5.3 Solutions
231
Case (i):
k
h 2 > 1
Let
k
h 2 − 1 = n 2
d 2 u
d θ 2 − n
2 u = 0
which has the solution u = Ae nθ + Be −nθ , where the constants A and B
depend on the initial conditions of projection. If these are such that either A
or B is zero then the path is an equiangular spiral
Case (ii):
k
h 2 = 1, the equation becomes
d 2 u
d θ 2 = 0, whose solution is v =
Aθ + B, a curve known as the reciprocal spiral curve.
Case (iii):
k
h 2 < 1. Let 1 −
k
h 2 = n 2 , the equation becomes
d 2 u
d θ 2 + n 2 u = 0
whose solution is u = A cos nθ + B sin nθ , a curve with infinite branches.
5.52 1
p 2 =
1
r 2 +
1
r 4
dr
d θ
2
(1)
r
2
= a
2 cos
2
θ
(2)
r
dr
d θ
= −a
2 sin
2
θ
(3)
∴
1
r 4
dr
d θ
2
=
a 4
r 6 sin
2 2θ =
a 4
r 6
1 −
r 4
a 4
=
a 4
r 6 −
1
r 2
(4)
From (1) and (4)
1
p 2 =
a 4
r 6
or p =
r 3
a 2
(5)
∴
d p
dr
=
3r 2
a 2
(6)
f = −
h 2
p 3
d p
dr
= −
3h 2 a 4
r 7
where we have used (5) and (6).
5.53 The polar equation of a circle with the origin on the circumference is r =
2a cos θ where a is the radius of the circle, Fig. 5.21.