Question

If $x={acos}^3\theta$ and $y={asin}^3\theta$, prove that $\frac{dy}{dx}=\sqrt[3]{3\frac{y}{x}}$

Answer 1

$x=a{\cos }^3\theta$

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

$=a3{\cos }^2\theta \frac{d}{d\theta }\left(\cos \theta \right)$

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

$y=a{\sin }^3\theta$

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

$=a3{\sin }^2\theta \frac{d}{d\theta }\left(\sin \theta \right)$

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

$\frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}$

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

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

$=-\tan \theta$ .....$\left(1\right)$

Again $\frac{y}{x}=\frac{a{\sin }^3\theta }{a{\cos }^3\theta }={\tan }^3\theta$

$\Rightarrow \tan \theta =\sqrt[3]{3\frac{y}{x}}$ .....$\left(2\right)$

$\therefore \frac{dy}{dx}=\sqrt[3]{3\frac{y}{x}}$ from $\left(1\right)$ and $\left(2\right)$