I9-24. |~r|= (x^2 +y^2 +z^2 )^1 /^2 = (~r·~r)^1 /^2 so that|~r|ν= (x^2 +y^2 +z^2 )ν/^2 =φ
∂φ
∂x=ν
2(x^2 +y^2 +z^2 )ν/^2 −^1 (2x) ∂φ
∂y=ν
2(x^2 +y^2 +z^2 )ν/^2 −^1 (2y ∂φ
∂z=ν
2(x^2 +y^2 +z^2 )ν/^2 −^1 (2z)so that
∇|~r|ν=∂φ
∂xˆe^1 +∂φ
∂yeˆ^2 +∂φ
∂zˆe^3 =ν|~r|ν− (^2) (xˆe 1 +yˆe 2 +zˆe 3 )
=ν|~r|ν−^2 ~r=ν|~r|ν−^2 |~r|~r
|~r|
=ν|~r|ν−^1 ˆe~r
I9-26. In problem 9-24 replace~rby~r−~r 0 and then use the results from problem
9-25 to obtain the result for part (b).
I9-27. LetM=∂U∂p andN=∂U∂V+P and show∂M∂V =∂p ∂v∂^2 U and∂N∂p=∂v ∂p∂^2 U + 1
Here ∂M∂v 6 =∂N∂p so that the line integral is path dependent.
I9-29. I=−^34 (5 +π)
I9-30. Potential energy^12 Kx^2
Kinetic +potential energy is constant or^12 mv^2 +^12 Kx^2 =constant
I9-32. (a) T=T(x) =c 1 x+c 2 ,
T(0) =c 2 =T 0 andT(L) =c 1 L+c 2 =T 1 =⇒ T=T(x) = (T^1 −LT^0 )x+T 0
I9-33. φ= 3x^2 z+ 4y^2
I9-34.
(a)F~ 1 =
Gm 1 m 2
|~r|^3 ~rwhere|~r|
(^2) =x (^2) +y (^2) +z 2
(b)F~ 1 =
Gm 1 m 2
|~r−~r 1 |^3 (~r−~r^1 )
where|~r−~r 1 |^2 = (x−x 1 )^2 + (y−y 1 )^2 + (z−z 1 )^2
I9-39. ~y=~yc+~yp=C~ 1 (1) +C~ 2 (t)−sinteˆ 1 + costˆe 2
I9-40. ~y 1 =C~ 1 sint−C~ 2 cost, ~y 2 =C~ 1 cost+C~ 2 sint
I9-41.
d~r
dt=r^0 e
θcotαω
︸ ︷︷ ︸
rω
ˆeθ+r 0 eθcotα
dθ
︸ ︷︷dtcotα︸
rωcotα
ˆer
Solutions Chapter 9