1
Test 1 solutions
Phz 3113 Fall 2007
1. Expand x/(ex − 1) to order x2 for x ¿ 1.
x2 /2!
1+x+
x
+ x3 /3! + · · · − 1
(1)
To order x2 ,
1
1
'
,
2
(1 + x/2! + x /3! + . . . )
1 + (x/2! + x2 /3!)
(2)
which is an alternating geometric series in (x/2! + x2 /3!), so
= 1 − (x/2! + x2 /3!) + (x/2! + x2 /3!)2 ' 1 − x/2 + x2 /12
2. The equation of state for a van der Waals gas is
³
a ´
p + 2 (V − b) = RT,
V
(3)
(4)
where a, b and R are constants. Consider two experiments on such a gas confined to a cylinder
where you may control p, V and/or T .
(a) Hold T constant and find dV /dp.
h³
i
a ´
d p + 2 (V − b) = d(RT ) = 0
V
so
dV
=
dp
b−V
a
V2
+
2(b−V )a
V3
+p
(5)
=
(b − V )V 3
pV 3 − aV + 2ab
(6)
=
RV 3
pV 3 − aV + 2ab
(7)
(b) Hold p constant and find dV /dT .
Similarly if dp = 0,
dV
=
dT
R
a
V2
+
2(b−V )a
V3
+p
3. Change variables x = u + v, y = u − v, to rewrite the differential equation
∂ 2w ∂ 2w
−
=1
∂x2
∂y 2
(8)
2
in terms of u and v (no need to solve the equation).
Sketch solution. First invert u = (x + y)/2, v = (x − y)/2. calculate partial derivatives:
·
¸
∂w ∂u ∂w ∂v
1 ∂w ∂w
∂w
=
+
=
+
∂x
∂u ∂x
∂v ∂x
2 ∂u
∂v
µ
¶
∂
1 ∂
∂
⇒
=
+
∂x
2 ∂u ∂v
(9)
and similarly for y
∂
1
=
∂y
2
µ
∂
∂
−
∂u ∂v
¶
(10)
The 2nd partials are e.g.
µ
¶µ
¶
∂2w
1 ∂
∂
∂
∂
=
+
+
w
∂x2
4 ∂u ∂v
∂u ∂v
·
¸
1 ∂ 2w ∂ 2w
∂ 2w
=
+
+2
4 ∂u2
∂v 2
∂u∂v
(11)
and similarly for y except the coefficient of the mixed partial derivative is negative. Constructing
2
∂2w
− ∂∂yw2 = 1, we find
∂x2
∂ 2w
= 1.
∂u∂v
(12)
4. Evaluate the integral
Z
Z
π
π
dy
y=0
dx
x=y
sin x
.
x
Reverse order of integrations:
Z π
Z
Z π
sin x x
dx
dy =
dx sin x = − cos x|π0 = 2.
x
0
0
0
(13)
(14)
~ ·A
~ = 0 and ∇
~ ·B
~ = 0, show that
5. If ∇
~ × (A
~ × B)
~ = (B
~ · ∇)
~ A
~ − (A
~ · ∇)
~ B.
~
∇
[Hint: ²ijk ²i`m = δj` δkm − δjm δk` ]
(15)
3
~ × (A
~ × B)]
~ i = ²ijk ∇j (A
~ × B)
~ k = ²ijk ²kmn Am Bn = ²kij ²kmn ∇j Am Bn
[∇
= (δim δjn − δin δjm )∇j Am Bn = ∇j Ai Bj − ∇j Aj Bi
~ · B) + (B
~ · ∇)A
~ i − Bi (∇
~ · A)
~ − (A
~ · ∇)B
~ i = (B
~ · ∇)A
~ i − (A
~ · ∇)B
~ i,
= Ai (∇
where in the last step I used the info given that both vectors had zero divergence, and the
product rule of differentiation.
6. Look for a minimum of the function 1/x + 4/y + 9/z for x, y, z > 0 and x + y + z = 12.
Lagrange multipliers:
F = 1/x + 4/y + 9/z + λ(x + y + z − 12),
(16)
so minimize:
∂F
1
=− 2 +λ=0 ;
∂x
x
∂F
4
=− 2 +λ=0 ;
∂y
y
∂F
9
= − 2 + λ = 0.
∂z
z
(17)
Together with the constraint equation x + y + z = 12 this system can be easily solved by noting
that the solutions should be positive. Therefore by eliminating λ we find x = y/2 = z/3. The
constraint equation is then x + 2x + 3x = 6, so x = 2, y = 4, z = 6 is a solution. Function takes
minimum value of 3 here.
~
7. (Extra
p credit) Consider the vector V = 4y î + xĵ + 2z k̂ and the scalar field ψ(x, y, z) =
1/ x2 + y 2 + z 2 .
~ × V~ = −3k̂
(a) show ∇
²ijk ∇j vk For i = 1, ²1jk ∇j vk = ∇2 v3 − ∇3 v2 = 0; for i = 2, ²2jk ∇j vk = ∇3 v1 − ∇1 v3 = 0;
for i = 3, ²3jk ∇j vk = ∇1 v2 − ∇2 v1 = 1 − 4 = −3. So answer is −3ẑ.
(b) evaluate
R
V~ · d~r from the origin (0,0,0) to (1,1,1) along the line x = t, y = t2 , z = t3 .
Since curl ~v 6= 0, integral depends on path in general.
Z Z Z 1,1,1
Z 1
(4y î + xĵ + 2z k̂) · (dxî + dy ĵ + dz k̂) =
(4t2 dt + t · 2tdt + 2t3 · 3t2 dt)
0,0,0
0
·
1
1
= 6 t3 + t6
3
6
¸1
=3
0
(18)
4
~ and ∇
~ × ∇ψ.
~
(c) evaluate ∇ψ
~ = −(x, y, z)/(x2 + y 2 + z 2 )3/2 = − ~r
∇ψ
r3
(19)
¯
¯
¯
¯ x̂
ŷ
ẑ
¯
¯
~
~
¯
∇ × ∇ψ = ¯ ∂x ∂y ∂z ¯¯ ,
¯ ∂ x ψ ∂ y ψ ∂z ψ ¯
(20)
and
which vanishes by equality of mixed partial derivatives. You can also do this problem with
²ijk notation if you like.
8. (Extra credit.) Calculate the radii of convergence of the following series:
(a)
∞
X
(nx)n
n=1
Use ratio test:
n!
¯
¯ ¯
¯
¯ an+1 ¯ ¯ (n + 1)n+1 xn+1 n! ¯
¯
¯
¯
¯
ρn = ¯
=
an ¯ ¯
(n + 1)!
nn xn ¯
µ
¶n
1
|x|(n + 1)n
= |x| 1 +
=
nn
n
µ
¶n
1
ρ = lim ρn = |x| lim 1 +
= |x|e
n→∞
n→∞
n
(21)
(22)
(23)
(24)
(b)
∞
X
n=1
Use ratio test:
xn
n2 + 1
¯ ¯
¯
¯
n+1
2
2
¯
¯ an+1 ¯ ¯
x
n
+
1
¯=¯
¯ = |x| n + 1
ρn = ¯¯
an ¯ ¯ n2 + 2n + 2 x2 ¯
n2 + 2n + 2
ρ = lim ρn = |x| < 1
n→∞
(25)
(26)
(27)