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Differentiability and Linearization · Tutorial 340 of 999

Relative and Percentage Error Estimates

Learn to convert small measurement uncertainties into first-order relative and percentage error estimates for common formulas.

Intermediate 9 min read

What You'll Learn

  • Distinguish relative error estimates from absolute error estimates
  • Derive the differential rule for products, quotients, and powers
  • Combine signed relative changes and estimate worst-case uncertainty
  • Apply the method to rectangular area and cylindrical volume
  • Recognize why a differential estimate is not automatically a rigorous error bound

Why Relative Error Is Useful

A small error in an input does not always produce a small error in the same units in the output. An uncertainty of \(0.1\) centimetres has a different significance for a length of \(2\) centimetres than for a length of \(200\) centimetres. Relative error compares a change with the size of the quantity itself, making it useful when the output is a product, quotient, power, area, or volume.

Differentials provide a first-order estimate of how a small change in measured inputs affects an output. For positive quantities, dividing the differential change by the quantity gives an estimated relative change. Multiplying that relative change by \(100\) expresses it as an estimated percentage change.

Definition: If \(Q>0\) is a quantity and \(dQ\) is its differential change, then \(dQ/Q\) is the differential relative change. The corresponding differential percentage change is \(100(dQ/Q)\%\). For small input changes, these estimate the actual relative and percentage errors.

The word “estimate” matters. A differential describes a first-order approximation, not necessarily an exact change or a guaranteed bound on the actual error. The accuracy of the estimate depends on the size of the input changes and on the behavior of the function.

The Relative-Change Rule for Products, Quotients, and Powers

Suppose a positive quantity is built from positive variables using powers:

$$ Q=Cx_1^{a_1}x_2^{a_2}\cdots x_n^{a_n}, $$

where \(C>0\) is constant, the \(x_i\) are positive, and the exponents \(a_i\) are fixed real numbers. Its differential relative change is

$$ \frac{dQ}{Q} = a_1\frac{dx_1}{x_1} +a_2\frac{dx_2}{x_2} +\cdots +a_n\frac{dx_n}{x_n}. $$

This formula includes the familiar cases. For a product \(Q=xy\), the relative changes add. For a quotient \(Q=x/y\), they subtract. For a power \(Q=x^a\), the relative change is \(a\,dx/x\). A power in a denominator has a negative exponent, so its contribution has a negative sign.

First-order error estimate: For small input changes, replace \(dx_i\) by the corresponding input uncertainty or change. The resulting value of \(dQ/Q\) estimates the signed relative change in \(Q\). If only maximum input uncertainties are known, add the absolute sizes of the individual contributions to estimate a worst-case first-order relative uncertainty.

Why the Rule Works

Here is a derivation for the stated product-of-powers form. Positivity ensures that the natural logarithm of each quantity is defined. Taking logarithms gives

$$ \ln Q = \ln C+\sum_{i=1}^{n}a_i\ln x_i. $$

Differentiate both sides. Since \(C\) and the exponents are constant, \(d(\ln C)=0\), and \(d(\ln x_i)=dx_i/x_i\). Also, \(d(\ln Q)=dQ/Q\). Therefore,

$$ \frac{dQ}{Q} = \sum_{i=1}^{n}a_i\frac{dx_i}{x_i}. $$

This is an exact identity between differentials. It becomes an error estimate when the differentials are used to approximate actual small changes. The signs encode whether a change increases or decreases the output: an increase in a numerator factor contributes positively, while an increase in a denominator factor contributes negatively.

1
Write the quantity in product-of-powers form.
Identify every measured input and its exponent, including negative exponents for denominator factors.
2
Write the relative differential.
For each input, multiply its relative change \(dx_i/x_i\) by its exponent.
3
Use signs or magnitudes as appropriate.
Keep signs to estimate a particular combination of changes; use absolute contributions for a worst-case first-order estimate.
4
Convert to an absolute or percentage estimate if needed.
Multiply the relative estimate by the nominal output for an absolute estimate, or by \(100\) for a percentage estimate.

Worked Example: Estimating Error in a Rectangle’s Area

Worked Example: Estimating Error in a Rectangle’s Area

A rectangle has measured length \(L=12.0\) cm and width \(W=8.0\) cm. Suppose the length may change by at most \(0.1\) cm and the width by at most \(0.2\) cm. Estimate the maximum first-order relative error, percentage error, and absolute error in the area.

The area is \(A=LW\). Differentiating gives

$$ dA=W\,dL+L\,dW. $$

Dividing by \(A=LW\), with \(L,W>0\), gives

$$ \frac{dA}{A} = \frac{W\,dL+L\,dW}{LW} = \frac{dL}{L}+\frac{dW}{W}. $$

For a worst-case estimate, use the maximum magnitudes of the input changes:

$$ \left|\frac{dA}{A}\right| \ \text{is estimated by}\ \frac{0.1}{12.0}+\frac{0.2}{8.0} = \frac{1}{120}+\frac{1}{40} = \frac{1}{30} \approx 0.03333. $$

The estimated relative uncertainty is \(0.03333\), or about \(3.33\%\). The nominal area is

$$ A=(12.0)(8.0)=96.0\text{ cm}^2. $$

Thus the estimated maximum absolute error is \(96.0(1/30)=3.2\text{ cm}^2\). As a check, the differential contributions are \(W(0.1)=0.8\text{ cm}^2\) and \(L(0.2)=2.4\text{ cm}^2\); their sum is \(3.2\text{ cm}^2\). The units are area units, as required.

Signed Changes and Worst-Case Estimates

A worst-case estimate adds the magnitudes of the contributions because the input changes could act in the same direction. If the directions of change are known, retain their signs instead. Some contributions may then cancel. Cancellation can make the estimated change smaller, but it should not be assumed when only uncertainty magnitudes are given.

Worked Example: Relative Error in a Quotient

Worked Example: Relative Error in a Quotient

Let \(q=x/y\), where \(x=10.0\) and \(y=4.00\). Suppose \(x\) increases by \(0.1\) while \(y\) decreases by \(0.04\). Estimate the signed relative change, percentage change, and absolute change in \(q\), then compare the estimate with the actual change for those inputs.

The quotient rule gives

$$ dq=\frac{1}{y}\,dx-\frac{x}{y^2}\,dy, \qquad \frac{dq}{q}=\frac{dx}{x}-\frac{dy}{y}. $$

The signed relative estimate is

$$ \frac{dq}{q} \approx \frac{0.1}{10.0}-\frac{-0.04}{4.00} = 0.01+0.01 = 0.02. $$

Thus the estimated relative change is \(2\%\). The starting quotient is \(q=10.0/4.00=2.5\), so the estimated absolute change is \(dq\approx 2.5(0.02)=0.05\). Direct substitution into the differential formula checks this value:

$$ dq = \frac{0.1}{4.00} - \frac{10.0}{(4.00)^2}(-0.04) = 0.025+0.025 = 0.05. $$

For comparison, the changed quotient is \(10.1/3.96\approx 2.550505\), so its actual change is about \(0.050505\), close to the differential estimate \(0.05\). The actual relative change is about \(0.050505/2.5\approx 0.020202\), or \(2.0202\%\). The small difference illustrates that the differential gives a first-order estimate, not an exact change.

Worked Example: Error Estimate for a Cylinder’s Volume

Worked Example: Error Estimate for a Cylinder’s Volume

A cylinder has measured radius \(r=3.00\) cm and height \(h=10.0\) cm. Suppose the radius may be in error by at most \(0.02\) cm and the height by at most \(0.1\) cm. Estimate the maximum first-order percentage and absolute errors in its volume.

The volume is \(V=\pi r^2h\). Applying the relative-change rule to the squared radius and the first power of height gives

$$ \frac{dV}{V} = 2\frac{dr}{r}+\frac{dh}{h}. $$

Using the maximum input uncertainties, the estimated maximum relative error is

$$ 2\frac{0.02}{3.00}+\frac{0.1}{10.0} = \frac{0.04}{3.00}+0.01 \approx 0.013333+0.010000 = 0.023333. $$

This is approximately \(2.3333\%\). The nominal volume is \(V=\pi(3.00)^2(10.0)=90\pi\text{ cm}^3\). Therefore, the estimated absolute error is

$$ dV_{\text{estimated}} = 90\pi(0.023333\ldots) = 2.1\pi\text{ cm}^3 \approx 6.597\text{ cm}^3. $$

Check by differentiating the volume formula directly:

$$ dV=2\pi rh\,dr+\pi r^2\,dh = 2\pi(3.00)(10.0)(0.02)+\pi(3.00)^2(0.1) = 1.2\pi+0.9\pi = 2.1\pi\text{ cm}^3. $$

The two methods agree, and the units are cubic centimetres. The factor of \(2\) on the radius contribution comes from the exponent \(2\) in \(r^2\).

Common Pitfall: Forgetting the Exponent

For a quantity proportional to the square of a measurement, the relative error contribution is twice the input’s relative error. For example, if \(A=\pi r^2\), then

$$ \frac{dA}{A}=2\frac{dr}{r}. $$

A common incorrect calculation drops the exponent and writes \(dA/A=dr/r\). With \(r=5.00\) cm and \(dr=0.02\) cm, this wrong calculation gives \(0.02/5.00=0.004\), or \(0.4\%\). The correction is

$$ \frac{dA}{A} = 2\frac{0.02}{5.00} = 0.008 = 0.8\%. $$

The same principle explains why a sphere’s volume, proportional to \(r^3\), has a relative error contribution of \(3\,dr/r\), not merely \(dr/r\).

Interpreting the Estimate Carefully

The relative-differential rule is useful because it makes the structure of an error calculation visible. In a product, relative changes add; in a quotient, denominator changes enter with the opposite sign; and an exponent scales the corresponding relative change. This makes the method especially effective for measurements whose formulas combine several inputs.

However, a differential estimate should not automatically be reported as a guaranteed maximum actual error. It omits higher-order effects, such as products of small input changes. If a rigorous bound on the actual error is required, additional information about the function over the full range of possible inputs is needed. For ordinary small measurement uncertainties, the first-order estimate is often informative, provided it is identified as an estimate and its assumptions are clear.

Key takeaway: For positive quantities formed from products, quotients, and powers, estimate relative error by adding each input’s relative change multiplied by its exponent. Preserve signs for a particular change; add magnitudes for a worst-case first-order estimate.

Check Your Understanding

Use differentials and show the relative-change calculation in each case.

  1. If \(Q=xy^3\), what is \(dQ/Q\) in terms of \(dx/x\) and \(dy/y\)?
  2. If \(R=x^2/y\), where \(x=5.0\) and \(y=2.0\), estimate the worst-case percentage error when the maximum uncertainties are \(0.1\) in \(x\) and \(0.02\) in \(y\).
  3. A measured radius has a relative uncertainty of \(0.5\%\). What is the first-order relative uncertainty in the area of a circle calculated from that radius?
  4. Why do signed relative changes subtract for a quotient, while worst-case uncertainty contributions are added in magnitude?
  5. Does a differential error estimate necessarily guarantee a bound on the actual error? Explain briefly.

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