Question

If $x=e^{\theta }(\sin \theta +\cos \theta )$ and $y=e^{\theta }(\sin \theta –\cos \theta ),$ where $\theta$ is a real parameter, then $\frac{d^{2}y}{d{x}^2}$ at $\theta =\frac{\pi }{6}$ is

• A. $\frac{\sqrt{3}}{2}$
• B. $\frac{4}{3\sqrt{3}}{e}^{-\pi /6}$
• C. $\frac{8}{3\sqrt{3}}{e}^{-\pi /6}$
• D. $\frac{4}{3}$

Answer 1

The correct option is B $\frac{4}{3\sqrt{3}}{e}^{-\pi /6}$

Given : $x=e^{\theta }(\sin \theta +\cos \theta ),y=e^{\theta }(\sin \theta -\cos \theta )$
$\frac{dx}{d\theta }=e^{\theta }(\sin \theta +\cos \theta )+e^{\theta }(\cos \theta -\sin \theta )\Rightarrow \frac{dx}{d\theta }=2e^{\theta }\cos \theta \frac{dy}{d\theta }=e^{\theta }(\sin \theta -\cos \theta )+e^{\theta }(\cos \theta +\sin \theta )\Rightarrow \frac{dy}{d\theta }=2e^{\theta }\sin \theta \Rightarrow \frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\tan \theta \Rightarrow \frac{dy}{dx}=\tan \theta \frac{d^{2}y}{d{x}^2}={\sec }^2\theta \cdot \frac{d\theta }{dx}=\frac{{\sec }^2\theta }{2e^{\theta }\cos \theta }\therefore \frac{d^{2}y}{d{x}^2}{|}_{\theta =\frac{\pi }{6}}=\frac{4}{3\sqrt{3}}{e}^{-\pi /6}$