If x = a cos^4 theta and y = a sin^4 theta,th | vedclass

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(A) Given $x = a \cos^4 	heta$ and $y = a \sin^4 	heta$. Differentiating $y$ with respect to $	heta$: $\frac{dy}{d	heta} = 4a \sin^3 	heta \cos \

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(A) Given $x = a \cos^4 	heta$ and $y = a \sin^4 	heta$. Differentiating $y$ with respect to $	heta$: $\frac{dy}{d	heta} = 4a \sin^3 	heta \cos 	heta$. Differentiating $x$ with respect to $	heta$: $\frac{dx}{d	heta} = 4a \cos^3 	heta (-\sin 	heta) = -4a \cos^3 	heta \sin 	heta$. Now,$\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{4a \sin^3 	heta \cos 	heta}{-4a \cos^3 	heta \sin 	heta} = -\frac{\sin^2 	heta}{\cos^2 	heta} = -	an^2 	heta$. At $	heta = \frac{3\pi}{4}$,$	an(\frac{3\pi}{4}) = -1$. Therefore,$\frac{dy}{dx} = -(	an(\frac{3\pi}{4}))^2 = -(-1)^2 = -1$.

If $x = a \cos^4 \theta$ and $y = a \sin^4 \theta$,then $\frac{dy}{dx}$ at $\theta = \frac{3\pi}{4}$ is

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