Question

The angle between Radius vector r=a(1-cos⁡θ)and tangent to the curve is ∅ given by _______

(a) ∅=\(\frac{π}{2}\)

(b) ∅=π

(c) ∅=\(-\frac{π}{2}\)

(d) ∅=0

This question was posed to me in an international level competition.

My question is taken from Polar Curves in section Differential Calculus of Engineering Mathematics

Answer 1

Right answer is (a) ∅=\(\frac{π}{2}\)

The explanation is: r= a(1-cos⁡θ)

taking logarithms on both sides we get,

log⁡r = log⁡a + log⁡(1-cos⁡θ)

differentiating w.r.t θ we get,

\( \frac{1}{r} \frac{dr}{dθ} = 0 + \frac{sin⁡θ}{1-cos⁡θ}\)

\(\frac{1}{r} \frac{dr}{dθ} = \frac{2 sin ⁡\frac{θ}{2} cos⁡\frac{θ}{2}}{2sin^2 \frac{θ}{2}} = cot⁡\frac{θ}{2}\) ..(1),

but \(cot⁡∅ = \frac{1}{r} \frac{dr}{dθ}\)….(2)

 From (1)&(2)

∅=\(\frac{π}{2}\).