Question

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

• A. ${\frac{dy}{dx}|}_{t=\frac{3\pi }{8}}=\frac{1}{2}$
• B. ${\frac{dy}{dx}|}_{t=\frac{\pi }{4}}=\frac{1}{2}$
• C. ${\frac{d^{2}y}{d{x}^2}|}_{t=\frac{\pi }{4}}=\frac{-1}{4}$
• D. ${\frac{d^{2}y}{d{x}^2}|}_{t=\frac{\pi }{6}}=-2$

Answer 1

Answer: (A), (C)

The correct options are
A ${\frac{dy}{dx}|}_{t=\frac{3\pi }{8}}=\frac{1}{2}$
C ${\frac{d^{2}y}{d{x}^2}|}_{t=\frac{\pi }{4}}=\frac{-1}{4}$

$\frac{dy}{dt}=2\cos 2t,\frac{dx}{dt}=-4\sin 2t$
$\frac{dy}{dx}=-\frac{1}{2}\cot 2t\Rightarrow {\frac{dy}{dx}|}_{t=\frac{3\pi }{8}}=-\frac{1}{2}(-1)=\frac{1}{2}$
$\frac{d^{2}y}{d{x}^2}={\csc }^2\left(2t\right)\times \frac{1}{-4\sin 2t}$
$\frac{d^{2}y}{d{x}^2}=-\frac{1}{4}{\csc }^{3}2t,$
${\frac{d^{2}y}{d{x}^2}|}_{t=\frac{\pi }{4}}=\frac{-1}{4}\times 1=-\frac{1}{4}$