Step by step
The question
Question: The sphere has radius 5 cm, measured with a possible error of 0.2 cm.
Estimate the percentage error in the volume.
1 Draw and name
The radius is r = 5 cm.
The possible error in it is dr = 0.2 cm: a small change.
2 Find dV
The volume of a sphere is V = 4/3 × π r³.
Take the derivative, times dr: dV = 4πr² dr.
The change in the volume is about the derivative times the change in the radius.
3 Put the numbers in
Put in r = 5 and dr = 0.2.
That is 20π, about 62.83 cm³.
Careful!
A small error in the radius makes a bigger error in the volume.
4 The error
The percentage error compares the error with the whole volume.
The whole volume is about 523.6 cm³.
So the percentage error is about 12%.
5 Check
Let's check with the real change: the volume at 5.2, minus the volume at 5, is 65.38 cm³.
Close to our 62.83.
Right!
Let's look back.
First: draw and name.
Then: find dV.
Then: put the numbers in.
Then: the error.
Last: check.