Step by step
The question
Question: The cube has side 5 cm, measured with a possible error of 0.2 cm.
Estimate the error in the volume.
1 Draw and name
The side is s = 5 cm.
s = 5 cm
The possible error in it is ds = 0.2 cm: a small change.
ds = 0.2 cm
2 Find dV
The volume of a cube is V = s³.
V = s3
Take the derivative, times ds: dV = 3s² ds.
dV = 3s2 ds
The change in the volume is about the derivative times the change in the side.
dV = (derivative) × ds
3 Put the numbers in
Put in s = 5 and ds = 0.2.
dV = 3(5)²(0.2)
So the error in the volume is about 15 cm³.
= 15 cm³
Careful!
A small error in the side makes a bigger error in the volume.
Careful: dV is not ds: the error grows
4 Check
Let's check with the real change: the volume at 5.2, minus the volume at 5, is 15.61 cm³.
Close to our 15.
Right!
V(5.2) − V(5) = 15.61 cm³
Let's look back.
First: draw and name.
Then: find dV.
Then: put the numbers in.
Last: check.
Answer
Careful: dV is not ds: the error grows