Question

If $x=a\sin 2t(1+\cos 2t)$ and $y=b\cos 2t(1-\cos 2t)$, then find $\frac{dy}{dx}$ at $t=\frac{\pi }{4}$.

Answer 1

$x=a\sin 2t(1+\cos 2t)$

Differentiating w.r.t $t$, we get

$\frac{dx}{dt}=2a\cos 2t(1+\cos 2t)+a\sin 2t(-2\sin 2t)$

$=2a\cos 2t(1+\cos 2t)-2a{\sin }^{2}2t$

${\left[\frac{dx}{dt}\right]}_{t=\frac{\pi }{4}}=2a\cos (2\times \frac{\pi }{4})\{1+\cos (2\times \frac{\pi }{4})\}-2a{\sin }^2(2\times \frac{\pi }{4})$

$=0-2a=-2a$

$y=b\cos 2t(1-\cos 2t)$

Differentiating w.r.t $t$, we get,

$\frac{dy}{dt}=-2b\sin 2t(1-\cos 2t)+b\cos 2t\left(2\sin 2t\right)$

$=-2b\sin 2t(1-\cos 2t)+2b\cos 2t\sin 2t$

${\left[\frac{dy}{dt}\right]}_{t=\frac{\pi }{4}}=-2b\sin \frac{\pi }{2}(1-\cos \frac{\pi }{2})+2b\cos \frac{\pi }{2}\sin \frac{\pi }{2}$

$=-2b$

Now, $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{-2b}{-2a}=\frac{b}{a}$