Question

Find $\frac{dy}{dx}$, if $x={\sin }^{3}t,y={\cos }^{3}t$.

Answer 1

Given,

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

Now differentiating $x$ and $y$ with respect to $t$ we get,

$\frac{dx}{dt}=3{\sin }^{2}t.\cos t$.....(1) and $\frac{dy}{dt}=-3{\sin }^{2}t.\cos t$.......(2).

Now,

$\frac{dy}{dx}$

$=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

$=\frac{-3{\cos }^{2}t.\sin t}{3{\sin }^{2}t.\cos t}$ [ Using (1) and (2)]

$=-\cot t.$