1
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ Did you use differentials? [result looks OK but just think differentials were asked to be used in process] $\endgroup$ – coffeemath May 28 '16 at 22:47
  • $\begingroup$ @coffeemath if you read $dV/dt$ instead of $V'$ (and the same for the other variable) it should be ok right? $\endgroup$ – bru1987 May 28 '16 at 22:51
  • 1
    $\begingroup$ 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 $dV$. $\endgroup$ – Semiclassical May 28 '16 at 22:54
  • 1
    $\begingroup$ 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 $s.$ $\endgroup$ – coffeemath May 28 '16 at 22:55
  • $\begingroup$ I see. What I actually meant was $dV/dl$. Thank you for the explanation. best regards! $\endgroup$ – bru1987 May 28 '16 at 22:57
1
$\begingroup$

You can also use the binomial theorem. If $|x| \ll 1$, then

$$(1 + x)^n = 1 + \binom{n}{1} x + \binom{n}{2} x^2 + \cdots + \binom{n}{n} x^n \approx 1 + n x$$

Hence,

$$\left(20 \cdot (1+0.02)\right)^3 = 20^3 \cdot (1+0.02)^3 \approx 20^3 \cdot (1 + 3 \cdot 0.02) = 20^3 \cdot (1 + 0.06)$$

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 $4\%$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.