If y=4 cos ^3(t) and x=4 sin ^3(t),then frac{ | vedclass

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(C) Given: $y=4 \cos ^3(t)$ and $x=4 \sin ^3(t)$.First,differentiate $y$ with respect to $t$:$\frac{d y}{d t} = 4 \cdot 3 \cos ^2(t) \cdot (-\sin(t))

Vedclass · Accepted Answer

$-\cot (t)$

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(C) Given: $y=4 \cos ^3(t)$ and $x=4 \sin ^3(t)$.First,differentiate $y$ with respect to $t$:$\frac{d y}{d t} = 4 \cdot 3 \cos ^2(t) \cdot (-\sin(t)) = -12 \sin(t) \cos ^2(t)$.Next,differentiate $x$ with respect to $t$:$\frac{d x}{d t} = 4 \cdot 3 \sin ^2(t) \cdot \cos(t) = 12 \sin ^2(t) \cos(t)$.Using the chain rule for parametric differentiation:$\frac{d y}{d x} = \frac{d y / d t}{d x / d t} = \frac{-12 \sin(t) \cos ^2(t)}{12 \sin ^2(t) \cos(t)}$.Simplifying the expression:$\frac{d y}{d x} = -\frac{\cos(t)}{\sin(t)} = -\cot(t)$.Thus,the correct option is $C$.

If $y=4 \cos ^3(t)$ and $x=4 \sin ^3(t)$,then $\frac{d y}{d x}$ is equal to

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