Question

If x = a (1 − cos³θ), y = a sin³θ, prove that \[\frac{d^2 y}{d x^2} = \frac{32}{27a} \text { at } \theta = \frac{\pi}{6}\]?

Answer 1

Here,

\[x = a\left( 1 - \cos^3 \theta \right), y = a \sin^3 \theta\]

\[\text { Differentiating w . r . t .} \theta, \text {we get }\]

\[\frac{d x}{d \theta} = 3a \cos^2 \theta \sin\theta and  \frac{d y}{d \theta} = 3a \sin^2 \theta \cos\theta\]

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

Differentiating w. r. t. x, we get

\[\frac{d^2 y}{d x^2} = \sec^2 \theta \frac{d \theta}{d x}\]

\[ = \frac{\sec^2 \theta}{3a \cos^2 \theta \sin\theta}\]

\[ = \frac{\sec^4 \theta}{3a \sin\theta}\]

\[ \therefore \frac{d^2 y}{d x^2} \text { at } \theta = \frac{\pi}{6}\]

\[ \Rightarrow \frac{d^2 y}{d x^2} = \frac{\left( \sec\frac{\pi}{6} \right)^4}{3a \sin\frac{\pi}{6}} = \frac{32}{27a}\]