# Homework Help: Differentials and tolerances

1. Dec 16, 2012

### OCTOPODES

1. The problem statement, all variables and given/known data

About how accurately must the interior diameter of a 10-m high cylindrical storage tank be measured to calculate the tank's volume to within 1% of its true value?

2. Relevant equations

$V=\frac{5}{2}\pi l^{2}$, where $V$ is volume and $l$ is diameter.
$dV=5\pi l \ dl$

3. The attempt at a solution

I'm really uncertain as to how to go about this problem. What follows is the textbook's method for a similar problem translated into this problem's terms.

--

We want any inaccuracy in our measurement to be small enough to make the corresponding increment $\Delta V$ in the volume satisfy the inequality

$|\Delta V|\leq\frac{1}{100}V=\dfrac{\pi l^{2}}{40}$.

We replace $\Delta V$ in this inequality by its approximation

$dV=\left(\dfrac{dV}{dl}\right)dl=5\pi l \ dl$.

This gives

$|5\pi l\ dl|\leq\dfrac{\pi l^{2}}{40}$, or $|dl|\leq\dfrac{1}{5\pi l}\cdot\dfrac{\pi l^{2}}{40}=\dfrac{1}{5}\cdot\dfrac{l}{40}=0.005l$.

We should measure $l$ with an error $dl$ that is no more than 0.5% of its true value.

--

I need some clarification for this solution. Could somebody annotate it, or perhaps write up a more intuitive one?

2. Dec 16, 2012

### haruspex

Your answer is correct. I would tend to start from the other end. V = πl2h/4. If there's a fractional error δ in l, the computed volume will be π(l(1+δ))2h/4 = V(1+δ)2 = V + 2Vδ + Vδ2 ≈ V + 2Vδ. So the fractional error in the volume will be about double that in the linear measurement.