Question

If $x = \sin\: t \: \cos \: 2t$ and $y = \cos\: t\: \sin\: 2t$, then at $ t = \frac{\pi}{4}$, the value of $ \frac{dy}{dx} $ is equal to :

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

Answer 1

Answer: (C)

Let $x = \sin \, t \cos 2 \, t$ and $y = \cos \:t. \sin \:2t$
Differentiate both w.r.t 't' $ \frac{dx}{dt} = \cos \: t \: \cos 2t = 2 \:\sin \:t . \sin\: 2t$ and $\frac{dy}{dt} = 2 \cos t .\cos 2t -\sin 2t .\sin t$
Now, $\frac{dy}{dt} = \frac{dy dt}{dx dt }= \frac{2 \cos t .\cos 2t -\sin 2t .\sin t}{\cos t.\cos 2t- 2\sin t. \sin 2t}$
Put $ t = \frac{\pi}{4} , \frac{dy}{dx} = \frac{2 \cos \frac{\pi}{4} .\cos \frac{\pi }{2} -\sin \frac{\pi }{2} .\sin \frac{\pi }{4}}{\cos \frac{\pi }{4}.\cos \frac{\pi }{2}- 2\sin \frac{\pi }{4}. \sin \frac{\pi }{2}}$
$\frac{\frac{-1}{\sqrt{2}}}{-2\left(\frac{1}{\sqrt{2}}\right)} = \frac{1}{2} \frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}=\frac{1}{2}$