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Question
Find the slope of the tangent and the normal to the following curve at the indicted point x = a (θ − sin θ), y = a(1 − cos θ) at θ = π/2 ?
Solution
\[ x = a\left( \theta - \sin \theta \right)\]
\[ \Rightarrow \frac{dx}{d\theta} = a\left( 1 - \cos \theta \right)\]
\[ y = a\left( 1 - \cos \theta \right)\]
\[ \Rightarrow \frac{dy}{d\theta} = a\left( \sin \theta \right)\]
\[ \therefore \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{a\left( \sin \theta \right)}{a\left( 1 - \cos \theta \right)} = \frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \sin^2 \frac{\theta}{2}} = \cot \frac{\theta}{2}\]
\[\text { Now,} \]
\[\text { Slope of the tangent }= \left( \frac{dy}{dx} \right)_{\theta = \frac{\pi}{2}} =cot\left( \frac{\frac{\pi}{2}}{2} \right)=cot\left( \frac{\pi}{4} \right)=1\]
\[\text { Slope of the normal }=\frac{- 1}{\left( \frac{dy}{dx} \right)_{\theta = \frac{\pi}{2}}}=\frac{- 1}{1}=-1\]
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