The energy of a system as a function of time t is given as E(t) = A^2 exp(-alpha t), where alpha = 0

Q: Why do we take the natural logarithm before differentiating?

Taking the logarithm transforms complex operations like multiplication and exponentiation into simple addition and subtraction. This makes it incredibly easy to apply the derivative and directly obtain the fractional error terms ($\frac{dE}{E}$, $\frac{dA}{A}$), which are exactly what we need for percentage error calculations.

Q: Why did the negative sign in front of $\alpha t$ become positive in the error equation?

In error analysis, we are always interested in the **maximum permissible error**. We must assume the worst-case scenario where all individual measurement errors compound and push the final result in the same direction. Therefore, we take the absolute value of each fractional error term, turning all negative signs into positive ones.

Q: How did we get the value $7.5$ for the $dt \times 100$ term?

The problem gives us the percentage error in time, which is mathematically written as $(\frac{dt}{t}) \times 100 = 1.50\%$. However, our differentiated equation contains the term $\alpha \cdot dt$. To find $dt \times 100$, we rearrange the percentage error formula: $dt \times 100 = 1.50 \times t$. Substituting the given time $t = 5 \text{ s}$, we get $1.50 \times 5 = 7.5$.

Q: Can we just plug the percentage errors directly into the derivative without multiplying by 100?

Yes, you can work entirely in fractional errors and multiply by 100 at the very end. However, multiplying the entire equation by 100 immediately after differentiating is a great way to avoid confusion, as it allows you to directly substitute the percentage values given in the question (like 1.25%) into the equation.

Q: Why does the final percentage error depend on the specific time $t = 5 \text{ s}$?

Because the time $t$ is in the exponent! The fractional error contributed by the exponential term is $\alpha \cdot dt$. Since the percentage error in time ($\frac{dt}{t}$) is constant, the absolute error $dt$ grows as $t$ increases. Therefore, the uncertainty contributed by the time measurement becomes larger as more time passes.

Question

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

Sigma Percentile · Accepted Answer

### The Anatomy of Error Propagation

Imagine you are tracking the energy of a decaying system. The mathematical model tells you that the energy follows a beautiful, smooth exponential curve: $E(t) = A^2 e^{-\alpha t}$. However, in the real world, our measuring instruments are never perfect. The initial amplitude $A$ has a slight uncertainty, and our stopwatch measuring the time $t$ is also slightly off.

These tiny imperfections don't just sit there; they propagate through the mathematical machinery of the equation, creating a 'band of uncertainty' around our perfect theoretical curve. Our mission in this problem is to calculate exactly how wide this error band is at a specific moment in time, $t = 5 	ext{ s}$.

### The Master Tool: Logarithmic Differentiation

When dealing with an equation that involves products, quotients, or exponents, calculating the error directly can be a nightmare. The most elegant and powerful tool in our arsenal is **logarithmic differentiation**. By taking the natural logarithm of both sides, we can shatter the complex multiplicative structure into simple, manageable linear terms.

Let's apply this to our energy equation:

$$\ln E = \ln(A^2 e^{-\alpha t})$$

Using the properties of logarithms, we can expand this:

$$\ln E = 2 \ln A - \alpha t$$

Now, we differentiate the entire equation. The derivative of $\ln x$ is $\frac{dx}{x}$, which perfectly represents the fractional error of a quantity.

$$\frac{dE}{E} = 2 \frac{dA}{A} - \alpha dt$$

### The Golden Rule of Maximum Error

Notice the negative sign in front of the $\alpha dt$ term? This is where many students fall into a trap. In error analysis, we are not trying to find the *exact* error (which is impossible to know); we are trying to find the **maximum permissible error**.

We must assume the absolute worst-case scenario: what if the error in $A$ and the error in $t$ both conspire to push the final energy value in the same direction? To account for this, we must take the absolute value of every individual error term. Errors always add up; they never cancel each other out.

$$\left(\frac{dE}{E}ight)_{	ext{max}} = 2 \left|\frac{dA}{A}ight| + \alpha |dt|$$

To convert this fractional error into a percentage error, we simply multiply the entire equation by 100:

$$\frac{dE}{E} 	imes 100 = 2 \left(\frac{dA}{A} 	imes 100ight) + \alpha (dt 	imes 100)$$

### Decoding the Time Trap

Now we need to substitute our given values. The problem states that the percentage error in $A$ is $1.25\%$. This means $\frac{dA}{A} 	imes 100 = 1.25$. That part is easy.

However, look closely at the time term in our differentiated equation. It is $\alpha (dt 	imes 100)$. It requires the **absolute error** in time ($dt$), not the fractional error.

The problem gives us the *percentage error* in time: $\frac{dt}{t} 	imes 100 = 1.50\%$.

To find the value of $dt 	imes 100$, we must rearrange this relationship and plug in the specific time $t = 5 	ext{ s}$:

$$dt 	imes 100 = 1.50 	imes t$$
$$dt 	imes 100 = 1.50 	imes 5 = 7.5 	ext{ s}$$

This is a critical insight: because time is in the exponent, the uncertainty it contributes to the final energy value grows larger as time goes on!

### The Final Calculation

With all our pieces ready, we can substitute them into our master error equation. We know $\alpha = 0.2 	ext{ s}^{-1}$.

$$\frac{dE}{E} 	imes 100 = 2(1.25) + 0.2(7.5)$$

$$\frac{dE}{E} 	imes 100 = 2.5 + 1.5$$

$$\frac{dE}{E} 	imes 100 = 4\%$$

And there we have it. The maximum percentage error in the energy at $t = 5 	ext{ s}$ is **$\pm 4\%$**. By carefully breaking down the equation and respecting the rules of error propagation, a seemingly complex problem unravels into a beautiful, clean result.

Enter Numerical Value:

Visualized Solution

\text{Understanding the Error Band}

\text{Logarithmic Differentiation}

\text{Differentiating for Error}

\text{Given Percentage Errors}

\text{Substituting Values}

\text{Final Calculation}

\text{Food for Thought}

The Sigma Insight: Errors in Measurement

The Anatomy of Error Propagation

The Master Tool

The Golden Rule of Maximum Error

Decoding the Time Trap

The Final Calculation

Similar Questions

A physical quantity $z$ depends on four observables $a, b, c$ and $d$ , as $z = \frac{a ^{2} b ^{2/3}}{c d ^{3}}$ . The percentages of error in the measurement of $a, b, c$ and $d$ are $2%$ , $1.5%$ , $4%$ and $2.5%$ respectively. The percentage of error in $z$ is

Two capacitors with capacitance values $C_{1} = 2000 \pm 10 pF$ and $C_{2} = 3000 \pm 15 pF$ are connected in series. The voltage applied across this combination is $V = 5.00 \pm 0.02 V$ . The percentage error in the calculation of the energy stored in this combination of capacitors is _______.

Comprehension Passage

Consider the ratio $r = \frac{( 1 - a )}{( 1 + a )}$ to be determined by measuring a dimensionless quantity $a$ . If the error in the measurement of $a$ is $Δ a$ ( $Δ a / a ≪ 1$ ), then what is the error $Δ r$ in determining $r$ ?

In an experiment, the initial number of radioactive nuclei is 3000. It is found that $1000 \pm 40$ nuclei decayed in the first 1.0 s. For $∣ x ∣ ≪ 1$ , $ln (1 + x) = x$ up to first power in $x$ . The error $Δ λ$ , in the determination of the decay constant $λ$ in $s^{- 1}$ , is

If the length of the pendulum in pendulum clock increases by $0.1%$ , then the error in time per day is

The resistance $R = \frac{V}{I}$ , where $V = (50 \pm 2) V$ and $I = (20 \pm 0.2) A$ . The percentage error in $R$ is $x$ %. The value of $x$ to the nearest integer is……… .

A student uses a simple pendulum of exactly $1 m$ length to determine $g$ , the acceleration due to gravity. He uses a stop watch with the least count of $1 s$ for this and records $40 s$ for $20$ oscillations. For this observation, which of the following statement(s) is/are true?

A simple pendulum is being used to determine the value of gravitational acceleration $g$ at a certain place. The length of the pendulum is $25.0 cm$ and a stop watch with $1 s$ resolution measures the time taken for $40$ oscillations to be $50 s$ . The accuracy in $g$ is