If \(x = a\sec^3\theta,\ y = a\tan^3\theta\), then \(\displaystyle\frac{d^2y}{dx^2}\) equals
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{\cos^5\theta}{3a\sin\theta}\)
In parametric form, x and y are both differentiated with respect to \(\theta\):
\( \displaystyle \frac{dx}{d\theta} = 3a\sec^2\theta \cdot \sec\theta\tan\theta = 3a\sec^3\theta\tan\theta \)
\( \displaystyle \frac{dy}{d\theta} = 3a\tan^2\theta\sec^2\theta \)
\( \displaystyle \frac{dy}{dx} = \frac{3a\tan^2\theta\sec^2\theta}{3a\sec^3\theta\tan\theta} = \frac{\tan\theta}{\sec\theta} = \sin\theta \)
Differentiating again with respect to x, by the chain rule:
\( \displaystyle \frac{d^2y}{dx^2} = \frac{d}{d\theta}(\sin\theta) \cdot \frac{d\theta}{dx} = \cos\theta \cdot \frac{1}{3a\sec^3\theta\tan\theta} \)
\( \displaystyle = \cos\theta \cdot \frac{\cos^3\theta \cdot \cos\theta}{3a\sin\theta} \)
\( \displaystyle = \frac{\cos^5\theta}{3a\sin\theta} \)
Tip: \(\frac{d^2y}{dx^2}\) is not \(\frac{d^2y/d\theta^2}{d^2x/d\theta^2}\). Differentiate \(\frac{dy}{dx}\) with respect to \(\theta\) and then multiply by \(\frac{d\theta}{dx}\).