Question

The energy of a system as a function of time $t$ is given as $E (t)= A ^{2} \exp (-\alpha t )$, where $\alpha=$ $0.4\, s ^{-1}$. 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 $t =6\, s$ is_________$\%$.

Answer 1

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