Estimating radius change $dr$ for a given volume change $dV$ of a sphere

Question

Note: Similar questions have been asked here and here and here, but this is slightly different.

The problem

Find the calculation error, of the radius, of a sphere, so that the error of the calculation of its volume is maximum $2\%$

The solution attempt

We know that the volume of a sphere is given by the equation $(1)$:

$$ V(r) = \frac{4}{3} \pi r^3 \quad (1) $$

Therefore the change of radius for a given volume error is given by the equation $(2)$:

$$ V(r) = \frac{4}{3} \pi r^3 \quad (2) \iff r = \sqrt[3]{\frac{3(V(r))}{4\pi}} \iff dr = \left(\sqrt[3]{\frac{3(V(r) )}{4\pi}}\right)' dV \iff $$

$$ dr = \frac{1}{3^{\frac{2}{3}}\cdot \:4^{\frac{1}{3}}\pi ^{\frac{1}{3}}V(r)^{\frac{2}{3}}} \quad dV \quad (2)$$

Therefore, if we plug $dV =\pm0.02$ into $(2)$, we get the error estimation of $r$ i.e:

$$ dr = \frac{\pm0.02}{3^{\frac{2}{3}}\cdot \:4^{\frac{1}{3}}\pi ^{\frac{1}{3}}V(r)^{\frac{2}{3}}} \iff $$

$$ dr \approx \pm (0.41\%) \frac{1}{V(r)^{\frac{2}{3}}} $$

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Is the syllogism correct and if not, why?

Answer 1

When we want that the volume error is $\>\leq2\%$ then we mean the relative error, i.e., we want $${|dV|\over V}\leq 2\%\ .$$ In your argument you required $\>|dV|\leq0.02$, i.e., referred to the absolute error. This "pure number" absolute error is somewhat strange in a zero context situation. For the volume of a virus (in mm${}^3$) it might be terribly large, whereas for the volume of the earth ball (in m${}^3$) it is acceptable.

Now, since $V=c\,r^3$ for some constant $c$ the relative error of $V$ is three times the (small) relative error of $r$, because we have $${dV\over dr}=3c\,r^2={3V\over r}\ ,$$ hence $${dV\over V}=3{dr\over r}\ .$$ It follows that we need $r$ to be determined with a maximal relative error of ${2\over3}\%$.

Answer 2

Let's apply the classic error analysis formula. If $$f:\mathbb{R}^n\to \mathbb{R} ~; f:(x_1,...,x_n)\mapsto f(x_1,...,x_n)$$ and we have some maximum error $\delta x_i$ on each argument, the maximum error in $f$ will be $$\delta f=\sqrt{\sum_{i=1}^{n} \left(\frac{\partial f}{\partial x_i}\delta x_i\right)^2}$$ Let's apply to our example. $$\delta V = 4\pi r^2 \delta r$$ A 2% error in $V$ means $\delta V=0.02 V$. $$0.02 V = 4\pi r^2 \delta r$$ $$0.02\left(\frac{4\pi}{3}r^3\right)=4\pi r^2 \delta r$$ $$\delta r = 0.02\frac{r}{3}$$ $$\delta r = 0.00666... \cdot r$$ The error in $r$ can be no more than $0.666...\%$.

Answer 3

Implicitly differentiating both sides, we have:

$$\frac{\mathrm{d}V}{\mathrm{d}r} = 4 \pi r^2 \Rightarrow \mathrm{d}V = 4 \pi r^2 \ \mathrm{d}r$$

As Christian Blatter explained, we are looking for the relative error which is $\frac{\mathrm{d}V}{V}$. However, for a given radius $r$, the volume $V = V(r) = \frac{4}{3} \pi r^3$:

$$\frac{\mathrm{d}V}{V} = \frac{4 \pi r^2 \ \mathrm{d} r}{V} = \frac{4 \pi r^2 \ \mathrm{d} r}{\frac{4}{3} \pi r^3} = 3 \frac{\mathrm{d} r}{r} \ $$

So it follows when $\left \lvert \frac{\mathrm{d}V}{V} \right \lvert ≤ 0.02$, $\left \lvert 3 \frac{\mathrm{d}r}{r} \right \lvert ≤ 0.02$ so $\left \lvert \frac{\mathrm{d} r}{r} \right \lvert ≤ \frac{2}{300}$. Rephrasing the statement, this means that the relative error of $r$ is no more than $\frac{2}{3} \%$.