# Using differentials to find change in surface area when given change in length

A metal cube is heated and the length of each edge is thereby increased by 0.1%. Use a differential to show that the surface area of the cube is then increased by p%. Find p.

What I have is the following:

Length: $x$

$\Delta x=0.001x$

Surface area: SA = $6x^2$

$$\Delta\text{SA}=p\cdot\text{SA}\implies \frac{\Delta \text{SA}}{\text{SA}}=p$$

$$\text{SA}=6x^2\implies\Delta \text{SA}=12x\Delta x\implies\Delta \text{SA}=12x(0.001x)\implies\Delta \text{SA}=0.012x^2$$

Then to find $p$,

$$p=\frac{\Delta \text{SA}}{\text{SA}}\implies p=\frac{0.012x^2}{6x^2}\implies p=0.002\implies p=0.2\%$$

But when I try some number, say 1000, for $x$, I get this:

$\text{SA}(x=1000):6000000$

$\text{SA}(x=1001): 6012006$

$\frac{\Delta\text{SA}}{\text{SA}}=\frac{12006}{6000000}=0.002001=0.2001\%$

Where did I mess up?

• When computing $\Delta SA$ you wrote by mistake $\Delta x=0.0001x$ instead of $\Delta x=0.001x$. Commented Dec 18, 2016 at 16:35
• Ah, my bad. The answers are a bit closer now, but why does the one with the values have the extra 0.0001%? Commented Dec 18, 2016 at 16:41
• You didn't mess up at all. When doing this stuff, physicists only take the linear approximation: if $h$ is very close to $0$ and $a\neq0$, $(a+h)^2=a^2+2a h+ h^2$. Now since $h$ is super close to $0$ and $a\neq0$, the term $h^2$ is negligible: so physicists neglect it: they gladly say $(a+h)^2\approx a^2+2ah$. In math we say $(a+h)^2=a^2+2ah+o(h)$ (or something similar). It should make sense: the $0.0001\%$ is irrelevant when compared to $0.2\%$. Commented Dec 18, 2016 at 16:43
• Ah, I see. Thanks a bunch! Commented Dec 18, 2016 at 16:45