Question

Find the slope of the tangent and the normal to the following curve at the indicted point  x = a cos³ θ, y = a sin³ θ at θ = π/4 ?

Answer 1

\[ x = a \cos^3 \theta\]

\[ \Rightarrow \frac{dx}{d\theta} = - 3a \cos^2 \theta \sin \theta\]

\[y = a \sin^3 \theta\]

\[ \Rightarrow \frac{dy}{d\theta} = 3a \sin^2 \theta \cos \theta\]

\[ \therefore \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{3a \sin^2 \theta \cos \theta}{- 3a \cos^2 \theta \sin \theta} = - \tan \theta\]

\[\text { Now, } \]

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

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