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2: Derivatives and Integrals of Vector Functions

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97817_13_ch13_p872-881.qk_97817_13_ch13_p872-881 11/8/10 11:32 AM Page 872

872

CHAPTER 13

VECTOR FUNCTIONS

if this limit exists. The geometric significance of this definition is shown in Figure 1. If the

l

points P and Q have position vectors r͑t͒ and r͑t ϩ h͒, then PQ represents the vector

r͑t ϩ h͒ Ϫ r͑t͒, which can therefore be regarded as a secant vector. If h Ͼ 0, the scalar

multiple ͑1͞h͒͑r͑t ϩ h͒ Ϫ r͑t͒͒ has the same direction as r͑t ϩ h͒ Ϫ r͑t͒. As h l 0, it

appears that this vector approaches a vector that lies on the tangent line. For this reason, the

vector rЈ͑t͒ is called the tangent vector to the curve defined by r at the point P, provided

that rЈ͑t͒ exists and rЈ͑t͒ 0. The tangent line to C at P is defined to be the line through P

parallel to the tangent vector rЈ͑t͒. We will also have occasion to consider the unit tangent

vector, which is

T͑t͒ ෇

z

Խ

rЈ͑t͒

rЈ͑t͒

Խ

z

r(t+h)-r(t)

P

TEC Visual 13.2 shows an animation

of Figure 1.

rª(t)

Q

r(t+h)-r(t)

h

P

Q

r(t)

r(t)

r(t+h)

r(t+h)

C

C

0

0

y

x

FIGURE 1

(a) The secant vector PQ

y

x

(b) The tangent vector rª(t)

The following theorem gives us a convenient method for computing the derivative of a

vector function r : just differentiate each component of r.

2 Theorem If r͑t͒ ෇ ͗ f ͑t͒, t͑t͒, h͑t͒͘ ෇ f ͑t͒ i ϩ t͑t͒ j ϩ h͑t͒ k, where f , t, and

h are differentiable functions, then

rЈ͑t͒ ෇ ͗ f Ј͑t͒, tЈ͑t͒, hЈ͑t͒͘ ෇ f Ј͑t͒ i ϩ tЈ͑t͒ j ϩ hЈ͑t͒ k

PROOF

rЈ͑t͒ ෇ lim

⌬t l 0

෇ lim

⌬t l 0

෇ lim

⌬t l 0

ͳ

1

͓r͑t ϩ ⌬t͒ Ϫ r͑t͔͒

⌬t

1

͓͗ f ͑t ϩ ⌬t͒, t͑t ϩ ⌬t͒, h͑t ϩ ⌬t͒͘ Ϫ ͗ f ͑t͒, t͑t͒, h͑t͔͒͘

⌬t

ͳ

lim

⌬t l 0

f ͑t ϩ ⌬t͒ Ϫ f ͑t͒ t͑t ϩ ⌬t͒ Ϫ t͑t͒ h͑t ϩ ⌬t͒ Ϫ h͑t͒

,

,

⌬t

⌬t

⌬t

ʹ

f ͑t ϩ ⌬t͒ Ϫ f ͑t͒

t͑t ϩ ⌬t͒ Ϫ t͑t͒

h͑t ϩ ⌬t͒ Ϫ h͑t͒

, lim

, lim

⌬t

l

0

⌬t

l

0

⌬t

⌬t

⌬t

ʹ

෇ ͗ f Ј͑t͒, tЈ͑t͒, hЈ͑t͒͘

Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).

Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

97817_13_ch13_p872-881.qk_97817_13_ch13_p872-881 11/8/10 11:32 AM Page 873

SECTION 13.2

v

DERIVATIVES AND INTEGRALS OF VECTOR FUNCTIONS

873

EXAMPLE 1

(a) Find the derivative of r͑t͒ ෇ ͑1 ϩ t 3 ͒ i ϩ teϪt j ϩ sin 2t k.

(b) Find the unit tangent vector at the point where t ෇ 0.

SOLUTION

(a) According to Theorem 2, we differentiate each component of r:

rЈ͑t͒ ෇ 3t 2 i ϩ ͑1 Ϫ t͒eϪt j ϩ 2 cos 2t k

(b) Since r͑0͒ ෇ i and rЈ͑0͒ ෇ j ϩ 2k, the unit tangent vector at the point ͑1, 0, 0͒ is

T͑0͒ ෇

Խ

EXAMPLE 2 For the curve r͑t͒ ෇ st i ϩ ͑2 Ϫ t͒ j, find rЈ͑t͒ and sketch the position

vector r͑1͒ and the tangent vector rЈ͑1͒.

y

2

SOLUTION We have

(1, 1)

r(1)

0

Խ

rЈ͑0͒

j ϩ 2k

1

2

k

rЈ͑0͒

s1 ϩ 4

s5

s5

rЈ͑t͒ ෇

rª(1)

1

x

FIGURE 2

Notice from Figure 2 that the tangent vector

points in the direction of increasing t. (See

Exercise 56.)

1

iϪj

2st

rЈ͑1͒ ෇

and

1

iϪj

2

The curve is a plane curve and elimination of the parameter from the equations

x ෇ st , y ෇ 2 Ϫ t gives y ෇ 2 Ϫ x 2, x ജ 0. In Figure 2 we draw the position vector

r͑1͒ ෇ i ϩ j starting at the origin and the tangent vector rЈ͑1͒ starting at the corresponding point ͑1, 1͒.

v

EXAMPLE 3 Find parametric equations for the tangent line to the helix with parametric equations

x ෇ 2 cos t

y ෇ sin t

z෇t

at the point ͑0, 1, ␲͞2͒.

SOLUTION The vector equation of the helix is r͑t͒ ෇ ͗2 cos t, sin t, t͘ , so

rЈ͑t͒ ෇ ͗ Ϫ2 sin t, cos t, 1͘

The parameter value corresponding to the point ͑0, 1, ␲͞2͒ is t ෇ ␲͞2, so the tangent

vector there is rЈ͑␲͞2͒ ෇ ͗Ϫ2, 0, 1͘ . The tangent line is the line through ͑0, 1, ␲͞2͒

parallel to the vector ͗ Ϫ2, 0, 1͘ , so by Equations 12.5.2 its parametric equations are

x ෇ Ϫ2t

y෇1

z෇

ϩt

2

12

The helix and the tangent line in Example 3 are

shown in Figure 3.

8

z

4

0

_1

FIGURE 3

_0.5

y 0

0.5

1

2

_2

0 x

Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).

Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

97817_13_ch13_p872-881.qk_97817_13_ch13_p872-881 11/8/10 11:32 AM Page 874

874

CHAPTER 13

VECTOR FUNCTIONS

In Section 13.4 we will see how rЈ͑t͒ and rЉ͑t͒

can be interpreted as the velocity and acceleration vectors of a particle moving through space

with position vector r͑t͒ at time t.

Just as for real-valued functions, the second derivative of a vector function r is the

derivative of rЈ, that is, rЉ ෇ ͑rЈ͒Ј. For instance, the second derivative of the function in

Example 3 is

rЉ͑t͒ ෇ ͗Ϫ2 cos t, Ϫsin t, 0͘

Differentiation Rules

The next theorem shows that the differentiation formulas for real-valued functions have

their counterparts for vector-valued functions.

3 Theorem Suppose u and v are differentiable vector functions, c is a scalar,

and f is a real-valued function. Then

d

1.

͓u͑t͒ ϩ v͑t͔͒ ෇ uЈ͑t͒ ϩ vЈ͑t͒

dt

d

͓cu͑t͔͒ ෇ cuЈ͑t͒

2.

dt

d

͓ f ͑t͒ u͑t͔͒ ෇ f Ј͑t͒ u͑t͒ ϩ f ͑t͒ uЈ͑t͒

3.

dt

d

͓u͑t͒ ؒ v͑t͔͒ ෇ uЈ͑t͒ ؒ v͑t͒ ϩ u͑t͒ ؒ vЈ͑t͒

4.

dt

d

͓u͑t͒ ϫ v͑t͔͒ ෇ uЈ͑t͒ ϫ v͑t͒ ϩ u͑t͒ ϫ vЈ͑t͒

5.

dt

d

͓u͑ f ͑t͔͒͒ ෇ f Ј͑t͒uЈ͑ f ͑t͒͒

(Chain Rule)

6.

dt

This theorem can be proved either directly from Definition 1 or by using Theorem 2 and

the corresponding differentiation formulas for real-valued functions. The proof of Formula 4

follows; the remaining formulas are left as exercises.

PROOF OF FORMULA 4 Let

u͑t͒ ෇ ͗ f1͑t͒, f2͑t͒, f3͑t͒͘

v͑t͒ ෇ ͗ t1͑t͒, t2͑t͒, t3͑t͒͘

3

Then

u͑t͒ ؒ v͑t͒ ෇ f1͑t͒ t1͑t͒ ϩ f2͑t͒ t2͑t͒ ϩ f3͑t͒ t3͑t͒ ෇

͚ f ͑t͒ t ͑t͒

i

i

i෇1

so the ordinary Product Rule gives

d

d

͓u͑t͒ ؒ v͑t͔͒ ෇

dt

dt

3

3

͚ f ͑t͒ t ͑t͒ ෇ ͚

i

i

i෇1

i෇1

d

͓ fi ͑t͒ ti ͑t͔͒

dt

3

͚ ͓ f Ј͑t͒ t ͑t͒ ϩ f ͑t͒ tЈ͑t͔͒

i

i

i

i

i෇1

3

3

͚ f Ј͑t͒ t ͑t͒ ϩ ͚ f ͑t͒ tЈ͑t͒

i

i෇1

i

i

i

i෇1

෇ uЈ͑t͒ ؒ v͑t͒ ϩ u͑t͒ ؒ vЈ͑t͒

v

Խ

Խ

EXAMPLE 4 Show that if r͑t͒ ෇ c (a constant), then rЈ͑t͒ is orthogonal to r͑t͒ for

all t.

Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).

Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

97817_13_ch13_p872-881.qk_97817_13_ch13_p872-881 11/8/10 11:33 AM Page 875

SECTION 13.2

DERIVATIVES AND INTEGRALS OF VECTOR FUNCTIONS

SOLUTION Since

Խ

r͑t͒ ؒ r͑t͒ ෇ r͑t͒

Խ

2

875

෇ c2

and c 2 is a constant, Formula 4 of Theorem 3 gives

0෇

d

͓r͑t͒ ؒ r͑t͔͒ ෇ rЈ͑t͒ ؒ r͑t͒ ϩ r͑t͒ ؒ rЈ͑t͒ ෇ 2rЈ͑t͒ ؒ r͑t͒

dt

Thus rЈ͑t͒ ؒ r͑t͒ ෇ 0, which says that rЈ͑t͒ is orthogonal to r͑t͒.

Geometrically, this result says that if a curve lies on a sphere with center the origin,

then the tangent vector rЈ͑t͒ is always perpendicular to the position vector r͑t͒.

Integrals

The definite integral of a continuous vector function r͑t͒ can be defined in much the same

way as for real-valued functions except that the integral is a vector. But then we can express

the integral of r in terms of the integrals of its component functions f, t, and h as follows.

(We use the notation of Chapter 4.)

y

b

a

n

r͑t͒ dt ෇ lim

͚ r͑t *͒ ⌬t

i

n l ϱ i෇1

ͫͩ ͚

ͪ ͚ͩ

n

෇ lim

nlϱ

ͪ ͚ͩ

n

f ͑ti*͒ ⌬t i ϩ

i෇1

ͪͬ

n

t͑ti*͒ ⌬t j ϩ

i෇1

h͑ti*͒ ⌬t k

i෇1

and so

y

b

a

r͑t͒ dt ෇

ͩy ͪ ͩy ͪ ͩy ͪ

b

a

f ͑t͒ dt i ϩ

b

a

t͑t͒ dt j ϩ

b

a

h͑t͒ dt k

This means that we can evaluate an integral of a vector function by integrating each component function.

We can extend the Fundamental Theorem of Calculus to continuous vector functions as

follows:

y

b

a

r͑t͒ dt ෇ R͑t͒]ba ෇ R͑b͒ Ϫ R͑a͒

where R is an antiderivative of r, that is, RЈ͑t͒ ෇ r͑t͒. We use the notation x r͑t͒ dt for indefinite integrals (antiderivatives).

EXAMPLE 5 If r͑t͒ ෇ 2 cos t i ϩ sin t j ϩ 2t k, then

ͩ

ͪ ͩy ͪ ͩy ͪ

y r͑t͒ dt ෇ y 2 cos t dt

sin t dt j ϩ

2t dt k

෇ 2 sin t i Ϫ cos t j ϩ t 2 k ϩ C

where C is a vector constant of integration, and

y

␲͞2

0

[

r͑t͒ dt ෇ 2 sin t i Ϫ cos t j ϩ t 2 k

]

␲͞2

0

෇ 2i ϩ j ϩ

␲2

k

4

Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).

Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

97817_13_ch13_p872-881.qk_97817_13_ch13_p872-881 11/8/10 11:33 AM Page 876

876

VECTOR FUNCTIONS

CHAPTER 13

13.2

Exercises

2

1. The figure shows a curve C given by a vector function r͑t͒.

13. r͑t͒ ෇ e t i Ϫ j ϩ ln͑1 ϩ 3t͒ k

(a) Draw the vectors r͑4.5͒ Ϫ r͑4͒ and r͑4.2͒ Ϫ r͑4͒.

(b) Draw the vectors

r͑4.5͒ Ϫ r͑4͒

0.5

14. r͑t͒ ෇ at cos 3t i ϩ b sin 3 t j ϩ c cos 3t k

15. r͑t͒ ෇ a ϩ t b ϩ t 2 c

r͑4.2͒ Ϫ r͑4͒

0.2

and

16. r͑t͒ ෇ t a ϫ ͑b ϩ t c͒

(c) Write expressions for rЈ͑4͒ and the unit tangent vector T(4).

(d) Draw the vector T(4).

17–20 Find the unit tangent vector T͑t͒ at the point with the given

value of the parameter t.

y

R

C

17. r͑t͒ ෇ ͗teϪt, 2 arctan t, 2e t ͘ ,

r(4.5)

1

Q

r(4.2)

t෇1

19. r͑t͒ ෇ cos t i ϩ 3t j ϩ 2 sin 2t k,

t෇0

20. r͑t͒ ෇ sin t i ϩ cos t j ϩ tan t k,

t ෇ ␲͞4

2

2

P

r(4)

0

23–26 Find parametric equations for the tangent line to the curve

with the given parametric equations at the specified point.

function r͑t͒ ෇ ͗t 2, t ͘ , 0 ഛ t ഛ 2, and draw the vectors

r(1), r(1.1), and r(1.1) Ϫ r(1).

(b) Draw the vector rЈ͑1͒ starting at (1, 1), and compare it with

the vector

r͑1.1͒ Ϫ r͑1͒

0.1

24. x ෇ e ,

t

Ϫt

25. x ෇ e

5. r͑t͒ ෇ sin t i ϩ 2 cos t j,

6. r͑t͒ ෇ e i ϩ e

t

Ϫt

j,

7. r͑t͒ ෇ e i ϩ e j,

2t

t

y ෇ eϪt sin t,

t෇0

y ෇ ln͑t 2 ϩ 3͒,

31. x ෇ t cos t, y ෇ t, z ෇ t sin t ;

2

Graphing calculator or computer required

(s3 , 1, 2)

͑Ϫ␲, ␲, 0͒

32. (a) Find the point of intersection of the tangent lines to the

9. r͑t͒ ෇ ͗t sin t, t 2, t cos 2t͘

t

t2

1

12. r͑t͒ ෇

k

1ϩt

1ϩt

1ϩt

͑0, 1, 0͒

30. x ෇ 2 cos t, y ෇ 2 sin t, z ෇ 4 cos 2t ;

t ෇ ␲͞6

11. r͑t͒ ෇ t i ϩ j ϩ 2st k

z ෇ t ; ͑2, ln 4, 1͒

29–31 Find parametric equations for the tangent line to the curve

with the given parametric equations at the specified point. Illustrate by graphing both the curve and the tangent line on a common

screen.

29. x ෇ t, y ෇ e Ϫt, z ෇ 2t Ϫ t 2 ;

t෇0

10. r͑t͒ ෇ ͗tan t, sec t, 1͞t ͘

z ෇ eϪt; ͑1, 0, 1͒

0 ഛ t ഛ ␲, where the tangent line is parallel to the plane

s3 x ϩ y ෇ 1.

9–16 Find the derivative of the vector function.

;

z ෇ te ; ͑1, 0, 0͒

28. Find the point on the curve r͑t͒ ෇ ͗2 cos t, 2 sin t, e t ͘ ,

t ෇ ␲͞4

8. r͑t͒ ෇ ͑1 ϩ cos t͒ i ϩ ͑2 ϩ sin t͒ j,

͑3, 0, 2͒

t2

section of the cylinders x 2 ϩ y 2 ෇ 25 and y 2 ϩ z 2 ෇ 20 at the

point ͑3, 4, 2͒.

CAS

t෇1

cos t,

z ෇ t3 ϩ t;

27. Find a vector equation for the tangent line to the curve of inter-

(a) Sketch the plane curve with the given vector equation.

(b) Find rЈ͑t͒.

(c) Sketch the position vector r͑t͒ and the tangent vector rЈ͑t͒ for

the given value of t.

4. r͑t͒ ෇ ͗t , t ͘,

y ෇ te ,

t

26. x ෇ st 2 ϩ 3 ,

3–8

t ෇ Ϫ1

y ෇ t 3 Ϫ t,

23. x ෇ 1 ϩ 2 st ,

Explain why these vectors are so close to each other in

length and direction.

3

2

22. If r͑t͒ ෇ ͗e 2t, eϪ2t, te 2t ͘ , find T͑0͒, rЉ͑0͒, and rЈ͑t͒ ؒ rЉ͑t͒.

2. (a) Make a large sketch of the curve described by the vector

2

2

21. If r͑t͒ ෇ ͗ t, t 2, t 3 ͘ , find rЈ͑t͒, T͑1͒, rЉ͑t͒, and rЈ͑t͒ ϫ rЉ͑t͒.

x

1

3. r͑t͒ ෇ ͗t Ϫ 2, t 2 ϩ 1 ͘ ,

t෇0

18. r͑t͒ ෇ ͗t ϩ 3t, t ϩ 1, 3t ϩ 4͘,

3

;

curve r͑t͒ ෇ ͗sin ␲ t, 2 sin ␲ t, cos ␲ t͘ at the points where

t ෇ 0 and t ෇ 0.5.

(b) Illustrate by graphing the curve and both tangent lines.

33. The curves r1͑t͒ ෇ ͗t, t 2, t 3 ͘ and r2͑t͒ ෇ ͗sin t, sin 2t, t ͘ inter-

sect at the origin. Find their angle of intersection correct to the

nearest degree.

CAS Computer algebra system required

1. Homework Hints available at stewartcalculus.com

Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).

Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

97817_13_ch13_p872-881.qk_97817_13_ch13_p872-881 11/8/10 11:33 AM Page 877

SECTION 13.3

34. At what point do the curves r1͑t͒ ෇ ͗t, 1 Ϫ t, 3 ϩ t 2 ͘ and

mula 5 of Theorem 3 to find

d

͓u͑t͒ ϫ v͑t͔͒

dt

35– 40 Evaluate the integral.

y

36.

y

37.

2

0

1

0

y

͑t i Ϫ t 3 j ϩ 3t 5 k͒ dt

ͩ

␲͞2

2

ͪ

50. If r͑t͒ ෇ u͑t͒ ϫ v͑t͒, where u and v are the vector functions in

Exercise 49, find rЈ͑2͒.

͑3 sin t cos t i ϩ 3 sin t cos t j ϩ 2 sin t cos t k͒ dt

y (t

39.

y ͑sec

ͩ

2

51. Show that if r is a vector function such that rЉ exists, then

i ϩ tst Ϫ 1 j ϩ t sin ␲ t k) dt

2

1

y

uЈ͑2͒ ෇ ͗3, 0, 4͘ , and v͑t͒ ෇ ͗ t, t 2, t 3 ͘ .

4

2t

k dt

1 ϩ t2

1 ϩ t2

38.

40.

49. Find f Ј͑2͒, where f ͑t͒ ෇ u͑t͒ ؒ v͑t͒, u͑2͒ ෇ ͗ 1, 2, Ϫ1 ͘ ,

2

0

2

877

48. If u and v are the vector functions in Exercise 47, use For-

r2͑s͒ ෇ ͗ 3 Ϫ s, s Ϫ 2, s 2 ͘ intersect? Find their angle of intersection correct to the nearest degree.

35.

ARC LENGTH AND CURVATURE

d

͓r͑t͒ ϫ rЈ͑t͔͒ ෇ r͑t͒ ϫ rЉ͑t͒

dt

t i ϩ t͑t 2 ϩ 1͒3 j ϩ t 2 ln t k͒ dt

ͪ

d

͓u͑t͒ ؒ ͑v͑t͒ ϫ w͑t͔͒͒.

dt

52. Find an expression for

1

t

te i ϩ

k dt

1Ϫt

s1 Ϫ t 2

2t

53. If r͑t͒

41. Find r͑t͒ if rЈ͑t͒ ෇ 2t i ϩ 3t 2 j ϩ st k and r͑1͒ ෇ i ϩ j.

0, show that

d

1

r͑t͒ ෇

r͑t͒ ؒ rЈ͑t͒.

dt

r͑t͒

Խ

[Hint: Խ r͑t͒ Խ2 ෇ r͑t͒ ؒ r͑t͒]

42. Find r͑t͒ if rЈ͑t͒ ෇ t i ϩ e t j ϩ te t k and r͑0͒ ෇ i ϩ j ϩ k.

Խ Խ

Խ

54. If a curve has the property that the position vector r͑t͒ is

43. Prove Formula 1 of Theorem 3.

always perpendicular to the tangent vector rЈ͑t͒, show that

the curve lies on a sphere with center the origin.

44. Prove Formula 3 of Theorem 3.

45. Prove Formula 5 of Theorem 3.

55. If u͑t͒ ෇ r͑t͒ ؒ ͓rЈ͑t͒ ϫ rЉ͑t͔͒, show that

46. Prove Formula 6 of Theorem 3.

uЈ͑t͒ ෇ r͑t͒ ؒ ͓rЈ͑t͒ ϫ rٞ͑t͔͒

47. If u͑t͒ ෇ ͗ sin t, cos t, t͘ and v͑t͒ ෇ ͗t, cos t, sin t͘ , use

Formula 4 of Theorem 3 to find

56. Show that the tangent vector to a curve defined by a vector

d

͓u͑t͒ ؒ v͑t͔͒

dt

function r͑t͒ points in the direction of increasing t. [Hint: Refer

to Figure 1 and consider the cases h Ͼ 0 and h Ͻ 0 separately.]

Arc Length and Curvature

13.3

In Section 10.2 we defined the length of a plane curve with parametric equations x ෇ f ͑t͒,

y ෇ t͑t͒, a ഛ t ഛ b, as the limit of lengths of inscribed polygons and, for the case where

f Ј and tЈ are continuous, we arrived at the formula

1

z

b

L ෇ y s͓ f Ј͑t͔͒ 2 ϩ ͓tЈ͑t͔͒ 2 dt ෇

a

y

b

a

ͱͩ ͪ ͩ ͪ

dx

dt

2

dy

dt

ϩ

2

dt

The length of a space curve is defined in exactly the same way (see Figure 1). Suppose

that the curve has the vector equation r͑t͒ ෇ ͗ f ͑t͒, t͑t͒, h͑t͒͘ , a ഛ t ഛ b, or, equivalently,

the parametric equations x ෇ f ͑t͒, y ෇ t͑t͒, z ෇ h͑t͒, where f Ј, tЈ, and hЈ are continuous. If

the curve is traversed exactly once as t increases from a to b, then it can be shown that its

length is

0

y

x

FIGURE 1

The length of a space curve is the limit

of lengths of inscribed polygons.

2

b

L ෇ y s͓ f Ј͑t͔͒ 2 ϩ ͓tЈ͑t͔͒ 2 ϩ ͓hЈ͑t͔͒ 2 dt

a

y

b

a

ͱͩ ͪ ͩ ͪ ͩ ͪ

dx

dt

2

ϩ

dy

dt

2

ϩ

dz

dt

2

dt

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97817_13_ch13_p872-881.qk_97817_13_ch13_p872-881 11/8/10 11:33 AM Page 878

878

CHAPTER 13

VECTOR FUNCTIONS

Notice that both of the arc length formulas 1 and 2 can be put into the more compact

form

L෇y

3

Խ rЈ͑t͒ Խ dt

b

a

because, for plane curves r͑t͒ ෇ f ͑t͒ i ϩ t͑t͒ j,

Խ rЈ͑t͒ Խ ෇ Խ f Ј͑t͒ i ϩ tЈ͑t͒ j Խ ෇ s͓ f Ј͑t͔͒

2

ϩ ͓ tЈ͑t͔͒ 2

2

ϩ ͓ tЈ͑t͔͒ 2 ϩ ͓hЈ͑t͔͒ 2

and for space curves r͑t͒ ෇ f ͑t͒ i ϩ t͑t͒ j ϩ h͑t͒ k,

Խ rЈ͑t͒ Խ ෇ Խ f Ј͑t͒ i ϩ tЈ͑t͒ j ϩ hЈ͑t͒ k Խ ෇ s͓ f Ј͑t͔͒

v EXAMPLE 1 Find the length of the arc of the circular helix with vector equation

r͑t͒ ෇ cos t i ϩ sin t j ϩ t k from the point ͑1, 0, 0͒ to the point ͑1, 0, 2␲͒.

Figure 2 shows the arc of the helix

whose length is computed in Example 1.

z

SOLUTION Since rЈ͑t͒ ෇ Ϫsin t i ϩ cos t j ϩ k, we have

Խ rЈ͑t͒ Խ ෇ s͑Ϫsin t͒

ϩ cos 2 t ϩ 1 ෇ s2

2

The arc from ͑1, 0, 0͒ to ͑1, 0, 2␲͒ is described by the parameter interval 0 ഛ t ഛ 2␲

and so, from Formula 3, we have

(1, 0, 2π)

(1, 0, 0)

L෇y

x

2␲

0

y

FIGURE 2

Խ rЈ͑t͒ Խ dt ෇ y

2␲

0

s2 dt ෇ 2s2 ␲

A single curve C can be represented by more than one vector function. For instance, the

twisted cubic

r1͑t͒ ෇ ͗t, t 2, t 3 ͘

4

1ഛtഛ2

could also be represented by the function

r2͑u͒ ෇ ͗e u, e 2u, e 3u ͘

5

0 ഛ u ഛ ln 2

where the connection between the parameters t and u is given by t ෇ e u. We say that Equations 4 and 5 are parametrizations of the curve C. If we were to use Equation 3 to compute the length of C using Equations 4 and 5, we would get the same answer. In general, it

can be shown that when Equation 3 is used to compute arc length, the answer is independent of the parametrization that is used.

Now we suppose that C is a curve given by a vector function

r͑t͒ ෇ f ͑t͒i ϩ t͑t͒j ϩ h͑t͒k

where rЈ is continuous and C is traversed exactly once as t increases from a to b. We define

its arc length function s by

z

s(t)

C

6

r(t)

r(a)

0

x

FIGURE 3

aഛtഛb

y

t

Խ

Խ

s͑t͒ ෇ y rЈ͑u͒ du ෇

a

y

t

a

ͱͩ ͪ ͩ ͪ ͩ ͪ

dx

du

2

ϩ

dy

du

2

ϩ

dz

du

2

du

Thus s͑t͒ is the length of the part of C between r͑a͒ and r͑t͒. (See Figure 3.) If we differentiate both sides of Equation 6 using Part 1 of the Fundamental Theorem of Calculus, we

obtain

ds

෇ rЈ͑t͒

7

dt

Խ

Խ

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