MA005: Calculus I

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 $46.3 \, \mathrm{ft} / \mathrm{min}$.