Question
If $\text{x}=10(\text{t}-\sin\text{t}),\text{y}=12(1-\cos\text{t}),$ find $\frac{\text{dy}}{\text{dx}}.$
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Answer
Here, $\text{x}=10(\text{t}-\sin\text{t}),\text{y}=12(1-\cos\text{t})$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=10(1-\cos\text{t})\ ...(\text{i})$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=12(\sin\text{t})\ ...(\text{ii})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{12(\sin\text{t})}{10(1-\cos\text{t})}$ From equation (i) and (ii)
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{12\sin\frac{\text{t}}{2}\cdot\cos\frac{\text{t}}{2}}{10\sin^2\frac{\text{t}}{2}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{6}{5}\cot\frac{\text{t}}{2}$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=10(1-\cos\text{t})\ ...(\text{i})$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=12(\sin\text{t})\ ...(\text{ii})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{12(\sin\text{t})}{10(1-\cos\text{t})}$ From equation (i) and (ii)
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{12\sin\frac{\text{t}}{2}\cdot\cos\frac{\text{t}}{2}}{10\sin^2\frac{\text{t}}{2}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{6}{5}\cot\frac{\text{t}}{2}$
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