Question

Find \(\frac{dy}{dx}\), if x=a^2 t^2 cotθ and y=at sin⁡θ.

(a) \(\frac{tan⁡θ \,sin⁡θ}{at}\)

(b) \(\frac{tan⁡θ \,sin⁡θ}{2at}\)

(c) \(\frac{tan⁡θ \,sin⁡θ}{2t}\)

(d) \(\frac{tan⁡θ \,sin⁡θ}{2a}\)

The question was asked during a job interview.

This interesting question is from Derivatives of Functions in Parametric Forms topic in division Continuity and Differentiability of Mathematics – Class 12

Answer 1

The correct option is (b) \(\frac{tan⁡θ \,sin⁡θ}{2at}\)

To explain: Given that, x=a^2 t^2 cotθ and y=at sin⁡θ

\(\frac{dx}{dt}\)=2ta^2 cot⁡θ

\(\frac{dy}{dt}\)=asin⁡θ

\(\frac{dy}{dx}\)=\(\frac{asin⁡θ}{2ta^2 \,cot⁡θ}=\frac{a sin⁡θ}{2a^2 t cos⁡θ}.sin⁡θ=\frac{tan⁡θ \,sin⁡θ}{2at}\)