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Question
If \[x = a \cos^3 \theta, y = a \sin^3 \theta, \text { then } \sqrt{1 + \left( \frac{dy}{dx} \right)^2} =\] ____________ .
Options
\[\tan^2 \theta\]
\[\sec^2 \theta\]
\[\sec \theta\]
\[\left| \sec \theta \right|\]
Solution
\[\left| \sec \theta \right|\]
\[\text { We have }, x = a \cos^3 \theta\]
\[ \Rightarrow \frac{dx}{d\theta} = a\frac{d}{d\theta}\left( \cos^3 \theta \right)\]
\[ \Rightarrow \frac{dx}{d\theta} = 3a \cos^2 \theta\frac{d}{d\theta}\left( \cos\theta \right)\]
\[ \Rightarrow \frac{dx}{d\theta} = - 3a \cos^2 \theta\sin\theta .......... \left( 1 \right)\]
\[\text { and }, \]
\[ y = a \sin^3 \theta\]
\[ \Rightarrow \frac{dy}{d\theta} = a\frac{d}{d\theta}\left( \sin^3 \theta \right)\]
\[ \Rightarrow \frac{dy}{d\theta} = 3a \sin^2 \theta\frac{d}{d\theta}\left( \sin\theta \right)\]
\[ \Rightarrow \frac{dy}{d\theta} = 3a \sin^2 \theta \cos\theta ............ \left( 2 \right)\]
\[\text { Dividing } \left( 2 \right) \text { by } \left( 1 \right), \text { we get }, \]
\[\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{3a \sin^2 \theta \cos\theta}{- 3a \cos^2 \theta\sin\theta}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\sin\theta}{- \cos\theta}\]
\[ \Rightarrow \frac{dy}{dx} = - \tan\theta\]
\[\text { Now }, \sqrt{1 + \left( \frac{dy}{dx} \right)^2} = \sqrt{1 + \tan^2 \theta} = \sqrt{\sec^2 \theta} = \left| \sec\theta \right|\]
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