Why Differentials Are Useful for Error
Measurements are never perfectly exact. If you measure the radius of a circle with a ruler, for example, your measurement might be off by a small amount. That small error in the input can produce an error in the quantity you calculate from it.
Differentials give us a fast way to estimate how much that measurement error will affect the final answer. The key idea is simple: a small change in the input produces an approximately predictable change in the output.
If , then a small change in the input produces an approximate change in the output.
For error propagation, represents the uncertainty or error in the measured input, while estimates the resulting error in the calculated output.
The Basic Error-Propagation Formula
Suppose and the measured value of has a small error of approximately . Since
the approximate error in is:
In practice, you will usually take the absolute value when reporting the size of the error:
This is an estimate, not an exact error. The approximation works well when the measurement error is small relative to the measured value.
Absolute Error vs. Relative Error
There are two different ways you might describe an error.
- Absolute error: the approximate size of the error in the calculated quantity.
- Relative error: the error compared with the size of the quantity itself.
- Percent error: relative error expressed as a percentage.
If is the approximate error in , then the relative and percent errors are approximately:
Notice that the denominator is the calculated quantity itself, .
A Step-by-Step Method
- Write the formula connecting the quantity with the measured variable.
- Differentiate the formula.
- Identify the input error, which is the uncertainty in the measured quantity.
- Use the differential relationship to estimate the resulting output error.
- Evaluate at the measured value, not at the unknown exact value.
- Take the absolute value if the question asks for the size or maximum approximate error.
- If requested, convert the error into relative or percent error.
Worked Examples
A circle has a measured radius of 10 cm. The measurement may be off by at most 0.1 cm. Estimate the maximum error in the calculated area.
Using the same circle, estimate the percent error in the calculated area.
A sphere has a measured radius of 6 inches, with a possible measurement error of 0.02 inches. Estimate the maximum error in the calculated volume.
A cube has side length cm. The volume must be accurate to within approximately cubic centimeters. About how accurately must the side length be measured?
Why the Differential Gives an Error Estimate
The differential comes from the tangent-line approximation. If , then near a point , the function can be approximated by its tangent line.
Therefore, a small change in produces an approximate change in given by:
That is why differential-based error propagation is especially useful when the measurement error is small. We are essentially replacing the actual curve with its tangent-line approximation near the measured value.
A Useful Shortcut for Percent Error
For many AP Calculus problems, you can make percent-error calculations especially efficient by starting with the relative differential.
For example, if , then:
Dividing both sides by gives:
This tells us something important: the relative error in the area is approximately twice the relative error in the radius.
Likewise, for the volume of a sphere, , so:
Therefore, the relative error in the volume is approximately three times the relative error in the radius.
Common Mistakes
If the radius is measured as 10 cm with an uncertainty of 0.1 cm, then and . They are not the same thing.
The differential comes from the derivative. For a circle, start with , differentiate to get , and only then substitute the measured radius and its error.
A negative differential can indicate that the output decreases when the input increases in a particular direction. But when a problem asks for the maximum error or size of the error, report the absolute value.
The relationship is an approximation. The actual change is . For sufficiently small measurement errors, however, the differential provides an excellent estimate.
Percent error is approximately , not simply the absolute error multiplied by 100. You must divide by the calculated quantity first.
Practice Problems
A circle has a measured radius of 8 cm with a possible measurement error of 0.05 cm. Estimate the maximum error in the calculated area.
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A sphere has a measured radius of 5 inches with a possible error of 0.01 inches. Estimate the maximum error in the calculated volume.
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The radius of a circular disk is measured as 12 cm with an uncertainty of 0.06 cm. Estimate the percent error in the calculated area.
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The side length of a cube is measured as 10 cm. If the measurement can be off by at most 0.02 cm, estimate the maximum error in the calculated volume.
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A circular region has area . If the radius has a relative error of approximately 1%, estimate the relative error in the calculated area.
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