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  • If x = asin 2t (1 + cos2t) and y = bcos2t, show that dy/dx at t =pi/4 = b/a

If x = a\sin 2t (1 + \cos2t) and y = b\cos2t(1 - \cos2t), show that \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\mathrm{at} \mathrm{t}=\frac{\pi}{4}}=\frac{\mathrm{b}}{\mathrm{a}}

Answers (1)

x = a\sin 2t (1 + \cos2t) $and y$ = b\cos2t(1 - \cos2t)

Differentiate w.r.t t

\\ x=a \sin 2 t(1+\cos 2 t) \\ \frac{d x}{d t}=\frac{d}{d t}[\operatorname{asin} 2 t(1+\cos 2 t)] \\ \frac{d x}{d t}=a \frac{d}{d t}[\sin 2 t(1+\cos 2 t)] \\ \frac{d x}{d t}=a\left[\sin 2 t \frac{d}{d t}(1+\cos 2 t)+(1+\cos 2 t) \frac{d}{d t}(\sin 2 t)\right] \\ \frac{d x}{d t}=a[\sin 2 t \cdot(-2 \sin 2 t)+(1+\cos 2 t) \cdot 2 \cos 2 t] \\ \frac{d x}{d t}=-2 a\left[\sin ^{2} 2 t-\cos 2 t(1+\cos 2 t)\right]

Also,

y = bcos2t

\\ \frac{\mathrm{dy}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{b} \cos 2 \mathrm{t}) \\ \frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{b} \frac{\mathrm{d}}{\mathrm{dt}}(\cos 2 \mathrm{t}) \\ \frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{b}\left[-\sin 2 \mathrm{t} \cdot \frac{\mathrm{d}}{\mathrm{dt}}(2 \mathrm{t})\right] \\ \frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{b}[-2 \sin 2 \mathrm{t}]

Now, for dy/dx


\\ \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}} \\ \frac{d y}{d x}=\frac{b[-2 \sin 2 t]}{-2 a\left[\sin ^{2} 2 t-\cos 2 t(1+\cos 2 t)\right]} \\ \frac{d y}{d x}=\frac{b[\sin 2 t]}{a\left[\sin ^{2} 2 t-\cos 2 t(1+\cos 2 t)\right]} \\ \frac{d y}{d x} \text { at } t=\frac{\pi}{4}

\\ \frac{d y}{d x}=\frac{b\left[\sin \frac{2 \pi}{4}\right]}{a\left[\sin ^{2} \frac{2 \pi}{4}-\cos \frac{2 \pi}{4}\left(1+\cos \frac{2 \pi}{4}\right)\right]} \\ \frac{d y}{d x}=\frac{b\left[\sin \frac{\pi}{2}\right]}{a\left[\sin ^{2} \frac{\pi}{2}-\cos \frac{\pi}{2}\left(1+\cos \frac{\pi}{2}\right)\right]} \\ \frac{d y}{d x}=\frac{b[1]}{a[1-0(1+0)]} \\ \frac{d y}{d x}=\frac{b}{a}

Hence Proved.

Posted by

infoexpert22

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