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

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(C) Given that $x=a \cos^{3} 	heta$ and $y=a \sin^{3} 	heta$. First,find the derivatives with respect to $	heta$: $\frac{dx}{d	heta} = a(3 \cos^{2

Vedclass · Accepted Answer

$\sec^{2} 	heta$

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(C) Given that $x=a \cos^{3} 	heta$ and $y=a \sin^{3} 	heta$. First,find the derivatives with respect to $	heta$: $\frac{dx}{d	heta} = a(3 \cos^{2} 	heta)(-\sin 	heta) = -3a \cos^{2} 	heta \sin 	heta$. $\frac{dy}{d	heta} = a(3 \sin^{2} 	heta)(\cos 	heta) = 3a \sin^{2} 	heta \cos 	heta$. Now,calculate $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{3a \sin^{2} 	heta \cos 	heta}{-3a \cos^{2} 	heta \sin 	heta} = -\frac{\sin 	heta}{\cos 	heta} = -	an 	heta$. Finally,compute $1 + (\frac{dy}{dx})^{2}$: $1 + (-	an 	heta)^{2} = 1 + 	an^{2} 	heta = \sec^{2} 	heta$.

If $x=a \cos^{3} \theta$ and $y=a \sin^{3} \theta$,then $1+\left(\frac{dy}{dx}\right)^{2}$ is

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