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 $x=\sin t\cos 2t$ 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, $\frac{dy}{dt}=\frac{dydt}{dxdt}=\frac{2\cos t.\cos 2t-\sin 2t.\sin t}{\cos t.\cos 2t-2\sin t.\sin 2t}$
Put $t=\frac{\pi }{4},\frac{dy}{dx}=\frac{2\cos \frac{\pi }{4}.\cos \frac{\pi }{2}-\sin \frac{\pi }{2}.\sin \frac{\pi }{4}}{\cos \frac{\pi }{4}.\cos \frac{\pi }{2}-2\sin \frac{\pi }{4}.\sin \frac{\pi }{2}}$
$\frac{\frac{-1}{\sqrt{2}}}{-2\left(\frac{1}{\sqrt{2}}\right)}=\frac{1}{2}\frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}=\frac{1}{2}$