Question

Which among the following is true for the curve r^n = a sin⁡nθ?

(a) Given family of curve is Self orthogonal

(b) Orthogonal trajectory is r^n=k cos⁡nθ  where k is an constant

(c) Orthogonal trajectory is r^n=k cosec⁡nθ  where k is an constant

(d) Orthogonal trajectory is r^n=k sin⁡nθ  where k is an constant

This question was posed to me during an interview for a job.

This intriguing question comes from Orthogonal Trajectories topic in chapter Ordinary Differential Equations – First Order & First Degree of Engineering Mathematics

Answer 1

The correct option is (b) Orthogonal trajectory is r^n=k cos⁡nθ  where k is an constant

The best explanation: Consider r^n = a sin⁡nθ..(1) –> n log r = log a + log(sin⁡nθ) …differentiating w.r.t θ

\( \frac{n}{r}  \frac{dr}{dθ} = \frac{n cos⁡nθ}{sin⁡nθ}  \,or\, \frac{1}{r}  \frac{dr}{dθ} = \frac{cos⁡nθ}{sin⁡nθ} = cot \,nθ\)

replacing \(\frac{1}{r}  \frac{dr}{dθ} \,by\, -r\frac{dθ}{dr} \,we\,\, get\, -r\frac{dθ}{dr} = cot \,nθ\)

separating the variables and integrating \(\int \frac{dr}{r} \int tan⁡\,nθ \,dθ = c \)

\( \rightarrow log \,r + \frac{log⁡(sec⁡nθ)}{n} = c\) or n log r + log⁡(sec nθ) = nc = log⁡k

log(r^n sec⁡nθ) = log k –> r^n  sec⁡ nθ=k or r^n=k cos⁡nθ ..(2) is the required orthogonal trajectory since (1)&(2) are not same the given family of curve is not self orthogonal.