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

Question

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

Professor PC Thomas Classes · Answer

(b): $E(t)=A^2 e^{-\alpha t}$ Taking natural logarithm on both sides, \ $ \ln (E)=2 \ln (A)+(-\alpha t) $ Differentiating both sides $ \frac{d E}{E}=2\left(\frac{d A}{A}ight)+(-\alpha d t) $ Errors always add up for maximum error. $ 	herefore \quad \frac{d E}{E}=2 \frac{d A}{A}+\alpha\left(\frac{d t}{t}ight) 	imes t \ $ Here, $\frac{d A}{A}=1.25 \%, \frac{d t}{t}=1.5 \%, t=5 \mathrm{~s}, \alpha=0.2 \mathrm{~s}^{-1}$ $\ 	herefore \quad \frac{d E}{E}=(2 	imes 1.25 \%)+(0.2) 	imes(1.5 \%) 	imes 5=4 \% $