Question

The energy of a system as a function of time $t$ is given as $E(t)=A^2$ exp$(-\alpha t)$, where $\alpha =0.2s^{-1}$. 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 $E(t)$ at $t=5s$ is

• A. $2\%$
• B. $4\%$
• C. $3\%$
• D. $5\%$

Answer 1

Answer: (B)

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