Practice Problems

1. Fig. 10 shows the tangent line to a function $\mathrm{g}$ at the point $(2,2)$ and a line segment $\Delta \mathrm{x}$ units long. On the figure, label the locations of 

(a) $2+\Delta x$ on the $x$-axis, (b) the point $(2+\Delta x, g(2+\Delta x))$, and

(c) the point $\left(2+\Delta x, g(2)+g^{\prime}(2) \cdot \Delta x\right)$.

(d) How large is the "error", $\left(g(2)+g^{\prime}(2) \cdot \Delta x\right)-(g(2+\Delta x))?$

In problem 3, find the equation of the tangent line $L$ to the given function $f$ at the given point $(\mathrm{a}, \mathrm{f}(\mathrm{a}))$. Use the value $\mathrm{L}(\mathrm{a}+\Delta \mathrm{x})$ to approximate the value of $\mathrm{f}(\mathrm{a}+\Delta \mathrm{x})$

3. (a) $\mathrm{f}(\mathrm{x})=\sqrt{\mathrm{x}}, \mathrm{a}=4, \Delta \mathrm{x}=0.2$

(b) $\mathrm{f}(\mathrm{x})=\sqrt{\mathrm{x}}, \mathrm{a}=81, \Delta \mathrm{x}=-1$

(c) $f(x)=\sin (x), a=0, \Delta x=0.3$

In problem 7, use the Linear Approximation Process to derive each approximation formula for $\mathrm{x}$ "close to" $\mathbf{0}$.

7. (a) $\ln (1+x) \approx x$

(b) $\cos (\mathrm{x}) \approx 1$

(c) $\tan (\mathrm{x}) \approx \mathrm{x}$

(d) $\sin (\pi / 2+x) \approx 1$

9. A rectangle has one side on the $\mathrm{x}$-axis, one side on the y-axis, and a corner on the graph of $y=x^{2}+1$ (Fig. 13).

(a) Use Linear Approximation of the area formula to estimate the increase in the area of the rectangle if the base grows from 2 to 2.3 inches.

(b) Calculate exactly the increase in the area of the rectangle as the base grows from 2 to 2.3 inches.

11. You are minting gold coins which must have a volume of $47.3 \pm 0.1 \mathrm{~cm}^{3}$. If you can manufacture the coins to be exactly $2 \mathrm{~cm}$ high, how much variation can you allow for the radius?

13. Your company is making dice (cubes) and the specifications require that their volume be $87 \pm 2 \mathrm{~cm}^{3}$. How long should each side be and how much variance can a side have in order to meet the specifications?

15. The period $\mathrm{P}$, in seconds, for a pendulum to make one complete swing and return to the release point is $\mathrm{P}=2 \pi$ $\sqrt{\mathrm{L} / \mathrm{g}}$ where $\mathrm{L}$ is the length of the pendulum in feet and $\mathrm{g}$ is $32 \mathrm{feet} / \mathrm{sec}^{2}$.

(a) If $L=2$ feet, what is the period of the pendulum?

(b) If $P=1$ second, how long is the pendulum?

(c) Estimate the change in $\mathrm{P}$ if $\mathrm{L}$ increases from $\mathrm{2}$ feet to $\mathrm{2.1}$ feet.

(d) The length of a $\mathrm{24}$ foot pendulum is increasing $\mathrm{2}$ inches per hour. Is the period getting longer or shorter? How fast is the period changing?

17. For the function in Fig. 14, estimate the value of $\mathbf{d f}$ when

(a) $\mathrm{x}=2$ and $\mathrm{dx}=1$

(b) $\mathrm{x}=4$ and $\mathrm{dx}=-1$

(c) $x=3$ and $d x=2$

19. Calculate the differentials df of the following functions:

(a) $f(x)=x^{2}-3 x$

(b) $f(x)=e^{x}$

(c) $f(x)=\sin (5 x)$

(d) $f(x)=x^{3}+2 x$ with $x=1$ and $d x=0.2$

(e) $\mathrm{f}(\mathrm{x})=\ln (\mathrm{x})$ with $\mathrm{x}=\mathrm{e}$ and $\mathrm{dx}=-0.1$

(f) $\mathrm{f}(\mathrm{x})=\sqrt{2 \mathrm{x}+5}$ with $\mathrm{x}=22$ and $\mathrm{dx}=3$.

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.

1. See Fig. 1.

3. (a) $\mathrm{f}(4)=2, \mathrm{f}^{\prime}(4)=\frac{1}{2 \sqrt{4}}=\frac{1}{4}$.

Then $y-2=\frac{1}{4}(x-4)$ so $y=\frac{1}{4} x+1$ $\sqrt{4.2}=\mathrm{f}(4.2) \approx \frac{1}{4}(4.2)+1=2.05$.

(b) $\mathrm{f}(81)=9, \mathrm{f}^{\prime}(81)=\frac{1}{18} \mathrm{so}$

$\begin{aligned} &\mathrm{y}-9=\frac{1}{18}(\mathrm{x}-81) \text { so } \mathrm{y}=\frac{1}{18}(\mathrm{x}-81)+9 \\ &\sqrt{80}=\mathrm{f}(80) \approx \frac{1}{18}(80-81)+9=9-\frac{1}{18} \approx 8.944\end{aligned}$

(c) $f(0)=0, f^{\prime}(0)=1$ so $y-0=1(x-0)$ and $y=x.$ Then $\sin (0.3)=f(0.3) \approx 0.3$.

5. $f(x)=(1+x)^{n}, f^{\prime}(x)=n(1+x)^{n-1}, f(0)=1$ and $f^{\prime}(0)=n$. Then $y-1=n(x-0)$ or $y=1+n x$. Therefore, $(1+x)^{n} \approx 1+n x$ (when $x$ is close to $\mathrm{0}$).

7. (a) $\mathrm{f}(\mathrm{x})=\ln (1+\mathrm{x}), \mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{1+\mathrm{x}}$, and $\mathrm{f}^{\prime}(0)=1$. Then $\mathrm{y}-0=1(\mathrm{x}-0)$ so $\mathrm{y}=\mathrm{x}$ and $\ln (1+\mathrm{x}) \approx \mathrm{x}$.

(b) $\mathrm{f}(\mathrm{x})=\cos (\mathrm{x}), \mathrm{f}^{\prime}(\mathrm{x})=-\sin (\mathrm{x})$, and $\mathrm{f}^{\prime}(0)=0$. Then $\mathrm{y}-1=0(\mathrm{x}-0)$ so $\mathrm{y}=1$ and $\cos (\mathrm{x}) \approx 1$.

(c) $f(x)=\tan (x), f^{\prime}(x)=\sec ^{2}(x)$, and $f^{\prime}(0)=1$. Then $y-0=1(x-0)$ so $y=x$ and $\tan (x) \approx x$.

(d) $f(x)=\sin \left(\frac{\pi}{2}+x\right), f^{\prime}(x)=\cos \left(\frac{\pi}{2}+x\right)$, and $f^{\prime}(0)=\cos \left(\frac{\pi}{2}+0\right)=0$. Then $y-1=0(x-0)$ so $y=1$ and $\sin \left(\frac{\pi}{2}+x\right) \approx 1$.

9. (a) Area $A(x)=($ base $)$ (height $)=x\left(x^{2}+1\right)=x^{3}+x$. Then $A^{\prime}(x)=3 x^{2}+1$ so $A^{\prime}(2)=13$ Then $\quad \Delta \mathrm{A} \approx \mathrm{A}^{\prime}(2) \cdot \Delta \mathrm{x}=(13)(2.3-2)=3.9$.

(b) (base)(height) $=(2.3)\left((2.3)^{2}+1\right)=14.467$. Then the actual difference $=14.467-(2)\left(2^{2}+1\right)=4.467$.

11. $\mathrm{V}=\pi r^{2} \mathrm{~h}=2 \pi \mathrm{r}^{2}$ and $\Delta \mathrm{V}=2 \pi 2 \mathrm{r} \Delta \mathrm{r}=4 \pi \mathrm{r} \Delta \mathrm{r}$. Since $\mathrm{V}=2 \pi \mathrm{r}^{2}=47.3$, we have $\mathrm{r}=\sqrt{47.3 /(2 \pi)} \approx 2.7437 \mathrm{~cm}$. We know $\Delta \mathrm{V}=\pm 0.1$ so, using $\Delta \mathrm{V}=4 \pi \mathrm{r} \Delta \mathrm{r}$, we have $\pm 0.1=4 \pi(2.7437) \Delta \mathrm{r}$ and $\Delta r=\frac{\pm 0.1}{4 \pi(2.7437)} \approx \pm 0.0029 \mathrm{~cm}$. The required tolerance is $\pm 0.0029 \mathrm{~cm}.$ (C.W. comment: "A coin $2 \mathrm{~cm}$ high is one hell of a coin! My eyeball estimate is that $47.3 \mathrm{~cm}^{3}$ of gold weighs around 2 pounds").

13. $\mathrm{V}=\mathrm{x}^{3}$. When $\mathrm{V}=87, \mathrm{x} \approx 4.431 \mathrm{~cm}. \Delta \mathrm{V}=3 \mathrm{x}^{2} \Delta \mathrm{x}$ so $\Delta \mathrm{x}=\frac{\Delta \mathrm{V}}{3 \mathrm{x}^{2}} \approx \frac{2}{3(4.431)^{2}}=0.034 \mathrm{~cm}$.

15. $\mathrm{P}=2 \pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}}}$ with $\mathrm{g}=32 \mathrm{ft} / \mathrm{sec}^{2}$.

(a) $\mathrm{P}=2 \pi \sqrt{\frac{2}{32}}=\frac{\pi}{2} \approx 1.57$ seconds.

(b) $1=2 \pi \sqrt{\frac{L}{32}}$ so $L=\frac{8}{\pi^{2}} \approx 0.81$ feet.

(c) $\mathrm{dP}=\frac{2 \pi}{\sqrt{32}} \frac{1}{2 \sqrt{2}} \mathrm{dL}=\frac{2 \pi}{4 \sqrt{2}} \frac{1}{2 \sqrt{2}}(0.1) \approx 0.039$ seconds.

(d) $2 \mathrm{in} / \mathrm{hr}=1 / 6 \mathrm{ft} / \mathrm{hr}=\frac{1}{21600} \mathrm{ft} / \mathrm{sec}. \frac{\mathrm{dP}}{\mathrm{dt}}=\frac{2 \pi}{4 \sqrt{2}} \frac{1}{2 \sqrt{24}} \frac{1}{21600} \approx 5.25 \times 10^{-6}$.

17. (a) $\mathrm{df}=\mathrm{f}^{\prime}(2) \mathrm{dx} \approx(0)(1)=0$

(b) $\mathrm{df}=\mathrm{f}^{\prime}(4) \mathrm{dx} \approx(0.3)(-1)=-0.3$

(c) $\mathrm{df}=\mathrm{f}^{\prime}(3) \mathrm{dx} \approx(0.5)(2)=1$

19. (a) $f(x)=x^{2}-3 x. f^{\prime}(x)=2 x-3. d f=(2 x-3) d x$.

(b) $f(x)=e^{x} \cdot f^{\prime}(x)=e^{x} \cdot d f=e^{x} d x$.

(c) $\mathrm{f}(\mathrm{x})=\sin (5 \mathrm{x}). \mathrm{f}^{\prime}(\mathrm{x})=5 \cos (5 \mathrm{x}) \cdot \mathrm{df}=5 \cos (5 \mathrm{x}) \mathrm{dx}$.

(d) $f(x)=x^{3}+2 x \cdot f^{\prime}(x)=3 x^{2}+2. d f=\left(3 x^{2}+2\right) d x.$ When $x=1$ and $d x=0.2, d f=\left(3 \cdot 1^{2}+2\right)(0.2)=1$.

(e) $\mathrm{f}^{\prime}(\mathrm{x})=1 / \mathrm{x} \cdot \mathrm{df}=\frac{1}{\mathrm{x}} \mathrm{dx}$. When $\mathrm{x}=\mathrm{e}$ and $\mathrm{dx}=-0.1, \mathrm{df}==\frac{1}{\mathrm{e}}(-0.1)=-\frac{1}{10 \mathrm{e}}$.

(f) $f(x)=\sqrt{2 x+5} \cdot f^{\prime}(x)=\frac{1}{\sqrt{2 x+5}} \cdot d f=\frac{1}{\sqrt{2 x+5}} d x.$ When $x=22$ and $d x=3, d f=\frac{1}{\sqrt{49}}(3)=\frac{3}{7}$.