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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»

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

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


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