If \(x = 2(\cos t + t\sin t)\) and \(y = 2(\sin t - t\cos t)\), \(0 \lt t \lt \displaystyle\frac{\pi}{2}\), then the value of \(\displaystyle\frac{d^{2}y}{dx^{2}}\) at \(t = \displaystyle\frac{\pi}{3}\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{12}{\pi}\)
Differentiating each with respect to t (product rule on \(t\sin t\) and \(t\cos t\)):
\( \displaystyle \frac{dx}{dt} = 2(-\sin t + \sin t + t\cos t) = 2t\cos t \)
\( \displaystyle \frac{dy}{dt} = 2(\cos t - \cos t + t\sin t) = 2t\sin t \)
In parametric form:
\( \displaystyle \frac{dy}{dx} = \frac{2t\sin t}{2t\cos t} = \tan t \)
For the second derivative, differentiate with respect to t and divide by dx/dt:
\( \displaystyle \frac{d^{2}y}{dx^{2}} = \frac{\frac{d}{dt}(\tan t)}{dx/dt} = \frac{\sec^{2}t}{2t\cos t} \)
\( \displaystyle = \frac{\sec^{3}t}{2t} \)
At \(t = \frac{\pi}{3}\), \(\sec t = 2\), so \(\sec^{3}t = 8\):
\( \displaystyle \frac{d^{2}y}{dx^{2}} = \frac{8}{2 \times \frac{\pi}{3}} = \frac{12}{\pi} \)