# % Error of Linear Approximations: Example Problem

I received the following question on my exam and got it right, although it was entirely a guess and I had absolutely no idea how to approach it. Any help with the logic or steps behind this would be greatly appreciated.

The side length of a square ice cube is claimed to be 1.0 inches, correct to within 0.001 inch. Use linear approximation to estimate the resulting error, measured in squared inches, in the surface area of the ice cube.

± 0.12

± 0.003

± 0.02

None of these

So, just to let you know, the correct answer is None of these. I drew a diagram thinking it would help (it didn't) and was pretty much stuck at that point. I know how Linear Approximation works and sort of understand percentage error. Does it just deal with the fact that (A)-(C) use the "±" sign? We were taught that percentage error is 100 • | (approximate - exact) / (exact) |, so I guess you can never have negatives?

Is there any algebraic way to solve this problem, though?

For a cube of side length $x$ the surface area is $6x^2$. Linear approximation says $f(x) \approx f(x_0) + f'(x_0)(x-x_0)$ for $x$ close to $x_0$. (This is the most important formula in differential calculus, in my opinion.) So in this case $6x^2 \approx 6x_0^2 + 12x_0(x-x_0)$. So if for instance $x=1.001,x_0=1$, the error in the area is approximately $12 \cdot 0.001=0.012$. If they were asking about percentages then this would be $100 \cdot \frac{0.012}{6} = 0.2\%$ but I do not think they are actually asking about percentages in this context.