# Using differentials to find the percentual error on calculating the volume of a cube

I am given the following problem:

The size of a cube measures $20 \, \rm{cm}$ with a percentage error of $\pm 2 \%$. Use differentials to estimate the error on calculating its volume.

What I have so far is

$$2 \% \cdot 20 \, \rm{cm} = 0.4 \, \rm{cm}$$

$$V = l^3 \Rightarrow V' = 3l^2 l'\\ V' = 3 \cdot 400 \cdot 0.4\\ V' = 480 \, \rm{cm}^3\\ E_{\%} = \frac{480}{20^3} = 0.06 = 6 \%$$

Is that correct or did I make a mistake somewhere?

Thank you.

• Did you use differentials? [result looks OK but just think differentials were asked to be used in process] – coffeemath May 28 '16 at 22:47
• @coffeemath if you read $dV/dt$ instead of $V'$ (and the same for the other variable) it should be ok right? – bru1987 May 28 '16 at 22:51
• It may be useful to include units in writing out the third line. Also, it's not really sensible to say that $V'=\dfrac{dV}{dt}=400$ cm^3: if $t$ is intended as a time then $\dfrac{dV}{dt}$ should have units of cm^3 per second. What you probably mean is not $\dfrac{dV}{dt}$ but just $dV$. – Semiclassical May 28 '16 at 22:54
• Yes, but maybe "move the dt to the other side" of equation like $dV=3s^2ds$ where $s$ is the cube's side length. Then the usual way is at that point replace $ds$ by the small difference in $s.$ – coffeemath May 28 '16 at 22:55
• I see. What I actually meant was $dV/dl$. Thank you for the explanation. best regards! – bru1987 May 28 '16 at 22:57

You can also use the binomial theorem. If $|x| \ll 1$, then
$$(1 + x)^n = 1 + \binom{n}{1} x + \binom{n}{2} x^2 + \cdots + \binom{n}{n} x^n \approx 1 + n x$$
$$\left(20 \cdot (1+0.02)\right)^3 = 20^3 \cdot (1+0.02)^3 \approx 20^3 \cdot (1 + 3 \cdot 0.02) = 20^3 \cdot (1 + 0.06)$$
A perturbation of $2\%$ in the side of the cube produces a perturbation of approximately $6\%$ in its volume. One can use the same argument to conclude that the perturbation in the surface area of the cube is approximately $4\%$.