# Why use differentials to compute error?

I get how to use differentials to compute error, but why is it a "good" method? For example, a standard problem is something like:

If the radius of a circle is $3 \pm 0.1$ cm, find the area with error.

What's wrong/undesirable with just finding the area for $r=2.9$ and $r=3.1$ to determine the error?

• In some cases, the maximum error might not occur at the maximum deviation from the given value. To give a pathological example, find the error of the function $$f(x)= \frac{1}{x-2}$$ on the interval $2.1\pm 0.2$. The biggest error (so to speak, I guess) occurs at $x-2$, which is not an endpoint of the error interval. – Ian Coley Mar 2 '14 at 21:24
• @IanColey I disagree: simply calculate the increment $\Delta f(x) = f(2.1) - f(2.1 \pm 0.2)$ and see which of the two errors is the largest. If the input is 2.3, then the error is $10-3.33=6.67$, and if the input is 1.9, the error is $10- (-10)=20$. Clearly, the latter yields the largest error. This has the additional benefit of avoiding figuring out what the derivative of $f(x)$ is. – Ius Klesar Jun 13 at 11:26

The nice thing about using differentials is that as you adjust the error, you don't need to compute the highest and lowest again; you just need to multiply some coefficient.

So, for your example, we might have a situation where we ask "if the radius $3 \pm 0.1$ cm, what's the error?" But the more useful question is "if we want the area to be within $.01$ of $3 \text{cm}^2$, how small does the error in the radius need to be?" In this case, we could solve for $\Delta r$, saying that if our radius is $3 \pm \Delta r$ cm, then $$\Delta A \approx 2 \pi (3 \text{cm})\Delta r \leq .01 \text{cm}$$ and solve for $\Delta r$.

On the other hand, doing things the other way, we would have to solve something like $$\Delta A = \pi (3\text{cm} + \Delta r)^2 - \pi (3\text{cm})^2 \leq .01 \text{cm}$$ In this situation, we simply have a quadratic equation, but in others we could wind up with something much more complicated.

If I understand your question correctly, there are a couple of reasons I can see. First, you can use a general formula instead of computing with specific numbers. Second, you can usually see how the error in measuring (say) $r$ affects the error in the computed value of $A(r)$.

For $A(r)=\pi r^2$, if the measured radius is $r\pm\epsilon$, then the computed area is $A(r\pm\epsilon)\approx A(r) \pm A'(r)\epsilon= \pi r^2\pm 2\pi r\epsilon$. This shows that the error in area is about $2\pi r$ times the error in measuring the radius.

I hope that's the sort of thing you were looking for.

Sometimes you cannot compute the error exactly. For instance, how do you compute the cubic root of 28? If you have a calculator, it is easy. Or you can do it by hand noting that $\sqrt{27}=3$. Let $f(x)=\sqrt{x}$. Then $$f'(x)=\frac{1}{3\sqrt{x^2}}$$ and $$\sqrt{28}=f(27+1)\approx f(27)+1\cdot f'(27)=3+\frac{1}{3\sqrt{27^2}}=3+\frac{1}{27}.$$