Chapter 13. CALCULUS OF VECTOR-VALUED FUNCTIONS
13.1 Chapter Introduction
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In this chapter, we study vector-
13.2 13.1: Vector–Valued Functions
Consider a particle moving in Rtag whose coordinates at time t are (x(t), y(t), z(t)). It is convenient to represent the particle’s path by the vector-
Functions f(x) (with real number values) are often called scalar-
Think of r(t) as a moving vector that points from the origin to the position of the particle at time t (Figure 1).
The parameter is often called t (for time), but we are free to use any other variable such as s or θ. It is best to avoid writing r(x) or r(y) to prevent confusion with the x-
More generally, a vector-
is a set of real numbers and whose range is a set of position vectors. The variable t is called a parameter, and the functions x(t), y(t), z(t) are called the components or coordinate functions. We usually take as domain the set of all values of t for which r(t) is defined—
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The terminal point of a vector-
We have already studied special cases of vector parametrizations. In Chapter 12, we described lines in R3 using vector parametrizations. Recall that
r(t) = 〈x0, y0, z0〉 + tv = 〈x0 + ta, y0 + tb, z0 + tc〉
parametrizes the line through P = (x0, y0, z0) in the direction of the vector v = 〈a, b, c〉.
In Chapter 11, we studied parametrized curves in the plane R2 in the form
c(t) = (x(t), y(t))
Such a curve is described equally well by the vector-
It is important to distinguish between the path parametrized by r(t) and the underlying curve 
traced by r(t). The curve
is the set of all points (x(t), y(t), z(t)) as t ranges over the domain of r(t). The path is a particular way of traversing the curve; it may traverse the curve several times, reverse direction, or move back and forth, etc.
EXAMPLE 1: The Path versus the Curve
Describe the path
How are the path and the curve
traced by r(t) different?
Solution As t varies from −∞ to ∞, the endpoint of the vector r(t) moves around a unit circle at height z = 1 infinitely many times in the counterclockwise direction when viewed from above (Figure 2). The underlying curve
traced by r(t) is the circle itself.
A curve in R3 is also referred to as a space curve (as opposed to a curve in R2, which is called a plane curve). Space curves can be quite complicated and difficult to sketch by hand. The most effective way to visualize a space curve is to plot it from different viewpoints using a computer (Figure 3). As an aid to visualization, we plot a “thickened” curve as in Figures 3 and 5, but keep in mind that space curves are one-
The projections onto the coordinate planes are another aid in visualizing space curves. The projection of a path r(t) = 〈x(t), y(t), z(t)〉 onto the xy-plane is the path p(t) = 〈x(t), y(t), 0〉 (Figure 4). Similarly, the projections onto the yz- and xz-planes are the paths 〈0, y(t), z(t)〉 and 〈x(t), 0, z(t)〉, respectively.
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EXAMPLE 2: Helix
Describe the curve traced by r(t) = 〈−sin t, cos t, t〉 for t ≥ 0 in terms of its projections onto the coordinate planes.
Solution The projections are as follows (Figure 4):
xy-plane (set z = 0): the path p(t) = 〈− sin t, cos t, 0〉, which describes a point moving counterclockwise around the unit circle starting at p(0) = (0, 1, 0).
xz-plane (set y = 0): the path 〈− sin t, 0,t〉, which is a wave in the z-direction.
yz-plane (set x = 0): the path 〈0, cos t, t〉, which is a wave in the z-direction.
The function r(t) describes a point moving above the unit circle in the xy-plane while its height z = t increases linearly, resulting in the helix of Figure 4.
Every curve can be parametrized in infinitely many ways (because there are infinitely many ways that a point can traverse a curve as a function of time). The next example describes two very different parametrizations of the same curve.
EXAMPLE 3: Parametrizing the Intersection of Surfaces
Parametrize the curve
obtained as the intersection of the surfaces x2 − y2 = z − 1 and x2 + y2 = 4 (Figure 5).
Solution We have to express the coordinates (x, y, z) of a point on the curve as functions of a parameter t. Here are two ways of doing this.
First method: Solve the given equations for y and z in terms of x. First, solve for y:
The equation x2 − y2 = z − 1 can be written z = x2 − y2 + 1. Thus, we can substitute y2 = 4 − x2 to solve for z:
z = x2 − y2 + 1 = x2 − (4 − x2) + 1 = 2x2 − 3
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Now use t = x as the parameter. Then 
, z = 2t2 − 3. The two signs of the square root correspond to the two halves of the curve where y > 0 and y < 0, as shown in Figure 6. Therefore, we need two vector-
Second method: Note that x2 + y2 = 4 has a trigonometric parametrization: x = 2 cos t, y = 2 sin t for 0 ≤ t < 2π. The equation x2 − y2 = z − 1 gives us
z = x2 − y2 + 1 = 4 cos2 t − 4 sin2 t + 1 = 4 cos 2t + 1
Thus, we may parametrize the entire curve by a single vector-
EXAMPLE 4
Parametrize the circle of radius 3 with center P = (2, 6, 8) located in a plane:
Parallel to the xy-plane
Parallel to the xz-plane
Solution
Acircle of radius R in the xy-plane centered at the origin has parametrization 〈R cos t, R sin t〉. To place the circle in a three-
dimensional coordinate system, we use the parametrization 〈R cos t, R sin t, 0〉. 727
Thus, the circle of radius 3 centered at (0, 0, 0) has parametrization 〈3 cos t, 3 sin t, 0〉. To move this circle in a parallel fashion so that its center lies at P = (2, 6, 8), we translate by the vector 〈2, 6, 8〉:
r1(t) = 〈2, 6, 8〉 + 〈3 cos t, 3 sin t, 0〉 = 〈2 + 3 cos t, 6 + 3 sin t, 8〉
The parametrization 〈3 cos t, 0, 3 sin t〉 gives us a circle of radius 3 centered at the origin in the xz-plane. To move the circle in a parallel fashion so that its center lies at (2, 6, 8), we translate by the vector 〈2, 6, 8〉:
r2(t) = 〈2, 6, 8〉 + 〈3 cos t, 0, 3 sin t〉 = 〈2 + 3 cos t, 6, 8 + 3 sin t〉
These two circles are shown in Figure 7.
Horizontal and vertical circles of radius 3 and center P = (2, 6, 8) obtained by translation.
13.3 13.1: SUMMARY
A vector-
valued function is a function of the form r(t) = 〈x(t), y(t), z(t)〉 = x(t)i + y(t)j + z(t)k
We often think of t as time and r(t) as a “moving vector” whose terminal point traces out a path as a function of time. We refer to r(t) as a vector parametrization of the path, or simply as a “path.”
The underlying curve
traced by r(t) is the set of all points (x(t), y(t), z(t)) in R3 for t in the domain of r(t). A curve in R3 is also called a space curve.Every curve
can be parametrized in infinitely many ways.The projection of r(t) onto the xy-plane is the curve traced by 〈x(t), y(t), 0〉. The projection onto the xz-plane is 〈x(t), 0, z(t)〉, and the projection onto the yz-plane is 〈0, y(t), z(t)〉.
13.4 13.1: EXERCISES
Preliminary Questions
Question 13.1
Which one of the following does not parametrize a line?
r1(t) = 〈8 − t, 2t, 3t〉
r2(t) = t3i − 7t3j + t3k
r3(t) = 〈8 − 4t3, 2 + 5t2, 9t3〉
Question 13.2
What is the projection of r(t) = ti + t4j + et k onto the xz-plane?
Question 13.3
Which projection of 〈cos t, cos 2t, sin t〉 is a circle?
Question 13.4
What is the center of the circle with parametrization
r(t) = (−2 + cos t)i + 2j + (3 − sin t)k?
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Question 13.5
How do the paths r1(t) = 〈cos t, sin t〉 and r2(t) = 〈sin t, cos t〉 around the unit circle differ?
Question 13.6
Which three of the following vector-
(−2 + cos t)i + 9j + (3 − sin t)k
(2 + cos t)i − 9j + (−3 − sin t)k
(−2 + cos 3t)i + 9j + (3 − sin 3t)k
(−2 − cos t)i + 9j + (3 + sin t)k
(2 + cos t)i + 9j + (3 + sin t)k
Exercises
Question 13.7
What is the domain of 
?
Question 13.8
What is the domain of 
?
Question 13.9
Evaluate r(2) and r(−1) for 
.
Question 13.10
Does either of P = (4, 11, 20) or Q = (−1, 6, 16) lie on the path r(t) = 〈1 + t, 2 + t2, t4〉?
Question 13.11
Find a vector parametrization of the line through P = (3, −5, 7) in the direction v = 〈3, 0, 1〉.
Question 13.12
Find a direction vector for the line with parametrization 
.
Question 13.13
Match the space curves in Figure 8 with their projections onto the xy-plane in Figure 9.
Question 13.14
Match the space curves in Figure 8 with the following vector-
r1(t) = 〈cos 2t, cos t, sin t〉
r2(t) = 〈t, cos 2t, sin 2t〉
r3(t) = 〈1, t, t〉
Question 13.15
Match the vector-
r(t) = 〈t + 15, e0.08t cos t, e0.08t sin t〉
r(t) = 〈cos t, sin t, sin 12t〉
r(t) = 〈cos3 t, sin3 t, sin 2t〉
r(t) = 〈t, t2, 2t〉
r(t) = 〈cos t, sin t, cos t sin 12t〉
Question 13.16
Which of the following curves have the same projection onto the xy-plane?
r1(t) = 〈t, t2, et〉
r2(t) = 〈et, t2, t〉
r3(t) = 〈t, t2, cos t〉
Question 13.17
Match the space curves (A)–(C) in Figure 11 with their projections (i)–(iii) onto the xy-plane.
Question 13.18
Describe the projections of the circle r(t) = 〈sin t, 0, 4 + cos t〉 onto the coordinate planes.
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In Exercises 13–
Question 13.19
r(t) = (9 cos t)i + (9 sin t)j
Question 13.20
r(t) = 7i + (12 cos t)j + (12 sin t)k
Question 13.21
r(t) = 〈sin t, 0, 4 + cos t〉
Question 13.22
r(t) = 〈6 + 3 sin t, 9, 4 + 3 cos t〈
Question 13.23
Let
be the curve r(t) = 〈t cos t, t sin t, t〉.
Show that
lies on the cone x2 + y2 = z2.Sketch the cone and make a rough sketch of
on the cone.
Question 13.24
Use a computer algebra system to plot the projections onto the xy- and xz-planes of the curve r(t) = 〈t cos t, t sin t, t〉 in Exercise 17.
In Exercises 19 and 20, let
r(t) = 〈sin t, cos t, sin t cos 2t〉
as shown in Figure 12.
Question 13.25
Find the points where r(t) intersects the xy-plane.
Question 13.26
Show that the projection of r(t) onto the xz-plane is the curve
Question 13.27
Parametrize the intersection of the surfaces
using t = y as the parameter (two vector functions are needed as in Example 3).
Question 13.28
Find a parametrization of the curve in Exercise 21 using trigonometric functions.
Question 13.29
Viviani’s Curve
is the intersection of the surfaces (Figure 13)
Parametrize each of the two parts of
corresponding to x ≥ 0 and x ≤ 0, taking t = z as parameter.Describe the projection of
onto the xy-plane.Show that
lies on the sphere of radius 1 with center (0, 1, 0). This curve looks like a figure eight lying on a sphere [Figure 13(B)].
Question 13.30
Show that any point on x2 + y2 = z2 can be written in the form (z cos θ, z sin θ, z) for some θ. Use this to find a parametrization of Viviani’s curve (Exercise 23) with θ as parameter.
Question 13.31
Use sine and cosine to parametrize the intersection of the cylinders x2 + y2 = 1 and x2 + z2 = 1 (use two vector-
Question 13.32
Use hyperbolic functions to parametrize the intersection of the surfaces x2 − y2 = 4 and z = xy.
Question 13.33
Use sine and cosine to parametrize the intersection of the surfaces x2 + y2 = 1 and z = 4x2 (Figure 14).
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In Exercises 28–
Question 13.34
Which of the following statements are true?
If r1 and r2 intersect, then they collide.
If r1 and r2 collide, then they intersect.
Intersection depends only on the underlying curves traced by r1 and r2, but collision depends on the actual parametrizations.
Question 13.35
Determine whether r1 and r2 collide or intersect:
Question 13.36
Determine whether r1 and r2 collide or intersect:
In Exercises 31–
Question 13.37
The vertical line passing through the point (3, 2, 0)
Question 13.38
The line passing through (1, 0, 4) and (4, 1, 2)
Question 13.39
The line through the origin whose projection on the xy-plane is a line of slope 3 and whose projection on the yz-plane is a line of slope 5 (i.e., Δz/Δy = 5)
Question 13.40
The horizontal circle of radius 1 with center (2, −1, 4)
Question 13.41
The circle of radius 2 with center (1, 2, 5) in a plane parallel to the yz-plane
Question 13.42
The ellipse 
in the xy-plane, translated to have center (9, −4, 0)
Question 13.43
The intersection of the plane 
with the sphere x2 + y2 + z2 = 1
Question 13.44
The intersection of the surfaces
Question 13.45
The ellipse 
in the xz-plane, translated to have center (3, 1, 5) [Figure 15(A)]
Question 13.46
The ellipse 
, translated to have center (3, 1, 5) [Figure 15(B)]
Further Insights and Challenges
Question 13.47
Sketch the curve parametrized by r(t) = 〈|t|+ t, |t|− t〉.
Question 13.48
Find the maximum height above the xy-plane of a point on r(t) = 〈et, sin t, t (4 − t)〉.
Question 13.49
Let
be the curve obtained by intersecting a cylinder of radius r and a plane. Insert two spheres of radius r into the cylinder above and below the plane, and let F1 and F2 be the points where the plane is tangent to the spheres [Figure 16(A)]. Let K be the vertical distance between the equators of the two spheres. Rediscover Archimedes’s proof that
is an ellipse by showing that every point P on
satisfies
Hint: If two lines through a point P are tangent to a sphere and intersect the sphere at Q1 and Q2 as in Figure 16(B), then the segments 
and 
have equal length. Use this to show that PF1 = PR1 and PF2 = PR2.
Question 13.50
Assume that the cylinder in Figure 16 has equation x2 + y2 = r2 and the plane has equation z = ax + by. Find a vector parametrization r(t) of the curve of intersection using the trigonometric functions cos t and sin t.
Question 13.51
Now reprove the result of Exercise 43 using vector geomentry. Assume that the cylinder has equation x2 + y2 = r2 and the plane has equation z = ax + by.
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Show that the upper and lower spheres in Figure 16 have centers
Show that the points where the plane is tangent to the sphere are
Hint: Show that 
and 
have length r and are orthogonal to the plane.
Verify, with the aid of a computer algebra system, that Eq. (2) holds with
To simplify the algebra, observe that since a and b are arbitrary, it suffices to verify Eq. (2) for the point P = (r, 0, ar).
13.5 13.2: Calculus of Vector-Valued Functions
In this section, we extend differentiation and integration to vector-
The first step is to define limits of vector-
DEFINITION Limit of a Vector-Valued Function
A vector-
. In this case, we write
We can visualize the limit of a vector-
THEOREM 1: Vector-Valued Limits Are Computed Componentwise
A vector-
The Limit Laws of scalar functions remain valid in the vector-
Proof
Let u = 〈a, b, c〉 and consider the square of the length
The term on the left approaches zero if and only if each term on the right approaches zero (because these terms are nonnegative). It follows that ∥r(t − u∥) approaches zero if and only if |x(t) − a|, |y(t) − b|, and |z(t) − c| tend to zero. Therefore, r(t) approaches a limit u as t → t0 if and only if x(t), y(t), and z(t) converge to the components a, b, and c.
EXAMPLE 1
Calculate 
, where r(t) = 〈t2, 1 − t, t-1〉.
Solution By Theorem 1,
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Continuity of vector-
By Theorem 1, r(t) is continuous at t0 if and only if the components x(t), y(t), z(t) are continuous at t0.
We define the derivative of r(t) as the limit of the difference quotient:
In Leibniz notation, the derivative is written dr/dt.
We say that r(t) is differentiable at t if the limit in Eq. (3) exists. Notice that the components of the difference quotient are difference quotients:
and by Theorem 1, r(t) is differentiable if and only if the components are differentiable. In this case, r′(t) is equal to the vector of derivatives 〈x′(t), y′(t), z′(t)〉.
By Theorems 1 and 2, vector-
THEOREM 2: Vector-Valued Derivatives Are Computed Componentwise
A vector-
Here are some vector-
Higher-
EXAMPLE 2
Calculate r″(3), where r(t) = 〈ln t, t, t2〉.
Solution We perform the differentiation componentwise:
Therefore, 
.
The differentiation rules of single-
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Differential Rules
Assume that r(t), r1(t), and r2(t) are differentiable. Then
Sum Rule:
Constant Multiple Rule: For any constant c, (c r(t))′ = c r′(t).
Product Rule: For any differentiable scalar-
valued function f(t), 
Chain Rule: For any differentiable scalar-
valued function g(t), 
Proof
Each rule is proved by applying the differentiation rules to the components. For example, to prove the Product Rule (we consider vector-
f(t)r(t) = f(t) 〈x(t), y(t)〉 = 〈f(t)x(t), f(t)y(t)〉
Now apply the Product Rule to each component:
The remaining proofs are left as exercises (Exercises 69–
EXAMPLE 3
Let r(t) = 〈t2, 5t, 1〉 and f(t) = e3t. Calculate:
Solution We have r′(t) = 〈2t, 5, 0〉 and f′(t) = 3e3t.
By the Product Rule,
By the Chain Rule,
There are three different Product Rules for vector-
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THEOREM 3: Product Rule for Dot and Cross Products
Assume that r1(t) and r2(t) are differentiable. Then
CAUTION
Order is important in the Product Rule for cross products. The first term in Eq. (5) must be written as
not 
. Similarly, the second term is 
. Why is order not a concern for dot products?
Proof
We verify Eq. (4) for vector-
The proof of Eq. (5) is left as an exercise (Exercise 71).
In the next example and throughout this chapter, all vector-
EXAMPLE 4
Prove the formula 
.
Solution By the Product Formula for cross products,
Here, r′(t) × r′(t) = 0 because the cross product of a vector with itself is zero.
The Derivative as a Tangent Vector
The derivative vector r′(t0) has an important geometric property: It points in the direction tangent to the path traced by r(t) at t = t0.
To understand why, consider the difference quotient, where Δr = r(t0 + h) − r(t0) and Δt = h with h ≠ 0:
Although it has been our convention to regard all vectors as based at the origin, the tangent vector r′(t) is an exception; we visualize it as a vector based at the terminal point of r(t). This makes sense because r′(t) then appears as a vector tangent to the curve (Figure 3).
The vector Δr points from the head of r(t0) to the head of r(t0 + h) as in Figure 2(A). The difference quotient Δr/Δt is a scalar multiple of Δr and therefore points in the same direction [Figure 2(B)].
As h = Δt tends to zero, Δr also tends to zero but the quotient Δr/Δt approaches a vector r′(t0), which, if nonzero, points in the direction tangent to the curve. Figure 3 illustrates the limiting process. We refer to r′(t0) as the tangent vector or the velocity vector at r(t0).
The tangent vector r′(t0) (if it is nonzero) is a direction vector for the tangent line to the curve. Therefore, the tangent line has vector parametrization:
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EXAMPLE 5: Plotting Tangent Vectors
Plot r(t) = 〈cos t, sin t, 4 cos2 t〉 together with its tangent vectors at 
and 
. Find a parametrization of the tangent line at 
.
Solution The derivative is r′(t) = 〈− sin t, cos t, −8 cos t sin t〉, and thus the tangent vectors at 
and 
are
Figure 4 shows a plot of r(t) with 
based at 
and 
based at 
.
r(t) = 〈cos t, sin t, 4 cos2 t〉 at
.At 
and thus the tangent line is parametrized by
There are some important differences between vector-
EXAMPLE 6: Horizontal Tangent Vectors on the Cycloid
The function
r(t) = 〈t − sin t, 1 − cos t〉
traces a cycloid. Find the points where:
r′(t) is horizontal and nonzero.
r′(t) is the zero vector.
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Solution The tangent vector is r′(t) = 〈1 − cos t, sin t〉. The y-component of r′(t) is zero if sin t = 0—
By periodicity, we conclude that r′(t) is nonzero and horizontal for t = π, 3π, 5π,…and r′(t) = 0 for t = 0, 2π, 4π,…(Figure 5).
CONCEPTUAL INSIGHT
The cycloid in Figure 5 has sharp points called cusps at points where x = 0, 2π, 4π,…. If we represent the cycloid as the graph of a function y = f(x), then f′(x) does not exist at these points. By contrast, the vector derivative r′(t) = 〈1 − cos t, sin t〉 exists for all t, but r′(t) = 0 at the cusps. In general, r′(t) is a direction vector for the tangent line whenever it exists, but we get no information about the tangent line (which may or may not exist) at points where r′(t) = 0.
The next example establishes an important property of vector-
EXAMPLE 7: Orthogonality of r and r′ When r Has Constant Length
Prove that if r(t) has constant length, then r(t) is orthogonal to r′(t).
Solution By the Product Rule for Dot Products,
This derivative is zero because ∥r(t)∥ is constant. Therefore r(t) · r′(t) = 0, and r(t) is orthogonal to r′(t) [or r′(t) = 0].
GRAPHICAL INSIGHT
The result of Example 7 has a geometric explanation. A vector parametrization r(t) consisting of vectors of constant length R traces a curve on the surface of a sphere of radius R with center at the origin (Figure 6). Thus r′(t) is tangent to this sphere. But any line that is tangent to a sphere at a point P is orthogonal to the radial vector through P, and thus r(t) is orthogonal to r′(t).
Vector-Valued Integration
The integral of a vector-
The integral exists if each of the components x(t), y(t), z(t) is integrable. For example,
Vector-
An antiderivative of r(t) is a vector-
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THEOREM 4
If R1(t) and R2(t) are differentiable and 
, then
R1(t) = R2(t) + c
for some constant vector c.
The general antiderivative of r(t) is written
∫r(t) dt = R(t) + c
where c = 〈c1, c2, c3〉 is an arbitrary constant vector. For example,
Fundamental Theorem of Calculus for Vector-Valued Functions
If r(t) is continuous on [a, b], and R(t) is an antiderivative of r(t), then
EXAMPLE 8: Finding Position via Vector-Valued Differential Equations
The path of a particle satisfies
Find the particle’s location at t = 4 if r(0) = 〈4, 1〉.
Solution The general solution is obtained by integration:
The initial condition r(0) = 〈4, 1〉 gives us
r(0) = 〈2, 0〉 + c = 〈4, 1〉
Therefore, c = 〈2, 1〉 and (Figure 7)
The particle’s position at t = 4 is
13.6 13.2: SUMMARY
Limits, differentiation, and integration of vector-
valued functions are performed componentwise. Differentation rules:
Sum Rule:
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Constant Multiple Rule: (c r(t))′ = c r′(t)
Chain Rule:
Product Rules:
The derivative r′(t0) is called the tangent vector or velocity vector.
If r′(t0) is nonzero, then it points in the direction tangent to the curve at r(t0). The tangent line has vector parametrization
L(t) = r(t0) + tr′(t0)
If
, then R1(t) = R2(t) + c for some constant vector c.
The Fundamental Theorem for vector-
valued functions: If r(t) is continuous and R(t) is an antiderivative of r(t), then 
13.7 13.2: EXERCISES
Preliminary Questions
Question 13.52
State the three forms of the Product Rule for vector-
In Questions 2–
Question 13.53
The derivative of a vector-
Question 13.54
There are two Chain Rules for vector-
Question 13.55
The terms “velocity vector” and “tangent vector” for a path r(t) mean one and the same thing.
Question 13.56
The derivative of a vector-
Question 13.57
The derivative of the cross product is the cross product of the derivatives.
Question 13.58
State whether the following derivatives of vector-
Exercises
In Exercises 1–
Question 13.59
Question 13.60
Question 13.61
Question 13.62
Question 13.63
Evaluate 
for 
.
Question 13.64
Evaluate 
for r(t) = 〈sin t, 1 − cos t, −2t〉.
In Exercises 7–
Question 13.65
r(t) = 〈t, t2, t3〉
Question 13.66
Question 13.67
r(t) = 〈e3s, e−s, s4〉
Question 13.68
Question 13.69
c(t) = t−1i − e2t k
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Question 13.70
a(θ) = (cos 3θ)i + (sin2 θ)j + (tan θ)k
Question 13.71
Calculate r′(t) and r″(t) for r(t) = 〈t, t2, t3〉.
Question 13.72
Stetch the curve r(t) = 〈1 − t2, t〉 for −1 ≤ t ≤ 1. Compute the tangent vector at t = 1 and add it to the sketch.
Question 13.73
Stetch the curve r1(t) = 〈t, t2〉 together with its tangent vector at t = 1. Then do the same for r2(t) = 〈t3, t6〉.
Question 13.74
Sketch the cycloid r(t) = 〈t − sin t, 1 − cos t〉 together with its tangent vectors at 
and 
.
In Exercises 17–
Question 13.75
Question 13.76
Question 13.77
Question 13.78
, assuming that
In Exercises 21 and 22, let
Question 13.79
Compute 
in two ways:
Calculate r1(t) · r2(t) and differentiate.
Use the Product Rule.
Question 13.80
Compute 
in two ways:
Calculate r1(t) × r2(t) and differentiate.
Use the Product Rule.
In Exercises 23–
using the Chain Rule.
Question 13.81
Question 13.82
Question 13.83
Question 13.84
Question 13.85
Let r(t) = 〈t2, 1 − t, 4t〉. Calculate the derivative of r(t) · a(t) at t = 2, assuming that a(2) = 〈1, 3, 3〉 and a′(2) = 〈−1, 4, 1〉.
Question 13.86
Let v(s) = s2i + 2sj + 9s−2k. Evaluate 
at s = 4, assuming that g(4) = 3 and g′(4) = −9.
In Exercises 29–
Question 13.87
Question 13.88
Question 13.89
Question 13.90
Question 13.91
Question 13.92
Question 13.93
Use Example 4 to calculate 
, where r(t) = 〈?t, t2, et〉.
Question 13.94
Let r(t) = 〈3 cos t, 5 sin t, 4 cos t〉. Show that ∥r(t)∥ is constant and conclude, using Example 7, that r(t) and r′(t) are orthogonal. Then compute r′(t) and verify directly that r′(t) is orthogonal to r(t).
Question 13.95
Show that the derivative of the norm is not equal to the norm of the derivative by verifying that ∥r(t)∥′ ≠ ∥r(t)′∥ for r(t) = 〈t, 1, 1〉.
Question 13.96
Show that 
for any constant vector a.
In Exercises 39–
Question 13.97
Question 13.98
Question 13.99
Question 13.100
Question 13.101
Question 13.102
Question 13.103
Question 13.104
In Exercises 47–
Question 13.105
Question 13.106
Question 13.107
Question 13.108
Question 13.109
Question 13.110
Question 13.111
Question 13.112
Question 13.113
Find the location at t = 3 of a particle whose path (Figure 8) satisfies
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Question 13.114
Find the location and velocity at t = 4 of a particle whose path satisfies
Question 13.115
A fighter plane, which can shoot a laser beam straight ahead, travels along the path r(t) = 〈5 − t, 21 − t2, 3 − t3/27〉. Show that there is precisely one time t at which the pilot can hit a target located at the origin.
Question 13.116
The fighter plane of Exercise 57 travels along the path r(t) = 〈t − t3, 12 − t2, 3 − t〉. Show that the pilot cannot hit any target on x-axis.
Question 13.117
Find all solutions to r′(t) = v with initial condition r(1) = w, where v and w are constant vectors in R3.
Question 13.118
Let u be a constant vector in R3. Find the solution of the equation r′(t) = (sin t)u satisfying r′(0) = 0.
Question 13.119
Find all solutions to r′(t) = 2r(t) where r(t) is a vector-
Question 13.120
Show that w(t) = 〈sin(3t + 4), sin(3t − 2), cos 3t〉 satisfies the differential equation w″(t) = −9w(t).
Question 13.121
Prove that the Bernoulli spiral (Figure 9) with parametrization r(t) = 〈et cos 4t, et sin 4t〉 has the property that the angle ψ between the position vector and the tangent vector is constant. Find the angle ψ in degrees.
Question 13.122
A curve in polar form r = f(θ) has parametrization
r(θ) = f(θ) 〈cos θ, sin θ〉
Let ψ be the angle between the radial and tangent vectors (Figure 10). Prove that
Hint: Compute r(θ) × r′(θ) and r(θ) · r′(θ).
Question 13.123
Prove that if ∥r(t)∥ takes on a local minimum or maximum value at t0, then r(t0) is orthogonal to r′(t0). Explain how this result is related to Figure 11. Hint: Observe that if ∥r(t0)∥ is a minimum, then r(t) is tangent at t0 to the sphere of radius ∥r(t0)∥ centered at the origin.
Question 13.124
Newton’s Second Law of Motion in vector form states that 
where F is the force acting on an object of mass m and p = mr′(t) is the object’s momentum. The analogs of force and momentum for rotational motion are the torque τ = r × F and angular momentum
J = r(t) × p(t)
Use the Second Law to prove that 
.
Further Insights and Challenges
Question 13.125
Let r(t) = 〈x(t), y(t)〉 trace a plane curve
. Assume that x′(t0) ≠ 0. Show that the slope of the tangent vector r′(t0) is equal to the slope dy/dx of the curve at r(t0).
Question 13.126
Prove that 
.
Question 13.127
Verify the Sum and Product Rules for derivatives of vector-
Question 13.128
Verify the Chain Rule for vector-
Question 13.129
Verify the Product Rule for cross products [Eq. (5)].
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Question 13.130
Verify the linearity properties
Question 13.131
Prove the Substitution Rule (where g(t) is a differentiable scalar function):
Question 13.132
Prove that if ∥r(t)∥ ≤ K for t ∈ [a, b], then
13.8 13.3: Arc Length and Speed
In Section 11.2, we derived a formula for the arc length of a plane curve given in parametric form. This discussion applies to paths in three-
REMINDER
The length of a path or curve is referred to as the arc length.
Recall that arc length is defined as the limit of the lengths of polygonal approximations.
we choose a partition a = t0 < t1 < t2 <⋯< tN = b and join the terminal points of the vectors r(tj) by segments, as in Figure 1. As in Section 11.2, we find that if r′(t) exists and is continuous on [a, b], then the lengths of the polygonal approximations approach a limit L as the maximum of the widths |tj − tj−1| tends to zero. This limit is the length s of the path which is computed by the integral in the next theorem.
THEOREM 1: Length of a Path
Assume that r(t) is differentiable and that r′(t) is continuous on [a, b]. Then the length s of the path r(t) for a ≤ t ≤ b is equal to
Keep in mind that the length s in Eq. (1) is the distance traveled by a particle following the path r(t). The path length s is not equal to the length of the underlying curve unless r(t) traverses the curve only once without reversing direction.
EXAMPLE 1
Find the arc length s of r(t) = 〈cos 3t, sin 3t, 3t〉 for 0 ≤ t ≤ 2π.
Solution The derivative is r′(t) = 〈−3 sin 3t, 3 cos 3t, 3〉, and
∥r′(t)∥2 = 9 sin2 3t + 9 cos2 3t + 9 = 9(sin2 3t + cos2 3t) + 9 = 18
Therefore, 
.
Speed, by definition, is the rate of change of distance traveled with respect to time t. To calculate the speed, we define the arc length function:
Thus s(t) is the distance traveled during the time interval [a, t]. By the Fundamental Theorem of Calculus,
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Now we can see why r′(t) is known as the velocity vector (and also as the tangent vector). It points in the direction of motion, and its magnitude is the speed (Figure 2). We often denote the velocity vector by v(t) and the speed by υ(t):
EXAMPLE 2
Find the speed at time t = 2 s of a particle whose position vector is
r(t) = t3i − etj + 4tk
Solution The velocity vector is v(t) = r′(t) = 3t2i − etj + 4k, and at t = 2,
v(2) = 12i − e2j + 4k
The particle’s speed is 
.
Arc Length Parametrization
Keep in mind that a parametrization r(t) describes not just a curve, but also how a particle traverses the curve, possibly speeding up, slowing down, or reversing direction along the way. Changing the parametrization amounts to describing a different way of traversing the same underlying curve.
We have seen that parametrizations are not unique. For example, r1(t) = 〈t, t2〉 and r2(s) = 〈s3, s6〉 both parametrize the parabola y = x2. Notice in this case that r2(s) is obtained by substituting t = s3 in r1(t).
In general, we obtain a new parametrization by making a substitution t = g(s)—that is, by replacing r(t) with r1(s) = r(g(s)) [Figure 3]. If t = g(s) increases from a to b as s varies from c to d, then the path r(t) for a ≤ t ≤ b is also parametrized by r1(s) for c ≤ s ≤ d.
EXAMPLE 3
Parametrize the path r(t) = (t2, sin t, t) for 3 ≤ t ≤ 9 using the parameter s, where t = g(s) = es.
Solution Substituting t = es in r(t), we obtain the parametrization
r1(s) = r(g(s)) = 〈e2s , sin es, es〉
Because s = ln t, the parameter t varies from 3 to 9 as s varies from ln 3 to ln 9. Therefore, the path is parametrized by r1(s) for ln 3 ≤ s ≤ ln 9.
One way of parametrizing a path is to choose a starting point and “walk along the path” at unit speed (say, 1 m/s). A parametrization of this type is called an arc length parametrization [Figure 4(A)]. It is defined by the property that the speed has constant value 1:
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In an arc length parametrization, the distance traveled over any time interval [a, b] is equal to the length of the interval:
Arc length parametrizations are also called unit speed parametrizations. We will use arc length parametrizations to define curvature in Section 13.4.
To find an arc length parametrization, start with any parametrization r(t) such that r′(t) ≠ 0 for all t, and form the arc length integral
The letter s is often used as the parameter in an arc length parametrization.
Because ∥r′(t)∥ ≠ 0, s(t) is an increasing function and therefore has an inverse t = g(s). By the formula for the derivative of an inverse (and since s′(t) = ∥r′(t)∥),
REMINDER
By Theorem 1 in Section 3.8, if g(x) is the inverse of f(x), then
Now we can show that the parametrization
r1(s) = r(g(s))
is an arc length parametrization. Indeed, by the Chain Rule,
In most cases we cannot evaluate the arc length integral s(t) explicitly, and we cannot find a formula for its inverse g(s) either. So although arc length parametrizations exist in general, we can find them explicitly only in special cases.
EXAMPLE 4: Finding an Arc Length Parametrization
Find the arc length parametrization of the helix r(t) = 〈cos 4t, sin 4t, 3t〉.
Solution First, we evaluate the arc length function
Then we observe that the inverse of s(t) = 5t is t = s/5; that is, g(s) = s/5. As shown above, an arc length parametrization is
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As a check, let’s verify that r1(s) has unit speed:
13.9 13.3: SUMMARY
The length s of a path r(t) = 〈x(t), y(t), z(t)〉 for a ≤ t ≤ b is
Arc length function:
Speed is the derivative of distance traveled with respect to time:
The velocity vector v(t) = r′(t) points in the direction of motion [provided that r′(t) ≠ 0] and its magnitude υ(t) = ∥r′(t)∥ is the object’s speed.
We say that r(s) is an arc length parametrization if ∥r′(s)∥ = 1 for all s. In this case, the length of the path for a ≤ s ≤ b is b − a.
If r(t) is any parametrization such that r′(t) ≠ 0 for all t, then
r1(s) = r(g(s))
is an arc length parametrization, where t = g(s) is the inverse of the arc length function.
13.10 13.3: EXERCISES
Preliminary Questions
Question 13.133
At a given instant, a car on a roller coaster has velocity vector r′ = 〈25, −35, 10〉 (in miles per hour). What would the velocity vector be if the speed were doubled? What would it be if the car’s direction were reversed but its speed remained unchanged?
Question 13.134
Two cars travel in the same direction along the same roller coaster (at different times). Which of the following statements about their velocity vectors at a given point P on the roller coaster is/are true?
The velocity vectors are identical.
The velocity vectors point in the same direction but may have different lengths.
The velocity vectors may point in opposite directions.
Question 13.135
A mosquito flies along a parabola with speed υ(t) = t2. Let L(t) be the total distance traveled at time t.
How fast is L(t) changing at t = 2?
Is L(t) equal to the mosquito’s distance from the origin?
Question 13.136
What is the length of the path traced by r(t) for 4 ≤ t ≤ 10 if r(t) is an arc length parametrization?
Exercises
In Exercises 1–
Question 13.137
Question 13.138
Question 13.139
Question 13.140
Question 13.141
Question 13.142
. Use the formula: 
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In Exercises 7 and 8, compute the arc length function 
for the given value of a.
Question 13.143
Question 13.144
In Exercises 9–
Question 13.145
Question 13.146
Question 13.147
Question 13.148
Question 13.149
What is the velocity vector of a particle traveling to the right along the hyperbola y = x−1 with constant speed 5 cm/s when the particle’s location is 
?
Question 13.150
A bee with velocity vector r′(t) starts out at the origin at t = 0 and flies around for T seconds. Where is the bee located at time T if 
? What does the quantity 
represent?
Question 13.151
Let
Show that r(t) parametrizes a helix of radius R and height h making N complete turns.
Guess which of the two springs in Figure 5 uses more wire.
Which spring uses more wire?Compute the lengths of the two springs and compare.
Question 13.152
Use Exercise 15 to find a general formula for the length of a helix of radius R and height h that makes N complete turns.
Question 13.153
The cycloid generated by the unit circle has parametrization
r(t) = 〈t − sin t, 1 − cos t〉
Find the value of t in [0, 2π] where the speed is at a maximum.
Show that one arch of the cycloid has length 8. Recall the identity sin2(t/2) = (1 − cos t)/2.
Question 13.154
Which of the following is an arc length parametrization of a circle of radius 4 centered at the origin?
r1(t) = 〈4 sin t, 4 cos t〉
r2(t) = 〈4 sin 4t, 4 cos 4t〉
Question 13.155
Let r(t) = 〈3t + 1, 4t − 5, 2t〉.
Evaluate the arc length integral
.Find the inverse g(s) of s(t).
Verify that r1(s) = r(g(s)) is an arc length parametrization.
Question 13.156
Find an arc length parametrization of the line y = 4x + 9.
Question 13.157
Let r(t) = w + tv be the parametrization of a line.
Show that the arc length function
is given by s(t) = t∥v∥. This shows that r(t) is an arc length parametrizaton if and only if v is a unit vector.Find an arc length parametrization of the line with w = 〈1, 2, 3〉 and v = 〈3, 4, 5〉.
Question 13.158
Find an arc length parametrization of the circle in the plane z = 9 with radius 4 and center (1, 4, 9).
Question 13.159
Find a path that traces the circle in the plane y = 10 with radius 4 and center (2, 10, −3) with constant speed 8.
Question 13.160
Find an arc length parametrization of r(t) = 〈et sin t, et cos t, et 〉.
Question 13.161
Find an arc length parametrization of r(t) = 〈t2, t3〉.
Question 13.162
Find an arc length parametrization of the cycloid with parametrization r(t) = 〈t − sin t, 1 − cos t〉.
Question 13.163
Find an arc length parametrization of the line y = mx for an arbitrary slope m.
Question 13.164
Express the arc length L of y = x3 for 0 ≤ x ≤ 8 as an integral in two ways, using the parametrizations r1(t) = 〈t, t3〉 and r2(t) = 〈t3, t9〉. Do not evaluate the integrals, but use substitution to show that they yield the same result.
Question 13.165
The curve known as the Bernoulli spiral (Figure 6) has parametrization r(t) = 〈et cos 4t, et sin 4t〉.
Evaluate
. It is convenient to take lower limit −∞ because r(−∞) = 〈0, 0〉.Use (a) to find an arc length parametrization of r(t).
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Further Insights and Challenges
Question 13.166
Prove that the length of a curve as computed using the arc length integral does not depend on its parametrization. More precisely, let
be the curve traced by r(t) for a ≤ t ≤ b. Let f(s) be a differentiable function such that f′(s) > 0 and that f(c) = a and f(d) = b. Then r1(s) = r(f(s)) parametrizes
for c ≤ s ≤ d. Verify that
Question 13.167
The unit circle with the point (−1, 0) removed has parametrization (see Exercise 73 in Section 11.1)
Use this parametrization to compute the length of the unit circle as an improper integral. Hint: The expression for ∥r′(t)∥ simplifies.
Question 13.168
The involute of a circle, traced by a point at the end of a thread unwinding from a circular spool of radius R, has parametrization (see Exercise 26 in Section 12.2)
r(θ) = 〈R(cos θ + θ sin θ), R(sin θ − θ cos θ)〉
Find an arc length parametrization of the involute.
Question 13.169
The curve r(t) = 〈t − tanh t, sech t〉 is called a tractrix (see Exercise 92 in Section 11.1).
Show that
is equal to s(t) = ln(cosh t).Show that
is an inverse of s(t) and verify that
is an arc length parametrization of the tractrix.
13.11 13.4: Curvature
Curvature is a measure of how much a curve bends. It is used to study geometric properties of curves and motion along curves, and has applications in diverse areas such as roller coaster design (Figure 1), optics, eye surgery (see Exercise 60), and biochemistry (Figure 2).
In Chapter 4, we used the second derivative f″(x) to measure the bending or concavity of the graph of y = f(x), so it might seem natural to take f″(x) as our definition of curvature. However, there are two reasons why this proposed definition will not work. First, f″(x) makes sense only for a graph y = f(x) in the plane, and our goal is to define curvature for curves in three-
does not capture the curvature of the circle, which by symmetry should be the same at all points.Consider a path with parametrization r(t) = 〈x(t), y(t), z(t)〉. We assume that r′(t) = 0 for all t in the domain of r(t). A parametrization with this property is called regular. At every point P along the path there is a unit tangent vector T = TP that points in the direction of motion of the parametrization. We write T(t) for the unit tangent vector at the terminal point of r(t):
For example, if r(t) = 〈t, t2, t3〉, then r′(t) = 〈1, 2t, 3t2〉, and the unit tangent vector at P = (1, 1, 1), which is the terminal point of r(1) = 〈1, 1, 1〉, is
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If we choose another parametrization, say r1(s), then we can also view T as function of s: T(s) is the unit tangent vector at the terminal point of r1(s).
Now imagine walking along a path and observing how the unit tangent vector T changes direction (Figure 4). A change in T indicates that the path is bending, and the more rapidly T changes, the more the path bends. Thus, 
would seem to be a good measure of curvature. However, 
depends on how fast you walk (when you walk faster, the unit tangent vector changes more quickly). Therefore, we assume that you walk at unit speed. In other words, curvature is the magnitude 
, where s is the parameter of an arc length parametrization. Recall that r(s) is an arc length parametrization if ∥r(s)∥ = 1 for all s.
DEFINITION Curvature
Let r(s) be an arc length parametrization and T the unit tangent vector. The curvature at r(s) is the quantity (denoted by a lowercase Greek letter “kappa”)
Our first two examples illustrate curvature in the case of lines and circles.
EXAMPLE 1: A Line Has Zero Curvature
Compute the curvature at each point on the line r(t) = 〈x0, y0, z0〉 + tu, where ∥u∥ = 1.
Solution First, we note that because u is a unit vector, r(t) is an arc length parametrization. Indeed, r′(t) = u and thus ∥r′(t)∥ = ∥u∥ = 1. Thus we have T(t) = r′(t)/∥r′(t)∥ = r′(t) and hence T′(t) = r″(t) = 0 (because r′(t) = u is constant). As expected, the curvature is zero at all points on a line:
EXAMPLE 2: The Curvature of a Circle of Radius R Is 1/R
Compute the curvature of a circle of radius R.
Solution Assume the circle is centered at the origin, so that it has parametrization r(θ) = 〈R cos θ, R sin θ〉 (Figure 5). This is not an arc length parametrization if R ≠ 1. To find an arc length parametrization, we compute the arc length function:
Example 2 shows that a circle of large radius R has small curvature 1/R. This makes sense because your direction of motion changes slowly when you walk at unit speed along a circle of large radius.
Thus s = Rθ, and the inverse of the arc length function is θ = g(s) = s/R. In Section 13.3, we showed that r1(s) = r(g(s)) is an arc length parametrization. In our case, we obtain
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The unit tangent vector and its derivative are
By definition of curvature,
This shows that the curvature is 1/R at all points on the circle.
In practice, it is often impossible to find an arc length parametrization explicitly. Fortunately, we can compute curvature using any regular parametrization r(t). To derive a formula, we need the following two results.
First is the fact that T(t) and T′(t) are orthogonal (see the marginal note). Second, arc length s is function s(t) of time t, so the derivatives of T with respect to t and s are related by the Chain Rule. Denoting the derivative with respect to t by a prime, we have
REMINDER
To prove that T(t) and T′(t) are orthogonal, note that T(t) is a unit vector, so T(t) · T(t) = 1. Differentiate using the Product Rule for Dot Products:
This shows that T(t) · T′(t) = 0
where 
is the speed of r(t). Since curvature is the magnitude 
, we obtain
THEOREM 1: Formula for Curvature
If r(t) is a regular parametrization, then the curvature at r(t) is
To apply Eq. (3) to plane curves, replace r(t) = 〈x(t), y(t)〉 by r(t) = 〈x(t), y(t), 0〉 and compute the cross product.
Proof
Since υ(t) = ∥r′(t)∥, we have r′(t) = υ(t)T(t). By the Product Rule,
r″(t) = υ′(t)T(t) + υ(t)T′(t)
Now compute the following cross product, using the fact that T(t) × T(t) = 0:
Because T(t) and T′(t) are orthogonal,
REMINDER
By Theorem 1 in Section 12.4,
∥v × w∥ = ∥v∥ ∥w∥ sin θ
where θ is the angle between v and w.
Eq. (4) yields ∥r′(t) × r″(t)∥ = υ(t)2∥T′(t)∥. Using Eq. (2), we obtain
∥r′(t) × r″(t)∥ = υ(t)2∥T′(t)∥ = υ(t)3κ(t) = ∥r′(t)∥3κ(t)
This yields the desired formula.
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EXAMPLE 3: Twisted Cubic Curve
Calculate the curvature κ(t) of the twisted cubic r(t) = 〈t, t2, t3〉. Then plot the graph of κ(t) and determine where the curvature is largest.
Solution The derivatives are
The parametrization is regular because r′(t) ≠ 0 for all t, so we may use Eq. (3):
The graph of κ(t) in Figure 6 shows that the curvature is largest at t = 0. The curve r(t) is illustrated in Figure 7. The plot is colored by curvature, with large curvature represented in blue, small curvature in green.
In the second paragraph of this section, we pointed out that the curvature of a graph y = f(x) must involve more than just the second derivative f″(x). We now show that the curvature can be expressed in terms of both f″(x) and f′(x).
THEOREM 2: Curvature of a Graph in the Plane
The curvature at the point (x, f (x)) on the graph of y = f(x) is equal to
Proof
The curve y = f(x) has parametrization r(x) = 〈x, f(x)〉. Therefore, r′(x) = 〈1, f′(x)〉 and r″(x) = 〈0, f″(x)〉. To apply Theorem 1, we treat r′(x) and r″(x) as vectors in R3 with z-component equal to zero. Then
Since 
, Eq. (3) yields
CONCEPTUAL INSIGHT
Curvature for plane curves has a geometric interpretation in terms of the angle of inclination, defined as the angle θ between the tangent vector and the horizonal (Figure 8). The angle θ changes as the curve bends, and we can show that the curvature κ is the rate of change of θ as you walk along the curve at unit speed (see Exercise 61).
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EXAMPLE 4
Compute the curvature of f(x) = x3 − 3x2 + 4 at x = 0, 1, 2, 3.
Solution We apply Eq. (5):
We obtain the following values:
Figure 9 shows that the graph bends more where the curvature is large.
Unit Normal Vector
We noted above that T′(t) and T(t) are orthogonal. The unit vector in the direction of T′(t), assuming it is nonzero, is called the unit normal vector and is denoted N(t) or simply N:
Furthermore, ∥T′(t)∥ = υ(t)κ(t) by Eq. (2), so we have
Intuitively, N points the direction in which the curve is turning (see Figure 11). This is particularly clear for a plane curve. In this case, there are two unit vectors orthogonal to T (Figure 10), and of these two, N is the vector that points to the “inside” of the curve.
EXAMPLE 5: Unit Normal to a Helix
Find the unit normal vector at 
to the helix r(t) = 〈cos t, sin t, t〉.
Solution The tangent vector r′(t) = 〈− sin t, cos t, 1〉 has length 
, so
Hence, 
(Figure 11).
on the helix in Example 5.We conclude by describing another interpretation of curvature in terms of the osculating or “best-
where the curvature κP is nonzero. The osculating circle, denoted OscP, is the circle of radius R = 1/κP through P whose center Q lies in the direction of the unit normal N (Figure 12). In other words, the center Q is determined by
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from P in the normal direction.Among all circles tangent to
at P, OscP “best fits” the curve (Figure 13; see also Exercise 71). We refer to R = 1/κP as the radius of curvature at P. The center Q of OscP is called the center of curvature at P.
EXAMPLE 6
Parametrize the osculating circle to y = x2 at 
.
Solution Let f(x) = x2. We use the parametrization
and proceed by the following steps.
Step 1. Find the radius.
Apply Eq. (5) to f(x) = x2 to compute the curvature:
The osculating circle has radius 
.
Step 2. Find N at 
.
For a plane curve, there is an easy way to find N without computing T′. The tangent vector is r′(x) = 〈1, 2x〉, and we know that 〈2x, −1〉 is orthogonal to r′(x) (because their dot product is zero). Therefore, N(x) is the unit vector in one of the two directions ± 〈2x, −1〉. Figure 14 shows that the unit normal vector points in the positive y-
has center Q and radius 
.Step 3. Find the center Q.
Apply Eq. (8) with 
:
Step 4. Parametrize the osculating circle.
The osculating circle has radius
, so it has parametrization
If a curve
lies in a plane, then this plane is the osculating plane. For a general curve in three-
To define the osculating circle at a point P on a space curve
, we must first specify the plane in which the circle lies. The osculating plane is the plane through P determined by the unit tangent TP and the unit normal NP at P (we assume that T′ ≠ 0, so N is defined). Intuitively, the osculating plane is the plane that “most nearly” contains the curve
near P (see Figure 15). The osculating circle is the circle in the osculating plane through P of radius R = 1/κP whose center is located in the normal direction NP from P. Equation (8) remains valid for space curves.
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13.12 13.4: SUMMARY
A parametrization r(t) is called regular if r′(t) ≠ 0 for all t. If r(t) is regular, we define the unit tangent vector
.Curvature is defined by
, where r(s) is an arc length parametrization.In practice, we compute curvature using the following formula, which is valid for arbitrary regular parametrizations:
The curvature at a point on a graph y = f(x) in the plane is
If ∥T′(t)∥ ≠ 0, we define the unit normal vector
.T′(t) = κ(t)υ(t)N(t)
The osculating plane at a point P on a curve
is the plane through P determined by the vectors TP and NP. It is defined only if the curvature κP at P is nonzero.The osculating circle OscP is the circle in the osculating plane through P of radius R = 1/κP whose center Q lies in the normal direction NP:
The center of OscP is called the center of curvature and R is called the radius of curvature.
13.13 13.4: EXERCISES
Preliminary Questions
Question 13.170
What is the unit tangent vector of a line with direction vector v = 〈2, 1, −2〉?
Question 13.171
What is the curvature of a circle of radius 4?
Question 13.172
Which has larger curvature, a circle of radius 2 or a circle of radius 4?
Question 13.173
What is the curvature of r(t) = 〈2 + 3t, 7t, 5 − t〉?
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Question 13.174
What is the curvature at a point where T′(s) = 〈1, 2, 3〉 in an arc length parametrization r(s)?
Question 13.175
What is the radius of curvature of a circle of radius 4?
Question 13.176
What is the radius of curvature at P if κP = 9?
Exercises
In Exercises 1–
Question 13.177
r(t) = 〈4t2, 9t〉
Question 13.178
r(t) = 〈et, t2〉
Question 13.179
r(t) = 〈3 + 4t, 3 − 5t, 9t〉
Question 13.180
r(t) = 〈1 + 2t, t2, 3 − t2〉
Question 13.181
r(t) = 〈cos πt, sin πt, t〉
Question 13.182
r(t) = 〈et, e−t, t2〉
In Exercises 7–
Question 13.183
r(t) = 〈1, et, t〉
Question 13.184
r(t) = 〈4 cos t, t, 4 sin t〉
Question 13.185
r(t) = 〈4t + 1, 4t − 3, 2t〉
Question 13.186
r(t) = 〈t−1, 1, t〉
In Exercises 11–
Question 13.187
Question 13.188
Question 13.189
Question 13.190
In Exercises 15–
Question 13.191
Question 13.192
Question 13.193
Question 13.194
Question 13.195
Find the curvature of r(t) = 〈2 sin t, cos 3t, t〉 at 
and 
(Figure 16).
Question 13.196
Find the curvature function κ(x) for y = sin x. Use a computer algebra system to plot κ(x) for 0 ≤ x ≤ 2π. Prove that the curvature takes its maximum at 
and 
. Hint: As a shortcut to finding the max, observe that the maximum of the numerator and the minimum of the denominator of κ(x) occur at the same points.
Question 13.197
Show that the tractrix r(t) = 〈t − tanh t, sech t〉 has the curvature function κ(t) = sech t.
Question 13.198
Show that curvature at an inflection point of a plane curve y = f(x) is zero.
Question 13.199
Find the value of α such that the curvature of y = e αx at x = 0 is as large as possible.
Question 13.200
Find the point of maximum curvature on y = ex.
Question 13.201
Show that the curvature function of the parametrization r(t) = 〈a cos t, b sin t〉 of the ellipse 
is
Question 13.202
Use a sketch to predict where the points of minimal and maximal curvature occur on an ellipse. Then use Eq. (9) to confirm or refute your prediction.
Question 13.203
In the notation of Exercise 25, assume that a ≥ b. Show that b/a2 ≤ κ(t) ≤ a/b2 for all t.
Question 13.204
Use Eq. (3) to prove that for a plane curve r(t) = 〈x(t), y(t)〉,
In Exercises 29–
Question 13.205
Question 13.206
Question 13.207
Question 13.208
Question 13.209
Let 
for the Bernoulli spiral r(t) = 〈et cos 4t, et sin 4t〉 (see Exercise 29 in Section 13.3). Show that the radius of curvature is proportional to s(t).
Question 13.210
The Cornu spiral is the plane curve r(t) = 〈x(t), y(t)〉, where
Verify that κ(t) = |t|. Since the curvature increases linearly, the Cornu spiral is used in highway design to create transitions between straight and curved road segments (Figure 17).
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Question 13.211
Plot and compute the curvature κ(t) of the clothoid r(t) = 〈x(t), y(t)〉, where
Question 13.212
Find the unit normal vector N(θ) to r(θ) = R 〈cos θ, sin θ〉, the circle of radius R. Does N(θ) point inside or outside the circle? Draw N(θ) at 
with R = 4.
Question 13.213
Find the unit normal vector N(t) to r(t) = 〈4, sin 2t, cos 2t〉.
Question 13.214
Stetch the graph of r(t) = 〈t, t3〉. Since r′(t) = 〈1, 3t2〉, the unit normal N(t) points in one of the two directions ±〈−3t2, 1〉. Which sign is correct at t = 1? Which is correct at t = −1?
Question 13.215
Find the normal vectors to r(t) = 〈t, cos t〉 at 
and 
.
Question 13.216
Find the unit normal to the Cornu spiral (Exercise 34) at 
.
Question 13.217
Find the unit normal to the clothoid (Exercise 35) at t = π1/3.
Question 13.218
Method for Computing N Let υ(t) = ∥r′(t)∥. Show that
Hint: N is the unit vector in the direction T′(t). Differentiate T(t) = r′(t)/υ(t) to show that υ(t)r″(t) − υ′(t)r′(t) is a positive multiple of T′(t).
In Exercises 43–
Question 13.219
Question 13.220
Question 13.221
Question 13.222
Question 13.223
Question 13.224
Question 13.225
Let f(x) = x2. Show that the center of the osculating circle at
is given by 
.
Question 13.226
Use Eq. (8) to find the center of curvature r(t) = 〈t2, t3〉 at t = 1.
In Exercises 51–
Question 13.227
Question 13.228
Question 13.229
Question 13.230
Question 13.231
Question 13.232
Question 13.233
Question 13.234
Question 13.235
Figure 18 shows the graph of the half-
, where r and p are positive constants. Show that the radius of curvature at the origin is equal to r. Hint: One way of proceeding is to write the ellipse in the form of Exercise 25 and apply Eq. (9).
and the osculating circle at the origin.Question 13.236
In a recent study of laser eye surgery by Gatinel, Hoang-
. During surgery, tissue is removed to a depth t(y) at height y for −S ≤ y ≤ S, where t(y) is given by Munnerlyn’s equation (for some R > r):
After surgery, the cross section of the cornea has the shape x = f(y) + t(y) (Figure 19). Show that after surgery, the radius of curvature at the point P (where y = 0) is R.
Question 13.237
The angle of inclination at a point P on a plane curve is the angle θ between the unit tangent vector T and the x-axis (Figure 20). Assume that r(s) is a arc length parametrization, and let θ = θ(s) be the angle of inclination at r(s). Prove that
Hint: Observe that T(s) = 〈cos θ(s), sin θ(s)〉.
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Question 13.238
A particle moves along the path y = x3 with unit speed. How fast is the tangent turning (i.e., how fast is the angle of inclination changing) when the particle passes through the point (2, 8)?
Question 13.239
Let θ(x) be the angle of inclination at a point on the graph y = f(x) (see Exercise 61).
Use the relation f′(x) = tan θ to prove that
.Use the arc length integral to show that
.Now give a proof of Eq. (5) using Eq. (12).
Question 13.240
Use the parametrization r(θ) = 〈f(θ) cos θ, f(θ) sin θ〉 to show that a curve r = f(θ) in polar coordinates has curvature
In Exercises 65–
Question 13.241
f(θ) = 2 cos θ
Question 13.242
f(θ) = θ
Question 13.243
f(θ) = eθ
Question 13.244
Use Eq. (13) to find the curvature of the general Bernoulli spiral r = aebθ in polar form (a and b are constants).
Question 13.245
Show that both r′(t) and r″(t) lie in the osculating plane for a vector function r(t). Hint: Differentiate r′(t) = υ(t)T(t).
Question 13.246
Show that
is an arc length parametrization of the osculating circle at r(t0).
Question 13.247
Two vector-
Let r(s) be an arc length parametrization of a path
, and let P be the terminal point of r(0). Let γ(s) be the arc length parametrization of the osculating circle given in Exercise 70. Show that r(s) and γ(s) agree to order 2 at s = 0 (in fact, the osculating circle is the unique circle that approximates
to order 2 at P).
Question 13.248
Let r(t) = 〈x(t), y(t), z(t)〉 be a path with curvature κ(t) and define the scaled path r1(t) = 〈λx(t), λy(t), λz(t)〉, where λ ≠ 0 is a constant. Prove that curvature varies inversely with the scale factor. That is, prove that the curvature κ1(t) of r1(t) is κ1(t) = λ−1κ(t). This explains why the curvature of a circle of radius R is proportional to 1/R (in fact, it is equal to 1/R). Hint: Use Eq. (3).
Further Insights and Challenges
Question 13.249
Show that the curvature of Viviani’s curve, given by r(t) = 〈1 + cos t, sin t, 2 sin(t/2)〉, is
Question 13.250
Let r(s) be an arc length parametrization of a closed curve
of length L. We call
an oval if dθ/ds > 0 (see Exercise 61). Observe that −N points to the outside of
. For k > 0, the curve
1 defined by r1(s) = r(s) − kN is called the expansion of c(s) in the normal direction.
Show that
.As P moves around the oval counterclockwise, θ increases by 2π [Figure 21(A)]. Use this and a change of variables to prove that
.
As P moves around the oval, θ increases by 2π.Show that
1 has length L + 2πk.
In Exercises 75–
, defined by B = T × N.
Question 13.251
Show that B is a unit vector.
Question 13.252
Follow steps (a)–(c) to prove that there is a number τ (lowercase Greek “tau”) called the torsion such that
Show that
and conclude that dB/ds is orthogonal to T.Differentiate B · B = 1 with respect to s to show that dB/ds is orthogonal to B.
Conclude that dB/ds is a multiple of N.
Question 13.253
Show that if
is contained in a plane 
, then B is a unit vector normal to
. Conclude that τ = 0 for a plane curve.
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Question 13.254
Torsion means “twisting.” Is this an appropriate name for τ? Explain by interpreting τ geometrically.
Question 13.255
Use the identity
a × (b × c) = (a · c)b − (a · b)c
to prove
Question 13.256
Follow steps (a)–(b) to prove
Show that dN/ds is orthogonal to N. Conclude that dN/ds lies in the plane spanned by T and B, and hence, dN/ds = aT + bB for some scalars a, b.
Use N · T = 0 to show that
and compute a. Compute b similarly. Equations (14) and (16) together with dT/dt = κN are called the Frenet formulas and were discovered by the French geometer Jean Frenet (1816–1900).
Question 13.257
Show that r′ × r″is a multiple of B. Conclude that
Question 13.258
The vector N can be computed using N = B × T [Eq. (15)] with B, as in Eq. (17). Use this method to find N in the following cases:
r(t) = 〈cos t, t, t2〉 at t = 0
r(t) = 〈t2, t−1, t〉 at t = 1
13.14 13.5: Motion in Three-Space
In this section, we study the motion of a particle traveling along a path r(t). Recall that
As we have seen, v(t) points in the direction of motion (if it is nonzero), and its magnitude υ(t) = ∥v(t)∥ is the particle’s speed. The acceleration vector is the second derivative r″(t), which we shall denote a(t). In summary,
EXAMPLE 1
Calculate and plot the velocity and acceleration vectors at t = 1 of 
. Then find the speed at t = 1 (Figure 2).
Solution
The speed at t = 1 is
If an object’s acceleration is given, we can solve for v(t) and r(t) by integrating twice:
with v0 and r0 determined by initial conditions.
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EXAMPLE 2
Find r(t) if
Solution We have
v(t) = ∫a(t) dt + v0 = 2ti + 6t2j + v0
The initial condition v(0) = v0 = 7i gives us v(t) = 2ti + 6t2j + 7i. Then we have
r(t) = ∫v(t) dt + r0 = t2i + 2t3j + 7t i + r0
The initial condition r(0) = r0 = 2i + 9k yields
r(t) = t2i + 2t3j + 7ti + (2i + 9k) = (t2 + 7t + 2)i + 2t3j + 9k
Newton’s Second Law of Motion is often stated in the scalar form F = ma, but a more general statement is the vector law F = ma, where F is the net force vector acting on the object and a is the acceleration vector. When the force varies from position to position, we write F(r(t)) for the force acting on a particle with position vector r(t) at time t. Then Newton’s Second Law reads
EXAMPLE 3
A bullet is fired from the ground at an angle of 60° above the horizontal. What initial speed υ0 must the bullet have in order to hit a point 150 m high on a tower located 250 m away (ignoring air resistance)?
Solution Place the gun at the origin, and let r(t) be the position vector of the bullet (Figure 3).
Step 1. Use Newton’s Law.
Gravity exerts a downward force of magnitude mg, where m is the mass of the bullet and g = 9.8 m/s2. In vector form,
F = 〈0, −mg〉 = m〈0, −g〉
Newton’s Second Law F = mr″(t) yields m〈0, −g〉 = mr″(t) or r″(t) = 〈0, −g〉. We determine r(t) by integrating twice:
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Step 2. Use the initial conditions.
By our choice of coordinates, r0 = 0. The initial velocity v0 has unknown magnitude υ0, but we know that it points in the direction of the unit vector 〈cos 60°, sin 60°〉. Therefore,
Step 3. Solve for υ0.
The bullet hits the point 〈250, 150〉 on the tower if there exists a time t such that r(t) = 〈250, 150〉; that is,
Equating components, we obtain
The first equation yields t = 500/υ0. Now substitute in the second equation and solve, using g = 9.8:
We obtain 
.
In linear motion, acceleration is the rate at which an object is speeding up or slowing down. The acceleration is zero if the speed is constant. By contrast, in two or three dimensions, the acceleration can be nonzero even when the object’s speed is constant. This happens when υ(t) = ∥v(t)∥ is constant but the direction of v(t) is changing. The simplest example is uniform circular motion, in which an object travels in a circular path at constant speed (Figure 4).
EXAMPLE 4: Uniform Circular Motion
Find a(t) and ∥a(t)∥ for motion around a circle of radius R with constant speed υ.
Solution Assume that the particle follows the circular path r(t) = R〈cos ωt, sin ωt〉 for some constant ω. Then the velocity and speed of the particle are
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Thus |ω| = υ/R, and accordingly,
The constant ω (lowercase Greek “omega”) is called the angular speed because the particle’s angle along the circle changes at a rate of ω radians per unit time.
The vector a(t) is called the centripetal acceleration: It has length υ2/R and points in toward the origin [because a(t) is a negative multiple of the position vector r(t)], as in Figure 4.
Understanding the Acceleration Vector
We have noted that v(t) can change in two ways: in magnitude and in direction. To understand how the acceleration vector a(t) “encodes” both types of change, we decompose a(t) into a sum of tangential and normal components.
Recall the definition of unit tangent and unit normal vectors:
Thus, v(t) = υ(t)T(t), where υ(t) = ∥v(t)∥, so by the Product Rule,
Furthermore, T′(t) = υ(t)κ(t)N(t) by Eq. (7) of Section 13.4, where κ(t) is the curvature. Thus we can write
When you make a left turn in an automobile at constant speed, your tangential acceleration is zero (because υ′(t) = 0) and you will not be pushed back against your seat. But the car seat (via friction) pushes you to the left toward the car door, causing you to accelerate in the normal direction. Due to inertia, you feel as if you are being pushed to the right toward the passenger’s seat. This force is proportional to κυ2, so a sharp turn (large κ) or high speed (large v) produces a strong normal force.
The coefficient aT(t) is called the tangential component and aN(t) the normal component of acceleration (Figure 5).
The normal component aN is often called the centripetal acceleration, especially in the case of circular motion where it is directed toward the center of the circle.
CONCEPTUAL INSIGHT
The tangential component aT = υ′(t) is the rate at which speed υ(t) changes, whereas the normal component aN = κ(t)υ(t)2 describes the change in v due to a change in direction. These interpretations become clear once we consider the following extreme cases:
A particle travels in a straight line. Then direction does not change [κ(t) = 0] and a(t) = υ′(t)T is parallel to the direction of motion.
A particle travels with constant speed along a curved path. Then υ′(t) = 0 and the acceleration vector a(t) = κ(t)υ(t)2N is normal to the direction of motion.
General motion combines both tangential and normal acceleration.
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EXAMPLE 5
The Giant Ferris Wheel in Vienna has radius R = 30 m (Figure 6). Assume that at time t = t0, the wheel rotates counterclockwise with a speed of 40 m/min and is slowing at a rate of 15 m/min2. Find the acceleration vector a for a person seated in a car at the lowest point of the wheel.
Solution At the bottom of the wheel, T = 〈1, 0〉 and N = 〈0, 1〉. We are told that aT = υ′ = −15 at time t0. The curvature of the wheel is κ = 1/R = 1/30, so the normal component is aN = κv2 = υ2/R = (40)2/30 ≈ 53.3. Therefore (Figure 7),
a ≈ −15T + 53.3N = 〈−15, 53.3〉 m/min2
The following theorem provides useful formulas for the tangential and normal components.
THEOREM 1: Tangential and Normal Components of Acceleration
In the decomposition a = aTT + aNN, we have
and
Proof
We have T · T = 1 and N · T = 0. Thus
and since 
, we have
and
Finally, the vectors aTT and aNN are the sides of a right triangle with hypotenuse a as in Figure 5, so by the Pythagorean Theorem,
Keep in mind that aN ≥ 0 but aT is positive or negative, depending on whether the object is speeding up or slowing down.
EXAMPLE 6
Decompose the acceleration vector a of r(t) = 〈t2, 2t, ln t〉 into tangential and normal components at 
(Figure 8).
on the curve r(t) = 〈t2, 2t, ln t〉.Solution First, we compute the tangential components T and aT. We have
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At 
,
Thus
and by Eq. (3),
Next, we use Eq. (4):
Summary of steps in Example 6:
This vector has length
and thus
Finally, we obtain the decomposition
a = 〈2, 0,-4〉 = aTT + aNN = −2T + 4N
EXAMPLE 7: Nonuniform Circular Motion
Figure 9 shows the acceleration vectors of three particles moving counterclockwise around a circle. In each case, state whether the particle’s speed υ is increasing, decreasing, or momentarily constant.
REMINDER
By Eq. (3), v′ = aT = a · T
v · w = ∥v∥ ∥w∥ cos θ
where θ is the angle between v and w.
Solution The rate of change of speed depends on the angle θ between a and T:
v′ = aT = a · T = ∥a∥ ∥T∥ cos θ = ∥a∥ cos θ
In (A), θ is obtuse so cos θ < 0 and υ′ < 0. The particle’s speed is decreasing.
In (B),
so cos θ = 0 and υ′ = 0. The particle’s speed is momentarily constant.In (C), θ is acute so cos θ > 0 and υ′ > 0. The particle’s speed is increasing.
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EXAMPLE 8
Find the curvature 
for the path r(t) = 〈t2, 2t, ln t〉 in Example 6.
Solution By Eq. (2), the normal component is
aN = κv2
In Example 6 we showed that aN = 4 and v = 〈1, 2, 2〉 at 
. Therefore, v2 = v · v = 9 and the curvature is 
.
13.15 13.5: SUMMARY
For an object whose path is described by a vector-
valued function r(t), 
The velocity vector v(t) points in the direction of motion. Its length υ(t) = ∥v(t)∥ is the object’s speed.
The acceleration vector a is the sum of a tangential component (reflecting change in speed) and a normal component (reflecting change in direction):
a(t) = aT(t)T(t) + aN(t)N(t)
13.16 13.5: EXERCISES
Preliminary Questions
Question 13.259
If a particle travels with constant speed, must its acceleration vector be zero? Explain.
Question 13.260
For a particle in uniform circular motion around a circle, which of the vectors v(t) or a(t) always points toward the center of the circle?
Question 13.261
Two objects travel to the right along the parabola y = x2 with nonzero speed. Which of the following statements must be true?
Their velocity vectors point in the same direction.
Their velocity vectors have the same length.
Their acceleration vectors point in the same direction.
Question 13.262
Use the decomposition of acceleration into tangential and normal components to explain the following statement: If the speed is constant, then the acceleration and velocity vectors are orthogonal.
Question 13.263
If a particle travels along a straight line, then the acceleration and velocity vectors are (choose the correct description):
Orthogonal
Parallel
Question 13.264
What is the length of the acceleration vector of a particle traveling around a circle of radius 2 cm with constant velocity 4 cm/s?
Question 13.265
Two cars are racing around a circular track. If, at a certain moment, both of their speedometers read 110 mph. then the two cars have the same (choose one):
aT
aN
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Exercises
Question 13.266
Use the table below to calculate the difference quotients 
for h = −0.2, −0.1, 0.1, 0.2. Then estimate the velocity and speed at t = 1.
|
r(0.8) |
〈1.557, 2.459,-1.970〉 |
|
r(0.9) |
〈1.559, 2.634,-1.740〉 |
|
r(1) |
〈1.540, 2.841,-1.443〉 |
|
r(1.1) |
〈1.499, 3.078,-1.035〉 |
|
r(1.2) |
〈1.435, 3.342,-0.428〉 |
Question 13.267
Draw the vectors r(2 + h) − r(2) and 
for h = 0.5 for the path in Figure 10. Draw v(2) (using a rough estimate for its length).
In Exercises 3–
Question 13.268
Question 13.269
Question 13.270
Question 13.271
Question 13.272
Find a(t) for a particle moving around a circle of radius 8 cm at a constant speed of υ = 4 cm/s (see Example 4). Draw the path and acceleration vector at 
.
Question 13.273
Sketch the path r(t) = 〈1 − t2, 1 − t〉 for −2 ≤ t ≤ 2, indicating the direction of motion. Draw the velocity and acceleration vectors at t = 0 and t = 1.
Question 13.274
Sketch the path r(t) = 〈t2, t3〉 together with the velocity and acceleration vectors at t = 1.
Question 13.275
The paths r(t) = 〈t2, t3〉 and r1(t) = 〈t4, t6〉 trace the same curve, and r1(1) = r(1). Do you expect either the velocity vectors or the acceleration vectors of these paths at t = 1 to point in the same direction? Compute these vectors and draw them on a single plot of the curve.
In Exercises 11–
Question 13.276
Question 13.277
Question 13.278
Question 13.279
In Exercises 15–
Question 13.280
Question 13.281
Question 13.282
Question 13.283
In Exercises 19–
Question 13.284
A bullet is fired from the ground at an angle of 45°. What initial speed must the bullet have in order to hit the top of a 120-
Question 13.285
Find the initial velocity vector v0 of a projectile released with initial speed 100 m/s that reaches a maximum height of 300 m.
Question 13.286
Show that a projectile fired at an angle θ with initial speed υ0 travels a total distance 
sin 2θ before hitting the ground. Conclude that the maximum distance (for a given υ0) is attained for θ = 45°.
Question 13.287
One player throws a baseball to another player standing 25 m away with initial speed 18 m/s. Use the result of Exercise 21 to find two angles θ at which the ball can be released. Which angle gets the ball there faster?
Question 13.288
A bullet is fired at an angle 
at a tower located d = 600 m away, with initial speed υ0 = 120 m/s. Find the height H at which the bullet hits the tower.
Question 13.289
Show that a bullet fired at an angle θ will hit the top of an h-meter tower located d meters away if its initial speed is
Question 13.290
A constant force F = 〈5, 2〉 (in newtons) acts on a 10-
Question 13.291
A force F = 〈24t, 16 − 8t〉 (in newtons) acts on a 4-
Question 13.292
A particle follows a path r(t) for 0 ≤ t ≤ T, beginning at the origin O. The vector 
is called the average velocity vector. Suppose that 
. Answer and explain the following:
Where is the particle located at time T if
?Is the particle’s average speed necessarily equal to zero?
Question 13.293
At a certain moment, a moving particle has velocity v = 〈2, 2, −1〉 and a = 〈0, 4, 3〉. Find T, N, and the decomposition of a into tangential and normal components.
Question 13.294
At a certain moment, a particle moving along a path has velocity v = 〈12, 20, 20〉 and acceleration a = 〈2, 1, −3〉. Is the particle speeding up or slowing down?
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In Exercises 30–
Question 13.295
r(t) = 〈t2, t3〉
Question 13.296
r(t) = 〈t, cos t, sin t〉
Question 13.297
Question 13.298
In Exercise 34–
Question 13.299
Question 13.300
Question 13.301
Question 13.302
Question 13.303
Question 13.304
Question 13.305
Question 13.306
Question 13.307
Let r(t) = 〈t2, 4t − 3〉. Find T(t) and N(t), and show that the decomposition of a(t) into tangential and normal components is
Question 13.308
Find the components aT and aN of the acceleration vector of a particle moving along a circular path of radius R = 100 cm with constant velocity υ0 = 5 cm/s.
Question 13.309
In the notation of Example 5, find the acceleration vector for a person seated in a car at (a) the highest point of the Ferris wheel and(b) the two points level with the center of the wheel.
Question 13.310
Suppose that the Ferris wheel in Example 5 is rotating clockwise and that the point P at angle 45° has acceleration vector a = 〈0, −50〉 m/min2 pointing down, as in Figure 11. Determine the speed and tangential acceleration of the Ferris wheel.
Question 13.311
At time t0, a moving particle has velocity vector v = 2i and acceleration vector a = 3i + 18k. Determine the curvature κ(t0) of the particle’s path at time t0.
Question 13.312
A space shuttle orbits the earth at an altitude 400 km above the earth’s surface, with constant speed υ = 28,000 km/h. Find the magnitude of the shuttle’s acceleration (in km/h2), assuming that the radius of the earth is 6378 km (Figure 12).
Question 13.313
A car proceeds along a circular path of radius R = 300 m centered at the origin. Starting at rest, its speed increases at a rate of t m/s2. Find the acceleration vector a at time t = 3 s and determine its decomposition into normal and tangential components.
Question 13.314
A runner runs along the helix r(t) = 〈cos t, sin t, t〉. When he is at position 
, his speed is 3 m/s and he is accelerating at a rate of 
. Find his acceleration vector a at this moment. Note: The runner’s acceleration vector does not coincide with the acceleration vector of r(t).
Question 13.315
Explain why the vector w in Figure 13 cannot be the acceleration vector of a particle moving along the circle. Hint: Consider the sign of w · N.
Question 13.316
Figure 14 shows acceleration vectors of a particle moving clockwise around a circle. In each case, state whether the particle is speeding up, slowing down, or momentarily at constant speed. Explain.
Question 13.317
Prove that 
.
Question 13.318
Suppose that r = r(t) lies on a sphere of radius R for all t. Let J = r × r′. Show that r′ = (J × r)/∥r∥2. Hint: Observe that r and r′ are perpendicular.
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Further Insights and Challenges
Question 13.319
The orbit of a planet is an ellipse with the sun at one focus. The sun’s gravitational force acts along the radial line from the planet to the sun (the dashed lines in Figure 15), and by Newton’s Second Law, the acceleration vector points in the same direction. Assuming that the orbit has positive eccentricity (the orbit is not a circle), explain why the planet must slow down in the upper half of the orbit (as it moves away from the sun) and speed up in the lower half. Kepler’s Second Law, discussed in the next section, is a precise version of this qualitative conclusion. Hint: Consider the decomposition of a into normal and tangential components.
In Exercises 55–
Question 13.320
Show that the car will not skid if the curvature κ of the road is such that (with R = 1/κ)
Note that braking (υ′ < 0) and speeding up (υ′ > 0) contribute equally to skidding.
Question 13.321
Suppose that the maximum radius of curvature along a curved highway is R = 180 m. How fast can an automobile travel (at constant speed) along the highway without skidding if the coefficient of friction is μ = 0.5?
Question 13.322
Beginning at rest, an automobile drives around a circular track of radius R = 300 m, accelerating at a rate of 0.3 m/s2. After how many seconds will the car begin to skid if the coefficient of friction is μ = 0.6?
Question 13.323
You want to reverse your direction in the shortest possible time by driving around a semicircular bend (Figure 16). If you travel at the maximum possible constant speed υ that will not cause skidding, is it faster to hug the inside curve (radius r) or the outside curb (radius R)? Hint: Use Eq. (5) to show that at maximum speed, the time required to drive around the semicircle is proportional to the square root of the radius.
Question 13.324
What is the smallest radius R about which an automobile can turn without skidding at 100 km/h if μ = 0.75 (a typical value)?
13.17 13.6: Planetary Motion According to Kepler and Newton
In this section, we derive Kepler’s laws of planetary motion, a feat first accomplished by Isaac Newton and published by him in 1687. No event was more emblematic of the scientific revolution. It demonstrated the power of mathematics to make the natural world comprehensible and it led succeeding generations of scientists to seek and discover mathematical laws governing other phenomena, such as electricity and magnetism, thermodynamics, and atomic processes.
According to Kepler, the planetary orbits are ellipses with the sun at one focus. Furthermore, if we imagine a radial vector r(t) pointing from the sun to the planet, as in Figure 1, then this radial vector sweeps out area at a constant rate or, as Kepler stated in his Second Law, the radial vector sweeps out equal areas in equal times (Figure 2). Kepler’s Third Law determines the period T of the orbit, defined as the time required to complete one full revolution. These laws are valid not just for planets orbiting the sun, but for any body orbiting about another body according to the inverse-
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Kepler’s Three Laws
Law of Ellipses: The orbit of a planet is an ellipse with the sun at one focus.
Law of Equal Area in Equal Time: The position vector pointing from the sun to the planet sweeps out equal areas in equal times.
Law of the Period of Motion:
, wherea is the semimajor axis of the ellipse (Figure 1).
G is the universal gravitational constant: 6.673 × 10−11 m3 kg−1 s−2.
M is the mass of the sun, approximately 1.989 × 1030 kg.
Kepler’s version of the Third Law stated only that T2 is proportional to a3. Newton discovered that the constant of proportionality is equal to 4π2/(GM), and he observed that if you can measure T and a through observation, then you can use the Third Law to solve for the mass M. This method is used by astronomers to find the masses of the planets (by measuring T and a for moons revolving around the planet) as well as the masses of binary stars and galaxies. See Exercises 2–
Our derivation makes a few simplifying assumptions. We treat the sun and planet as point masses and ignore the gravitational attraction of the planets on each other. And although both the sun and the planet revolve around their mutual center of mass, we ignore the sun’s motion and assume that the planet revolves around the center of the sun. This is justified because the sun is much more massive than the planet.
We place the sun at the origin of the coordinate system. Let r = r(t) be the position vector of a planet of mass m, as in Figure 1, and let (Figure 3)
be the unit radial vector at time t (er is the unit vector that points to the planet as it moves around the sun). By Newton’s Universal Law of Gravitation (the inverse-
where k = GM (Figure 3). Combining the Law of Gravitation with Newton’s Second Law of Motion F(r(t)) = mr″(t), we obtain
Kepler’s Laws are a consequence of this differential equation.
Kepler’s Second Law
The key to Kepler’s Second Law is the fact that the following cross product is a constant vector (even though both r(t) and r′(t) are changing in time):
THEOREM 1
The vector J is constant—
In physics, mJ is called the angular momentum vector. In situations where J is constant, we say that angular momentum is conserved. This conservation law is valid whenever the force acts in the radial direction.
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Proof
By the Product Rule for cross products (Theorem 3 in Section 13.2)
The cross product of parallel vectors is zero, so the second term is certainly zero. The first term is also zero because r″(t) is a multiple of er by Eq. (1), and hence also of r(t).
REMINDER
a × b is orthogonal to both a and b
a × b = 0 if a and b are parallel, that is, one is a multiple of the other.
How can we use Eq. (2)? First of all, the cross product J is orthogonal to both r(t) and r′(t). Because J is constant, r(t) and r′(t) are confined to the fixed plane orthogonal to J. This proves that the motion of a planet around the sun takes place in a plane.
We can choose coordinates so that the sun is at the origin and the planet moves in the counterclockwise direction (Figure 4). Let (r, θ) be the polar coordinates of the planet, where r = r(t) and θ = θ(t) are functions of time. Note that r(t) = ∥r(t)∥.
Recall from Section 11.4 (Theorem 1) that the area swept out by the planet’s radial vector is
Kepler’s Second Law states that this area is swept out at a constant rate. But this rate is simply dA/dt. By the Fundamental Theorem of Calculus, 
, and by the Chain Rule,
Thus, Kepler’s Second Law follows from the next theorem, which tells us that dA/dt has the constant value 
.
THEOREM 2
Let J = ∥J∥ (J is constant by Theorem 1). Then
Proof
We note that in polar coordinates, er = 〈cos θ, sin θ〉. We also define the unit vector eθ = 〈−sin θ, cos θ〉 that is orthogonal to er (Figure 5). In summary,
We see directly that the derivatives of er and eθ with respect to θ are
The time derivative of er is computed using the Chain Rule:
Now apply the Product Rule to r = rer:
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Using er × er = 0, we obtain
J = r × r′ = rer × (r′er + rθ′eθ) = r2θ′(er × eθ)
To compute cross products of vectors in the plane, such as r, er, and eθ, we treat them as vectors in three-
It is straightforward to check that er × eθ = k, and since k is a unit vector, J = ∥J∥ = |r2θ′|. However, θ′ > 0 because the planet moves in the counterclockwise direction, so J = r2θ′. This proves Theorem 2.
Proof of the Law of Ellipses
Let v = r′(t) be the velocity vector. Then r″ = v′ and Eq. (1) may be written
REMINDER
Eq. (1) states:
Where r(t) = ∥r(t)∥.
On the other hand, by the Chain Rule and the relation r(t)2θ′(t) = J of Eq. (3),
Together with Eq. (6), this yields 
, or
This is a first-
where c is an arbitrary constant vector.
We are still free to rotate our coordinate system in the plane of motion, so we may assume that c points along the y-axis. We can then write c = 〈0, (k/J)e〉 for some constant e. We finish the proof by computing J = r × v:
Direct calculation yields
so our equation becomes 
. Since k is a unit vector,
Solving for r, we obtain the polar equation of a conic section of eccentricity e (an ellipse, parabola, or hyperbola):
REMINDER
The equation of a conic section in polar coordinates is discussed in Section 11.5.
This result shows that if a planet travels around the sun in a bounded orbit, then the orbit must be an ellipse. There are also “open orbits” that are either parabolic and hyperbolic. They describe comets that pass by the sun and then continue into space, never to return. In our derivation, we assumed implicitly that J ≠ 0. If J = 0, then θ′(t) = 0. In this case, the orbit is a straight line, and the planet falls directly into the sun.
Kepler’s Third Law is verified in Exercises 23 and 24.
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CONCEPTUAL INSIGHT
We exploited the fact that J is constant to prove the law of ellipses without ever finding a formula for the position vector r(t) of the planet as a function of time t. In fact, r(t) cannot be expressed in terms of elementary functions. This illustrates an important principle: Sometimes it is possible to describe solutions of a differential equation even if we cannot write them down explicitly.
The Hubble Space Telescope produced this image of the Antenna galaxies, a pair of spiral galaxies that began to collide hundreds of millions of years ago.
HISTORICAL PERSPECTIVE
The astronomers of the ancient world (Babylon, Egypt, and Greece) mapped out the nighttime sky with impressive accuracy, but their models of planetary motion were based on the erroneous assumption that the planets revolve around the earth. Although the Greek astronomer Aristarchus (310–
Constants:
Gravitational constant:
G ≈ 6.673 × 10−11 m3 kg−1 s−2
Mass of the sun:
M ≈ 1.989 × 1030 kg
k = GM ≈ 1.327 × 1020
The German astronomer Johannes Kepler was the son of a mercenary soldier who apparently left his family when Johannes was 5 and may have died at war. He was raised by his mother in his grandfather’s inn. Kepler’s mathematical brilliance earned him a scholarship at the University of Tübingen and at age of 29, he went to work for the Danish astronomer Tycho Brahe, who had compiled the most complete and accurate data on planetary motion then available. When Brahe died in 1601, Kepler succeeded him as “Imperial Mathematician” to the Holy Roman Emperor, and in 1609 he formulated the first two of his laws of planetary motion in a work entitled Astronomia Nova (New Astronomy).
In the centuries since Kepler’s death, as observational data improved, astronomers found that the planetary orbits are not exactly elliptical. Furthermore, the perihelion (the point on the orbit closest to the sun) shifts slowly over time (Figure 6). Most of these deviations can be explained by the mutual pull of the planets, but the perihelion shift of Mercury is larger than can be accounted for by Newton’s Laws. On November 18, 1915, Albert Einstein made a discovery about which he later wrote to a friend, “I was beside myself with ecstasy for days.” He had been working for a decade on his famous General Theory of Relativity, a theory that would replace Newton’s law of gravitation with a new set of much more complicated equations called the Einstein Field Equations. On that 18th of November, Einstein showed that Mercury’s perihelion shift was accurately explained by his new theory. At the time, this was the only substantial piece of evidence that the General Theory of Relativity was correct.
13.18 13.6: SUMMARY
Kepler’s three laws of planetary motion:
Law of Ellipses
Law of Equal Area in Equal Time
Law of the Period
, where T is the period (time to complete one full revolution) and a is the semimajor axis (Figure 7).
Planetary orbit.
According to Newton’s Universal Law of Gravitation and Second Law of Motion, the position vector r(t) of a planet satisfies the differential equation
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Properties of J = r(t) × r′(t):
J is a constant of planetary motion.
Let J = ∥J∥. Then J = r(t)2θ′(t).
The planet sweeps out area at the rate
.
A planetary orbit has polar equation
, where e is the eccentricity of the orbit.
13.19 13.6: EXERCISES
Preliminary Questions
Question 13.325
Describe the relation between the vector J = r × r′ and the rate at which the radial vector sweeps out area.
Question 13.326
Equation (1) shows that r″ is proportional to r. Explain how this fact is used to prove Kepler’s Second Law.
Question 13.327
How is the period T affected if the semimajor axis a is increased four-
Exercises
Question 13.328
Kepler’s Third Law states that T2/a3 has the same value for each planetary orbit. Do the data in the following table support this conclusion? Estimate the length of Jupiter’s period, assuming that a = 77.8 × 1010 m.
|
Planet |
Mercury |
Venus |
Earth |
Mars |
|---|---|---|---|---|
|
a (1010 m) |
5.79 |
10.8 |
15.0 |
22.8 |
|
T (years) |
0.241 |
0.615 |
1.00 |
1.88 |
Question 13.329
Finding the Mass of a Star Using Kepler’s Third Law, show that if a planet revolves around a star with period T and semimajor axis a, then the mass of the star is 
.
Question 13.330
Ganymede, one of Jupiter’s moons discovered by Galileo, has an orbital period of 7.154 days and a semimajor axis of 1.07 × 109 m. Use Exercise 2 to estimate the mass of jupiter.
Question 13.331
An astronomer observes a planet orbiting a star with a period of 9.5 years and a semimajor axis of 3 × 108 km. Find the mass of the star using Exercise 2.
Question 13.332
Mass of the Milky Way The sun revolves around the center of mass of the Milky Way galaxy in an orbit that is approximately circular, of radius a ≈ 2.8 × 1017 km and velocity υ ≈ 250 km/s. Use the result of Exercise 2 to estimate the mass of the portion of the Milky Way inside the sun’s orbit (place all of this mass at the center of the orbit).
Question 13.333
A satellite orbiting above the equator of the earth is geosynchronous if the period is T = 24 hours (in this case, the satellite stays over a fixed point on the equator). Use Kepler’s Third Law to show that in a circular geosynchronous orbit, the distance from the center of the earth is R ≈ 42,246 km. Then compute the altitude h of the orbit above the earth’s surface. The earth has mass M ≈ 5.974 × 1024 kg and radius R ≈ 6371 km.
Question 13.334
Show that a planet in a circular orbit travels at constant speed. Hint: Use that J is constant and that r(t) is orthogonal to r′(t) for a circular orbit.
Question 13.335
Verify that the circular orbit
r(t) = 〈R cos ωt, R sin ωt〉
satisfies the differential equation, Eq. (1), provided that ω2 = kR−3.
Then deduce Kepler’s Third Law 
for this orbit.
Question 13.336
Prove that if a planetary orbit is circular of radius R, then υT = 2πR, where υ is the planet’s speed (constant by Exercise 7) and T is the period. Then use Kepler’s Third Law to prove that 
.
Question 13.337
Find the velocity of a satellite in geosynchronous orbit about the earth. Hint: Use Exercises 6 and 9.
Question 13.338
A communications satellite orbiting the earth has initial position r = 〈29,000, 20,000, 0〉 (in km) and initial velocity r′ = 〈1, 1, 1〉 (in km/s), where the origin is the earth’s center. Find the equation of the plane containing the satellite’s orbit. Hint: This plane is orthogonal to J.
Question 13.339
Assume that the earth’s orbit is circular of radius R = 150 × 106 km (it is nearly circular with eccentricity e = 0.017). Find the rate at which the earth’s radial vector sweeps out area in units of km2/s. What is the magnitude of the vector J = r × r′ for the earth (in units of km2 per second)?
Exercises 13–
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Question 13.340
Use the polar equation of an ellipse
to show that rper = a(1 − e) and rap = a(1 + e). Hint: Use the fact that rper + rap = 2a.
Question 13.341
Use the result of Exercise 13 to prove the formulas
Question 13.342
Use the fact that J = r × r′ is constant to prove
υper(1 − e) = υap(1 + e)
Hint: r is perpendicular to r′ at the perihelion and aphelion.
Question 13.343
Compute rper and rap for the orbit of Mercury, which has eccentricity e = 0.244 (see the table in Exercise 1 for the semimajor axis).
Question 13.344
Conservation of Energy The total mechanical energy (kinetic energy plus potential energy) of a planet of mass m orbiting a sun of mass M with position r and speed υ = ∥r′∥ is
Prove the equations
Then use Newton’s Law F = ma and Eq. (1) to prove that energy is conserved—
that is, 
.
Question 13.345
Show that the total energy [Eq. (8)] of a planet in a circular orbit of radius R is 
. Hint: Use Exercise 9.
Question 13.346
Prove that 
as follows:
Use Conservation of Energy (Exercise 17) to show that
Show that
using Exercise 13.Show that
using Exercise 15. Then solve for υper using (a) and (b).
Question 13.347
Show that a planet in an elliptical orbit has total mechanical energy 
, where a is the semimajor axis. Hint: Use Exercise 19 to compute the total energy at the perihelion.
Question 13.348
Prove that 
at any point on an elliptical orbit, where r = ∥r∥, υ is the velocity, and a is the semimajor axis of the orbit.
Question 13.349
Two space shuttles A and B orbit the earth along the solid trajectory in Figure 9. Hoping to catch up to B, the pilot of A applies a forward thrust to increase her shuttle’s kinetic energy. Use Exercise 20 to show that shuttle A will move off into a larger orbit as shown in the figure. Then use Kepler’s Third Law to show that A’s orbital period T will increase (and she will fall farther and farther behind B)!
Further Insights and Challenges
Exercises 23 and 24 prove Kepler’s Third Law. Figure 10 shows an elliptical orbit with polar equation
where p = J2/k. The origin of the polar coordinates is at F1. Let a and b be the semimajor and semiminor axes, respectively.
Question 13.350
This exercise shows that 
.
Show that CF1 = ae. Hint: rper = a(1 − e) by Exercise 13.
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Show that
.Show that F1A + F2A = 2a. Conclude that F1B + F2B = 2a and hence F1B = F2B = a.
Use the Pythagorean Theorem to prove that
.
Question 13.351
The area A of the ellipse is A = πab
Prove, using Kepler’s First Law, that
, where T is the period of the orbit.Use Exercise 23 to show that
.Deduce Kepler’s Third Law:
.
Question 13.352
According to Eq. (7) the velocity vector of a planet as a function of the angle θ is
Use this to explain the following statement: As a planet revolves around the sun, its velocity vector traces out a circle of radius k/J with center at the terminal point of c (Figure 11). This beautiful but hidden property of orbits was discovered by William Rowan Hamilton in 1847.
13.20 CHAPTER REVIEW EXERCISES
Question 13.353
Determine the domains of the vector-
r1(t) = 〈t−1, (t + 1)−1, sin−1 t〉
Question 13.354
Sketch the paths r1(θ) = 〈θ, cos θ〉 and r2(θ) = 〈cos θ, θ〉 in the xy-plane.
Question 13.355
Find a vector parametrization of the intersection of the surfaces x2 + y4 + 2z3 = 6 and x = y2 in R3.
Question 13.356
Find a vector parametrization using trigonometric functions of the intersection of the plane x + y + z = 1 and the elliptical cylinder 
in R3.
In Exercises 5–
Question 13.357
Question 13.358
Question 13.359
Question 13.360
Question 13.361
Question 13.362
In Exercises 11–
Question 13.363
Question 13.364
Question 13.365
Question 13.366
Question 13.367
Calculate
Question 13.368
Calculate
Question 13.369
A particle located at (1, 1, 0) at time t = 0 follows a path whose velocity vector is v(t) = 〈1, t, 2t2〉. Find the particle’s location at t = 2.
Question 13.370
Find the vector-
Question 13.371
Calculate r(t) assuming that
Question 13.372
Solve r″(t) = 〈t2 − 1, t + 1, t3〉 subject to the initial conditions r(0) = 〈1, 0, 0〉 and r′(0) = 〈−1, 1, 0〉
Question 13.373
Compute the length of the path
Question 13.374
Express the length of the path r(t) = 〈ln t, t, et 〉 for 1 ≤ t ≤ 2 as a definite integral, and use a computer algebra system to find its value to two decimal places.
Question 13.375
Find an arc length parametrization of a helix of height 20 cm that makes four full rotations over a circle of radius 5 cm.
Question 13.376
Find the minimum speed of a particle with trajectory r(t) = 〈t, et−3,e4−t〉.
Question 13.377
A projectile fired at an angle of 60° lands 400 m away. What was its initial speed?
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Question 13.378
A specially trained mouse runs counterclockwise in a circle of radius 0.6 m on the floor of an elevator with speed 0.3 m/s while the elevator ascends from ground level (along the z-axis) at a speed of 12 m/s. Find the mouse’s acceleration vector as a function of time. Assume that the circle is centered at the origin of the xy-plane and the mouse is at (2, 0, 0) at t = 0.
Question 13.379
During a short time interval [0.5, 1.5], the path of an unmanned spy plane is described by
A laser is fired (in the tangential direction) toward the yz-plane at time t = 1. Which point in the yz-plane does the laser beam hit?
Question 13.380
A force F = 〈12t + 4, 8 − 24t〉 (in newtons) acts on a 2-
Question 13.381
Find the unit tangent vector to r(t) = 〈sin t, t, cos t〉 at t = π.
Question 13.382
Find the unit tangent vector to r(t) = 〈t2, tan−1 t, t〉 at t = 1.
Question 13.383
Calculate κ(1) for r(t) = 〈ln t, t〉.
Question 13.384
Calculate 
for r(t) = 〈tan t, sec t, cos t〉.
In Exercises 33 and 34, write the acceleration vector a at the point indicated as a sum of tangential and normal components.
Question 13.385
Question 13.386
Question 13.387
At a certain time t0, the path of a moving particle is tangent to the y-axis. The particle’s speed at time t0 is 4 m/s, and its acceleration vector is a = 〈5, 4, 12〉. Determine the curvature of the path at t0.
Question 13.388
Parametrize the osculating circle to y = x2 − x3 at x = 1.
Question 13.389
Parametrize the osculating circle to 
at x = 4.
Question 13.390
If a planet has zero mass (m = 0), then Newton’s laws of motion reduce to r″(t) = 0 and the orbit is a straight line r(t) = r0 + tv0, where r0 = r(0) and v0 = r′(0) (Figure 1). Show that the area swept out by the radial vector at time t is 
and thus Kepler’s Second Law continues to hold (the rate is constant).
Question 13.391
Suppose the orbit of a planet is an ellipse of eccentricity e = c/a and period T (Figure 2). Use Kepler’s Second Law to show that the time required to travel from A′ to B′ is equal to
Question 13.392
The period of Mercury is approximately 88 days, and its orbit has eccentricity 0.205. How much longer does it take Mercury to travel from A′ to B′ than from B′ to A (Figure 2)?