Question 305: Units , Dimensions and Error Analysis

Question

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

Accepted Answer

$$E(t) = {A^2}{e^{ - \alpha t}}$$ ...... (i)
$$\alpha$$ = 0.2 s
$$-$$1
$$\left( {{{dA} \over A}} ight) 	imes 100 = 1.25\% $$
$$\left( {{{dt} \over t}} ight) 	imes 100 = 1.50$$
$$ \Rightarrow (dt 	imes 100) = 1.5t = 1.5 	imes 5 = 7.5$$
Differentiating on both sides of equation (i), we get
$$
d \mathrm{E}=(2 \mathrm{~A} d \mathrm{~A}) e^{-\alpha t}+\mathrm{A}^2 e^{-\alpha t}(-\alpha d t)
$$
Dividing throughout by $\mathrm{E}=\mathrm{A}^2 e^{-\alpha t}$
$$
\frac{d \mathrm{E}}{\mathrm{E}}=\frac{2}{\mathrm{~A}} d \mathrm{~A}+\alpha d t
$$
(Considering worst possible case)
$$	herefore$$ $$\left( {{{dE} \over E}} ight) 	imes 100 =   2\left( {{{dA} \over A}} ight) 	imes 100 + \alpha (dt 	imes 100)$$
$$ =  2(1.25) + 0.2(7.5)$$
$$ =  2.5 + 1.5$$
$$ =  4\% $$

Units , Dimensions and Error Analysis

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

Integer / Numeric Answer Type