Linear Approximation and Differentials

Most scientific experiments involve using instruments to take measurements, but the instruments are not perfect, and the measurements we get from them are only accurate up to a certain level. If we know the level of accuracy of our instruments and measurements, we can use the idea of linear approximation to estimate the level of accuracy of results we calculate from our measurements.

If we measure the side $x$ of a square to be 8 inches, then, of course, we would calculate its area to be $A(x)=x^{2}=64$ square inches. Suppose, as is reasonable in a real measurement, that our measuring instrument could only measure or be read to the nearest $0.05$ inches. Then our measurement of 8 inches would really mean some number between $8-0.05$ $=7.95$ inches and $8+0.05=8.05$ inches, and the true area of the square would be between $\mathrm{A}(7.95)=63.2025$ and $\mathrm{A}(8.05)=64.8025$ square inches. Our possible "error" or "uncertainty", because of the limitations of the instrument, could be as much as $64.8025-64=0.8025$ square inches so we could report the area of the square to be $64 \pm 0.8025$ square inches. We can also use the linear approximation method to estimate the "error" or uncertainty of the area. (For a function as simple as the area of a square, this linear approximation method really isn't needed, but it is used to illustrate the idea).

For a square with side $x$, the area is $A(x)=x^{2}$ and $A^{\prime}(x)=2 x.$ If $\Delta x$ represents the "error" or uncertainty of our measurement of the side, then, using the linear approximation technique for $\mathrm{A}(\mathrm{x})$,

$\mathrm{A}(\mathrm{x}) \approx \mathrm{A}(\mathrm{a})+\mathrm{A}^{\prime}(\mathrm{a}) \cdot \Delta \mathrm{x}$ so the uncertainty of our calculated area is $\mathrm{A}(\mathrm{x})-\mathrm{A}(\mathrm{a}) \approx \mathrm{A}^{\prime}(\mathrm{x}) \cdot \Delta \mathrm{x}.$ In this example, $\mathrm{a}=8$ inches and $\Delta \mathrm{x}=0.05$ inches so

$\mathrm{A}(8.05) \approx \mathrm{A}(8)+\mathrm{A}^{\prime}(8) \cdot(0.05)=64+2(8) \cdot(0.05)=64.8$ square inches

and the uncertainty in our calculated area is approximately

$\mathrm{A}(8+0.05)-\mathrm{A}(8) \approx \mathrm{A}^{\prime}(8) \cdot \Delta \mathrm{x}=2(8$ inches $)(0.05$ inches $)=0.8$ square inches

This process can be summarized as:

Practice 5: If we measure the side of a cube to be $4 \mathrm{~cm}$ with an uncertainty of $0.1 \mathrm{~cm}$, what is the volume of the cube and the uncertainty of our calculation of the volume? (Use linear approximation).

Example 4: We are using a tracking telescope to follow a small rocket. Suppose we are 3000 meters from the launch point of the rocket, and, 2 seconds after the launch, we measure the angle of the inclination of the rocket to be $64^{\circ}$ with a possible "error" of $2^{\circ}$ (Fig. 5). How high is the rocket and what is the possible error in this calculated height?

Solution: Our measured angle is $x=1.1170$ radians and $\Delta x=0.0349$ radians (all of our trigonometric work is in radians), and the height of the rocket at an angle $x$ is $\mathrm{f}(\mathrm{x})=3000 \cdot \tan (\mathrm{x})$ so $\mathrm{f}(1.1170) \approx 6151 \mathrm{~m}.$ Our uncertainty in the height is 

$\Delta \mathrm{f}(\mathrm{x}) \approx \mathrm{f}^{\prime}(\mathrm{x}) \cdot \Delta \mathrm{x} \approx 3000 \cdot \sec ^{2}(\mathrm{x}) \cdot \Delta \mathrm{x}=3000 \sec ^{2}(1.1170) \cdot(0.0349)=545 \mathrm{~m}$.

If our measured angle of $64^{\circ}$ can be in error by as much as $2^{\circ}$, then our calculated height of $6151 \mathrm{~m}$ can be in error by as much as $545 \mathrm{~m}$. The height is $6151 \pm 545$ meters.

In some scientific and engineering applications, the calculated result must be within some given specification. You might need to determine how accurate the initial measurements must be in order to guarantee the final calculation is within the specification. Added precision usually costs time and money, so it is important to choose a measuring instrument which is good enough for the job but not too good or too expensive.

Example 5: Your company produces ball bearings (spheres) with a volume of $10 \mathrm{~cm}^{3}$, and the volume must be accurate to within $0.1 \mathrm{~cm}^{3}$. What radius should the bearings have and what error can you tolerate in the radius measurement to meet the accuracy specification for the volume? $\left(\mathrm{V}=\frac{4}{3} \pi \mathrm{r}^{3}\right)$

Solution: Since we want $\mathrm{V}=10$, we can solve $10=\frac{4}{3} \pi \mathrm{r}^{3}$ for $\mathrm{r}$ to get $\mathrm{r}=1.3365 \mathrm{~cm}$. $\mathrm{V}(\mathrm{r})=\frac{4}{3} \pi \mathrm{r}^{3}$ and $\mathrm{V}^{\prime}(\mathrm{r})=4 \pi \mathrm{r}^{2} \quad$ so $\Delta \mathrm{V} \approx \mathrm{V}^{\prime}(\mathrm{r}) \cdot \Delta \mathrm{r}$. In this case we have been given that $\Delta \mathrm{V}=0.1 \mathrm{~cm}^{3}$, and we have calculated $\mathrm{r}=1.3365 \mathrm{~cm}$ so $0.1 \mathrm{~cm}^{3}=\mathrm{V}$ '(1.3365 $\mathrm{cm}) \cdot \Delta \mathrm{r}=\left(22.45 \mathrm{~cm}^{2}\right) \cdot \Delta \mathrm{r}$.

Solving for $\Delta \mathrm{r}$, we get $\Delta \mathrm{r} \approx 0.0045 \mathrm{~cm}$. To meet the specifications for the allowable error in the volume, we must allow no more than $0.0045 \mathrm{~cm}$ variation in the radius. If we measure the diameter of the sphere rather than the radius, then we want $\mathrm{d}=2 \mathrm{r}=2(1.3365 \pm 0.0045)=2.673 \pm 0.009 \mathrm{~cm}$.

Practice 7: You want to determine the height of the rocket to within 10 meters when it is 4000 meters high (Fig. 6). How accurate must your angle measurement be? (Do your calculations in radians).