Question

If $x=2sin\theta -sin2\theta$ and $y=2cos\theta -cos2\theta ,\theta ϵ\left[0,2\pi \right],\frac{d^{2}y}{d{x}^2}$ at $\theta =\pi$ is :

• A. $\frac{3}{8}$
• B. $\frac{3}{4}$
• C. $\frac{3}{2}$
• D. $-\frac{3}{4}$

Answer 1

Answer: (A)

$x=2sin\theta -sin2\theta$
$\Rightarrow \frac{dx}{d\theta }=2cos\theta -2cos2\theta =4sin\left(\frac{\theta }{2}\right)sin\left(\frac{3\theta }{2}\right)$
$y=2cos\theta -cos2\theta$
$\Rightarrow \frac{dy}{d\theta }=2sin\theta +2sin2\theta =4sin\frac{\theta }{2}cos\frac{3\theta }{2}$
$\Rightarrow \frac{dy}{d\theta }=cot\left(\frac{3\theta }{2}\right)\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{-\frac{3}{2}cose{c}^2\left(\frac{3\theta }{2}\right)}{4sin\left(\frac{\theta }{2}\right)sin\frac{3\theta }{2}}$
$\Rightarrow {\left(\frac{d^{2}y}{d{x}^2}\right)}_{\theta =\pi }=\frac{3}{8}$
Alternate :-
$\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{-2sin\theta +2sin2\theta }{2sin\theta -2sin2\theta }=\frac{sin\theta -sin2\theta }{-cos\theta +cos2\theta }$
$\frac{d^{2}y}{d{x}^2}.\frac{dx}{d\theta }=\frac{\left(-cos\theta +cos2\theta \right)\left(cos\theta -2cos2\theta \right)-\left(sin\theta -2sin2\theta \right)\left(sin\theta -sin2\theta \right)}{{\left(-cos\theta +cos2\theta \right)}^2}$
$\frac{d^{2}y}{d{x}^2}.\left(-2-2\right)=\frac{\left(+1+1\right)\left(-1-2\right)-\left(0\right)}{{\left(1+1\right)}^2}$
$\frac{d^{2}y}{d{x}^2}\left(-4\right)=\frac{2\times -3}{4}=-\frac{3}{2}$
$\frac{d^{2}y}{d{x}^2}=\frac{3}{8}$