Question

If $x=a{\cos }^3\theta$, $y=a{\sin }^3\theta$, prove that $y^1=-\sqrt[3]{3\frac{y}{x}}$

Answer 1

Given,

$x=a{\cos }^3\theta$ …….. (1)

$y=a{\sin }^3\theta$ …….. (2)

On differentiating both equations w.r.t $\theta$, we get

$\frac{dx}{d\theta }=a\left(3{\cos }^2\theta \right)(-\sin \theta )$

$\frac{dx}{d\theta }=-3a\sin \theta {\cos }^2\theta$

$\frac{dy}{d\theta }=a\left(3{\sin }^2\theta \right)\left(\cos \theta \right)$

$\frac{dy}{d\theta }=3a{\sin }^2\theta \cos \theta$

Therefore,

$\frac{dy}{dx}=-\frac{\sin \theta }{\cos \theta }$

$\frac{dy}{dx}=-\tan \theta$

Hence, proved.