James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Section 13.3 Arc Length and Curvature 861
48. If u and v are the vector functions in Exercise 47, use For-
34. At what point do the curves r 1std − kt, 1 2 t, 3 1 t 2 l and
d
dt fustd ? vstdg [Hint: Refer to Figure 1 and consider the cases h . 0 and
h , 0 separately.]
r 2ssd − k3 2 s, s 2 2, s 2 l intersect? Find their angle of
mula 5 of Theorem 3 to find
intersection correct to the nearest degree.
d
fustd 3 vstdg
dt
35–40 Evaluate the integral.
35. y 2
st i 2 t 3 j 1 3t 5 49. Find f 9s2d, where f std − ustd ? vstd, us2d − k1, 2, 21l,
kd dt
0
u9s2d − k3, 0, 4l, and vstd − kt, t 2 , t 3 l.
36. y 4
(2t 3y2 i 1 st 1 1dst k) dt
50. If rstd − ustd 3 vstd, where u and v are the vector functions
1
in Exercise 49, find r9s2d.
37. y S D 1 0 t 1 1 i 1 1
t 2 1 1 j 1 t
t 2 1 1 k 51. If rstd − a cos t 1 b sin t, where a and b are constant
dt
vectors, show that rstd 3 r9std − a 3 b.
38. y y4
52. If r is the vector function in Exercise 51, show that
ssec t tan t i 1 t cos 2t j 1 sin 2 2t cos 2t kd dt
0
r0std 1 2 rstd − 0.
39. y ssec 2 t i 1 tst 2 1 1d 3 j 1 t 2 ln t kd dt
53. Show that if r is a vector function such that r0 exists, then
40.
d
y Ste 2t i 1 t
frstd 3 r9stdg − rstd 3 r0std
1 2 t j 1 1
D
s1 2 t k dt
dt
2
54. Find an expression for d fustd ? svstd 3 wstddg.
dt
41. Find rstd if r9std − 2t i 1 3t 2 j 1 st k and rs1d − i 1 j.
55. If rstd ± 0, show that d | dt rstd | − 1 rstd ? r9std.
| rstd |
42. Find rstd if r9std − t i 1 e t j 1 te t k and rs0d − i 1 j 1 k.
[Hint: | rstd | 2 − rstd ? rstd]
43. Prove Formula 1 of Theorem 3.
56. If a curve has the property that the position vector rstd is
44. Prove Formula 3 of Theorem 3.
always perpendicular to the tangent vector r9std, show that
the curve lies on a sphere with center the origin.
45. Prove Formula 5 of Theorem 3.
57. If ustd − rstd ? fr9std 3 r0stdg, show that
46. Prove Formula 6 of Theorem 3.
u9std − rstd ? fr9std 3 r-stdg
47. If ustd − ksin t, cos t, tl and vstd − kt, cos t, sin tl, use
58. Show that the tangent vector to a curve defined by a vector
Formula 4 of Theorem 3 to find
function rstd points in the direction of increasing t.
z
0
y
x
FIGURE 1
The length of a space curve is the limit
of lengths of inscribed polygons.
Length of a Curve
In Section 10.2 we defined the length of a plane curve with parametric equations x − f std,
y − tstd, a < t < b, as the limit of lengths of inscribed polygons and, for the case where
f 9 and t9 are continuous, we arrived at the formula
1 L − y b
sf f 9stdg 2 1 ft9stdg 2 b
dt − y
a
a
ÎS
dtD
dx 2
1S
dy
dtD2
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 rstd − k f std, tstd, hstdl, a < t < b, or, equivalently,
the parametric equations x − f std, y − tstd, z − hstd, where f 9, t9, and h9 are continuous.
If the curve is traversed exactly once as t increases from a to b, then it can be shown
Copyright 2016 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.