Question

A cube has side length measured to be 12 cm. Find the (linearly) approximate error in volume if the error in side length measurement is Δs.

Answer 1

\[
V=r^3
\]So \[
dV = \frac{dV}{dr} dr=3r^2dr
\]

Answer 2

Change those \(r\)s into \(s\)s...

Answer 3

So we know \(s=12cm\).\[
dV=3(12)^2dr = 3(144)dr=432dr
\]

Answer 4

Thanks! Is that the final answer?

Answer 5

Nope

Answer 6

Mm, what am I supposed to do after?

Answer 7

You only have how 'off' it is.

Answer 8

The problem is that I'm not quite sure what kind of error they're looking for.

Answer 9

If they want relative error, then we want\[
dV/V = 432\Delta s/(12)^3
\]

Answer 10

Ahh, okay, thanks.

Answer 11

That gets us\[
\frac{\Delta V}{V} \approx \frac{dV}{V} =\frac{3\Delta s}{12}=\frac{\Delta s}{4}
\]

Answer 12

Okay thanks! :)

Answer 13

got a way of checking the answer?

Answer 14

Nope, but I'm sure it's right.