Question

If $x=e^t\sin t,y=e^t\cos t$, then $\frac{d^{2}y}{d{x}^2}$ at $t=\mathrm{\pi }$ is equal to

• A. $2e^{\mathrm{\pi }}$
• B. $\frac{1}{2}{e}^{\mathrm{\pi }}$
• C. $\frac{1}{2e^{\mathrm{\pi }}}$
• D. $\frac{2}{e^{\mathrm{\pi }}}$

Answer 1

The correct option is D

$\frac{2}{e^{\mathrm{\pi }}}$

Explanation for the correct option:

Step 1. Find $\frac{dx}{dt}$ and $\frac{dy}{dt}$.

Differentiate $x=e^t\sin t$ with respect to $t$.

$\begin{array}{rcl}\frac{dx}{dt}& =& \frac{d}{dt}\left(e^t\sin t\right)\\ & =& e^t\sin t+e^t\cos t\\ & =& e^t\left(\sin t+\cos t\right)\end{array}$

Differentiate $y=e^t\cos t$ with respect to $t$.

$\begin{array}{rcl}\frac{dy}{dt}& =& \frac{d}{dt}\left(e^t\cos t\right)\\ & =& e^t\cos t+e^t\left(-\sin t\right)\\ & =& e^t\left(\cos t-\sin t\right)\end{array}$

Step 2. Find the value of $\frac{dy}{dx}$.

The value of $\frac{dy}{dx}$ is given as: $\frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}$.

Substitute the found values.

$\begin{array}{rcl}\frac{dy}{dx}& =& e^t\left(\cos t-\sin t\right)\times \frac{1}{e^t\left(\sin t+\cos t\right)}\\ & =& \frac{\cos t-\sin t}{\sin t+\cos t}\end{array}$

Step 3. Find the value of $\frac{d^{2}y}{d{x}^2}$.

The value of $\frac{d^{2}y}{d{x}^2}$ is given as: $\frac{d^{2}y}{d{x}^2}=\frac{d}{dt}\left(\frac{dy}{dx}\right)\frac{dt}{dx}$

Substitute the values.

$\begin{array}{rcl}\frac{d^{2}y}{d{x}^2}& =& \frac{d}{dt}\left(\frac{\cos t-\sin t}{\cos t+\sin t}\right)\frac{1}{e^t\left(\sin t+\cos t\right)}\\ & =& \frac{\left(-\sin t-\cos t\right)\left(\cos t+\sin t\right)-\left(\cos t-\sin t\right)\left(-\sin t+\cos t\right)}{{\left(\cos t+\sin t\right)}^2}\frac{1}{e^t\left(\sin t+\cos t\right)}\\ & =& \frac{-{\sin }^{2}t-{\cos }^{2}t-2\sin t\cos t-{\cos }^{2}t-{\sin }^{2}t+2\sin t\cos t}{e^t{\left(\cos t+\sin t\right)}^3}\\ & =& \frac{-1-1}{e^t{\left(\cos t+\sin t\right)}^3}\\ & =& \frac{-2}{e^t{\left(\cos t+\sin t\right)}^3}\end{array}$

Step 4. Find the value of $\frac{d^{2}y}{d{x}^2}$ at $t=\mathrm{\pi }$.

In $\frac{d^{2}y}{d{x}^2}=\begin{array}{rcl}& & \frac{-2}{e^t{\left(\cos t+\sin t\right)}^3}\end{array}$ substitute $\mathrm{\pi }$ for $t$ to get the value of $\frac{d^{2}y}{d{x}^2}$ at $t=\mathrm{\pi }$.

$\begin{array}{rcl}{\overline{)\frac{d^{2}y}{d{x}^2}}}_{t=\mathrm{\pi }}& =& \frac{-2}{e^{\mathrm{\pi }}{\left(\mathrm{cos\pi }+\mathrm{sin\pi }\right)}^3}\\ & =& \frac{-2}{e^{\mathrm{\pi }}{\left(-1+0\right)}^3}\left[\mathrm{cos\pi }=-1,\mathrm{sin\pi }=0\right]\\ & =& \frac{-2}{e^{\mathrm{\pi }}(-1)}\\ & =& \frac{2}{e^{\mathrm{\pi }}}\end{array}$

Hence, the correct option is (D).