Practice Problems

1. What is the slope of the line through $(3,9)$ and $(x, y)$ for $y=x^{2}$ and $x=2.97$? $x=3.001$? $\mathrm{x}=3+\mathrm{h}$? What happens to this last slope when $\mathrm{h}$ is very small (close to 0)? Sketch the graph of $y=x^{2}$ for $x$ near $3$.

3. What is the slope of the line through $(2,4)$ and $(x, y)$ for $y=x^{2}+x-2$ and $x=1.99$? $\mathrm{x}=2.004$? $\mathrm{x}=2+\mathrm{h}$? What happens to this last slope when $\mathrm{h}$ is very small? Sketch the graph of $y=x^{2}+x-2$ for $x$ near $2$.

5. Fig. 9 shows the temperature during a day in Ames.

(a) What was the average change in temperature from 9 am to 1 pm?

(b) Estimate how fast the temperature was rising at 10 am and at 7 pm?

Problem 9 defines new functions $A(x)$ in terms of AREAS bounded by the functions $y=3$ and $y=x+1$. This may seem a strange way to define a functions $A(x)$, but this idea will become important later in calculus. We are just getting an early start here.

9. Define $A(x)$ to be the area bounded by the $x$ and y axes, the horizontal line $\mathrm{y}=3$, and the vertical line at $\mathrm{x}$ (Fig. 13). For example, $\mathrm{A}(4)=12$ is the area of the 4 by 3 rectangle.

1. $m=\frac{y-9}{x-3}. \quad$ If $x=2.97$, then $m=\frac{-0.1791}{-0.03}=5.97.$ If $x=3.001$, then $m=\frac{0.006001}{0.001}=6.001$.

If $x=3+h$, then $m=\frac{(3+h)^{2}-9}{(3+h)-3}=\frac{9+6 h+h^{2}-9}{h}=6+h$. When $h$ is very small (close to 0$)$, $6+\mathrm{h}$ is very close to 6.

3. $\mathrm{m}=\frac{\mathrm{y}-\mathbf{4}}{\mathrm{x}-2}. \quad$ If $\mathrm{x}=1.99$, then $\mathrm{m}=\frac{-0.0499}{-0.01}=4.99.$ If $\mathrm{x}=2.004$, then $\mathrm{m}=\frac{0.020016}{0.004}=5.004$.

If $x=2+h$, then $m=\frac{\left\{(2+h)^{2}+(2+h)-2\right\}-4}{(2+h)-2}=\frac{4+4 h+h^{2}+2+h-2-4}{h}=5+h.$ When $\mathrm{h}$ is very small, $5+\mathrm{h}$ is very close to 5.

5. All of these answers are approximate. Your answers should be close to these numbers.

(a) average rate of temperature change $\approx \frac{80^{\circ}-64^{\circ}}{1 \mathrm{pm}-9 \mathrm{am}}=\frac{16^{\mathrm{o}}}{4 \text { hours }}=4^{\mathrm{o}}$ per hour.

(b) at 10am, temperature was rising about $5^{\circ}$ per hour.

at 7pm, temperature was rising about $-10^{\circ}$ per hour (falling about $10^{\circ}$ per hour).

7. All of these answers are approximate. Your answers should be close to these numbers.

(a) average velocity $\approx \frac{300 \mathrm{ft}-0 \mathrm{ft}}{20 \sec -0 \mathrm{sec}}=15$ feet per second.

(b) average velocity $\approx \frac{100 \mathrm{ft}-200 \mathrm{ft}}{30 \sec -10 \mathrm{sec}}=-5$ feet per second.

(c) at $\mathrm{t}=10$ seconds, velocity $\approx 30$ feet per second (between 20 and $35 \mathrm{ft} / \mathrm{s})$.

at $\mathrm{t}=20$ seconds, velocity $\approx-1$ feet per second.

at $\mathrm{t}=30$ seconds, velocity $\approx-40$ feet per second.

9. (a) $\mathrm{A}(0)=0, \mathrm{~A}(1)=3, \mathrm{~A}(2)=6, \mathrm{~A}(2.5)=7.5, \mathrm{~A}(3)=9$.

(b) the area of the rectangle bounded below by the $\mathrm{x}$-axis, above by the line $\mathrm{y}=3$, on the left by the vertical line $\mathrm{x}=1$, and on the right by the vertical line $\mathrm{x}=4$.

(c) Graph of $\mathrm{y}=\mathrm{A}(\mathrm{x})=3 \mathrm{x}$.