Thomas Calculus 13th [Solutions]

Chapter 13 Additional and Advanced Exercises 969
CHAPTER 13 ADDITIONAL AND ADVANCED EXERCISES
1. (a) r( ) ( a cos ) i ( a sin ) j b k dr
( a sin ) i ( a cos ) j bk
d ;
dt
dt
v
2 2
2gz 2gb 4 gb gb
2gz dr
a b d d d
2
dt dt dt a b a b dt 2 a b a b
(b)
2 2 2 2 2 2
2 2 2 2 2 2 2 2
d 2gb d 2gb 1 2 2gb
dt 2 t C;
dt t 1 2 2
0 0 C gb
0 2
t 2 2
a b a b a b
a b
2 a
gbt
2
2 2
b
2 2
gb t
; z b z
2 a b
2 2
gbt
(c) v( t) dr
( asin ) i ( a cos ) j bk d ( asin ) i ( a cos ) j bk , from part (b)
dt dt 2 2
a b
( asin ) i ( a cos ) j bk
gbt gbt
v( t) T;
2 2 2 2 2 2
a b a b a b
2 2
d r
2
( a cos ) i ( asin ) j d ( a sin ) i ( a cos ) j bk
d
2 dt
2
dt
dt
2
gbt
gb
2 2 ( a cos ) i ( a sin ) j [( a sin ) i ( a cos ) j bk
]
2 2
a b a b
2
( a sin ) i ( a cos ) j b k gb a
gbt ( cos ) i (sin ) j gb T a
gbt N (there is
2 2 2 2 2 2 2 2
2 2
a b a b a b a b a b
no component in the direction of B).
2. (a) r( ) a cos i a sin j b k dr
a cos a sin i asin a cos j bk
d ;
dt
dt
2 2 2 2
1 2 2gb
v 2gz dr
a a b d d
dt dt dt 2 2 2 2
a a b
t t 2 2 2 2
1 2 t 2 2 2 2
1 2
2 2 2 2
1 2
(b) s v dt a a b d dt a a b d a a u b du
0 0 dt 0 0
2 2
2 2 2
a a b u du a a c u du , where
0
2
a
0
a 2 2 2 2 2 2
ln
ln
2
c c c c c
2 2 2
a b 2 2 2 2
c s a u c u c ln u c u
a
2 2
0
(1 e) r dr (1 e) r ( esin ) dr (1 e) r ( esin )
0 0 (1 e) r
1 ecos d 2
0( esin ) 0 sin 0
(1 ecos ) d
(1 ecos )
0 or . Note that dr 0 when sin 0 and dr 0 when sin 0. Since sin 0 on 0
d
d
(1 e)
r0
and sin 0 on 0 , r is a minimum when 0 and r(0)
r
1 ecos0
0
0 0
0
3. r ;
2
4 (a) f ( x) x 1 1 sin x 0 f (0) 1 and f (2) 2 1 1 sin 2 1 since sin 2 1; since f is continuous
2
2 2
on [0, 2], the Intermediate Value Theorem implies there is a root between 0 and 2
(b) Root 1.4987011335179
5. (a) v x i y j and v r ur
r u r (cos ) i (sin ) j r ( sin ) i (cos ) j v i x and
v i r cos r sin x r cos r sin ; v j y and
v j r sin r cos y r sin r cos
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