Practice Problems

1. An expandable sphere is being filled with liquid at a constant rate from a tap (imagine a water balloon connected to a faucet). When the radius of the sphere is 3 inches, the radius is increasing at 2 inches per minute. How fast is the liquid coming out of the tap? $\left(\mathrm{V}=\frac{4}{3} \pi \mathrm{r}^{3}\right)$

3. One hour later the right triangle in Problem 2 is 15 inches long and 13 inches high

(Fig. 9), and the base and height are changing at the same rate as in Problem 2.

(a) Is the area increasing or decreasing now?

(b) Is the hypotenuse increasing or decreasing now?

(c) Is the perimeter increasing or decreasing now?

5. The length of a 12 foot by 8 foot rectangle is increasing at a rate of 3 feet per second and the width is decreasing at 2 feet per second (Fig. 11). 

(a) How fast is the perimeter changing?

(b) How fast is the area changing?

7. An oil tanker in Puget Sound has sprung a leak, and a circular oil slick is forming (Fig. 13). The oil slick is 4 inches thick everywhere, is 100 feet in diameter, and the diameter is increasing at 12 feet per hour. Your job, as the Coast Guard commander or the tanker's captain, is to determine how fast the oil is leaking from the tanker.

9. Lava flowing from a hole at the top of a hill is forming a conical mountain whose height is always the same as the width of its base (Fig. 15). If the mountain is increasing in height at 2 feet per hour when it is 500 feet high, how fast is the lava flowing (how fast is the volume of the mountain increasing)? $\left(\mathrm{V}=\frac{1}{3} \pi \, \mathrm{r}^{2} \mathrm{~h}\right.$)

11. Answer parts (a) and (b) in Problem 10 for when the person is 20 feet from the lamp post.

13. The string of a kite is perfectly taut and always makes an angle of $\mathrm{}^{\circ}$ above horizontal (Fig. 18).

(a) If the kite flyer has let out 500 feet of string, how high is the kite?

(b) If the string is let out at a rate of 10 feet per second, how fast is the kite's height increasing?

15. The 8 foot diameter of a spherical gas bubble is increasing at 2 feet per hour, and the 12 foot long edges of a cube containing the bubble are increasing at 3 feet per hour. Is the volume contained between the spherical bubble and the cube increasing or decreasing? At what rate?

17. The snow in a hemispherical pile melts at a rate proportional to its exposed surface area (the surface area of the hemisphere). Show that the height of the snow pile is decreasing at a constant rate.

19. Define $\mathrm{A}(\mathrm{x})$ to be the area bounded by the $\mathrm{x}$ and y axes, the horizontal line $\mathrm{y}=5$, and a vertical line at $\mathrm{x}$ (Fig. 20).

(a) Find a formula for $\mathrm{A}$ as a function of $x$.

(b) Determine $\frac{\mathrm{d} \mathrm{A}(\mathrm{X})}{\mathrm{dx}}$ when $\mathrm{x}=\mathrm{1, 2, 4, \, and \, 9}$.

21. You are walking along a sidewalk toward a 40 foot wide sign which is adjacent to the sidewalk and perpendicular to it (Fig. 22).

(a) If your viewing angle $\theta$ is $10^{\circ}$, then how far are you from the nearest corner of the sign?

(b) If your viewing angle is $10^{\circ}$ and you are walking at 25 feet per minute, then how fast is your viewing angle changing?

(c) If your viewing angle is $10^{\circ}$ and is increasing at $2^{0}$ per minute, then how fast are you walking?

Source: Dale Hoffman, https://s3.amazonaws.com/saylordotorg-resources/wwwresources/site/wp-content/uploads/2012/12/MA005-3.7-Related-Rates.pdf
This work is licensed under a Creative Commons Attribution 3.0 License.

1. $\mathrm{V}=\frac{4}{3} \pi \mathrm{r}^{3}(\mathrm{r}=\mathrm{r}(\mathrm{t}))$ so $\frac{\mathrm{d} \mathrm{V}}{\mathrm{dt}}=4 \pi \mathrm{r}^{2} \frac{\mathrm{d} \mathrm{r}}{\mathrm{dt}}$.

When $\mathrm{r}=3$ in.,$\frac{\mathrm{d} \mathrm{r}}{\mathrm{dt}}=2 \, \mathrm{in} / \mathrm{min}$, so $\frac{\mathrm{d} \mathrm{V}}{\mathrm{dtt}}=4 \pi(3 \mathrm{in})^{2}(2 \, \mathrm{in} / \mathrm{min})=72 \pi \mathrm{in}^{3} / \mathrm{min} \approx 226.19 \, \mathrm{in}^{3} / \mathrm{min}$.

3. $\mathrm{b}=15$ in., $\mathrm{h}=13$ in., $\frac{\mathrm{d} \mathrm{b}}{\mathrm{dt}}=3 \, \mathrm{in} / \mathrm{hr}, \frac{\mathrm{d} \mathrm{h}}{\mathrm{dt}}=-3 \, \mathrm{in} / \mathrm{hr}$.

(a) $\mathrm{A}=\frac{1}{2}$ bh so $\frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}=\frac{1}{2}\left\{\mathrm{~b} \frac{\mathrm{d} \mathrm{h}}{\mathrm{dt}}+\mathrm{h} \frac{\mathrm{d} \mathrm{b}}{\mathrm{dt}}\right\}=\frac{1}{2}\{(15 \, \mathrm{in})(-3 \, \mathrm{in} / \mathrm{hr})+(13 \, \mathrm{in})(3 \, \mathrm{in} / \mathrm{hr})\} < 0$ so $\mathrm{A}$ is decreasing.

(b) Hypotenuse $\mathrm{C}=\sqrt{\mathrm{b}^{2}+\mathrm{h}^{2}}$ so $\frac{\mathrm{d} \mathrm{C}}{\mathrm{dt}}=\frac{\mathrm{b} \frac{\mathrm{d} \mathrm{b}}{\mathrm{dt}}+\mathrm{h} \frac{\mathrm{d} \mathrm{h}}{\mathrm{dt}}}{\sqrt{\mathrm{b}^{2}+\mathrm{h}^{2}}}=\frac{15(3)+13(-3)}{\sqrt{15^{2}+13^{2}}} > 0$ so $\mathrm{C}$ is increasing.

(c) Perimeter $\mathrm{P}=\mathrm{b}+\mathrm{h}+\mathrm{C}$ so $\frac{\mathrm{d} \mathrm{P}}{\mathrm{dt}}=\frac{\mathrm{d} \mathrm{b}}{\mathrm{dt}}+\frac{\mathrm{d} \mathrm{h}}{\mathrm{dt}}+\frac{\mathrm{d} \mathrm{C}}{\mathrm{dt}}=(3)+(-3)+\frac{6}{\sqrt{394}} > 0$ so $\mathrm{P}$ is increasing.

5. (a) $\mathrm{P}=2 \mathrm{x}+2 \mathrm{y}$ so $\frac{\mathrm{d} \mathrm{P}}{\mathrm{dt}}=2 \frac{\mathrm{d} \mathrm{x}}{\mathrm{dt}}+2 \frac{\mathrm{dy}}{\mathrm{dt}}=2(3 \, \mathrm{ft} / \mathrm{sec})+2(-2 \, \mathrm{ft} / \mathrm{sec})=2 \, \mathrm{ft} / \mathrm{sec}$.

(b) $A=x y$ so $\frac{\mathrm{d} A}{d t}=x \frac{d y}{d t}+y \frac{d x}{d t}=(12 \, \mathrm{ft})(-2 \, \mathrm{ft} / \mathrm{sec})+(8 \mathrm{ft})(3 \, \mathrm{ft} / \mathrm{sec})=0 \, \mathrm{ft}^{2} / \mathrm{sec}$.

7. $\mathrm{V}=\pi \mathrm{r}^{2} \mathrm{~h}=\pi \mathrm{r}^{2}(1 / 3)$ so $\frac{\mathrm{d} \mathrm{V}}{\mathrm{dt}}=\frac{2 \pi}{3} \mathrm{r} \frac{\mathrm{dr}}{\mathrm{dt}}$.

When $\mathrm{r}=50 \mathrm{ft}$. and $\frac{\mathrm{dr}}{\mathrm{dt}}=6 \, \mathrm{ft} / \mathrm{hr}$, then $\frac{\mathrm{d} \mathrm{V}}{\mathrm{dt}}=\frac{2 \pi}{3}(50 \mathrm{ft})(6 \, \mathrm{ft} / \mathrm{hr})=200 \pi \, \mathrm{ft}^{3} / \mathrm{hr} \approx 628.32 \mathrm{ft}^{3} / \mathrm{hr}$.

9. $\mathrm{w}(\mathrm{t})=\mathrm{h}(\mathrm{t})$ for all $\mathrm{t}$ so $\frac{\mathrm{d} \mathrm{w}}{\mathrm{dt}}=\frac{\mathrm{d} \mathrm{h}}{\mathrm{dt}} \cdot \mathrm{V}=\frac{1}{3} \pi r^{2} \mathrm{~h}$ and $\mathrm{r}=\mathrm{w} / 2=\mathrm{h} / 2$ so $\mathrm{V}=\frac{1}{3} \pi(\mathrm{h} / 2)^{2} \mathrm{~h}=\frac{1}{12} \pi \mathrm{h}^{3}$ and $\frac{\mathrm{d} \mathrm{V}}{\mathrm{dt}}=\frac{1}{4} \pi \mathrm{h}^{2} \frac{\mathrm{dh}}{\mathrm{dt}}.$ When $\mathrm{h}=500 \, \mathrm{ft}$ and $\frac{\mathrm{dh}}{\mathrm{dt}}=2 \, \mathrm{ft} / \mathrm{hr}$, then $\frac{\mathrm{d} \mathrm{V}}{\mathrm{dt}}=\frac{1}{4} \pi(500)^{2}(2)=125,000 \, \pi \, \mathrm{ft}^{3} / \mathrm{hr}$.

11. Let $x$ be the distance from the lamp post to the person, and $L$ be the length of the shadow, both in feet. By similar triangles, $\frac{\mathrm{L}}{6}=\frac{\mathrm{x}}{8} \quad$ so $\mathrm{L}=\frac{3}{4} \mathrm{x} \cdot \frac{\mathrm{d} \mathrm{x}}{\mathrm{dt}}=3 \, \mathrm{ft} / \mathrm{sec}$.

(a) $\frac{\mathrm{d} \mathrm{L}}{\mathrm{dt}}=\frac{3}{4} \frac{\mathrm{dx}}{\mathrm{dt}}=\frac{3}{4}(3 \, \mathrm{ft} / \mathrm{sec})=2.25 \, \mathrm{ft} / \mathrm{sec}$.

(The value of $\mathrm{x}$ does not enter into the calculations).

(b) $\frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{x}+\mathrm{L})=\frac{\mathrm{dx}}{\mathrm{dt}}+\frac{\mathrm{d} \mathrm{L}}{\mathrm{dt}}=5.25 \, \mathrm{ft} / \mathrm{sec}$.

13. (a) $\sin \left(35^{\circ}\right)=\frac{h}{500} \quad$ so $\mathrm{h}=500 \cdot \sin \left(35^{\circ}\right) \approx 287 \, \mathrm{ft}$.

(b) $\mathrm{L}=$ length of the string so $\mathrm{h}=\mathrm{L} \cdot \sin \left(35^{\circ}\right)$ and $\frac{\mathrm{d} \mathrm{h}}{\mathrm{dt}}=\sin \left(35^{\mathrm{O}}\right) \frac{\mathrm{d} \mathrm{L}}{\mathrm{dt}}=\sin \left(35^{\mathrm{O}}\right)(10 \, \mathrm{ft} / \mathrm{sec}) \approx 5.7 \, \mathrm{ft} / \mathrm{sec}$.

15. $\mathrm{V}=\mathrm{s}^{3}-\frac{4}{3} \pi \mathrm{r}^{3} \cdot \mathrm{r}=\frac{1}{2}($ diameter $)=4 \, \mathrm{ft}, \frac{\mathrm{dr}}{\mathrm{dt}}=1 \, \mathrm{ft} / \mathrm{hr}, \mathrm{s}=12 \, \mathrm{ft}, \frac{\mathrm{ds}}{\mathrm{dt}}=3 \, \mathrm{ft} / \mathrm{hr}$.

$\frac{\mathrm{d} \mathrm{V}}{\mathrm{dt}}=3 \mathrm{~s}^{2} \frac{\mathrm{d} \mathrm{s}}{\mathrm{dt}}-4 \pi \mathrm{r}^{2} \frac{\mathrm{dr}}{\mathrm{dt}}=3(12 \, \mathrm{ft})^{2}$ (3 $\left.\mathrm{ft} / \mathrm{hr}\right)-4 \pi(4 \, \mathrm{ft})^{2}(1 \, \mathrm{ft} / \mathrm{hr}) \approx 1094.94 \, \mathrm{ft}^{3} / \mathrm{hr}$. The volume is increasing at about $1094.94 \, \mathrm{ft}^{3} / \mathrm{hr}$.

17. Given: $\quad \frac{\mathrm{d} \mathrm{V}}{\mathrm{dt}}=\mathrm{k} \cdot 2 \pi \mathrm{r}^{2}$ with $\mathrm{k}$ constant. We also have $\mathrm{V}=\frac{2}{3} \pi \mathrm{r}^{3}$ so $\frac{\mathrm{dV}}{\mathrm{dt}}=2 \pi \mathrm{r}^{2} \frac{\mathrm{dr}}{\mathrm{dt}}$.

Therefore, $\mathrm{k} \cdot 2 \pi \mathrm{r}^{2}=2 \pi \mathrm{r}^{2} \frac{\mathrm{dr}}{\mathrm{dt}}$ so $\frac{\mathrm{dr}}{\mathrm{dt}}=\mathrm{k}$. The radius $\mathrm{r}$ is changing at a constant rate.

19. (a) $\mathrm{A}=5 \mathrm{x}$

(b) $\frac{\mathrm{d} \mathrm{A}}{\mathrm{dx}}=5$ for all $\mathrm{x} > 0$.

(c) $\mathrm{A}=5 \mathrm{t}^{2}$

(d) $\frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}=10 \mathrm{t}.$ When $\mathrm{t}=1, \frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}=10 ;$ when $\mathrm{t}=2, \frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}=20 ;$ when $\mathrm{t}=3, \frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}=30$.

(e) $\mathrm{A}=10+5 \cdot \sin (\mathrm{t}) \cdot \frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}=5 \cdot \cos (\mathrm{t})$.

21. (a) $\tan \left(10^{\circ}\right)=\frac{40}{x}$ so $x=\frac{40}{\tan \left(10^{\circ}\right)} \approx 226.9 \mathrm{ft}$.

For parts (b) and (c) we need to work in radians since our formulas for the derivatives of the trigonometric functions assume that the angles are measured in radians: $360^{\circ} \approx 2 \pi$ radians so $10^{\circ} \approx 0.1745$ radians and $2^{\mathrm{O}} \approx 0.0349$ radians,

(b) $\mathrm{x}=\frac{40}{\tan (\theta)}=40 \cot (\theta)$ so $\frac{\mathrm{dx}}{\mathrm{dt}}=-40 \csc ^{2}(\theta) \frac{\mathrm{d} \theta}{\mathrm{dt}}$ and

$\frac{\mathrm{d} \theta}{\mathrm{dt}}=\frac{\sin ^{2}(\theta) \frac{\mathrm{d} \mathrm{x}}{\mathrm{dt}}}{-40}=\frac{\sin ^{2}(0.1745)(-25)}{-40} \approx \frac{(0.1736)^{2}(-25)}{-40} \approx 0.0188$ radians/min $\approx 1.079^{\circ} / \mathrm{min}.$

$\frac{\mathrm{d} \mathrm{x}}{\mathrm{dt}}-40 \csc ^{2}(\theta) \frac{\mathrm{d} \theta}{\mathrm{dt}}=-40 \frac{1}{\sin ^{2}(\theta)} \frac{\mathrm{d} \theta}{\mathrm{dt}} \approx-40 \frac{1}{(0.1736)^{2}}(0.0349) \approx-46.3 \, \mathrm{ft} / \mathrm{min}.$ (The " - " indicates the distance to the sign is decreasing: you are approaching the sign). Your speed is $46.3 \, \mathrm{ft} / \mathrm{min}$.