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  4. Let x(t)=2 √2 cos t √ sin 2 t and y(t)=2 √2 sin t √ sin 2 t, t ∈(0, (π/2)). Then (1+((d y/d x))2/(d2 y)d x2) at t=(π/4) is equal to

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Q. Let $x(t)=2 \sqrt{2} \cos t \sqrt{\sin 2 t}$ and $y(t)=2 \sqrt{2} \sin t \sqrt{\sin 2 t}, t \in\left(0, \frac{\pi}{2}\right)$. Then $\frac{1+\left(\frac{d y}{d x}\right)^2}{\frac{d^2 y}{d x^2}}$ at $t=\frac{\pi}{4}$ is equal to

JEE MainJEE Main 2022Differential Equations

Solution:

$ x =2 \sqrt{2} \cos t \sqrt{\sin 2 t } $
$ \frac{ dx }{ dt }=\frac{2 \sqrt{2} \cos 3 t }{\sqrt{\sin 2 t}}$
$ y ( t )=2 \sqrt{2} \sin t \sqrt{\sin 2 t } $
$ \frac{ dy }{ dt }=\frac{2 \sqrt{2} \sin 3 t }{\sqrt{\sin 2 t}}$
$ \frac{ dy }{ dx }=\tan 3 t $
$ \frac{ dy }{ dx }=-1 \text { at } t =\frac{\pi}{4}$
$ \frac{d^2 y}{d x^2}=\frac{3}{2 \sqrt{2}} \sec ^3 3 t \cdot \sqrt{\sin 2 t}=-3 \text { at } t=\frac{\pi}{4} $
$ \therefore \frac{1+\left(\frac{d y}{d x}\right)^2}{\frac{d^2 y}{d x^2}}=\frac{1+1}{-3}=-\frac{2}{3}$