I'm looking at a simple problem involving the following question:
The dimensions of a rectangular box are $5$, $10$, and $20$ cm, with a possible measurement error in each side of $\pm .1$ cm. Use differentials to find what possible error should be attached to its volume.
This is simple enough given that $$dV = \frac{\partial V}{\partial y}dy + \frac{\partial V}{\partial x}dx + \frac{\partial V}{\partial z}dz$$
we can simply plug in $\frac{\partial V}{\partial x}=200, \frac{\partial V}{\partial y} = 100, \frac{\partial V}{\partial z} = 50$, and $dx=dy=dz=.1$.
I also understand that $\Delta V \approx dV$. Thus the solution comes out to be that the error is such that $\Delta V \approx \pm 35 \text{cm}^3$.
But the solution given goes a bit further than this to claim $$\vert \Delta V \vert \leq 35 \text{cm}^3.$$
How is it that the claim above holds? If $dV$ is a linear approximation of $\Delta V$, might it be the case that $\Delta V$ increases more quickly than $dV$?
And indeed $20.1 \times 10.1 \times 5.1 = 20\times 10\times 5 + 35.351$
So at the upper bound, this formula does indeed under-estimate the error. And this is because the estimate is a linear function and the volume calculation is convex (cubic).
As a liner estimate of some error term, we are making the assumption that this error follows some distribution over $(-0.1,0.1)$ and our upper bound of this linear estimate is $35 cm$ even if the true upper bound is a little bit higher.