Question

If $x=a{\cos }^{3}t,y=a{\sin }^{3}t$,
show that $\frac{dy}{dx}=-{\left(\frac{y}{x}\right)}^{\frac{1}{3}}$.

Answer 1

Given,

$x=a{\cos }^{3}t,y=a{\sin }^{3}t$

$\frac{dx}{dt}=-3a\sin t{\cos }^{2}t$

$\frac{dy}{dt}=3a\cos t{\sin }^{2}t$

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{3a\cos t{\sin }^{2}t}{-3a\sin t{\cos }^{2}t}=\frac{-\sin t}{\cos t}=-\sqrt[3]{3\frac{y}{x}}$