Question

If $x=a\sin 2t(1+\cos 2t)$ and $y=b\cos 2t(1-\cos 2t)$, find the values of $\frac{dy}{dx}$ at $t=\frac{\pi }{4}$ and $t=\frac{\pi }{3}$
OR
If $y=x^x$, prove that $\frac{d^{2}y}{d{x}^2}-\frac{1}{y}{\left(\frac{dy}{dx}\right)}^2-\frac{y}{x}=0$

Answer 1

$x=a\sin 2t(1+\cos 2t),y=b\cos 2t(1-\cos 2t)$

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

$=b[-2\sin 2t(1-\cos 2t)+2\cos 2t\sin 2t]\times \left[\frac{1}{-2a{\sin }^{2}2t+2a\cos 2t(1+\cos 2t)}\right]$

Substituting $t=\frac{\pi }{4}$, we get

$\frac{dy}{dx}=\frac{b[-2]}{a[-2]}=\frac{b}{a}$

At $t=\frac{\pi }{3}$, we get

$\frac{dy}{dx}=\frac{b[\frac{3\sqrt{3}}{2}-\frac{\sqrt{3}}{2}]}{a[\frac{-3}{2}-\frac{1}{2}]}$ $=\frac{b\sqrt{3}}{-2a}$