PHZ3113–Introduction to Theoretical Physics
Fall 2008
Problem Set 2 Solutions
Monday Sept. 15, 2008
1. (a) As x → ∞, applying L’Hospital we find
xn
nxn−1
const
→
→ . . . ... x → 0
x
x
e
e
e
(1)
1
ln x
→ p.
p
x
px
(2)
(b) similarly
If p > 0 this → 0.
R ∞ −ax
k
m sin kx = a2 +k
2 to find
0 e
2. Take derivative wrt a of
−
So
R∞
0
Z ∞
0
xe−ax sin kxdx =
Z ∞
0
xe−ax sin kx = −
2ak
.
(a2 +k2 )2
e−ax x cos kx =
(a2
2ak
.
+ k 2 )2
(3)
Differentiate wrt k to obtain
1
2k 2
a2 − k 2
−
=
a2 + k 2 (a2 + k 2 )2
(a2 + k 2 )2
(4)
3. First let’s calculate the total differential
2 −q 2
dr = e−p
(−2pdp − 2qdq)
(5)
as well as the differentials dp = es ds and dq = −e−s ds. Substitute
2
2
dr = e−p −q (−2pes + 2qe−s )ds
dr
2
2
⇒
= 2e−p −q (e−2s − e2s )
ds
∂2u
∂ y
= ∂x
e sin x
∂x∂y
∂2u
∂2u
so ∂x∂y = ∂x∂y as must
4. (a)
= −ey cos x, while
∂2u
∂y∂x
=
∂
∂y
(6)
(7)
− ey cos x = −ey cos x,
be true for any smooth function.
(b)
∂ 2u
∂ 2u
y
=
−e
sin
x
and
= ey sin x
∂x2
∂y 2
so indeed the sum
∂2u
∂x2
+
∂2u
∂y 2
= 0 for this case.
1
(8)
5. (a) First, solve for y explicitly:
1
y=
2
Ã
√
!
± 12x2 + x + 20
√
+1
x
(9)
so derivative is
dy
5 − 3x2
= ± 3/2 √
dx
x
12x2 + x + 20
(10)
Now do implicitly:
dx y 2 + 2xy dy − 6x dx = y dx + x dy
dy
y + 6x − y 2
⇒
=
.
dx
2xy − x
(11)
(12)
To check that this agrees with the explicit method, substitute back in for y as
above:
dy
=
dx
1
2
= ±
³ √
2
± 12x
√ +x+20
x
2x
x3/2
√
´
+ 1 + 6x −
³ ³ √
³ ³ √
1
2
2
2
± 12x
√ +x+20
x
2
± 12x
√ +x+20
x
1
2
+1
´´
+1
−x
5 − 3x
12x2 + x + 20
´´2
(13)
(14)
it works!
Solution is y = − 2(x+2)
, so dy/dx =
(b) x = 3y−4
y+2
x−3
derivative, we find the differential
(y + 2) dx + x dy = 3 dy
⇒
10
.
(x−3)2
To take the implicit
dy
y+2
=−
.
dx
x−3
(15)
Check again by substituting for y:
2(x+2)
2 − x−3
dy
10
=−
=
dx
x−3
(x − 3)2
2
OK
(16)