Thomas Calculus 13th [Solutions]

950 Chapter 13 Vector-Valued Functions and Motion in Space
(1) 2 1 1
2
2 2
2. The circle of curvature is tangent to the curve at P 0, 2 circle has same
tangent as the curve v(1) 2i is tangent to the circle the center lies on the y -axis. If t 1( t 0), then
2 2 2
2
t 1
t 1 0 t 2t 1 0 t 1 2t 2 since 0 1 2 1 2 2
t
t t t
t t
y on
both sides of 0, 2 the curve is concave down center of circle of curvature is 0, 4
2 2
x y 4 4 is an equation of the circle of curvature
23.
2
y x f ( x) 2x and f ( x) 2
2 2
2
3 2
2
3 2
1 (2 x) 1 4x
24.
4
3
y x f ( x)
x and
4
1
2
3x
2
3 2
6
3 2
3
x
1
2
3x
x
2
f ( x) 3x
25. y sin x f ( x) cos x and f ( x) sin x
sin x
sin x
2
3 2
2
3 2
1 cos x 1 cos x
x
x
x
26. y e f ( x)
e and f ( x)
e
1
x
| e |
e
2
3 2
2
3 2
x
x
e
1 e
x
1 d T
2
27. We will use the formula
( a)
to find the curvature at the point a,
a .
v( a)
dt
By Example 4 in Section 13.4,
2 dT
2
3/2
v( t) 1 4t
and 2 1 4t
2 ti
j .
dt
1 d T 2 2
3/2
2 2
At t a this gives
( a) 1 4a 1 4a
. Thus the radius of
v( a) dt 2 2
3/2
1 4a 1 4a
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