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Question
If \[x = 10 \left( t - \sin t \right), y = 12 \left( 1 - \cos t \right), \text { find } \frac{dy}{dx} .\] ?
Solution
\[\text { We have, x }= 10\left( t - \sin t \right) \text { and y } = 12\left( 1 - \cos t \right)\]
\[ \Rightarrow \frac{dx}{dt} = \frac{d}{dt}\left[ 10\left( t - \sin t \right) \right] \text { and }\frac{dy}{dt} = \frac{d}{dt}\left[ 12\left( 1 - \cos t \right) \right]\]
\[ \Rightarrow \frac{dx}{dt} = 10\frac{d}{dt}\left( t - \sin t \right) \text { and } \frac{dy}{dt} = 12\frac{d}{dt}\left( 1 - \cos t \right)\]
\[ \Rightarrow \frac{dx}{dt} = 10\left( 1 - \cos t \right) \text{ and } \frac{dy}{dt} = 12\left[ 0 - \left( - \sin t \right) \right] = 12 \sin t\]
\[ \therefore \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{12 \sin t}{10\left( 1 - \cos t \right)}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{12 \times 2\sin\frac{t}{2}\cos\frac{t}{2}}{10 \times 2 \sin^2 \frac{t}{2}}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{6}{5}\cot\frac{t}{2}\]
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