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.

Question

anonymous · Answer

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

anonymous · Answer

Change those $r$s into $s$s...

anonymous · Answer

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

anonymous · Answer

Thanks! Is that the final answer?

anonymous · Answer

Nope

anonymous · Answer

Mm, what am I supposed to do after?

anonymous · Answer

You only have how 'off' it is.

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

Ahh, okay, thanks.

anonymous · Answer

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

anonymous · Answer

Okay thanks! :)

anonymous · Answer

got a way of checking the answer?

anonymous · Answer

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