Question

2. Show that in the equiangular spiral r = ae cot a, the tangent is inclined at a constant angle to the radius vector.​

Answer 1

Angle between radius vector and tangent:

Let P (r, θ) be any point on the curve r = f(θ) and Q (r + Δr, θ + Δθ) be any other point on the curve, where Δ indicates very small quantity. If φ be the angle between the radius vector and the tangent to the curve f(θ) at T, then

tanφ = r / (dr/dθ) = r dθ/dr

Solution:

The given curve is

r = a e^(θ cotα) ..... (1)

Differentiating both sides of (1) with respect to θ, we get

dr/dθ = a e^(θ cotα) cotα

or, dr/dθ = r cotα

or, 1 / (dr/dθ) = 1 / (r cotα)

or, r / (dr/dθ) = r / (r cotα)

or, tanφ = tanα

or, φ = α = constant

Hence proved.