Why Measurement Uncertainty Propagates
A measured input is rarely exact. If a length is recorded as \(x=10.0\text{ cm}\) with uncertainty \(0.1\text{ cm}\), the actual length is being represented as close to the recorded value, not known to equal it exactly. A quantity calculated from that measurement inherits some uncertainty. Differentials give a first-order estimate of how large that output uncertainty is.
The central idea is simple: near a known input \(x\), a small input change produces an output change estimated by the derivative times that input change. For uncertainty, we want the size of the output change, so we use the magnitude of the derivative.
From a Measurement to a Differential
Suppose the measured input is \(x\), and the true input differs from it by a small amount \(dx\). The corresponding output is \(y=f(x)\). Its actual change is \(f(x+dx)-f(x)\), while the differential predicts the change \(dy=f'(x)\,dx\). Thus, for a small input uncertainty, the estimated output uncertainty is the magnitude \(|dy|=|f'(x)|\,|dx|\).
When a measurement is reported as \(x\pm\delta x\), the symbol \(\delta x\) denotes a nonnegative size of uncertainty. The differential \(dx\) may have either sign, depending on whether the input is above or below the reported value. Since uncertainty is a size, set \(|dx|=\delta x\) in the estimate. The result \(\delta y\approx |f'(x)|\delta x\) has output units: derivative units multiplied by input units.
Why the Derivative Gives the Estimate
The derivative is defined by the limiting ratio of output change to input change. At a point where \(f\) is differentiable, for a small nonzero \(h\), the actual output change can be written as
As \(h\) approaches zero, the ratio in parentheses approaches \(f'(x)\). Therefore, when \(h\) is small, the output change is approximately \(f'(x)h\). Taking magnitudes gives the uncertainty estimate. This reasoning explains both the method and its limitation: the estimate is best when the input uncertainty is small enough that the derivative is representative across the nearby input values.
Write the output as \(y=f(x)\) and record the measured input and its nonnegative uncertainty.
Find \(f'(x)\), then evaluate it at the measured input.
Compute \(\delta y\approx |f'(x)|\delta x\), retaining the units.
Report the output value and the estimated uncertainty, and state that the uncertainty is a first-order approximation.
Worked Examples
Worked Example: Uncertainty in the Area of a Circle
A circle has measured radius \(r=10.0\text{ cm}\) with uncertainty \(\delta r=0.1\text{ cm}\). Estimate the uncertainty in its area \(A=\pi r^2\).
First differentiate the area formula with respect to \(r\):
At \(r=10.0\text{ cm}\), the derivative is \(2\pi(10.0)=20\pi\text{ cm}\). Its units are area divided by length. The differential uncertainty estimate is
The estimated area itself is \(A=\pi(10.0)^2=100\pi\text{ cm}^2\approx314.16\text{ cm}^2\). Thus the estimated result is \(314.16\text{ cm}^2\) with absolute uncertainty about \(6.28\text{ cm}^2\).
Check: Differentiating \(A=\pi r^2\) gives \(dA/dr=2\pi r\), as used. For an increase of \(0.1\text{ cm}\), the exact area change is \(\pi(10.1)^2-\pi(10.0)^2=2.01\pi\text{ cm}^2\), close to the differential estimate \(2\pi\text{ cm}^2\). The small difference is why the result is an estimate.
Worked Example: Percentage Uncertainty in a Volume
A cube has measured side length \(s=4.00\text{ cm}\) with uncertainty \(\delta s=0.02\text{ cm}\). Estimate the uncertainty in its volume \(V=s^3\), and find the percentage uncertainty.
Differentiate and evaluate at the measured side length:
The absolute output uncertainty is therefore
The estimated volume is \(V=(4.00)^3=64.0\text{ cm}^3\). The relative uncertainty is approximately \(0.96/64.0=0.015\), so the percentage uncertainty is \(0.015(100\%)=1.5\%\).
Check: The derivative of \(s^3\) is \(3s^2\), and multiplying its value at \(4.00\) by \(0.02\) gives \(0.96\). Directly, the volume change for an increase to \(4.02\text{ cm}\) is \((4.02)^3-4^3=64.964808-64=0.964808\text{ cm}^3\), close to the differential estimate. The percentage estimate also agrees with the power rule: the side’s percentage uncertainty is \(0.02/4.00\times100\%=0.5\%\), and cubing makes the estimated percentage uncertainty three times as large, or \(1.5\%\).
Percentage Uncertainty for Power Functions
For a power function \(y=Cx^n\), where \(C\ne0\), \(x>0\), and \(n\) is a real constant, differentiate to obtain \(y'=Cnx^{n-1}\). The differential estimate is \(|dy|\approx |Cnx^{n-1}|\delta x\). Since \(|y|=|C|x^n\), divide by \(|y|\) to compare the estimated output uncertainty with the output’s magnitude:
The coefficient \(C\) cancels, and the exponent determines the multiplier. For example, a square (\(n=2\)) has about twice the input’s percentage uncertainty; a cube (\(n=3\)) has about three times the input’s percentage uncertainty. For a reciprocal square (\(n=-2\)), the multiplier is \(|-2|=2\): the negative exponent affects the direction of change, but uncertainty size remains nonnegative. The rule is a first-order estimate, not an exact percentage relationship for a finite input interval.
Worked Example: A Reciprocal-Square Model
Suppose \(y=120/x^2\), where \(x=3.00\) has uncertainty \(\delta x=0.03\). Estimate the absolute and percentage uncertainty in \(y\).
Differentiate the model:
At \(x=3.00\), the output and derivative are
Thus the absolute uncertainty estimate is
The relative uncertainty is \((4/15)/(40/3)=1/50=0.02\), giving a percentage uncertainty of \(2\%\). This also follows from the power rule: the input percentage uncertainty is \((0.03/3.00)\times100\%=1\%\), and the exponent has magnitude \(2\).
Check: Differentiating \(120/x^2\) gives \(-240/x^3\), and its value at \(3.00\) is \(-80/9\). For an increase to \(x=3.03\), the exact change is \(120/(3.03)^2-120/9\approx-0.263\), close in magnitude to \(0.267\). The negative change indicates the output decreases; the uncertainty estimate is its positive size.
Signed Change Is Not Uncertainty Size
A derivative can be negative. That sign tells whether the output increases or decreases as the input increases; it does not make an uncertainty negative. For example, if \(f(x)=1/x\) and \(x=2\) with input uncertainty \(0.1\), then \(f'(2)=-1/4\). Multiplying the signed derivative by \(0.1\) gives \(dy\approx-0.025\). Calling this a “negative uncertainty” is incorrect. The estimated uncertainty magnitude is \(|dy|\approx0.025\). Indeed, increasing \(x\) to \(2.1\) changes the output by \(1/2.1-1/2=-1/42\approx-0.02381\): the change is negative, but its size is positive.
When to Use Caution
Differential propagation estimates local change. It is appropriate when the model is differentiable at the measured value and the uncertainty is small enough for the local linear approximation to be useful. If the derivative varies substantially across the possible input range, or the uncertainty is not small, the estimate may not describe the full output range well. Also, a zero derivative makes the first-order estimate zero; it does not by itself prove that the output has no uncertainty. The conclusion should always be described as an estimate unless a separate argument establishes an exact bound.
Check Your Understanding
Use differentials and show the units and interpretation in each response.
- If \(y=x^4\), \(x=2.00\text{ m}\), and \(\delta x=0.01\text{ m}\), what is the estimated absolute uncertainty in \(y\)?
- A measured radius has a \(0.4\%\) uncertainty. What percentage uncertainty does the differential rule estimate for the area of a circle?
- For \(y=5/x\), why should an uncertainty estimate use the magnitude of \(dy\) rather than its sign?
- State the assumptions and the local nature of the percentage-uncertainty rule for \(y=Cx^n\).
- What does it mean if the differential estimate is zero, and why might that not settle the actual output uncertainty?
