2: Derivatives and Integrals of Vector Functions
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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
jϩ
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
zt
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 rЈ͑͞2͒ ͗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
y1
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.
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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
i1
so the ordinary Product Rule gives
d
d
͓u͑t͒ ؒ v͑t͔͒
dt
dt
3
3
͚ f ͑t͒ t ͑t͒ ͚
i
i
i1
i1
d
͓ fi ͑t͒ ti ͑t͔͒
dt
3
͚ ͓ f Ј͑t͒ t ͑t͒ ϩ f ͑t͒ tЈ͑t͔͒
i
i
i
i
i1
3
3
͚ f Ј͑t͒ t ͑t͒ ϩ ͚ f ͑t͒ tЈ͑t͒
i
i1
i
i
i
i1
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.
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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 i1
ͫͩ ͚
ͪ ͚ͩ
n
lim
nlϱ
ͪ ͚ͩ
n
f ͑ti*͒ ⌬t i ϩ
i1
ͪͬ
n
t͑ti*͒ ⌬t j ϩ
i1
h͑ti*͒ ⌬t k
i1
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
iϩ
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)
t1
19. r͑t͒ cos t i ϩ 3t j ϩ 2 sin 2t k,
t0
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,
t0
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
iϩ
jϩ
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 ;
t0
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
t1
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 ͘ ,
t0
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
jϩ
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
jϩ
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
Ly
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, 2͒.
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, 2͒ is described by the parameter interval 0 ഛ t ഛ 2
and so, from Formula 3, we have
(1, 0, 2π)
(1, 0, 0)
Ly
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
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.