Question

If $x=\cos t(3-2{\cos }^{2}t)$ and $y=\sin t(3-2{\sin }^{2}t)$, find the value of $\frac{dy}{dx}$ at $t=\frac{\pi }{4}$

Answer 1

$x=\cos t(3-2{\cos }^{2}t)$

$=3\cos t-2{\cos }^{3}t$ $\because \cos 3t=4{\cos }^{3}t-3\cos t$

$x=2{\cos }^3-\cos 3t$

$dx=(-6{\cos }^{2}t\sin t+3\sin 3t)dt$

$y=\sin t(3-2{\sin }^{2}t)$

$=3\sin t-2{\sin }^{3}t$ $\because \sin 3t=3\sin t-4{\sin }^{3}t$

$y=\sin 3t+2{\sin }^{3}t$

$dy=(3\cos 3t+6{\sin }^{2}t\cos t)dt$

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

$\frac{dy}{dx}{|}_{t=\frac{\pi }{4}}=\frac{-\frac{1}{\sqrt{2}}+\frac{2}{2\sqrt{2}}}{\frac{1}{\sqrt{2}}-\frac{2}{2\sqrt{2}}}=\frac{0}{0}$

$\therefore \frac{dy}{dx}{|}_{t=\frac{\pi }{4}}$ is not defined.