Question

The energy of a system as a function of time $t$ is given as $E(t)=A^2\exp (-\alpha t)$, where $\alpha =$ $0.4s^{-1}$. The measurement of $A$ has an error of $1.7\%$. If the error in the measurement of time is $1.50\%$, the percentage error in the value of $E(t)$ at $t=6s$ is_________$\%$.

Answer 1

Answer: 7

$E(t)=A^2{e}^{-\alpha t}$
Taking natural logarithm on both sides,
$\ln (E)=2\ln (A)+(-\alpha t)$
Differentiating both sides,
$\frac{dE}{E}=2\left(\frac{dA}{A}\right)+(-\alpha dt)$
As errors always add up for maximum error,
$\therefore \frac{dE}{E}=2\frac{dA}{A}+\alpha \left(\frac{dt}{t}\right)\times t$
Here, $\frac{dA}{A}=1.7\%,\frac{dt}{t}=1.5\%,t=6s$,
$\alpha =0.4s^{-1}$
$\therefore \%\frac{dE}{E}=(2\times 1.7\%)+(0.4)\times (1.5\%)\times 6$
$=3.4\%+3.6\%$
$=7\%$