Error Estimation with Differentials
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 4.2: “Linear Approximations and Differentials” |
| Book URL | https://openstax.org/details/books/calculus-volume-1 |
Freely available and openly licensed.
Try This First: Imprecise Measurement, Precise Consequence
A sphere is measured to have radius $r = 10$ cm, but the measurement could be off by up to $\pm 0.1$ cm.
Before computing: how much could the computed volume $V = \frac{4}{3}\pi r^3$ be off?
Predict: would you expect the volume error to be large or small relative to the volume itself?
The volume is about 4189 cm$^3$. The absolute error turns out to be about $\pm 125.7$ cm$^3$ -- only about $3\%$ of the total. Errors in a measured quantity propagate to computed quantities, but in a controlled, predictable way via the differential.
Quantity-First Framing
In any physical measurement, there is uncertainty. If a length is measured as $x$ but the true value could be anywhere in $[x - \Delta x, x + \Delta x]$, then a quantity $y = f(x)$ computed from that measurement has uncertainty approximately $|dy| = |f'(x)|\,|\Delta x|$.
This “propagated error” quantifies how imprecision in input leads to imprecision in output. The size of the derivative $|f'(x)|$ controls how much the error is amplified or reduced.
Prerequisite Check
Quick Reference
Propagated error (absolute): \[ |\Delta y| \approx |dy| = |f'(x)|\,|\Delta x|. \]
Relative (fractional) error: \[ \frac{|\Delta y|}{|y|} \approx \frac{|dy|}{|f(x)|}. \]
Percentage error: multiply relative error by 100%.
Key Concepts
1. Propagated Error: Setting Up
Example 1. A sphere’s radius is measured as $r = 10$ cm with a possible error of $|\Delta r| \leq 0.1$ cm. Estimate the propagated error in the volume $V = \frac{4}{3}\pi r^3$.
$\frac{dV}{dr} = 4\pi r^2$.
Propagated error: \[ |\Delta V| \approx |dV| = 4\pi r^2\,|\Delta r| = 4\pi(100)(0.1) = 40\pi \approx 125.7 \text{ cm}^3. \]
Relative error: \[ \frac{|\Delta V|}{V} = \frac{40\pi}{\frac{4}{3}\pi(1000)} = \frac{40\pi}{\frac{4000\pi}{3}} = \frac{40 \cdot 3}{4000} = \frac{120}{4000} = 0.03 = 3\%. \]
A 1% error in radius ($\Delta r / r = 0.1/10 = 1\%$) leads to a 3% error in volume. This factor of 3 comes from the exponent: $V \propto r^3$, so relative errors are 3 times amplified.
2. Two Representations
Algebraic: $\frac{|\Delta V|}{V} \approx \frac{|dV/dr|\,|\Delta r|}{V} = \frac{4\pi r^2\,|\Delta r|}{\frac{4}{3}\pi r^3} = \frac{3|\Delta r|}{r}$.
This shows directly: the relative error in volume is 3 times the relative error in radius. For any quantity $f(x) = cx^n$, the relative error is amplified by the exponent $n$.
Contextual: If a manufacturer allows a 0.5% tolerance on the radius of a sphere, the volume tolerance is 1.5%. Knowing this lets the engineer set appropriate tolerances.
3. Amplification Factor: The Derivative
The propagated error formula $|\Delta y| \approx |f'(x)|\,|\Delta x|$ shows that $|f'(x)|$ is the amplification factor. A large derivative means a small input error leads to a large output error; a small derivative means the output is relatively insensitive to input errors.
Example 2. Compare the sensitivity of $f(x) = x^{10}$ and $g(x) = \sqrt{x}$ near $x = 1$.
$f'(1) = 10$. A 1% error in $x$ near 1 leads to approximately 10% error in $f(x)$. High sensitivity.
$g'(1) = 0.5$. A 1% error in $x$ near 1 leads to approximately 0.5% error in $g(x)$. Low sensitivity.
4. Percentage Error Problems
Example 3. A cube’s side is measured as $s = 5$ cm with $|\Delta s| = 0.02$ cm. Estimate the percentage error in the volume $V = s^3$.
$dV = 3s^2\,ds = 3(25)(0.02) = 1.5$ cm$^3$.
$V = 125$ cm$^3$. Relative error: $1.5/125 = 0.012 = 1.2\%$.
Again, the relative error in volume is $3 \times$ the relative error in side ($0.02/5 = 0.4\%$, and $3 \times 0.4\% = 1.2\%$).
5. Ask Why: Why Does the Exponent Multiply the Relative Error?
For $y = cx^n$: $\ln y = \ln c + n\ln x$. Differentiating both sides: \[ \frac{dy}{y} = n\,\frac{dx}{x}. \]
The relative change in $y$ is exactly $n$ times the relative change in $x$. This is why:
- Squaring a quantity doubles the relative error.
- Cubing triples the relative error.
- Taking a square root halves the relative error.
This is a logarithmic-differentiation result: the exponent multiplies the relative change.
Named Misconception: average-rate-as-arithmetic-mean
A common error: computing the error by averaging instead of using the differential. If the volume at $r = 10$ is $V_1$ and at $r = 10.1$ is $V_2$, the error is NOT $(V_1 + V_2)/2 - V_1$; it is $V_2 - V_1 \approx dV$. The differential gives the error directly without evaluating the function at two points.
Common Errors
| Error | Specific example | Correction |
|---|---|---|
| Not using the derivative | Computing $V(10.1) - V(10)$ exactly | Use $dV = 4\pi r^2 |dr|$ -- the point of differentials is to avoid computing $f$ twice |
| Confusing absolute and relative error | “The relative error is 125.7 cm$^3$” | $125.7$ cm$^3$ is the absolute error; relative error is dimensionless (3%) |
| Wrong amplification | “1% error in radius gives 1% error in volume” | The exponent (3 for $r^3$) multiplies the relative error; it is 3% |
Common Misconceptions
a 1% error in the input produces a 1% error in the output regardless of the formula.
This is the rate-as-fixed-number error. The relative error formula $\frac{|dy|}{|y|} = n\frac{|dx|}{|x|}$ shows that the exponent $n$ in a power function amplifies or dampens the relative error. A 1% error in the radius of a sphere ($r^3$) produces a 3% error in the volume, while a 1% error in the argument of a square root produces only a 0.5% error in the output. The derivative, not the formula itself, controls how errors propagate.
the propagated error in area or volume can be estimated by averaging the function values at the minimum and maximum of the measured quantity.
This is the average-rate-as-arithmetic-mean error. Averaging $V(r - \Delta r)$ and $V(r + \Delta r)$ does not yield the propagated error; it yields a value close to $V(r)$ itself. The propagated absolute error is $|dV| = |V'(r)||\Delta r| = 4\pi r^2 |\Delta r|$, which comes from the derivative at the measured value, not from averaging function values at the two endpoints.
Leveled Practice
Level 1 -- Computing Propagated Error
Problem 1. The area of a circle is computed from $A = \pi r^2$. If $r = 8$ cm with error $|\Delta r| = 0.05$ cm, estimate $|\Delta A|$ and the percentage error.
Show answer
$dA = 2\pi r\,dr = 2\pi(8)(0.05) = 0.8\pi \approx 2.51$ cm$^2$.
$A = 64\pi \approx 201.1$ cm$^2$. Relative error: $0.8\pi/(64\pi) = 0.8/64 = 1.25\%$.
Problem 2. A cylindrical can has radius 5 cm and height 20 cm. The radius is measured with error $\pm 0.1$ cm and the height with error $\pm 0.2$ cm. Estimate the total propagated error in volume $V = \pi r^2 h$.
Show answer
$dV = 2\pi rh\,dr + \pi r^2\,dh$.
From radius error: $|dV_r| = 2\pi(5)(20)(0.1) = 20\pi$.
From height error: $|dV_h| = \pi(25)(0.2) = 5\pi$.
Total: $|dV| \leq 25\pi \approx 78.5$ cm$^3$.
$V = \pi(25)(20) = 500\pi \approx 1571$ cm$^3$. Relative error: $25\pi/(500\pi) = 5\%$.
Level 2 -- Relative Error
Problem 3. Show that for $y = x^n$, the relative error satisfies $|dy/y| = n|dx/x|$. Then use this to find the percentage error in $A = \pi r^2$ if $r$ has a 2% relative error.
Show answer
$y = x^n$, $dy = nx^{n-1}dx$. $dy/y = nx^{n-1}dx/(x^n) = n(dx/x)$. So $|dy/y| = n|dx/x|$.
For $A = \pi r^2$: $n = 2$, so $|dA/A| = 2|dr/r| = 2(0.02) = 4\%$.
Level 3 -- Low-Floor-High-Ceiling Extension
Problem 4 (Extension).
(a) (Floor) A square of side $s$ is measured as 6 cm with error 0.03 cm. Estimate the absolute and percentage error in the area.
(b) (Mid) For $f(x) = \ln x$ near $x = 10$ with $|\Delta x| = 0.05$, estimate the error in $f$. Why is the relative error of $\ln x$ not the same as the relative error of $x$?
(c) (Ceiling) A quality-control engineer measures a sphere’s radius with a 0.5% relative error. She wants the relative error in the surface area $S = 4\pi r^2$ to be no more than 2%. Is this guaranteed? What is the largest allowable relative error in radius if the surface area must be within 2% relative error?
Show answer
(a) $A = s^2$. $|dA| = 2s|ds| = 2(6)(0.03) = 0.36$ cm$^2$. $A = 36$ cm$^2$. Relative error: $0.36/36 = 1\%$.
(b) $f'(x) = 1/x$. $|df| = |dx|/x = 0.05/10 = 0.005$. $f(10) = \ln 10 \approx 2.303$. Relative error in $\ln x$: $0.005/2.303 \approx 0.22\%$. This is different from $|\Delta x|/x = 0.05/10 = 0.5\%$ because the relative error in $\ln x$ is not the same as the relative error in $x$; they are related by $|d(\ln x)/\ln x| = |dx|/(x\ln x)$.
(c) $S = 4\pi r^2$. $|dS/S| = 2|dr/r|$. With $|dr/r| = 0.5\% = 0.005$: $|dS/S| = 1\% < 2\%$. Yes, guaranteed.
Largest allowable $|dr/r|$: $2|dr/r| \leq 2\% \Rightarrow |dr/r| \leq 1\%$.
Mastery Checklist
You have mastered this skill when you can do all of the following without referring to notes:
Mental Model
The derivative is a local amplifier. Near a given $x$, the function $f$ scales input changes by $|f'(x)|$. A large derivative means the function amplifies errors; a small derivative means the function dampens errors.
Error estimation with differentials is exactly this: take the input error $|\Delta x|$, multiply by the amplification factor $|f'(x)|$, and the result is the approximate output error $|\Delta y|$.
The relative-error formula for power functions ($|dy/y| = n\,|dx/x|$) is the key fact to internalize: the exponent tells you how many times the relative error is amplified.
Connections
Within MATH161
- Differentials: This lesson directly applies the differential $dy = f'(x)\,dx$ from the previous lesson to a practical estimation problem.
- Logarithmic differentiation: The formula $d(\ln y) = n\,d(\ln x)$ for $y = x^n$ is logarithmic differentiation -- the same technique used to differentiate complex products and powers.
- Integration: Total error over an interval (propagation through a process) is an integral. The differential $dy$ gives the local version of the same idea.