Test Index

Differentiation

Section: Mathematics

© examsnet.com

Question : 6 of 34

Marks: +1, -0

If

x=\sqrt{2} e^t(\sin t-\cos t)

y=\sqrt{2} e^t(\sin t+\cos t)

\left(\frac{d^2 y}{d x^2}\right)_{t=\frac{\pi}{4}}=

[AP EAPCET 23 May 2025 S1]

$- e^{-\pi/4}$
$\sqrt{2} e^{\pi/4}$
$\sqrt{2} e^{-\pi/4}$
$e^{-\pi/4}$

Solution:

Given,

x=\sqrt{2} e^t(\sin t-\cos t)

and

y=\sqrt{2} e^t(\sin t+\cos t)

On differentiating both side, we get

\therefore \;\; \frac{dx}{dt}=\sqrt{2} e^t[2\sin t] \;\text{ and }\; \frac{dy}{dt}=\sqrt{2} e^t(2\cos t)

\therefore \;\; \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \cot t

Again, differentiating both side,

\frac{d^2 y}{d x^2} = \frac{d \cot t}{d x} = \frac{d \cot t}{d t} \times \frac{dt}{dx}

= -\csc^2 t \times \frac{1}{\sqrt{2} e^t (2\sin t)} = \frac{-1}{2 \sqrt{2} e^t \cdot \sin^3 t}

\therefore \left. \frac{d^2 y}{d x^2} \right|_{t=\frac{\pi}{4}} = \frac{-1}{2 \sqrt{2} \times e^{\frac{\pi}{4}} \cdot \left( \frac{1}{\sqrt{2}} \right)^3} = -e^{\frac{-\pi}{4}}

© examsnet.com

Go to Question:

Prev Question

Next Question