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(A) Given the parametric equations of the curve: $x = a \cos^3 	heta$ and $y = a \sin^3 	heta$. First,find the derivative $\frac{dy}{dx}$ using the

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(A) Given the parametric equations of the curve: $x = a \cos^3 	heta$ and $y = a \sin^3 	heta$. First,find the derivative $\frac{dy}{dx}$ using the chain rule: $\frac{dx}{d	heta} = 3a \cos^2 	heta (-\sin 	heta) = -3a \cos^2 	heta \sin 	heta$ $\frac{dy}{d	heta} = 3a \sin^2 	heta (\cos 	heta) = 3a \sin^2 	heta \cos 	heta$ $\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{3a \sin^2 	heta \cos 	heta}{-3a \cos^2 	heta \sin 	heta} = -	an 	heta$ At $	heta = \pi / 4$,the slope of the tangent is $m_t = -	an(\pi / 4) = -1$. The slope of the normal $m_n$ is given by $-\frac{1}{m_t}$. $m_n = -\frac{1}{-1} = 1$.

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

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