Find the slope of the normal to the curve x=a | vedclass

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(A) Given the parametric equations of the curve are $x=a \cos ^{3} 	heta$ and $y=a \sin ^{3} 	heta$. First,we find the derivatives with respect to $

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(A) Given the parametric equations of the curve are $x=a \cos ^{3} 	heta$ and $y=a \sin ^{3} 	heta$. First,we find the derivatives with respect to $	heta$: $\frac{dx}{d	heta} = 3a \cos^{2} 	heta (-\sin 	heta) = -3a \cos^{2} 	heta \sin 	heta$ $\frac{dy}{d	heta} = 3a \sin^{2} 	heta (\cos 	heta) = 3a \sin^{2} 	heta \cos 	heta$ The slope of the tangent is given by $\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{3a \sin^{2} 	heta \cos 	heta}{-3a \cos^{2} 	heta \sin 	heta} = -\frac{\sin 	heta}{\cos 	heta} = -	an 	heta$. At $	heta = \frac{\pi}{4}$,the slope of the tangent is $m_{T} = -	an(\frac{\pi}{4}) = -1$. The slope of the normal $m_{N}$ is given by $-\frac{1}{m_{T}}$. Therefore,$m_{N} = -\frac{1}{-1} = 1$.

Find the slope of the normal to the curve $x=a \cos ^{3} \theta, y=a \sin ^{3} \theta$ at $\theta=\frac{\pi}{4}$.

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