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 ?

Answer 1

\[x = a\left( \theta - \sin \theta \right)\]

\[ \Rightarrow \frac{dx}{d\theta} = a\left( 1 - \cos \theta \right)\]

\[ y = a\left( 1 + \cos \theta \right) \]

\[ \Rightarrow \frac{dy}{d\theta} = a\left( - \sin \theta \right)\]

\[ \therefore \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{a\left( - \sin \theta \right)}{a\left( 1 - \cos \theta \right)} = \frac{- 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \sin^2 \frac{\theta}{2}} = - \text { cot } \frac{\theta}{2}\]

\[\text { Now,} \]

\[\text { Slope of the tangent }= \left( \frac{dy}{dx} \right)_{\theta = \frac{- \pi}{2}} =-\text { cot }\left( \frac{\frac{- \pi}{2}}{2} \right)=-\text { cot }\left( \frac{- \pi}{4} \right)=1\]

\[\text { Slope of the normal }=\frac{- 1}{\left( \frac{dy}{dx} \right)_{\theta = \frac{- \pi}{2}}}=\frac{- 1}{1}=-1\]