Question

For the curve in polar form \(\sqrt{\frac{r}{a}} = sec⁡(\frac{θ}{2}) \,the\, \,value\, \,of\, \frac{ds}{dθ}\) is _____

(a) r sec θ

(b) r sec \((\frac{θ}{2})\)

(c) r sec (2θ)

(d) r cosec \((\frac{θ}{2})\)

This question was addressed to me at a job interview.

The above asked question is from Derivative of Arc Length topic in chapter Differential Calculus of Engineering Mathematics

Answer 1

Right choice is (c) r sec (2θ)

For explanation I would say: Squaring the given curve on both side i.e \(r=a sec ^2 (\frac{θ}{2})\)…(1)

\(\frac{dr}{dθ} = a.2 sec⁡(\frac{θ}{2}) sec⁡(\frac{θ}{2}) tan (\frac{θ}{2}).1/2 = a sec^2 (\frac{θ}{2}) tan(\frac{θ}{2}) \)

from (1)

\(\frac{dr}{dθ} = r tan (\frac{θ}{2}) \,the\, \,equation\, \,for\, \frac{ds}{dθ} \,is\, = \sqrt{(r^2+(\frac{dr}{dθ})^2)} = \sqrt{r^2(1+ tan^2 (\frac{θ}{2}))} \)

\(\frac{ds}{dθ} = r sec (\frac{θ}{2}).\)