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प्रश्न
If x = a (1 − cos3θ), y = a sin3θ, prove that \[\frac{d^2 y}{d x^2} = \frac{32}{27a} \text { at } \theta = \frac{\pi}{6}\]?
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उत्तर
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
\[\frac{d^2 y}{d x^2} = \sec^2 \theta \frac{d \theta}{d x}\]
\[ = \frac{\sec^2 \theta}{3a \cos^2 \theta \sin\theta}\]
\[ = \frac{\sec^4 \theta}{3a \sin\theta}\]
\[ \therefore \frac{d^2 y}{d x^2} \text { at } \theta = \frac{\pi}{6}\]
\[ \Rightarrow \frac{d^2 y}{d x^2} = \frac{\left( \sec\frac{\pi}{6} \right)^4}{3a \sin\frac{\pi}{6}} = \frac{32}{27a}\]
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