Percentage Error in Time Dependent Energy Function

The energy of the system as a function of time $t$ is given by: $E(t) = A^2 \exp(-\alpha t)$

Taking the natural logarithm on both sides, we get: $\ln E = 2 \ln A - \alpha t$

Differentiating both sides to find the relative error: $\frac{dE}{E} = 2 \frac{dA}{A} - \alpha dt$

To calculate the maximum fractional error, we take the absolute sum of the individual contributions: $\frac{\Delta E}{E} = 2 \left( \frac{\Delta A}{A} \right) + \alpha \Delta t$

Rewriting $\Delta t$ in terms of the fractional error in time $\frac{\Delta t}{t}$: $\frac{\Delta E}{E} = 2 \left( \frac{\Delta A}{A} \right) + \alpha t \left( \frac{\Delta t}{t} \right)$

Multiplying the equation by $100$ to express the relation in terms of percentage errors: $\left( \frac{\Delta E}{E} \times 100\% \right) = 2 \left( \frac{\Delta A}{A} \times 100\% \right) + \alpha t \left( \frac{\Delta t}{t} \times 100\% \right)$

We are given the following values:

• $\alpha = 0.2 \text{ s}^{-1}$
• $t = 5 \text{ s}$
• Percentage error in $A$: $\frac{\Delta A}{A} \times 100\% = 1.25\%$
• Percentage error in $t$: $\frac{\Delta t}{t} \times 100\% = 1.50\%$

First, evaluate the term $\alpha t$: $\alpha t = 0.2 \text{ s}^{-1} \times 5 \text{ s} = 1$

Now, substituting these values into the percentage error formula: $\text{Percentage error in } E = 2 \times (1.25\%) + 1 \times (1.50\%)$ $\text{Percentage error in } E = 2.50\% + 1.50\% = 4\%$

Thus, the percentage error in the value of $E(t)$ evaluated at $t = 5\text{ s}$ is $4$.