Question

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

Answer 1

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

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

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

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

$[\because {\cos }^{2}A-{\sin }^{2}A=\cos 2A]$

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

$\Rightarrow \frac{dy}{dy}=2b[-\sin 2t+2\sin 2t\cos 2t]=2b(-\sin 2t+\sin 4t)$

$[\because 2\sin A\cos A=\sin 2A]$

Now, $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{2b[-\sin 2t+\sin 4t]}{2a[\cos 2t+\cos 4t]}$

At $t=\frac{\pi }{4},\frac{dy}{dx}=\frac{b}{a}\left[\frac{-\sin \frac{x}{2}+\sin \pi }{\cos \frac{x}{2}+\cos \pi }\right]=\frac{b}{a}\left[\frac{-1+0}{0-1}\right]=\frac{b}{a}$

$\Rightarrow \frac{dy}{dx}=\frac{b}{a}(att=\frac{\pi }{4})$