet	Semimajor axis $𝑎$ ( $1010$ m)	Period $𝑇$ (years)
Mercury	5.79	0.241
Venus	10.81	0.615
Mars	22.78	1.881
Saturn	142.70	29.457

Earth’s major axis Estimate the length of the major axis of Earth’s orbit if its orbital period is 365.256 days.
Estimate the length of the major axis of the orbit of Uranus if its orbital period is 84 years.
The eccentricity of Earth’s orbit is $𝑒 : = : 0.0167$ , so the orbit is nearly circular, with radius approximately $150 \times 106$ km. Find, in units of km $2 / s,$ the rate $𝑑 𝐴 / 𝑑 𝑡$ satisfying Kepler’s second law.
Jupiter’s orbital period Estimate the orbital period of Jupiter, assuming that $𝑎 = 77.8 \times 1010 m$
Mass of Jupiter Io is one of the moons of Jupiter. It has a semimajor axis of $0.042 \times 1010$ m and an orbital period of 1.769 days. Use these data to estimate the mass of Jupiter.
Distance from Earth to the moon The period of the moon’s rotation around Earth is $2.36055 \times 106$ s. Estimate the distance to the moon.

CHAPTER 12 Questions to Guide Your Review

State the rules for differentiating and integrating vector functions. Give examples.
How do you define and calculate the velocity, speed, direction of motion, and acceleration of a body moving along a sufficiently differentiable space curve? Give an example.
What is special about the derivatives of vector functions of constant length? Give an example.
What are the vector and parametric equations for ideal projectile motion? How do you find a projectile’s maximum height, flight time, and range? Give examples.
How do you define and calculate the length of a segment of a smooth space curve? Give an example. What mathematical assumptions are involved in the definition?
How do you measure distance along a smooth curve in space from a preselected base point? Give an example.
What is a differentiable curve’s unit tangent vector? Give an example.
Define curvature, circle of curvature (osculating circle), center of curvature, and radius of curvature for twice-differentiable curves in the plane. Give examples. What curves have zero curvature? Constant curvature?
What is a plane curve’s principal normal vector? When is it defined? Which way does it point? Give an example.
How do you define N and κ for curves in space? How are these quantities related? Give examples.
What is a curve’s binormal vector? Give an example. How is this vector related to the curve’s torsion? Give an example.
What formulas are available for writing a moving object’s acceleration as a sum of its tangential and normal components? Give an example. Why might one want to write the acceleration this way? What if the object moves at a constant speed? At a constant speed around a circle?
State Kepler’s laws.

CHAPTER 12 Practice Exercises

Motion in the Plane

In Exercises 1 and 2, graph the curves and sketch their velocity and acceleration vectors at the given values of t. Then write a in the form $𝐚 = 𝑎 T 𝐓 + 𝑎 N 𝐍$ without finding T and N, and find the value of κ at the given values of t.

r i j ( ) 4 cos 2 sin , 0 and 4 t t t t = + = ( ) ( ) π
r i j ( ) 3 sec 3 tan , 0 t t t t = + = ( ) ( )
The position of a particle in the plane at time t is

𝐫 = 1 \sqrt 1 + 𝑡 2 𝐢 + 𝑡 \sqrt 1 + 𝑡 2 𝐣 .

Find the particle’s highest speed.

Suppose $𝐫 (𝑡) = (𝑒 𝑡 c o s 𝑡) 𝐢 + (𝑒 𝑡 s i n 𝑡) 𝐣$ . Show that the angle between r and a never changes. What is the angle?
Finding curvature At point P, the velocity and acceleration of a particle moving in the plane are $𝐯 = 3 𝐢 + 4 𝐣$ and $𝐚 = 5 𝐢 + 15 𝐣 .$ Find the curvature of the particle’s path at P.
Find the point on the curve $𝑦 = 𝑒 𝑥$ where the curvature is greatest.
A particle moves around the unit circle in the xy-plane. Its position at time t is $𝐫 = 𝑥 𝐢 + 𝑦 𝐣 .$ , where x and y are differentiable functions of t. Find $𝑑 𝑦 / 𝑑 𝑡 i f 𝐯 \cdot 𝐢 = \nabla 𝑦 .$ Is the motion clockwise or counterclockwise?
You send a message through a pneumatic tube that follows the curve $9 𝑦 = 𝑥 3$ (distance in meters). At the point $(3, 3), 𝐯 \cdot 𝐢 = 4$ and $𝐚 \cdot 𝐢 = - 2$ . Find the values of v j⋅ and a j⋅ at (3, 3 .)
Characterizing circular motion A particle moves in the plane so that its velocity and position vectors are always orthogonal. Show that the particle moves in a circle centered at the origin.
Speed along a cycloid A circular wheel with radius 1 m and center C rolls to the right along the x-axis at a half-turn per second. (See the accompanying figure.) At time t seconds, the position vector of the point P on the wheel’s circumference is

𝐫 = (𝜋 𝑡 - s i n 𝜋 𝑡) 𝐢 + (1 - c o s 𝜋 𝑡) 𝐣 .

a. Sketch the curve traced by P during the interval $0 \leq 𝑡 \leq 3$

b. Find v and a at $𝑡 = 0, 1, 2 .$ , and 3 and add these vectors to your sketch.

c. At any given time, what is the forward speed of the topmost point of the wheel? Of C?

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Projectile Motion

Shot put A shot leaves the thrower’s hand 2 m above the ground at $a 4 5 \circ$ angle at 14 m s. Where is it 3 s later?
Javelin A javelin leaves the thrower’s hand 2.5 m above the ground at $a 4 5 \circ$ angle at $24 m / s .$ . How high does it go?
A golf ball is hit with an initial speed $𝑣 0$ at an angle α to the horizontal from a point that lies at the foot of a straight-sided hill that is inclined at an angle φ to the horizontal, where

0 < 𝜙 < 𝛼 < 𝜋 2 .

Show that the ball lands at a distance

2 𝑣 20 c o s 𝛼 𝑔 c o s 2 𝜙 s i n (𝛼 - 𝜙),

measured up the face of the hill. Hence, show that the greatest range that can be achieved for a given $𝑣 0$ occurs when $𝛼 = (𝜙 / 2) + (𝜋 / 4)$ , that is, when the initial velocity vector bisects the angle between the vertical and the hill.

Javelin In Potsdam in 1988, Petra Felke of (then) East GermanyT set a women’s world record by throwing a javelin 80 m.

a. Assuming that Felke launched the javelin at $a 4 0 \circ$ angle to the horizontal 2 m above the ground, what was the javelin’s initial speed?

b. How high did the javelin go?

Motion in Space

Find the lengths of the curves in Exercises 15 and 16.

$𝐫 (𝑡) = (2 c o s 𝑡) 𝐢 + (2 s i n 𝑡) 𝐣 + 𝑡 2 𝐤, 0 \leq 𝑡 \leq 𝜋 / 4$
$𝐫 (𝑡) = (3 c o s 𝑡) 𝐢 + (3 s i n 𝑡) 𝐣 + 2 𝑡 3 / 2 𝐤, 0 \leq 𝑡 \leq 3$

In Exercises 17–20, find T, N, B, κ, and τ at the given value of t.

𝐫 (𝑡) = 49 (1 + 𝑡) 3 / 2 𝐢 + 49 (1 - 𝑡) 3 / 2 𝐣 + 13 𝑡 𝐤, 𝑡 = 0

𝟏 𝟖 . 𝐫 (𝑡) = (𝑒 𝑡 s i n 2 𝑡) 𝐢 + (𝑒 𝑡 c o s 2 𝑡) 𝐣 + 2 𝑒 𝑡 𝐤, 𝑡 = 0

$𝐫 (𝑡) = 𝑡 𝐢 + 12 𝑒 2 𝑡 𝐣, 𝑡 = l n 2$

𝟐 𝟎 . 𝐫 (𝑡) = (3 c o s h 2 𝑡) 𝐢 + (3 s i n h 2 𝑡) 𝐣 + 6 𝑡 𝐤, 𝑡 = l n 2

In Exercises 21 and 22, write a in the form a $𝚷 = 𝑎 T 𝐓 + 𝑎 N 𝐍 𝐚 𝐭 𝑡 = 𝟎$ without finding T and N.

$𝐫 (𝑡) = (2 + 3 𝑡 + 3 𝑡 2) 𝐢 + (4 𝑡 + 4 𝑡 2) 𝐣 - (6 c o s 𝑡) 𝐤$
$𝐫 (𝑡) = (2 + 𝑡) 𝐢 + (𝑡 + 2 𝑡 2) 𝐣 + (1 + 𝑡 2) 𝐤$
Find T, N, B, κ, and τ as functions of t if

𝐫 (𝑡) = (s i n 𝑡) 𝐢 + (\sqrt 2 c o s 𝑡) 𝐣 + (s i n 𝑡) 𝐤 .

At what times in the interval $0 \leq 𝑡 \leq 𝜋$ are the velocity and acceleration vectors of the motion $𝐫 (𝑡) = 𝐢 + (5 c o s 𝑡) 𝐣 + (3 s i n 𝑡) 𝐤$ orthogonal?
The position of a particle moving in space at time $𝑡 \geq 0$ is

𝐫 (𝑡) = 2 𝐢 + (4 s i n 𝑡 2) 𝐣 + (3 - 𝑡 𝜋) 𝐤 .

Find the first time r is orthogonal to the vector $𝐢 - 𝐣 .$

Find equations for the osculating, normal, and rectifying planes of the curve $𝐫 (𝑡) = 𝑡 𝐢 + 𝑡 2 𝐣 + 𝑡 3 𝐤$ at the point (1, 1, 1 .)
Find parametric equations for the line that is tangent to the curve $𝐫 (𝑡) = 𝑒 𝑡 𝐢 + (s i n 𝑡) 𝐣 + l n (1 - 𝑡) 𝐤 a t 𝑡 = 0 .$
Find parametric equations for the line that is tangent to the helix $𝐫 (𝑡) {˙ = (\sqrt 2 c o s 𝑡) ˆ 𝐢 + (\sqrt 2 s i n 𝑡) 𝐣 + 𝑡 𝐤$ at the point where $𝑡 = 𝜋 / 4$

Theory and Examples

Synchronous curves By eliminating α from the ideal projectile equations

𝑥 = (𝑣 0 c o s 𝛼) 𝑡, 𝑦 = (𝑣 0 s i n 𝛼) 𝑡 - 12 𝑔 𝑡 2,

show that $𝑥 2 + (𝑦 + 𝑔 𝑡 2 / 2) 2 = 𝑣 0 2 𝑡 2$ . This shows that projectiles launched simultaneously from the origin at the same initial speed will, at any given instant, all lie on the circle of radius $𝑣 0 𝑡$ centered at $(0, - 𝑔 𝑡 2 / 2)$ , regardless of their launch angle. These circles are the synchronous curves of the launching.

Radius of curvature Show that the radius of curvature of a twice-differentiable plane curve $𝐫 (𝑡) = 𝑓 (𝑡) 𝐢 + 𝑔 (𝑡) 𝐣$ is given by the formula

𝜌 = ˙ 𝑥 2 + ˙ 𝑦 2 \sqrt ¨ 𝑥 2 + ¨ 𝑦 2 - ¨ 𝑠 2, w h e r e ¨ 𝑠 = 𝑑 𝑑 𝑡 \sqrt ˙ 𝑥 2 + ˙ 𝑦 2 .

An alternative definition of curvature in the plane An alternative definition gives the curvature of a sufficiently differentiable plane curve to be $| 𝑑 𝜙 / 𝑑 𝑠 |$ , where φ is the angle between T and i (Figure 12.37a). Figure 12.37b shows the distance s measured counterclockwise around the circle $𝑥 2 + 𝑦 2 = 𝑎 2$ from the point $(𝑎, 0)$ to a point P, along with the angle φ at P. Calculate the circle’s curvature using the alternative definition. (Hint: $𝜙 = 𝜃 + 𝜋 / 2 .)$ 1

教材插图

FIGURE 12.37 Figures for Exercise 31.

The view from Skylab 4 What percentage of Earth’s surface area could the astronauts see when Skylab 4 was at its apogee height, 437 km above the surface? To find out, model the visible surface as the surface generated by revolving the circular arc GT, shown here, about the y-axis. Then carry out these steps:

Step 1. Use similar triangles in the figure to show that $𝑦 0 / 6380 = 6380 / (6380 + 437)$ . Solve for $𝑦 0$

Step 2. To four significant digits, calculate the visible area as

𝑉 𝐴 = \int 6380 𝑦 0 2 𝜋 𝑥 \sqrt 1 + (𝑑 𝑥 𝑑 𝑦) 2 𝑑 𝑦 .

Step 3. Express the result as a percentage of Earth’s surface area.

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CHAPTER 12 Additional and Advanced Exercises

Applications

A frictionless particle $𝑃,$ starting from rest at time $𝑡 = 0$ at the point $(𝑎, 0, 0)$ , slides down the helix

𝐫 (𝜃) = (𝑎 c o s 𝜃) 𝐢 + (𝑎 s i n 𝜃) 𝐣 + 𝑏 𝜃 𝐤 (𝑎, 𝑏 > 0)

under the influence of gravity, as in the accompanying figure. The θ in this equation is the cylindrical coordinate θ, and the helix is the curve $𝑟 = 𝑎, 𝑧 = 𝑏 𝜃, 𝜃 \geq 0 .$ , in cylindrical coordinates. We assume θ to be a differentiable function of t for the motion. The law of conservation of energy tells us that the particle’s speed after it has fallen straight down a distance z is $\sqrt 2 𝑔 𝑧$ , where $𝑔$ is the constant acceleration of gravity.

a. Find the angular velocity $𝑑 𝜃 / 𝑑 𝑡$ when $𝜃 = 2 𝜋$

b. Express the particle’s θ- and z-coordinates as functions of t.

c. Express the tangential and normal components of the velocity $𝑑 𝐫 / 𝑑 𝑡$ and acceleration $𝑑 2 𝐫 / 𝑑 𝑡 2$ as functions of t. Does the acceleration have any nonzero component in the direction of the binormal vector B?

教材插图

Suppose the curve in Exercise 1 is replaced by the conical helix $𝑟 = 𝑎 𝜃, 𝑧 = 𝑏 𝜃$ shown in the accompanying figure.

a. Express the angular velocity dθ dt as a function of θ.

b. Express the distance the particle travels along the helix as a function of θ.

教材插图

Motion in Polar and Cylindrical Coordinates

Deduce from the orbit equation

𝑟 = (1 + 𝑒) 𝑟 0 1 + 𝑒 c o s 𝜃

that a planet is closest to its sun when $𝜃 = 0,$ , and show that $𝑟 = 𝑟 0$ at that time.

A Kepler equation The problem of locating a planet in its orbitT at a given time and date eventually leads to solving “Kepler” equations of the form

𝑓 (𝑥) = 𝑥 - 1 - 12 s i n 𝑥 = 0 .

a. Show that this particular equation has a solution between $𝑥 = 0$ and $𝑥 = 2$

b. With your computer or calculator in radian mode, use Newton’s method to find the solution to as many places as you can.

In Section $12.6,$ we found the velocity of a particle moving in the plane to be

𝐯 = ˙ 𝑥 𝐢 + ˙ 𝑦 𝐣 = ˙ 𝑟 𝐮 𝑟 + 𝑟 ˙ 𝜃 𝐮 𝜃 .

_a. _Express _x _and _y _in _terms _of _r _and _rθ _by _evaluating _the _dot products v ⋅ i and $𝐯 \cdot 𝐣 .$

b. Express $˙ 𝑟$ _and _rθ _in _terms _of _x _and $˙ 𝑦$ by evaluating the dot products $𝐕 \cdot 𝐮 𝑗$ and $𝐯 \cdot 𝐮 𝜃 .$

Express the curvature of a twice-differentiable curve $𝑟 = 𝑓 (𝜃)$ in the polar coordinate plane in terms of f and its derivatives.
A slender rod through the origin of the polar coordinate plane rotates (in the plane) about the origin at the rate of 3 rad min. A beetle starting from the point (2, 0 crawls along the rod toward) the origin at the rate of 1 cm min.

a. Find the beetle’s acceleration and velocity in polar form when it is halfway to (1 cm from) the origin.

b. To the nearest millimeter, what will be the length of the pathT the beetle has traveled by the time it reaches the origin?

Arc length in cylindrical coordinates

a. Show that when you express $𝑑 𝑠 2 = 𝑑 𝑥 2 + 𝑑 𝑦 2 + 𝑑 𝑧 2$ in terms of cylindrical coordinates, you get ds dr r d dz= + +θ .2 2 2 2 2

b. Interpret this result geometrically in terms of the edges and a diagonal of a box. Sketch the box.

c. Use the result in part (a) to find the length of the curve

𝑟 = 𝑒 𝜃, 𝑧 = 𝑒 𝜃, 0 \leq 𝜃 \leq l n 8 .

Unit vectors for position and motion in cylindrical coordinates When the position of a particle moving in space is given in cylindrical coordinates, the unit vectors we use to describe its position and motion are

𝐮 𝑟 = (c o s 𝜃) 𝐢 + (s i n 𝜃) 𝐣, 𝐮 𝜃 = - (s i n 𝜃) 𝐢 + (c o s 𝜃) 𝐣,

and k (see accompanying figure). The particle’s position vector is then $𝐫 = 𝑟 𝐮 𝑟 + 𝑧 𝐤$ , where r is the positive polar distance coordinate of the particle’s position.

教材插图

a. Show that $𝐮 𝑟, 𝐮 𝜃,$ and k, in this order, form a right-handed frame of unit vectors.

b. Show that

𝑑 𝐮 𝑟 𝑑 𝜃 = 𝐮 𝜃 a n d 𝑑 𝐮 𝜃 𝑑 𝜃 = - 𝐮 𝑟 .

c. Assuming that the necessary derivatives with respect to t exist, express $𝐯 = ˙ 𝐫$ and $𝐚 = ¨ 𝐫$ in terms of $𝐮 𝑟, 𝐮 𝜃, 𝐤, ˙ 𝑟,$ _and θ_.

d. Conservation of angular momentum Let r( ) denote thet position in space of a moving object at time t. Suppose the force acting on the object at time t is

𝐅 (𝑡) = - 𝑐 | 𝐫 (𝑡) | 3 𝐫 (𝑡),

where c is a constant. In physics the angular momentum of an object at time t is defined to be $𝐋 (𝑡) = 𝐫 (𝑡) \times 𝑚 𝐯 (𝑡)$ where m is the mass of the object and v( ) is the velocity.t Prove that angular momentum is a conserved quantity; that is, prove that $𝐋 (𝑡)$ is a constant vector, independent of time. Remember Newton’s law $𝐅 = 𝑚 𝐚 .$ . (This is a calculus problem, not a physics problem.)

CHAPTER 12 Technology Application Projects

Mathematica/Maple Projects

Projects can be found within MyLab Math.

• Radar Tracking of a Moving Object

Visualize position, velocity, and acceleration vectors to analyze motion.

• Parametric and Polar Equations with a Figure Skater

Visualize position, velocity, and acceleration vectors to analyze motion.

• Moving in Three Dimensions

Compute distance traveled, speed, curvature, and torsion for motion along a space curve. Visualize and compute the tangential, normal, and binormal vectors associated with motion along a space curve.

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OVERVIEW The volume of a right circular cylinder is a function $𝑉 = 𝜋 𝑟 2 ℎ$ of its radius and its height, so it is a function $𝑉 (𝑟, ℎ)$ of two variables r and h. The speed of sound through seawater is primarily a function of salinity S and temperature T. The monthly payment on a home mortgage is a function of the principal borrowed $𝑃,$ the interest rate i, and the term t of the loan. These are examples of functions that depend on more than one independent variable. In this chapter we extend the ideas of single-variable differential calculus to functions of several variables.