If x = a sin 2t (1 + cos 2t) and y = b cos 2t | vedclass

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(A) Given $x = a \sin 2t (1 + \cos 2t) = a \sin 2t + a \sin 2t \cos 2t = a \sin 2t + \frac{a}{2} \sin 4t$. Differentiating with respect to $t$: $\frac

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$\frac{b}{a} 	an t$

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(A) Given $x = a \sin 2t (1 + \cos 2t) = a \sin 2t + a \sin 2t \cos 2t = a \sin 2t + \frac{a}{2} \sin 4t$. Differentiating with respect to $t$: $\frac{dx}{dt} = 2a \cos 2t + 2a \cos 4t = 2a (\cos 2t + \cos 4t)$. Given $y = b \cos 2t (1 - \cos 2t) = b \cos 2t - b \cos^2 2t$. Using $\cos^2 2t = \frac{1 + \cos 4t}{2}$,we have $y = b \cos 2t - \frac{b}{2} (1 + \cos 4t)$. Differentiating with respect to $t$: $\frac{dy}{dt} = -2b \sin 2t + b \sin 4t = -2b \sin 2t + 2b \sin 2t \cos 2t = -2b \sin 2t (1 - \cos 2t)$. Using $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-2b \sin 2t (1 - \cos 2t)}{2a (\cos 2t + \cos 4t)} = \frac{-b \sin 2t (1 - \cos 2t)}{a \cos 2t (1 + 2 \cos 2t - 1)} = \frac{-b \sin 2t (1 - \cos 2t)}{a \cos 2t (2 \cos 2t)} = \frac{-b (2 \sin t \cos t) (2 \sin^2 t)}{a (2 \cos^2 t - 1) (2 \cos 2t)} = \frac{b}{a} 	an t$.

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

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