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Q.
If $x=\sin t \cos 2 t$ and $y=\cos t \sin 2 t$, then at $t =\frac{\pi}{4}$, the value of $\frac{ dy }{ dx }$ is equal to:
Continuity and Differentiability
Solution:
Let $x=\sin t \cos 2 t$ and $y=\cos t \cdot \sin 2 t$
Differentiate both w.r.t ' $t $'
$\frac{d x}{d t}=\cos t \cos 2 t-2 \sin t .\sin 2 t$
and $\frac{d y}{d t}=2 \cos t \cos 2 t-\sin 2 t .\sin t$
Now, $\frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{2 \cos t .\cos 2 t-\sin 2 t .\sin t}{\cos t \cdot \cos 2 t-2 \sin t .\sin 2 t}$
Put $t =\frac{\pi}{4}$
$\frac{ dy }{ dx }=\frac{2 \cos \frac{\pi}{4} .\cos \frac{\pi}{2}-\sin \frac{\pi}{2} \sin \frac{\pi}{4}}{\cos \frac{\pi}{4} \cos \frac{\pi}{2}-2 \sin \frac{\pi}{4} \sin \frac{\pi}{2}}=\frac{\frac{-1}{\sqrt{2}}}{-2\left(\frac{1}{\sqrt{2}}\right)}=\frac{1}{2}$