Question

The energy of a system as a function of time t is given as $E\left(t\right)=A^{2}exp(-\alpha t)$, where $\alpha =0.2s^{-1}$. The measurement of A has an error of 1.25%. If the error in the measurement of time is 1.50%, the percentage error in the value of $E\left(t\right)$ at t = 5 s is:

• A. 2%
• B. 4%
• C. 3%
• D. 5%

Answer 1

The correct option is B 4%

$E\left(t\right)=A^2{e}^{-\alpha t}$
Taking natural logarithm on both sides,
$ln\left(E\right)=2ln\left(A\right)+(-\alpha t)$
Differentiating both sides
$\frac{dE}{E}=2\frac{dA}{A}+\left(\alpha dt\right)$
Errors always add up for maximum error.
$\therefore \frac{dE}{E}=2\frac{dA}{A}+\alpha \left(\frac{dt}{t}\right)\times t$
Here, $\frac{dA}{A}=1.25$ %, $\frac{dt}{t}=1.5$%, $t=5s$, $\alpha =0.2s^{-1}$
$\therefore \frac{dE}{E}=(2\times 1.25$%$)+(0.2)\times (1.5$%$)\times 5=4$%