Why Products and Quotients Need Their Own Error Formulas
When a measured quantity is calculated from two measurements, uncertainty in either input affects the result. For a product, a small change in either factor changes the product. For a quotient, a small change in the denominator matters as well as a change in the numerator, and it affects the result with the opposite sign.
Differentials give a first-order estimate of these effects. They are useful when the measurement errors are small relative to the measured values and the calculation stays in a region where the formula is defined. The estimates describe local changes; they are not exact error formulas in general.
Differentials for a Product
Let \(P=xy\), where \(x\) and \(y\) are measured quantities. Write \(dx\) and \(dy\) for small changes in the measurements. The differential of \(P\) is
This is a signed estimate: if the input changes have known signs, their contributions may reinforce or partly cancel. If instead we know only bounds \(|dx|\leq \Delta x\) and \(|dy|\leq \Delta y\), then the triangle inequality gives the first-order worst-case estimate
When \(x\) and \(y\) are both nonzero, divide the differential formula by \(P=xy\). This gives the relative-error relation
The signed relative changes add. For bounded uncertainties, the corresponding first-order bound is
Each term on the right is an input's relative uncertainty. The formula says that the product's worst-case relative uncertainty is estimated by adding the input relative uncertainties.
Differentials for a Quotient
Let \(Q=x/y\), with \(y\ne 0\). Its differential is
The negative sign on the denominator contribution is important: increasing the denominator while holding the numerator fixed decreases the quotient. For bounded errors \(|dx|\leq\Delta x\) and \(|dy|\leq\Delta y\), the first-order worst-case estimate is
If \(x\ne 0\) as well as \(y\ne 0\), then \(Q\ne 0\), so we can divide by \(Q\) to obtain
For bounded uncertainties, the first-order relative bound is therefore
Although the signed denominator contribution has a minus sign, its uncertainty contributes positively to the worst-case bound. The bounds above treat the input errors as independently capable of taking either sign; they are worst-case estimates, not statistical formulas.
Why These Formulas Work
The product formula follows directly by comparing the exact change with its first-order part. If the inputs change to \(x+dx\) and \(y+dy\), the product changes by
The differential \(dP=y\,dx+x\,dy\) keeps the terms linear in the input changes. The omitted term \(dx\,dy\) is a product of two small changes, so it is of second order when both changes are small. This explains why the differential estimates the actual change well locally, while not generally being exact.
For the quotient, the exact change is defined whenever \(y+dy\ne 0\). Subtracting the original quotient gives
The differential can be written \(dQ=(y\,dx-x\,dy)/y^2\). To see how it relates to the exact change, subtract it:
When \(y+dy\) stays away from zero, this remainder consists of terms involving products of small input changes. Thus the differential gives the first-order change. If \(y+dy\) is close to zero, that reasoning is not reliable; the quotient may change dramatically or cease to be defined.
Worked Examples
Worked Example: Uncertainty in a Rectangular Area
A rectangle has measured side lengths \(x=12.0\text{ cm}\) and \(y=8.0\text{ cm}\), with uncertainty bounds \(\Delta x=0.1\text{ cm}\) and \(\Delta y=0.2\text{ cm}\). Estimate the area and its worst-case absolute and relative uncertainty using differentials.
The area is \(A=xy\), so its estimated value is
For a product, \(dA=y\,dx+x\,dy\). The worst-case estimate uses the absolute contributions:
The relative uncertainty estimate is
Checking with the input relative uncertainties gives \(0.1/12.0+0.2/8.0=0.00833+0.025=0.03333\), the same estimate. Thus the area is approximately \(96.0\text{ cm}^2\), with first-order worst-case uncertainty \(3.2\text{ cm}^2\), or about \(3.33\%\).
Worked Example: Uncertainty in a Density
A sample has measured mass \(m=250\text{ g}\) with uncertainty bound \(\Delta m=2\text{ g}\), and measured volume \(V=50\text{ cm}^3\) with uncertainty bound \(\Delta V=1\text{ cm}^3\). Estimate the density and its worst-case uncertainty.
Density is the quotient \(\rho=m/V\), so
The quotient differential is \(d\rho=dm/V-m\,dV/V^2\). Applying the uncertainty bounds gives
The relative bound is \(0.14/5=0.028=2.8\%\). It also follows by adding the input relative bounds: \(2/250+1/50=0.008+0.020=0.028\). The denominator term has a negative sign in the signed differential, but for a worst-case bound its magnitude is added. The estimated density is \(5\text{ g/cm}^3\), with first-order worst-case uncertainty \(0.14\text{ g/cm}^3\), or \(2.8\%\).
Worked Example: Signed Product Change Versus a Worst-Case Bound
Suppose \(x=40\) and \(y=5\). In one particular measurement, the changes are \(dx=0.4\) and \(dy=-0.05\). Compare the differential estimate with the exact product change, then find the worst-case estimate if the input uncertainty bounds are \(\Delta x=0.4\) and \(\Delta y=0.05\).
The original product is \(P=40(5)=200\). The signed differential estimate is
The two first-order contributions cancel. The exact new product is
so the exact change is \(199.98-200=-0.02\). This agrees with the product expansion: \(\Delta P=y\,dx+x\,dy+dx\,dy=0+(0.4)(-0.05)=-0.02\).
If the signs of the errors are not known, the worst-case differential bound is instead
The signed estimate of zero describes these particular opposing changes; it is not a guarantee that any possible pair of errors produces zero change. The bound \(4\) is the first-order worst-case estimate. The exact cross term in this particular case is \(-0.02\), consistent with the differential approximation being first order.
Common Pitfall: Losing the Quotient's Minus Sign
A frequent error is to treat the numerator and denominator as if their relative changes simply add with the same sign. For example, for \(Q=x/y\), writing \(dQ/Q=dx/x+dy/y\) is incorrect. Increasing \(y\) with \(x\) fixed makes \(Q\) smaller, not larger.
The correction follows from differentiating the quotient with respect to each input, or from the exact-change calculation above:
For example, if \(x=10\), \(y=2\), \(dx=0\), and \(dy=0.1\), then \(Q=5\) and the differential predicts \(dQ=-(10/2^2)(0.1)=-0.25\). The exact new quotient is \(10/2.1\approx4.762\), a decrease of about \(0.238\), consistent with the negative differential estimate. The incorrect plus sign would predict an increase.
Using the Estimates Carefully
Before applying a formula, identify whether the errors are signed changes or uncertainty bounds. Signed changes go into the differential with their signs. Bounds go into an absolute-value estimate, where each possible contribution is added. Do not use a signed cancellation as a worst-case guarantee.
Relative-error formulas also have specific requirements. The product formula \(dP/P=dx/x+dy/y\) requires both factors to be nonzero. The quotient relative formula requires both numerator and denominator to be nonzero. If a quantity used as a relative-error denominator is zero, use the absolute differential formula instead. In every case, the differential is a local, first-order estimate; large measurement errors or a denominator close to zero can make it unreliable.
Check Your Understanding
Use the differential formulas and state any conditions needed for your answers.
- For \(P=xy\), write the differential and the first-order worst-case absolute-error bound when \(|dx|\leq\Delta x\) and \(|dy|\leq\Delta y\).
- If both factors in a product are nonzero, how is the product's signed relative change related to the factors' signed relative changes?
- For \(Q=x/y\), explain why the denominator term has a negative sign in the differential.
- When is the relative-error formula for a quotient undefined, even if the quotient itself is defined?
- Why can a signed differential estimate of zero fail to mean that the exact change is zero?
