Question

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