Question

The energy of a system as a function of time t is given as $E(t) = A^2 \exp(-\alpha t)$, where $\alpha = 0.2 s^{-1}$. The measurement of A has an error of 1.25%. If the error in the measurement of time is 1.50%, find the maximum percentage error in the calculation of value of $E(t)$ at $t = 4$ s.

Answer 1

3.7%

Answer 2

The energy of the system is given by the equation $E(t) = A^2 \exp(-\alpha t)$.
To find the maximum percentage error in $E(t)$, we use the method of error propagation.
Take the natural logarithm of both sides of the equation:
$\ln E = \ln(A^2) + \ln(e^{-\alpha t})$
$\ln E = 2 \ln A - \alpha t$

To find the maximum fractional error, we differentiate the logarithmic equation and sum the absolute values of the individual terms:
$\frac{\Delta E}{E} = \left| \frac{\partial(\ln E)}{\partial A} \Delta A \right| + \left| \frac{\partial(\ln E)}{\partial t} \Delta t \right|$
$\frac{\Delta E}{E} = \left| \frac{2}{A} \Delta A \right| + \left| (-\alpha) \Delta t \right|$
$\frac{\Delta E}{E} = 2 \frac{\Delta A}{A} + \alpha \Delta t$

Now, let's substitute the given values:

1. Percentage error in the measurement of A: $\frac{\Delta A}{A} \times 100\% = 1.25\%$.
   So, $\frac{\Delta A}{A} = 0.0125$.
2. Percentage error in the measurement of time: $\frac{\Delta t}{t} \times 100\% = 1.50\%$.
   So, $\frac{\Delta t}{t} = 0.015$.
3. Given $\alpha = 0.2 \text{ s}^{-1}$.
4. The calculation is at $t = 4 \text{ s}$.

Calculate the individual error contributions:
First term: $2 \frac{\Delta A}{A} = 2 \times 0.0125 = 0.025$.
In percentage: $0.025 \times 100\% = 2.5\%$.

Second term: $\alpha \Delta t$.
First, find $\Delta t$ using the given percentage error in time and the value of $t$:
$\Delta t = \left(\frac{\Delta t}{t}\right) \times t = 0.015 \times 4 \text{ s} = 0.06 \text{ s}$.
Now, calculate $\alpha \Delta t$:
$\alpha \Delta t = (0.2 \text{ s}^{-1}) \times (0.06 \text{ s}) = 0.012$.
In percentage: $0.012 \times 100\% = 1.2\%$.

Finally, sum the percentage errors to find the maximum percentage error in $E(t)$:
Maximum percentage error in $E(t) = (2.5\% + 1.2\%) = 3.7\%$.