Step by step
The question
Question: The sphere has radius 5 cm, measured with a possible error of 0.1 cm.
Estimate the error in the volume.
1 Draw and name
The radius is r = 5 cm.
r = 5 cm
The possible error in it is dr = 0.1 cm: a small change.
dr = 0.1 cm
2 Find dV
The volume of a sphere is V = 4/3 × π r³.
V = 43πr3
Take the derivative, times dr: dV = 4πr² dr.
dV = 4πr2 dr
The change in the volume is about the derivative times the change in the radius.
dV = (derivative) × dr
3 Put the numbers in
Put in r = 5 and dr = 0.1.
dV = 4π(5)²(0.1)
That is 10π, about 31.42 cm³.
= 10π ≈ 31.42 cm³
Careful!
A small error in the radius makes a bigger error in the volume.
Careful: dV is not dr: the error grows
4 Check
Let's check with the real change: the volume at 5.1, minus the volume at 5, is 32.05 cm³.
Close to our 31.42.
Right!
V(5.1) − V(5) = 32.05 cm³
Let's look back.
First: draw and name.
Then: find dV.
Then: put the numbers in.
Last: check.
Answer
Careful: dV is not dr: the error grows