
6-26 Calculus – Differentiation and Integration
= =
∴ =
=
2
2 2
2
2
2
2 2
2
sin cos
sin
cot
cot cot .
θ θ
θ
θ
φ
θ
φ
θ
or
Example 6.25 Show that the radius vector is inclined at a constant angle to the tangent at any point on
the equiangular spiral r = ae
bq
.
Solution: Given
r ae
b
=
q
. (1)
Differentiating Eq. (1) w.r.t q, we get
1 1
1
1
1
r
dr
d
b r
d
dr b
b
b
θ
θ
φ
φ
= =
∴ =
=
−
or
ortan
tan .
This is a constant. This property of equiangular spiral justifies the adjective “Equiangular”.
Angle of Intersection of Two Polar Curves
Assume that the two curves r = f(q) and r = F(q) intersect at P and let the values of f at the point P
for the two curves be f
1
and f
2
, respectively ...