Linear Approximation and Differentials

Newton's method used tangent lines to "point toward" a root of the function. In this section we examine and use another geometric characteristic of tangent lines:

This idea is used to approximate the values of some commonly used functions and to predict the "error" or uncertainty in a final calculation if we know the "error" or uncertainty in our original data. Finally, we define and give some examples of a related concept called the differential of a function.

Source: Dale Hoffman, https://s3.amazonaws.com/saylordotorg-resources/wwwresources/site/wp-content/uploads/2012/12/MA005-3.9-Linear-Approximation.pdf
This work is licensed under a Creative Commons Attribution 3.0 License.

Since this section uses tangent lines frequently, it is worthwhile to recall how we find the equation of the line tangent to $f$ at a point $x=a$. The line tangent to $f$ at $x=a$ goes through the point $(a, f(a))$ and has slope $f^{\prime}(a)$ so, using the point-slope form $y-y_{0}=m\left(x-x_{0}\right)$ for linear equations, we have

$y-f(a)=f^{\prime}(a) \cdot(x-a) \text { and } y=f(a)+f^{\prime}(a) \cdot(x-a)$

This final result is the equation of the line tangent to $\mathrm{f}$ at $\mathrm{x}=\mathrm{a}$.

Example 1: Find the equation of the line $\mathrm{L}(\mathrm{x})$ which is tangent to the graph of $\mathrm{f}(\mathrm{x})=\sqrt{\mathrm{x}}$ at the point $(9,3)$. Evaluate $\mathrm{L}(9.1)$ and $\mathrm{L}(8.88)$ to approximate $\sqrt{9.1}$ and $\sqrt{8.88}$.

Solution: $f(x)=\sqrt{x}=x^{1 / 2}$ and $f^{\prime}(x)=\frac{1}{2} x^{-1 / 2}=\frac{1}{2 \sqrt{x}}$ so $f(9)=3$ and $f^{\prime}(9)=\frac{1}{2 \sqrt{9}}=\frac{1}{6}$. Then

$\mathrm{L}(\mathrm{x})=\mathrm{f}(9)+\mathrm{f}^{\prime}(9) \cdot(\mathrm{x}-9)=3+\frac{1}{6}(\mathrm{x}-9)$. If $\mathrm{x}$ is close to $\mathrm{9}$, then the value of $\mathrm{L}(\mathrm{x})$ is a good approximation of the value of $\sqrt{x}$ (Fig. 2). The number $9.1$ is close to 9 so $\sqrt{9.1}=\mathrm{f}(9.1) \approx \mathrm{L}(9.1)=3+\frac{1}{6}(9.1-9)=3.016666$

Similarly, $\sqrt{8.88}=\mathrm{f}(8.88) \approx \mathrm{L}(8.88)=3+\frac{1}{6}(8.88-9)=2.98.$ In fact, $\sqrt{9.1} \approx 3.016621$, so our estimate, using $\mathrm{L(9.1)}$, is within $0.000045$ of the exact answer. $\sqrt{8.88} \approx 2.979933$ (accurate to 6 decimal places) and our estimate is within $0.00007$ of the exact answer.

In each example, we got a good estimate of a square root with very little work. The graph indicates the tangent line $\mathrm{L}(\mathrm{x})$ is slightly above $\mathrm{f}(\mathrm{x})$, and each estimate is slightly larger than the exact value.

Practice 1: Find the equation of the line $\mathrm{L}(\mathrm{x})$ tangent to the graph of $\mathrm{f}(\mathrm{x})=\sqrt{\mathrm{x}}$ at the point $(16,4)$ (Fig. 3). Evaluate $\mathrm{L}(16.1)$ and $\mathrm{L(15.92)}$ to approximate $\sqrt{16.1}$ and $\sqrt{15.92}$. Are your approximations using $\mathrm{L}$  larger or smaller than the exact values of the square roots?

Practice 2: Find the equation of the line $\mathrm{L}(\mathrm{x})$ tangent to the graph of $\mathrm{f}(\mathrm{x})=\mathrm{x}^{3}$ at the point $(1,1)$ and use $\mathrm{L}(\mathrm{x})$ to approximate $(1.02)^{3}$ and $(0.97)^{3}$. Do you think your approximations using $\mathrm{L}$ are larger or smaller than the exact values?

The process we have used to approximate square roots and cubics can be used to approximate any differentiable function, and the main result about the linear approximation follows from the two statements in the boxes. Putting these two statements together, we have the process for Linear Approximation.

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).

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?

In Fig. 7, the change in value of the function $\mathrm{f}$ near the point $(x, f(x))$ is $\Delta f=f(x+\Delta x)-f(x)$ and the change along the tangent line is $\mathrm{f}^{\prime}(\mathrm{x}) \cdot \Delta \mathrm{x}$. If $\Delta \mathrm{x}$ is small, then we have used the approximation that $\Delta \mathrm{f} \approx \mathrm{f}^{\prime}(\mathrm{x}) \cdot \Delta \mathrm{x}$. This leads to the definition of a new quantity, $\mathrm{df}$, called the differential of $\mathrm{f}$.

The differential of $f$ represents the change in $f$, as $x$ changes from $x$ to $x+d x$, along the tangent line to the graph of $\mathrm{f}$ at the point $(\mathrm{x}, \mathrm{f}(\mathrm{x}))$. If we take $\mathrm{dx}$ to be the number $\Delta \mathrm{x}$, then the differential is an approximation of $\Delta f: \Delta f \approx f^{\prime}(x) \cdot \Delta x=f^{\prime}(x) \cdot d x=d f$.

Example 7: Determine the differential df of each of $\mathrm{f}(\mathrm{x})=\mathrm{x}^{3}-7 \mathrm{x}, \mathrm{g}(\mathrm{x})=\sin (\mathrm{x})$, and $\mathrm{h}(\mathrm{r})=\pi \mathrm{r}^{2}$. Solution: $\quad \mathrm{df}=\mathrm{f}^{\prime}(\mathrm{x}) \cdot \mathrm{d} \mathrm{x}=\left(3 \mathrm{x}^{2}-7\right) \mathrm{dx}, \mathrm{dg}=\mathrm{g}^{\prime}(\mathrm{x}) \cdot \mathrm{dx}=\cos (\mathrm{x}) \mathrm{dx}$, and $\mathrm{dh}=\mathrm{h}^{\prime}(\mathrm{r}) \mathrm{dr}=2 \pi \mathrm{r} \mathrm{dr}$.

Practice 9: Determine the differentials of $\mathrm{f}(\mathrm{x})=\ln (\mathrm{x}), \mathrm{u}=\sqrt{1-3 \mathrm{x}}$, and $\mathrm{r}=3 \cos (\theta)$.

We will do little with differentials for a while, but are used extensively in integral calculus.

An approximation is most valuable if we also have have some measure of the size of the "error", the distance between the approximate value and the value being approximated. Typically, we will not know the exact value of the error (why not?), but it is useful to know that the error must be less than some number. For example, if one scale gives the weight of a gold pendant as $10.64$ grams with an error less than $.3$ grams $(10.64 \pm.3$ grams) and another scale gives the weight of the same pendant as $10.53$ grams with an error less than $.02$ grams $(10.53 \pm.02$ grams), then we can have more faith in the second approximate weight because of the smaller "error" guarantee. Before finding a guarantee on the size of the error of the linear approximation process, we will check how well the linear approximation process approximates some functions we can compute exactly. Then we will prove one bound on the possible error and state a somewhat stronger bound.

Example 8: Let $\mathrm{f}(\mathrm{x})=\mathrm{x}^{2}$. Evaluate $\mathrm{f}(2+\Delta \mathrm{x}), \mathrm{L}(2+\Delta \mathrm{x})$ and $|\mathrm{f}(2+\Delta \mathrm{x})-\mathrm{L}(2+\Delta \mathrm{x})|$ for
                  $\Delta \mathrm{x}=0.1,0.05,0.01,0.001$ and for a general value of $\Delta \mathrm{x}$.

Solution: $\quad \mathrm{f}(2+\Delta \mathrm{x})=(2+\Delta \mathrm{x})^{2}=2^{2}+4 \Delta \mathrm{x}+(\Delta \mathrm{x})^{2}$ and $\mathrm{L}(2+\Delta \mathrm{x})=\mathrm{f}(2)+\mathrm{f}^{\prime}(2) \cdot \Delta \mathrm{x}=2^{2}+4 \cdot \Delta \mathrm{x}$. Then

$\begin{array}{l|l|l|l} \Delta \mathrm{x} & \mathrm{f}(2+\Delta \mathrm{x}) & \mathrm{L}(2+\Delta \mathrm{x}) & |\mathrm{f}(2+\Delta \mathrm{x})-\mathrm{L}(2+\Delta \mathrm{x})| \\ \hline 0.1 & 4.41 & 4.4 & 0.01 \\ 0.05 & 4.2025 & 4.2 & 0.0025 \\ 0.01 & 4.0401 & 4.04 & 0.0001 \\ 0.001 & 4.004001 & 4.004 & 0.000001 \end{array}$

Cutting the value of $\Delta x$ in half makes the error $1 / 4$ as large. Cutting $\Delta x$ to $1 / 10$ as large makes the error $1 / 100$ as large. In general, $|\mathrm{f}(2+\Delta \mathrm{x})-\mathrm{L}(2+\Delta \mathrm{x})|=\left|\left(2^{2}+4 \cdot \Delta \mathrm{x}+(\Delta \mathrm{x})^{2}\right)-\left(2^{2}+4 \cdot \Delta \mathrm{x}\right)\right|=(\Delta \mathrm{x})^{2}$.

This function and error also have a nice geometric interpretation (Fig. 8): $f(x)=x^{2}$ is the area of a square of side $x$ so $f(2+\Delta x)$ is the area of a square of side $2+\Delta x$, and that area is the sum of the pieces with areas $2^{2}, 2 \cdot \Delta x, 2 \cdot \Delta x$, and $(\Delta x)^{2}$. The linear approximation $\mathrm{L}(2+\Delta x)=2^{2}+4 \cdot \Delta x$ to the area of the square includes the 3 largest pieces $2^{2}, 2 \cdot \Delta x$ and $2 \cdot \Delta x$, but it omits the small square with area $(\Delta x)^{2}$ so the approximation is in error by the amount $(\Delta x)^{2}$.

Practice 10: Let $\mathrm{f}(\mathrm{x})=\mathrm{x}^{3}$. Evaluate $\mathrm{f}(4+\Delta \mathrm{x}), \mathrm{L}(4+\Delta \mathrm{x})$ and

$|\mathrm{f}(4+\Delta \mathrm{x})-\mathrm{L}(4+\Delta \mathrm{x})|$ for $\Delta \mathrm{x}=0.1,0.05,0.01,0.001$ and for a general value of $\Delta x$. Use Fig. 9 to give a geometric interpretation of $f(4+\Delta x)$, $\mathrm{L}(4+\Delta \mathrm{x})$ and $|\mathrm{f}(4+\Delta \mathrm{x})-\mathrm{L}(4+\Delta \mathrm{x})|$.

In both the example and practice problem, the error $|\mathrm{f}(\mathrm{a}+\Delta \mathrm{x})-\mathrm{L}(\mathrm{a}+\Delta \mathrm{x})|$ turned out to be very small, proportional to $(\Delta x)^{2}$, when $\Delta x$ was small. In general, the error approaches $\mathrm{0}$ as $\Delta x$ approaches $\mathrm{0}$.

Theorem :      If $\quad \mathrm{f}(\mathrm{x})$ is differentiable at a and $\mathrm{L}(\mathrm{a}+\Delta \mathrm{x})=\mathrm{f}(\mathrm{a})+\mathrm{f}^{\prime}(\mathrm{a}) \cdot \Delta \mathrm{x}$

Proof: $\quad|\mathrm{f}(\mathrm{a}+\Delta \mathrm{x})-\mathrm{L}(\mathrm{a}+\Delta \mathrm{x})|=\left|\mathrm{f}(\mathrm{a}+\Delta \mathrm{x})-\mathrm{f}(\mathrm{a})-\mathrm{f}^{\prime}(\mathrm{a}) \cdot \Delta \mathrm{x}\right|=\left\{\frac{\mathrm{f}(\mathrm{a}+\Delta \mathrm{x})-\mathrm{f}(\mathrm{a})}{\Delta \mathrm{x}}-\mathrm{f}^{\prime}(\mathrm{a})\right\} \cdot \Delta \mathrm{x}.$ But $\mathrm{f}$ is differentiable

at $x=\mathrm{a}$ so $\lim _{\Delta x \rightarrow 0} \frac{f(a+\Delta x)-f(a)}{\Delta x}=\mathrm{f}^{\prime}(\mathrm{a})$ and $\lim _{\Delta x \rightarrow 0}\left\{\frac{f(a+\Delta x)-f(a)}{\Delta x}-f^{\prime}(a)\right\}=0$.

Then $\lim _{\Delta x \rightarrow 0}|f(a+\Delta x)-L(a+\Delta x)|=\lim _{\Delta x \rightarrow 0}\left\{\frac{f(a+\Delta x)-f(a)}{\Delta x}-f^{\prime}(a)\right\} \cdot \lim _{\Delta x \rightarrow 0} \Delta x=0 \cdot 0=0$

Not only does the difference $f(a+\Delta x)-L(a+\Delta x)$ approach $\mathrm{0}$, but this difference approaches $\mathrm{0}$ so fast that we can divide it by $\Delta \mathrm{x}$, another quantity approaching $\mathrm{0}$, and the quotient still approaches $\mathrm{0}$.

In the next chapter we can prove that the error of the linear approximation process is proportional to $(\Delta \mathrm{x})^{2}$. For now we just state the result.

Theorem: If $\mathrm{f}$ is differentiable at $\mathrm{a}$ and $\left|\mathrm{f}^{\prime \prime}(\mathrm{x})\right| \leq \mathrm{M}$ for all $\mathrm{x}$ between $\mathrm{a}$ and $\mathrm{a}+\Delta \mathrm{x}$

Practice 1: $\mathrm{f}(\mathrm{x})=\mathrm{x}^{1 / 2}$ so $\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2 \sqrt{\mathrm{x}}}$. At the point $(16,4)$ on the graph of $\mathrm{f}$, the slope of the tangent line is $f^{\prime}(16)=\frac{1}{2 \sqrt{16}}=\frac{1}{8}$. The equation of the tangent line is

$y-4=\frac{1}{8}(x-16)$ or $y=\frac{1}{8} x+2: L(x)=\frac{1}{8} x+2$. Then

$\sqrt{16.1} \approx \mathrm{L}(16.1)=\frac{1}{8}(16.1)+2=\mathbf{4.0125}$ and $\sqrt{15.92} \approx \mathrm{L}(15.92)=\frac{1}{8}(15.92)+2=\mathbf{3.99}$

Practice 2: $\quad \mathrm{f}(\mathrm{x})=\mathrm{x}^{3}$ so $\mathrm{f}^{\prime}(\mathrm{x})=3 \mathrm{x}^{2}$. At $(1,1)$, the slope of the tangent line is $\mathrm{f}^{\prime}(1)=3$. The equation of the tangent line is $y-1=3(x-1)$ or $y=3 x-2: L(x)=3 x-2$. Then $(1.02)^{3} \approx \mathrm{L}(1.02)=3(1.02)-2=\mathbf{1.06}$ and $(0.97)^{3} \approx \mathrm{L}(0.97)=3(0.97)-2=\mathbf{0.91}$.

Practice 3: $f(x)=x^{4}$ so $f^{\prime}(x)=4 x^{3}. a=1$ and $\Delta x=0.06$.

$(1.06)^{4}=\mathrm{f}(1.06) \approx \mathrm{L}(1.06)=\mathrm{f}(1)+\mathrm{f}^{\prime}(1) \cdot(0.06)=1^{4}+4(1)^{3} \cdot(0.06)=1.24$

Practice 4: $\begin{aligned} &\mathrm{f}(1.1) \approx \mathrm{f}(1)+\mathrm{f}^{\prime}(1) \cdot(0.1)=0.7854+(0.5) \cdot(0.1)=\mathbf{0.8354}. \\ &\mathrm{f}(1.23) \approx \mathrm{f}(1.2)+\mathrm{f}^{\prime}(1.2) \cdot(0.03)=0.8761+(0.4098) \cdot(0.03)=\mathbf{0.888394}. \\ &\mathrm{f}(1.38) \approx \mathrm{f}(1.4)+\mathrm{f}^{\prime}(1.4) \cdot(-0.02)=0.9505+(0.3378) \cdot(-0.02)=\mathbf{0.943744}. \end{aligned}$

Practice 5: $\quad x=4 \mathrm{~cm}$ and $\Delta x=0.1 \mathrm{~cm}. f(x)=x^{3}$ so $f^{\prime}(x)=3 x^{2}$ and $f(4)=4^{3}=64 \mathrm{~cm}^{3}$. Then "error" $\Delta \mathrm{f} \approx \mathrm{f}^{\prime}(\mathrm{x}) \Delta \mathrm{x}=3 \mathrm{x}^{2} \cdot \Delta \mathrm{x}$. When $\mathrm{x}=4$ and $\Delta \mathrm{x}=0.1$, "error" $\Delta \mathrm{f} \approx 3(4)^{2}(0.1)=\mathbf{4. 8} \mathrm{cm}^{3}$.

Practice 6: $43 \pm 1^{\circ}$ is $0.75049 \pm 0.01745$ radians. $f(x)=3000 \cdot \tan (x)$ so $\mathrm{f}(0.75049)=3000 \cdot \tan (0.75049) \approx \mathbf{2 7 9 7. 5} \mathrm{m}. \quad \mathrm{f}^{\prime}(\mathrm{x})=3000 \sec ^{2}(\mathrm{x}) \mathrm{so}$ 

$\Delta f(x) \approx f^{\prime}(x) \cdot \Delta x=3000 \cdot \sec ^{2}(x) \cdot \Delta x=3000 \cdot \sec ^{2}(0.75049) \cdot(0.01745)=97.9 \mathrm{~m}$

The height of the rocket is $\mathbf{2797.5} \pm \mathbf{97.9} \mathbf{~ m}$.

Practice 7: $\mathrm{f}(\theta)=2000 \cdot \tan (\theta)$ so $\mathrm{f}^{\prime}(\theta)=2000 \cdot \sec ^{2}(\theta).$ We know $4000=2000 \cdot \tan (\theta)$ so

$\tan (\theta)=2$ and $\theta \approx 1.10715$ (radians). $\mathrm{f}^{\prime}(\theta)=2000 \cdot \sec ^{2}(\theta)$ so $\mathrm{f}^{\prime}(1.10715)=2000 \cdot \sec ^{2}(1.10715) \approx 10,000$.

Finally, "error" $\Delta \mathrm{f} \approx \mathrm{f}^{\prime}(\theta) \cdot \Delta \theta$ so $10 \approx 10,000 \cdot \Delta \theta$ and $\Delta \theta \approx 10 / 10,000=\mathbf{0. 0 0 1}($ radians $) \approx 0.057^{\circ}$.

Practice 8: $\mathrm{A}(\mathrm{r})=6 \mathrm{r}^{2}$ so $\mathrm{A}^{\prime}(\mathrm{r})=12 \mathrm{r}$ and $\Delta \mathrm{A} \approx \mathrm{A}^{\prime}(\mathrm{r}) \cdot \Delta \mathrm{r}=12 \mathrm{r} \cdot \Delta \mathrm{r}.$ We are also told that $\Delta \mathrm{r} / \mathrm{r} < 0.03$ Percentage error is $\frac{\Delta \mathrm{A}}{\mathrm{A}} \cdot 100=\frac{12 \mathrm{r} \cdot \Delta \mathrm{r}}{6 \mathrm{r}^{2}} \cdot 100=\frac{2 \cdot \Delta \mathrm{r}}{\mathrm{r}} \cdot 100 < 200 \cdot(0.03)=\mathbf{6}$.

Practice 9: $\begin{aligned} &\mathrm{f}(\mathrm{x})=\ln (\mathrm{x}) \qquad \mathrm{df}=\mathrm{f}^{\prime}(\mathrm{x}) \cdot \mathrm{dx}=\frac{1}{\mathbf{x}} \mathrm{d} \mathrm{x} \\ &\mathrm{u}=\sqrt{1-3 \mathrm{x}} \qquad \mathrm{du}=\frac{\mathrm{du}}{\mathrm{dx}} \cdot \mathrm{dx}=\frac{-3}{2 \sqrt{1-3 \mathrm{x}}} \cdot \mathrm{dx} \\ &\mathrm{r}=3 \cos (\theta) \mathrm{dr}=\frac{\mathrm{dr}}{\mathrm{d} \theta} \cdot \mathrm{d} \theta=-\mathbf{3} \sin (\boldsymbol{\theta}) \cdot \mathrm{d} \theta \end{aligned}$

Practice 10: $f(x)=x^{3}, f^{\prime}(x)=3 x^{2}$, and $L(4+\Delta x)=f(4)+f^{\prime}(4) \Delta x=4^{3}+3(4)^{2} \Delta x=64+48 \cdot \Delta x$.

$\begin{array}{l|l|l|l} { }_­ \Delta \mathrm{x} & \mathrm{f}(4+\Delta \mathrm{x}) & \mathrm{L}(4+\Delta \mathrm{x}) & \mathrm{f}(4+\Delta \mathrm{x})-\mathrm{L}(4+\Delta \mathrm{x}) \\ \hline 0.1 & 68.921 & 68.8 & 0.121 \\ 0.05 & 66.430125 & 66.4 & 0.030125 \\ 0.01 & 64.481201 & 64.48 & 0.001201 \\ 0.001 & 64.048012 & 64.048 & 0.000012 \end{array}$

$\mathrm{f}(4+\Delta \mathrm{x})$ is the actual volume of the cube with side length $4+\Delta \mathrm{x}$.

$\mathrm{L}(4+\Delta \mathrm{x})$ is the volume of the cube with side length $4(\mathrm{v}=64)$ plus the volume of the 3 "slabs" $\left(\mathrm{v}=3 \cdot 4^{2} \cdot \Delta \mathrm{x}\right)$

$|f(4+\Delta x)-L(4+\Delta x)|$ is the volume of the "leftover" pieces from $L$ : the 3 "rods" $\left(v=3 \cdot 4 \cdot(\Delta x)^{2}\right)$ and the tiny cube $\left(\mathrm{v}=(\Delta \mathrm{x})^{3}\right)$.