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KEAM 2019 Math Paper

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Question : 3 of 121

Marks: +1, -0

If

x=\sin t \cos 2 t

y=\cos t \sin 2 t

t=\;\frac{\pi}{4}

\;\frac{d y}{d x}

is equal to

$\frac{1}{2}$
0
- 1
$\;-\frac{1}{2}$
1

Solution:

We have,

x=\sin t \cos 2 t

and

y=\cos t \sin 2 t

Differentiating both, we get,

\;\frac{d x}{d t} = \cos t \cos 2 t - 2 \sin t \sin 2 t

\;\frac{d y}{d t} = 2 \cos t \cos 2 t - \sin t \sin 2 t

Dividing both, we get.

\;\frac{d y}{d x} = \;\frac{2 \cos t \cos 2 t - sint \sin 2 t}{\cos t \cos 2 t - 2 \sin t \sin 2 t}

\therefore \;\; \left(\;\frac{d y}{d x}\right)_{\;\text{at }\; t=\;\frac{\pi}{4}} = \;\; \;\frac{2 \cos \frac{\pi}{4} \cos \frac{\pi}{2} - \sin \frac{\pi}{4} \sin \frac{\pi}{2}}{\cos \;\frac{\pi}{4} \cos \;\frac{\pi}{2} - 2 \sin \;\frac{\pi}{4} \sin \;\frac{\pi}{2}}

\;\; = \;\frac{0 - \frac{1}{\sqrt{2}}}{0 - 2 \times \;\frac{1}{\sqrt{2}}}

\Rightarrow \left(\;\frac{d y}{d x}\right)_{a t} = \;\frac{\pi}{4} \;\; = \;\frac{1}{2}

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