Thomas Calculus 13th [Solutions]

948 Chapter 13 Vector-Valued Functions and Motion in Space
13.
r
3 2
t 2 4 2 2
i t j, t 0 v t i t j v t t t t
3 2
1, since
2 2
d T
2
1 t d T
i j
1 t 1 t 1
dt 2
3 2
2
3 2 dt 2
3/2
2
3/2
2
3 2
t 1 t 1 t 1 t 1 t 1 t 1
t
0 T v t i 1 j
v 2 2
t 1 t 1
d T
dt
N 1 i t j;
1 d T 1 1 1 .
d T 2 2
2 2 3 2
t 1 t 1 v dt
t t 1 t 1 2
t t 1
dt
14.
3 3 2 2
r cos t i sin t j, 0 t v 3cos t sin t i 3sin t cos t j
2
2
2
2
2
4 2 4 2
v 3cos sin 3sin cos 9cos sin 9sin cos 3cos sin , since
t t t t t t t t t t
2 2
cos t sin t d T
d
sin cos sin cos 1
dt
t t T
T v i j i j
v
dt
t t
d T
dt
N sin t i cos t j;
1 d T 1 1 1 .
d T
v dt 3cost sin t 3cost sin t
dt
0 t
2
15.
2 2
r ti a cosh t j, a 0 v i sinh t j v 1 sinh t cosh t cosh t
a a a a a
1 1 2
sech t tanh t d T
T v i j sech t tanh t i sech t j
v a a dt a a a a a
d T
T 1 2 2 1 4 1
dt
dt 2 a a 2 a a a d T
a
a
dt
a a
d
sech t tanh t sech t sech t N tanh t i sech t j;
1 d T 1 1 1 2
sech t sech t .
v dt cosh
t a a a a
a
16.
2
2
r cosh i sinh j k v sinh i cosh j k v sinh cosh 1 2 cosh
1 tanh 1 1 sech d T 1 2
T v
t i j t k sech t i 1 sech t tanh t k
v 2 2 2 dt 2 2
d T
dt
t t t t t t t t
1 4 1 2 2 1
dt
sech t sech t tanh t sech t N sech t i tanh t k;
2 2 2
d T
1 d T 1 1 sech 1 2
t
dt 2cosh 2 2
sech t
v
.
t
d T
dt
17.
2
2a
2 2
3 2
y ax y 2ax y 2 a ; from Exercise 5(a), ( x) 2a 1 4a x
2 2
3 2
1 4a x
2 2
5 2
3
2
( x) 2a 1 4a x 8 a x ; thus, ( x) 0 x 0. Now, ( x ) 0 for x 0 and ( x ) 0 for
2
x 0 so that ( x ) has an absolute maximum at x 0 which is the vertex of the parabola. Since x 0 is the
only critical point for ( x ), the curvature has no minimum value.
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