Question

If x = 2sint - sin2t, y= 2cost-cos2t, then the value of $\frac{d^{2}y}{d{x}^2}$ at $t=\frac{\pi }{2}$ is

Answer 1

$x=2\sin t-\sin 2t$

$y=2\cos t-\cos 2t$

$\frac{dx}{dt}=2\cot t-2\cos 1t$, at $x=\frac{\pi }{2},\frac{dx}{dt}=\cos 2$

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

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

$\frac{d^{2}y}{d{x}^2}=\frac{(\cos t-\cos 2t)(-\cos t+2\cos 2t)\frac{dt}{dx}-(\sin t+\sin 2t)}{(\cos t-\cos 2t{)}^2}$

$(\sin t+2\sin 2t)\frac{dt}{dx}$

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

at $x=\frac{\pi }{2}$

Put $x=\frac{\pi }{2}$

$=2\left[\frac{(0+1)(-0+2)-(-1+0)(-1+0)}{(0+1{)}^2}\right]$

$=2\left[\frac{+2-1}{1}\right]=2$