Math · Calculus I · Worked example
Estimate an error with differentials
The radius of a ball is measured as 10 cm, with a possible error of 0.1 cm. Estimate the possible error in the computed volume, and the relative error.
Differentiate the formula
dV/dr = 4πr², so a small change dr in the radius changes the volume by about dV = 4πr² dr.
Substitute the measurement
With r = 10 and dr = 0.1, dV = 4π(100)(0.1) = 40π ≈ 126 cm³.
Find the relative error
The volume is V = (4/3)π(1000) ≈ 4189 cm³, so the relative error is dV/V = 40π ÷ (4000π/3) = 0.03, or 3%. A 1% error in the radius becomes about 3% in the volume, because V depends on r³.
Compare with the exact change
The exact change from r = 10 to r = 10.1 is (4/3)π(10.1³ − 10³) ≈ 126.9 cm³, close to the differential’s 125.7 cm³.
Result
The possible error in the volume is about 40π ≈ 126 cm³, a relative error of about 3%.
Your turn
A square’s side is measured as 5.0 cm with a possible error of 0.02 cm. Use differentials to estimate the possible error in its area.
Show the answer and explanation
About 0.2 cm².
A = s², so dA = 2s ds = 2(5.0)(0.02) = 0.2 cm². The relative error is 0.2/25 = 0.8%, twice the 0.4% error in the side.
Keep exploring
Halve the measurement error to 0.05 cm: the volume error halves too, to 20π ≈ 63 cm³, and the relative error to 1.5%.
Return to the concept →Sources and scope
Authored study material. Tool results depend on the stated inputs and model assumptions.
Try in the workspace
Open the example inputs, change a value and keep a useful result on your board.
Check the derivative in Math Open worked example on a board Linearization in Math ReferenceYour existing work stays on this device. Examples open as editable copies.