Question

Angle of intersection between two polar curves given by r=a(1+sin⁡θ) & r=a(1-sin⁡θ) is given by ________

(a) \(\frac{π}{4}\)

(b) \(\frac{π}{2}\)

(c) Π

(d) 0

I have been asked this question in an interview.

The query is from Polar Curves in division Differential Calculus of Engineering Mathematics

Answer 1

Correct choice is (b) \(\frac{π}{2}\)

For explanation: r=a(1+sin⁡θ) : r=a(1-sin⁡θ)

taking logarithm on both the equations

log⁡r = log⁡a + log⁡(1+sin⁡θ) : log⁡r = log⁡a + log⁡(1-sin⁡θ)

differentiating on both side we get

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

\(cot⁡∅1 = \frac{cos⁡θ}{1+sin⁡θ} : cot⁡∅2 = \frac{-cos⁡θ}{1-sin⁡θ}\)

where ∅1&∅2 are the angle between tangent & the vector respectively

\(tan⁡∅1 = \frac{1+sin⁡θ}{cos⁡θ} : tan⁡∅2 = \frac{1-sin⁡θ}{cos⁡θ}\)

\(tan⁡∅1 . tan⁡∅2 = \frac{1+sin⁡θ}{cos⁡θ} . \frac{1-sin⁡θ}{-cos⁡θ} = \frac{1-sin^2 θ}{-cos^2 θ} = -\frac{cos^2 θ}{cos^2 θ} = -1\)

above is the condition of orthogonality of two polar curves thus

|∅1-∅2|=\(\frac{π}{2}\).