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प्रश्न
If x = cos θ, y = sin3 θ, prove that \[y\frac{d^2 y}{d x^2} + \left( \frac{dy}{dx} \right)^2 = 3 \sin^2 \theta\left( 5 \cos^2 \theta - 1 \right)\] ?
उत्तर
Here,
\[x = \cos\theta \text { and y } = \sin^3 \theta\]
\[\text { Differentiating w . r . t }. \theta, \text { we get }\]
\[\frac{d x}{d \theta} = - \sin\theta \text { and } \frac{d y}{d \theta} = 3 \sin^2 \theta \cos\theta\]
\[ \therefore \frac{d y}{d x} = \frac{3 \sin^2 \theta \cos\theta}{- \sin\theta} = - 3\sin\theta \cos\theta\]
\[\text { Differentiating w . r . t . x, we get }\]
\[\frac{d^2 y}{d x^2} = \left( - 3 \cos^2 \theta + 3 \sin^2 \theta \right)\frac{d \theta}{d x} = \frac{\left( - 3 \cos^2 \theta + 3 \sin^2 \theta \right)}{- \sin\theta}\]
\[\text { Now,} \]
\[\text { LHS = y }\frac{d^2 y}{d x^2} + \left( \frac{d y}{d x} \right)^2 \]
\[ = \sin^3 \theta \times \frac{\left( - 3 \cos^2 \theta + 3 \sin^2 \theta \right)}{- \sin\theta} + \left( - 3\sin\theta \cos\theta \right)^2 \]
\[ = 3 \sin^2 \theta \cos^2 \theta - 3 \sin^4 \theta + 9 \sin^2 \theta \cos^2 \theta \]
\[ = 12 \sin^2 \theta \cos^2 \theta - 3 \sin^4 \theta \]
\[ = 3 \sin^2 \theta\left( 4 \cos^2 \theta - \sin^2 \theta \right) \]
\[ = 3 \sin^2 \theta\left( 5 \cos^2 \theta - 1 \right) [ \because \cos^2 + \sin^2 \theta = 1]\]
\[ = \text { RHS }\]
Hence proved.
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