Find frac{dy}{dx},if y=12(1-cos t) and x=10(t | vedclass

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(A) Given $y=12(1-\cos t)$ and $x=10(t-\sin t)$.Differentiating $x$ with respect to $t$:$\frac{dx}{dt} = \frac{d}{dt}[10(t-\sin t)] = 10(1-\cos t)$.Di

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$\frac{6}{5} \cot \frac{t}{2}$

Vedclass · Answer

(A) Given $y=12(1-\cos t)$ and $x=10(t-\sin t)$.Differentiating $x$ with respect to $t$:$\frac{dx}{dt} = \frac{d}{dt}[10(t-\sin t)] = 10(1-\cos t)$.Differentiating $y$ with respect to $t$:$\frac{dy}{dt} = \frac{d}{dt}[12(1-\cos t)] = 12(0 - (-\sin t)) = 12\sin t$.Using the chain rule for parametric differentiation:$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{12\sin t}{10(1-\cos t)}$.Using trigonometric identities $\sin t = 2\sin \frac{t}{2} \cos \frac{t}{2}$ and $1-\cos t = 2\sin^2 \frac{t}{2}$:$\frac{dy}{dx} = \frac{12(2\sin \frac{t}{2} \cos \frac{t}{2})}{10(2\sin^2 \frac{t}{2})} = \frac{12}{10} \cot \frac{t}{2} = \frac{6}{5} \cot \frac{t}{2}$.

Find $\frac{dy}{dx}$,if $y=12(1-\cos t)$ and $x=10(t-\sin t)$.

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