Question

if $x=2\sin t-\sin 2t,y=2\cos t-\cos 2t$, then the value of $\frac{d^{2}y}{d{x}^2}$ at $t=\frac{\pi }{2}$ is

• A. $2$
• B. $-1/2$
• C. $-3/4$
• D. $-3/2$

Answer 1

Answer: (B)

$-1/2$

Given, $x=2\sin t-\sin 2t$ and $y=2\cos t-\cos 2t$

$\left(\frac{dx}{dt}\right)=2\cos t-2\cos 2t=\left[2\sin \left(\frac{3t}{2}\right)\sin \left(\frac{t}{2}\right)\right]$
and $\left(\frac{dy}{dt}\right)=-2\sin t+2\sin 2t=2\left[\sin \frac{3t}{2}\cos \frac{t}{2}\right]$
Hence $\left(\frac{dy}{dx}\right)=\cot \left(\frac{t}{2}\right)$ and $\left(\frac{d^{2}y}{{dx}^2}\right)=-\frac{1}{2}{\csc }^2\left(\frac{t}{2}\right)\times \frac{dt}{dx}$
Therefore $\left(\frac{d^{2}y}{{dx}^2}\right)=-\frac{1}{2}{\csc }^2\left(\frac{t}{2}\right)\times \frac{1}{4\sin \frac{3t}{2}\sin \frac{t}{2}}$
And at $t=\frac{\pi }{2}$
$\frac{d^{2}y}{{dx}^2}=(-\frac{1}{2})={\left(\sqrt{2}\right)}^2\times \frac{1}{4\times \left(\frac{t}{2}\right)}=-\frac{1}{2}$