Question

For the below mentioned figure the angle between radius vector (op) ⃗ and tangent to the polar curve where r=f(θ) has the one among the following relation?

(a) \(tan\,⁡ω=\frac{tan⁡∅+tan⁡θ}{1-tan⁡∅tan⁡θ}\)

(b) \(tan⁡\,ω=\frac{tan⁡∅-tan⁡θ}{1+tan⁡∅tan⁡θ}\)

(c) \(tan\,⁡∅=r(\frac{dr}{dθ})\)

(d) \(tan⁡\,ω=tan⁡∅+tan⁡θ \)

I have been asked this question in an interview.

This intriguing question comes from Polar Curves topic in chapter Differential Calculus of Engineering Mathematics

Answer 1

Right choice is (a) \(tan\,⁡ω=\frac{tan⁡∅+tan⁡θ}{1-tan⁡∅tan⁡θ}\)

The best explanation: From the fig (1) ω=∅+θ…….. (triangle rule & \(\hat{opt}\) = ∅)

tan ⁡ω=tan(⁡θ+∅)

\(tan⁡\,ω=\frac{tan⁡∅+tan⁡θ}{1-tan⁡∅tan⁡θ}\) ……(1) further more w.k.t x=rcos⁡θ, y=rsin⁡θ

But \(tan\,⁡ω = \frac{dy}{dx}\)……(slope of the tangent TT’)

\(tan\,ω = \frac{\frac{dy}{dθ}} {\frac{dx}{dθ}}\)……..since x & y are functions of θ

i.e \(\displaystyle tan⁡\,ω = \frac{\frac{d(r sin⁡θ)}{dθ}}{\frac{d(r cos⁡θ)}{dθ}} = \frac{r cos⁡θ + \frac{dr}{dθ} sin⁡θ}{-r sin⁡θ + \frac{dr}{dθ} cos⁡θ}\)

\(\displaystyle tan⁡\,ω = \frac{\frac{r cos⁡θ} {\frac{dr}{dθ} cos⁡θ} + \frac{\frac{dr}{dθ} sin⁡θ}{\frac{dr}{dθ} cos⁡θ}} {\frac{-r sin⁡θ}{\frac{dr}{dθ} cos⁡θ}+\frac{\frac{dr}{dθ} cos⁡θ}{\frac{dr}{dθ} cos⁡θ}} = \frac{\frac{r}{\frac{dr}{dθ}}+tan⁡θ}{\frac{r}{\frac{dr}{dθ}}tan⁡θ}\)…………(2)

from (1) &(2)

\(tan⁡∅ = \frac{r}{\frac{dr}{dθ}} = r (\frac{dθ}{dr}).\)