Step by step
The question
Question: The sphere has radius 3 m, measured with a possible error of 0.05 m.
Estimate the error in the volume.
1 Draw and name
The radius is r = 3 m.
r = 3 m
The possible error in it is dr = 0.05 m: a small change.
dr = 0.05 m
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 = 3 and dr = 0.05.
dV = 4π(3)²(0.05)
That is 1.8π, about 5.65 m³.
= 1.8π ≈ 5.65 m³
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 3.05, minus the volume at 3, is 5.75 m³.
Close to our 5.65.
Right!
V(3.05) − V(3) = 5.75 m³
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