If $x = a\sin 2t(1 + \cos 2t)$ and $y = b\cos 2t(1 - \cos 2t)$, then $(\frac{dy}{dx})_{\text{at } x=\frac{\pi}{4}}$ is equal to

Given $x = a\sin 2t(1 + \cos 2t)$ and $y = b\cos 2t(1 - \cos 2t)$, find $\frac{dy}{dx}$ at $t = \frac{\pi}{4}$.

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Since both $x$ and $y$ are functions of $t$:

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

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For $x = a\sin 2t(1 + \cos 2t)$:

$\frac{dx}{dt} = a[(2\cos 2t)(1 + \cos 2t) + \sin 2t(-2\sin 2t)]$

$= 2a[\cos 2t(1 + \cos 2t) - \sin^2 2t]$

$= 2a[\cos 2t + \cos^2 2t - \sin^2 2t]$

Using $\cos^2 2t - \sin^2 2t = \cos 4t$:

$\frac{dx}{dt} = 2a[\cos 2t + \cos 4t]$

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For $y = b\cos 2t(1 - \cos 2t)$:

$\frac{dy}{dt} = b[(-2\sin 2t)(1 - \cos 2t) + \cos 2t(2\sin 2t)]$

$= 2b[-\sin 2t(1 - \cos 2t) + \cos 2t \sin 2t]$

$= 2b[-\sin 2t + \sin 2t\cos 2t + \cos 2t \sin 2t]$

$= 2b[-\sin 2t + 2\sin 2t\cos 2t]$

Using $2\sin 2t\cos 2t = \sin 4t$:

$\frac{dy}{dt} = 2b[-\sin 2t + \sin 4t]$

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$\frac{dy}{dx} = \frac{2b[-\sin 2t + \sin 4t]}{2a[\cos 2t + \cos 4t]}$

$= \frac{b[-\sin 2t + \sin 4t]}{a[\cos 2t + \cos 4t]}$

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At $t = \frac{\pi}{4}$:

$2t = \frac{\pi}{2}$ and $4t = \pi$

$\sin 2t = \sin \frac{\pi}{2} = 1$

$\cos 2t = \cos \frac{\pi}{2} = 0$

$\sin 4t = \sin \pi = 0$

$\cos 4t = \cos \pi = -1$

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$\frac{dy}{dx}\bigg|_{t=\frac{\pi}{4}} = \frac{b[-1 + 0]}{a[0 + (-1)]}$

$= \frac{b(-1)}{a(-1)}$

$= \frac{b}{a}$

Therefore, $\frac{dy}{dx}$ at $t = \frac{\pi}{4}$ is $\frac{b}{a}$.