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AP Calculus AB & BC Intermediate

Error Propagation Using Differentials

How differentials turn measurement uncertainty in an input into an estimate of the resulting error in an output, with step-by-step examples involving area, volume, and percent error.

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.

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

  1. Write the formula connecting the quantity with the measured variable.
  2. Differentiate the formula.
  3. Identify the input error, which is the uncertainty in the measured quantity.
  4. Use the differential relationship to estimate the resulting output error.
  5. Evaluate at the measured value, not at the unknown exact value.
  6. Take the absolute value if the question asks for the size or maximum approximate error.
  7. If requested, convert the error into relative or percent error.

Worked Examples

Example 1: Error in the Area of a Circle

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.

Step 1: Write the area formula:
Step 2: Differentiate:
Step 3: The measured radius is , and the possible measurement error is .
Step 4: Substitute into the differential:
Result: The maximum approximate error in the area is .
Example 2: Percent Error in Area

Using the same circle, estimate the percent error in the calculated area.

Step 1: The calculated area is:
Step 2: From the previous example, the approximate area error is .
Step 3: Compute the relative error:
Step 4: Convert to a percentage:
Result: The calculated area has an approximate percent error of .
Example 3: Error in the Volume of a Sphere

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.

Step 1: Write the volume formula:
Step 2: Differentiate:
Step 3: Use and .
Step 4: Substitute:
Result: The maximum approximate error in the calculated volume is cubic inches.
Example 4: Using Differentials to Estimate Measurement Precision

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?

Step 1: Write the volume formula:
Step 2: Differentiate:
Step 3: We know the allowable volume error , so solve the differential equation for .
Step 4: Substitute :
Result: The side length should be measured to within approximately cm.

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

Confusing the measured value with the measurement error

If the radius is measured as 10 cm with an uncertainty of 0.1 cm, then and . They are not the same thing.

Forgetting to differentiate before substituting

The differential comes from the derivative. For a circle, start with , differentiate to get , and only then substitute the measured radius and its error.

Reporting a signed error when the question asks for error size

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.

Treating the differential as an exact error

The relationship is an approximation. The actual change is . For sufficiently small measurement errors, however, the differential provides an excellent estimate.

Using the wrong denominator for percent error

Percent error is approximately , not simply the absolute error multiplied by 100. You must divide by the calculated quantity first.

Practice Problems

Question 1

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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Question 2

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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Question 3

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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Question 4

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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Question 5

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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