Question

If $x=a\sin 2t(1+\cos 2t)$ and $y=b2t(1-\cos 2t)$, find the values of $\frac{dy}{dx}$ at $t=\frac{\pi }{4}$ and $t=\frac{\pi }{3}$.

Answer 1

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

$dy=\left(2b\right(1-\cos 2t)+4bt\sin 2t)dt$

$\frac{dy}{dx}$ at $t=\frac{\pi }{4}$

$\frac{\left(2b\right(1-\cos 2t)+4bt\sin 2t)}{\left(2a\cos 2t\right(1+\cos 2t)-2a{\sin }^{2}2t)}$

$⟹\frac{2b+b\pi }{-2a}$

At $t=\frac{\pi }{3}$

$\frac{3b+\frac{4b\pi \sqrt{3}}{6}}{\frac{a}{4}-\frac{3a}{2}}$