Question

The value of $\frac{d^{2}y}{d{x}^2}$ at the point where $t=0$ is

• A. $1$
• B. $2$
• C. $-2$
• D. $3$

Answer 1

Answer: (B)

$2$

$x=e^t\cos t$ and $y=e^{t}sint$

$\frac{dx}{dt}=\frac{d}{dt}\left(e^t\cos t\right)$

$=-e^t\sin t+\cos t{e}^t$

$=e^t(\cos t-\sin t)$

$\frac{dy}{dt}=\frac{d}{dt}\left(e^t\sin t\right)$

$=e^t\cos t+\sin t{e}^t$

$=e^t(\cos t+\sin t)$

$\frac{dy}{dx}=\frac{\cos t+\sin t}{\cos t-\sin t}$

$\frac{d^{2}y}{d{x}^2}=\frac{\left(\cos t-\sin t\right)(-\sin t+\cos t)-\left(\cos t+\sin t\right)(-\sin t-\cos t)}{{\left(\cos t-\sin t\right)}^2}$

$\frac{d^{2}y}{d{x}^2}=\frac{{\left(\cos t-\sin t\right)}^2+{\left(\cos t+\sin t\right)}^2}{{\left(\cos t-\sin t\right)}^2}$

$\frac{d^{2}y}{d{x}^2}{|}_{att=0}=\frac{{(1-0)}^2+{(1+0)}^2}{{(1-0)}^2}=2$