If x = a cos theta and y = b sin theta,then f | vedclass

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(C) Given $x = a \cos 	heta$ and $y = b \sin 	heta$. First,find $\frac{dy}{dx}$ using the chain rule: $\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta}

Vedclass · Accepted Answer

$-\frac{3b}{a^3} \csc^4 	heta \cot 	heta$

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(C) Given $x = a \cos 	heta$ and $y = b \sin 	heta$. First,find $\frac{dy}{dx}$ using the chain rule: $\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{b \cos 	heta}{-a \sin 	heta} = -\frac{b}{a} \cot 	heta$. Next,find the second derivative $\frac{d^2y}{dx^2}$: $\frac{d^2y}{dx^2} = \frac{d}{dx} \left( -\frac{b}{a} \cot 	heta ight) = -\frac{b}{a} (-\csc^2 	heta) \frac{d	heta}{dx} = \frac{b}{a} \csc^2 	heta \left( \frac{1}{-a \sin 	heta} ight) = -\frac{b}{a^2} \csc^3 	heta$. Finally,find the third derivative $\frac{d^3y}{dx^3}$: $\frac{d^3y}{dx^3} = \frac{d}{dx} \left( -\frac{b}{a^2} \csc^3 	heta ight) = -\frac{b}{a^2} (3 \csc^2 	heta \cdot -\csc 	heta \cot 	heta) \frac{d	heta}{dx} = \frac{3b}{a^2} \csc^3 	heta \cot 	heta \left( \frac{1}{-a \sin 	heta} ight) = -\frac{3b}{a^3} \csc^4 	heta \cot 	heta$.

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

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