The energy of a system as a function of time $t$ is given as $E (t) = e^{- α t} A^{2}$ , where $α = 0.2 s^{- 1}$ . 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 $t = 5 s is$
(Approximate your answer in integer value)

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Solution

We have,
$E (t) = A^{2} e^{- α t}$
$E (t + Δ t) = (A + Δ A)^{2} e^{- α (t + Δ t)}$
$E (t + Δ t) = (A^{2} + Δ A^{2} + 2 A Δ A) e^{- α (t + Δ t)}$
Here, $Δ A^{2}$ will be very small and hence neglected.

$E (t + Δ t) = (A^{2} + 2 A Δ A) e^{- α (t + Δ t)}$

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

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

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