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

$x=\cos t(3-2{\cos }^{2}t)$
$\frac{dx}{dt}=\cos t[-4\cos t(-\sin t\left)\right]+(3-2{\cos }^{2}t)(-\sin t)$
$=4\sin t{\cos }^{2}t-3\sin t+2\sin t{\cos }^{2}t$
$=6\sin t{\cos }^{2}t-3\sin t$

$y=\sin t(3-2{\sin }^{2}t)$
$\frac{dy}{dx}=\cos t(3-2{\sin }^{2}t)+\sin t(-4\sin t\cos t)$
$=3\cos t-2{\sin }^{2}t\cos t-4{\sin }^{2}t\cos t$
$=3\cos t-6{\sin }^{2}t\cos t$

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{3\cos t(1-2{\sin }^{2}t)}{3\sin t(2{\cos }^{2}t-1)}=\frac{\cos t.\cos 2t}{\sin t\cos 2t}=\cot t$
At $t=\frac{\pi }{4},\frac{dy}{dx}=\cot \pi /4=1$