# Using Differentials to Estimate Errors

Suppose that we measured some quantity $x$ and know error $\Delta{y}$ in measurements. If we have function $y={f{{\left({x}\right)}}}$, how can we estimate error $\Delta{y}$ in measurement of ${y}$?

Since error is very small we can write that $\Delta{y}\approx{d}{y}$, so error in measurement is differential of the function. Since ${d}{x}=\Delta{x}$, then error in measurement of $y$ can be caluclated using formula ${d}{y}={f{'}}{\left({x}\right)}{d}{x}$.

Example. The radius of a sphere was measured and found to be 20 cm with a possible error in measurement of at most 0.01 cm. What is the maximum error in using this value of the radius to compute the volume of the sphere? Find the relative and percentage error in both radius and volume.

Volume of sphere is ${V}=\frac{{4}}{{3}}\pi{{r}}^{{3}}$.

Differentiate with respect to ${r}$: ${d}{V}=\frac{{4}}{{3}}\pi\cdot{3}{{r}}^{{2}}{d}{r}$ or ${d}{V}={4}\pi{{r}}^{{2}}{d}{r}$.

We have ${d}{r}={0.01}$ and ${r}={20}$, so ${d}{V}={4}\cdot\pi\cdot{{20}}^{{2}}\cdot{0.01}\approx{50.27}$.

So, the maximum error in the calculated volume is about ${50.27}\ {c}{{m}}^{{3}}$.

Relative error in the volume is calculated by dividing the error by the total volume. Same with radius.

Relative error in the radius is $\frac{{{d}{r}}}{{r}}=\frac{{0.01}}{{{20}}}={0.0005}$.

Percentage error in the radius is $\frac{{{d}{r}}}{{r}}\cdot{100}\%={0.05}\%$.

Relative error in the volume is $\frac{{{d}{V}}}{{V}}=\frac{{{4}\pi{{r}}^{{2}}{d}{r}}}{{\frac{{4}}{{3}}\pi{{r}}^{{3}}}}={3}\frac{{{d}{r}}}{{r}}={3}\cdot{0.0005}={0.0015}$.

Percentage error in the volume is $\frac{{{d}{V}}}{{V}}\cdot{100}\%={0.15}\%$.

Note, since $\frac{{{d}{V}}}{{V}}={3}\frac{{{d}{r}}}{{r}}$, then the relative error in the volume is approximately three times the relative error in the radius.