If x=a cos ^{3} theta and y=a sin ^{3} theta, | vedclass

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(D) Given $x=a \cos ^{3} 	heta$ and $y=a \sin ^{3} 	heta$. Differentiating with respect to $	heta$: $\frac{dx}{d	heta} = 3a \cos^{2} 	heta (-\sin

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$-\sqrt[3]{\frac{y}{x}}$

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(D) Given $x=a \cos ^{3} 	heta$ and $y=a \sin ^{3} 	heta$. Differentiating with respect to $	heta$: $\frac{dx}{d	heta} = 3a \cos^{2} 	heta (-\sin 	heta) = -3a \cos^{2} 	heta \sin 	heta$. $\frac{dy}{d	heta} = 3a \sin^{2} 	heta (\cos 	heta) = 3a \sin^{2} 	heta \cos 	heta$. Now,$\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{3a \sin^{2} 	heta \cos 	heta}{-3a \cos^{2} 	heta \sin 	heta} = -\frac{\sin 	heta}{\cos 	heta} = -	an 	heta$. From the given equations: $\frac{y}{a} = \sin^{3} 	heta \Rightarrow \sin 	heta = (\frac{y}{a})^{1/3}$. $\frac{x}{a} = \cos^{3} 	heta \Rightarrow \cos 	heta = (\frac{x}{a})^{1/3}$. Substituting these into the expression for $\frac{dy}{dx}$: $\frac{dy}{dx} = -\frac{(y/a)^{1/3}}{(x/a)^{1/3}} = -(\frac{y}{x})^{1/3} = -\sqrt[3]{\frac{y}{x}}$.

If $x=a \cos ^{3} \theta$ and $y=a \sin ^{3} \theta$,then $\frac{d y}{d x}$ is

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