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

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

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$|\sec 	heta|$

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(C) Given $x = a \cos^3 	heta$ and $y = a \sin^3 	heta$. First,we find the derivatives 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,$\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{3a \sin^2 	heta \cos 	heta}{-3a \cos^2 	heta \sin 	heta} = -\frac{\sin 	heta}{\cos 	heta} = -	an 	heta$. Then,$\left(\frac{dy}{dx}ight)^2 = (-	an 	heta)^2 = 	an^2 	heta$. Finally,$\sqrt{1 + \left(\frac{dy}{dx}ight)^2} = \sqrt{1 + 	an^2 	heta} = \sqrt{\sec^2 	heta} = |\sec 	heta|$.

If $x = a \cos^3 \theta$ and $y = a \sin^3 \theta$,then find the value of $\sqrt{1 + \left(\frac{dy}{dx}\right)^2}$.

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