MA005: Calculus I

The "error" we have been examining is called the absolute error to distinguish it from two other commonly used terms, the relative error and the percentage error which compare the absolute error with the magnitude of the number being measured. An "error" of 6 inches in measuring the circumference of the earth would be extremely small, but a 6 inch error in measuring your head for a hat would result in a very bad fit.

Example 6: If the relative error in the calculation of the area of a circle must be less than $0.4$, then what relative error can we tolerate in the measurement of the radius?

Solution: $\mathrm{A}(\mathrm{r})=\pi \mathrm{r}^{2}$ so $\mathrm{A}^{\prime}(\mathrm{r})=2 \pi \mathrm{r}$ and $\Delta \mathrm{A} \approx \mathrm{A}^{\prime}(\mathrm{r}) \Delta \mathrm{r}=2 \pi \mathrm{r} \Delta \mathrm{r}$. The Relative Error of $\mathrm{A}$ is 

$\frac{\Delta \mathrm{A}}{\mathrm{A}} \approx \frac{2 \pi \mathrm{r} \Delta \mathrm{r}}{\pi \mathrm{r}^{2}}=2 \frac{\Delta \mathrm{r}}{\mathrm{r}}$. We can guarantee that the Relative Error of $\mathrm{A}, \frac{\Delta \mathrm{A}}{\mathrm{A}}$, is less than $0.4$ if the Relative Error of $\mathrm{r}, \frac{\Delta \mathrm{r}}{\mathrm{r}}=\frac{1}{2} \frac{\Delta \mathrm{A}}{\mathrm{A}}$, is less than $\frac{1}{2}(0.4)=0.2$.

Practice 8: If you can measure the side of a cube with a percentage error less than 3%, then what will the percentage error for your calculation of the surface area of the cube be?