Question

If $x=2\cos t-\cos 2t$, $y=2\sin t-\sin 2t$

then $\frac{d^{2}y}{d{x}^2}$ at $t=\frac{\pi }{2}$ is?

• A. $1$
• B. $\frac{-3}{2}$
• C. $\frac{3}{2}$
• D. $\frac{5}{2}$

Answer 1

The correct option is A $1$

Now, $\frac{dx}{dt}=-2sint+2sin2t$

Also, $\frac{dy}{dt}=2cost-2cos2t$

Thus, $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

Thus, $\frac{dy}{dx}=\frac{2cost-2cos2t}{-2sint+2sin2t}$

Thus, $\frac{d^{2}y}{d{x}^2}=\frac{d}{dt}\frac{dy}{dx}.\frac{dt}{dx}$

Thus, $\frac{d^{2}y}{d{x}^2}=\frac{d}{dt}\times \frac{2cost-2cos2t}{-2sint+2sin2t}\times \frac{1}{(-2sint+2sin2t)}$

Thus, $\frac{d^{2}y}{d{x}^2}=\frac{d}{dt}\left(\frac{2cost-2cos2t}{(-2sint+2sin2t{)}^2}\right)$

Or, $\frac{d^{2}y}{d{x}^2}=\left(\frac{(-2sint+2sin2t{)}^2(-2sint+4sin2t)-(2cost-2cos2t)\times 2(-2sint+4cos2t)}{(-2sint+2sin2t{)}^4}\right)$

Thus, $\frac{d^{2}y}{d{x}^2}$ at $\frac{\pi }{2}$ is given by,

$\frac{d^{2}y}{d{x}^2}=\frac{(-2{)}^2(-2)-(2\left)\right(2\left)\right(-2-4)}{(-2{)}^4}$

Or, $\frac{d^{2}y}{d{x}^2}=1$