Question

If $x=2\cos t-\cos 2t,y=2\sin t-\sin 2t$, then the value of $\frac{d^{2}y}{d{x}^2}{|}_{t=\pi {/}_2}$ is

• A. $\frac{3}{2}$
• B. $\frac{-5}{2}$
• C. $\frac{5}{2}$
• D. $\frac{-3}{2}$

Answer 1

The correct option is D $\frac{-3}{2}$

$\frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}=\frac{2cost-2cos2t}{-2sint+2sin2t}$
$\frac{d}{dt}\left(\frac{dy}{dx}\right)\frac{dt}{dx}=\frac{d^{2}y}{d{x}^2}$
$\frac{d^{2}y}{d{x}^2}=\frac{d}{dt}\left(\frac{cost-cos2t}{sin2t-sint}\right)\left(\frac{1}{-2sint+2sin2t}\right)$
$\frac{d^{2}y}{d{x}^2}{|}_{t=\pi {/}_2}=\frac{(-sint+2sin2t)(sin2t-sint)-(cost-cos2t)(2cos2t-cost)}{[sin2t-sint{]}^2(2sin2t-2sint)}$
$=\frac{(-1)(-1)-\left(1\right)(-2)}{(-1{)}^2(-2)}$
$=-\frac{3}{2}$