Question

If $x=2{\cos }^4\left(t+3\right),y=3{\sin }^4\left(t+3\right)$ show that $\frac{dy}{dx}=-\sqrt{\frac{3y}{2x}}$

Answer 1

We have,

$x=2{\cos }^4(t+3)$ ……. (1)

On differentiating both sides w.r.t $t$, we get

$\frac{dx}{dt}=2\left(4{\cos }^3(t+3)(-\sin (t+3))\right)$

$\frac{dx}{dt}=-8\sin (t+3){\cos }^3(t+3)$ ……. (2)

Since,

$y=3{\sin }^4(t+3)$ …….. (3)

On differentiating both sides w.r.t $t$, we get

$\frac{dy}{dt}=3\left(4{\sin }^3(t+3)\cos (t+3)\right)$

$\frac{dy}{dt}=12{\sin }^3(t+3)\cos (t+3)$ …….. (4)

On dividing equation (4) by (2), we get

$\frac{dy}{dx}=\frac{12{\sin }^3(t+3)\cos (t+3)}{-8\sin (t+3){\cos }^3(t+3)}$

$\frac{dy}{dx}=-\frac{3{\sin }^2(t+3)}{2{\cos }^2(t+3)}$

From equation (1) and (3), we get

$\frac{dy}{dx}=-\frac{3\sqrt{\frac{y}{3}}}{2\sqrt{\frac{x}{2}}}$

$\frac{dy}{dx}=-\sqrt{\frac{3y}{2x}}$

Hence, proved.