Question

The energy of a system as a function of time $t$ is given as $E\left(t\right)=e^{-\alpha t}{A}^2$, 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 $t=5s\text{is}$
(Approximate your answer in integer value)

Answer 1

We have,
$E\left(t\right)=A^2{e}^{-\alpha t}$
$E(t+\Delta t)=(A+\Delta A{)}^2{e}^{-\alpha (t+\Delta t)}$
$E(t+\Delta t)=(A^2+\Delta A^2+2A\Delta A)e^{-\alpha (t+\Delta t)}$
Here, $\Delta A^2$ will be very small and hence neglected.

$E(t+\Delta t)=(A^2+2A\Delta A)e^{-\alpha (t+\Delta t)}$

Now if you divide whole equation with $E\left(t\right)$, we get
$\frac{E(t+\Delta t)}{E\left(t\right)}=\frac{(A^2+2A\Delta A)e^{-\alpha (t+\Delta t)}}{A^2{e}^{-\alpha t}}$
$\frac{E(t+\Delta t)}{E\left(t\right)}=\frac{(A^2+2A\Delta A)e^{-\Delta t}}{A^2}$
$\frac{E(t+\Delta t)}{E\left(t\right)}=(1+\frac{2\Delta A}{A})e^{-\Delta t}$

Again, $e^{-\Delta t}$ is very small value and hence neglected.
$\frac{E(t+\Delta t)}{E\left(t\right)}=(1+\frac{2\Delta A}{A})$
$\frac{E(t+\Delta t)}{E\left(t\right)}-1=\left(\frac{2\Delta A}{A}\right)$
$\%$ Error in $E\left(t\right)=\left(\frac{2\Delta A}{A}\right)\times 100$
$\%$ Error in $E\left(t\right)=2\times 1.25\%$
$\%$ Error in $E\left(t\right)=2.5\%\approx 3\%$