I6-35. (a) t 22 ˆe 1 +tˆe 2 −t 33 ˆe 3 +~c

I6-36.

~a=d~vdt= costˆe 1 + sintˆe 2

~v=~v(t) = sintˆe 1 −costˆe 2 +~c

~v(0) =−ˆe 2 +~c= 2ˆe 3 =⇒ ~c= 2ˆe 3 +ˆe 2

~v=d~r

dt

= sintˆe 1 + (1−cost)ˆe 2 + 2ˆe 3

~r=~r(t) =−costeˆ 1 + (t−sint)ˆe 2 + 2tˆe 3

I6-39. (a)ω= 5ˆe 3

(b)~v=ω×~r=−5 sin5tˆe 1 + 5 cos 5tˆe 2

I6-40. ~r=etˆe 1 + costˆe 2 + sintˆe 3 with~v=d~rdt and~a=d~vdt=ddt^2 ~r 2

I6-41.

C~=C~(x) =~r(x) +αN~(x) =xˆe 1 +yeˆ 2 +(1 + (y

′) (^2) )
y′′
[−y′ˆe 1 +ˆe 2 ]
Ifx=x(t)andy=y(t), then
y′=
dy
dx=
dy
dt
dx
dt

y ̇
x ̇
y′′=d
(^2) y
dx^2
= d
dx
(
dy
dx
)
= d
dx
(
y ̇
x ̇
)

d
dx
(y ̇
x ̇
)
dx
dt
=x ̇y ̈−y ̇ ̈x
( ̇x)^3
Substitute these derivatives intoC~=C~(x)and simplify.
I6-42. (b) C~=C~(x) =xˆe 1 +exˆe 2 +
(1 +e^2 x)
ex [−e
xˆe 1 +ˆe 2 ]
I6-43. When eˆ=ˆeA=
1
|A~|
A~, then ˆeA·A~=|A~|
I6-47.
∂U
∂x= (4xy+y
(^2) )ˆe 1 + (y+ 6xy)ˆe 2 and ∂^2 U
∂x^2 = 4y
ˆe 1 + 6yˆe 2
I6-48. If~v=d~rdt=ω×~r, then
dx
dt
ˆe 1 +dy
dt
ˆe 2 +dz
dt
ˆe 3 = (zω 2 −yω 3 )ˆe 1 + (xω 3 −zω 1 )ˆe 2 + (yω 1 −xω 2 )ˆe 3
Equate like components to show result.
Solutions Chapter 6