- An average rate uses a total change over a time interval. - An instantaneous rate describes what is happening at one moment. - Keep the direction as part of each position rate.

1. The average rate of change of position is \(\frac{120\,\text{mi}}{2\,\text{h}}=60\,\text{mph}\) east. 2. The stated velocity \(55\,\text{mph}\) east describes the rate of change of position at one instant. 3. Therefore, \(60\,\text{mph}\) east is the average rate and \(55\,\text{mph}\) east is the instantaneous rate.

Average rate of change of position: \(60\,\text{mph}\) east. Instantaneous rate of change of position: \(55\,\text{mph}\) east.

- Average rate of change is change in output divided by change in input. - Use the later height minus the earlier height. - Use hours for the elapsed time so the rate has units of centimeters per hour.

1. The change in height is \(30-18=12\,\text{cm}\). 2. The elapsed time is \(2\,\text{h}\). 3. The average rate of change is \(\frac{12\,\text{cm}}{2\,\text{h}}=6\,\text{cm/h}\).

\(6\,\text{cm/h}\)

- Which temperature was measured first, and which was measured last? - Think about how a negative temperature change affects the sign of the rate.

1. Divide the temperature change by the time: \(\frac{-14-(-8)}{3}=\frac{-6}{3}=-2\). 2. The average rate is \(-2\,^\circ\text{C}\) per hour. The negative sign means the temperature fell.

\(-2\,^\circ\text{C}\) per hour; the temperature decreased.

- A secant's slope is found just like the slope of any line through two points. - Keep the order of the coordinates consistent in the numerator and denominator.

1. A secant joins the two points on the curve. 2. Its slope is \(\frac{11-3}{5-1}=\frac{8}{4}=2\).

The secant slope is \(2\).

- How can you find the two function values needed for an average rate? - Use the change in output divided by the change in input.

1. Evaluate the endpoints: \(f(1)=1\) and \(f(3)=9\). 2. The average rate is \(\frac{f(3)-f(1)}{3-1}=\frac{9-1}{2}=4\).

\(4\)

- Rearrange the average-rate relationship to recover total change. - The interval length is the difference between the two times. - Interpret the sign using the direction of the volume change.

1. The time interval has length \(7-4=3\,\text{min}\). 2. Total change equals average rate times elapsed time: \((-2.5\,\text{gal/min})(3\,\text{min})=-7.5\,\text{gal}\). 3. The negative sign means the tank contains less water at \(t=7\) than at \(t=4\); the volume decreased by \(7.5\,\text{gal}\).

a) \(-7.5\,\text{gal}\) b) The tank's volume decreased by \(7.5\,\text{gal}\) over the interval.

- Compute the hour-long average from total distance and total time. - Ask which information an interval average ignores about what happens inside the interval. - Distinguish a whole-interval quantity from a one-moment quantity.

1. Each cyclist's average speed over the hour is \(\frac{12\,\text{mi}}{1\,\text{h}}=12\,\text{mph}\). 2. An average speed depends only on the total distance and total elapsed time over the interval. 3. An instantaneous speed describes one moment, so the cyclists can have different values at \(1{:}30\) p.m. even though their one-hour averages are equal.

Both average \(12\,\text{mph}\) over the hour, but average speed does not determine the speed at a particular instant; the instantaneous speeds can therefore be \(8\,\text{mph}\) and \(16\,\text{mph}\).

- Compare how each line meets the function graph rather than using its label as a clue. - The average rate over an interval comes from a line through the two endpoint values. - The instantaneous rate at a point comes from the tangent slope there.

1. Line \(p\) passes through the two function points \((0,0)\) and \((2,4)\), so it is the secant for the interval and represents the average rate of change. 2. Its slope is \(\frac{4-0}{2-0}=2\). 3. Line \(q\) touches the graph at \((2,4)\) with the same local direction as the curve, so it is the tangent and represents the instantaneous rate at \(x=2\). 4. Using \((1,0)\) and \((2,4)\) on \(q\), its slope is \(\frac{4-0}{2-1}=4\).

a) \(p\), because it is the secant through the two interval endpoints. b) \(q\), because it is tangent to the graph at \(x=2\). c) Average rate: \(2\); instantaneous rate at \(x=2\): \(4\)

- Write the average-rate formula with the unknown endpoint value still in place. - Use the full input change from \(2\) to \(5\). - Solve the resulting linear equation for the unknown output.

1. Use the average-rate equation \(\frac{R(5)-R(2)}{5-2}=-3\). 2. Substitute \(R(2)=14\): \(\frac{R(5)-14}{3}=-3\). 3. Thus \(R(5)-14=-9\), so \(R(5)=5\).

\(R(5)=5\)

- Use the two columns at the ends of each interval. - As the intervals shorten, what pattern do the rates suggest? - Can a few calculated rates tell you what happens for every still-smaller interval?

1. On \([1,1.5]\), the average rate is \(\frac{4.25-3}{1.5-1}=2.5\). 2. On \([1,1.1]\), the average rate is \(\frac{3.21-3}{1.1-1}=2.1\). 3. On \([1,1.01]\), the rate is \(\frac{3.0201-3}{1.01-1}=2.01\). 4. The slopes approach \(2\), so \(f'(1)\) is estimated as \(2\). Three secant slopes do not prove the exact derivative.

\([1,1.5]\): \(2.5\); \([1,1.1]\): \(2.1\); \([1,1.01]\): \(2.01\). Estimate: \(f'(1)\approx2\); not an exact proof.

- What total volume change corresponds to the stated rate and time? - Decide whether to add or subtract the change from the starting amount.

1. The volume changes by \((-3.5)(4)=-14\,\text{gal}\). 2. Add this change to the starting volume: \(25-14=11\,\text{gal}\).

\(11\,\text{gal}\)

- Different intervals can have different lengths; use each interval's own width. - How should you handle the full interval if the smaller intervals have different lengths?

1. On \([0,1]\), the rate is \(\frac{8-2}{1}=6\). 2. On \([1,3]\), the rate is \(\frac{12-8}{2}=2\). 3. On \([0,3]\), the rate is \(\frac{12-2}{3}=\frac{10}{3}\). The first interval has the larger average rate.

\([0,1]\): \(6\); \([1,3]\): \(2\); \([0,3]\): \(\frac{10}{3}\). The rate was greater on \([0,1]\).

- Which straight line connects the points at the two ends of the interval? - Which straight line just touches the curve at the requested input? - Read slopes as rises divided by runs using the grid.

1. The secant \(s\) joins \((0,0)\) and \((2,4)\), giving average rate \(\frac{4-0}{2-0}=2\). 2. The tangent \(t\) goes through \((2,4)\) and \((1,0)\); its slope is \(\frac{4-0}{2-1}=4\). This is the instantaneous rate at \(x=2\).

Average rate: \(2\). Instantaneous rate at \(x=2\): \(4\).

- What does landing at the nozzle's height tell you about the water's height at the basin? - Where is the highest point of a downward-opening parabola? - What happens to the average slope over a shorter and shorter starting interval?

1. The basin condition is \(h_c(4)=0\): \(-0.5(4)^2+4c=0\). Thus, \(-8+4c=0\), so \(c=2\). 2. For \(c=2.5\), the vertex occurs at \(x=2.5\). The maximum height is \(h_{2.5}(2.5)=3.125\,\text{m}\). 3. For \(c=1.5\), the average slope on \([0,t]\) is \(\frac{h_{1.5}(t)-h_{1.5}(0)}{t}=\frac{-0.5t^2+1.5t}{t}=-0.5t+1.5\), for \(t>0\). 4. As \(t\) approaches \(0\) from the right, the average slope approaches \(1.5\). This is the initial slope. 5. The angle \(\alpha\) satisfies \(\tan(\alpha)=1.5\), so \(\alpha=\arctan(1.5)\approx56.3^\circ\).

a) \(c=2\) b) \(3.125\,\text{m}\) c) The average slope is \(-0.5t+1.5\), which approaches \(1.5\). The launch angle is approximately \(56.3^\circ\).

- Read each list as rates over intervals that are becoming shorter. - Ask which value each list seems to approach. - Distinguish a numerical estimate from a proof about an infinite limiting process.

1. The listed secant slopes suggest that the right-hand rate of \(f\) is near \(6\). 2. For \(g\), the slopes suggest a right-hand rate near \(4\). 3. These are estimates, not proved limits: a finite list alone cannot establish what happens for all smaller intervals. 4. The average rate over \([2,4]\) uses only the two endpoint values. It need not match a local rate at \(t=3\).

Estimated right-hand rates: \(f\): about \(6\); \(g\): about \(4\). The finite data do not prove exact limits, and an interval average does not determine the local rate.

- Use endpoint position change over endpoint time change for part a). - For part b), use the two times that bracket \(t=2\) most closely. - Keep the time interval used for each rate clear when comparing them.

1. The average velocity from \(t=1\) to \(t=3\) is \(\frac{27-1}{3-1}=13\,\text{m/s}\). 2. A local secant estimate around \(t=2\) is \(\frac{9.261-6.859}{2.1-1.9}=12.01\,\text{m/s}\). 3. The local estimate is slightly smaller than the longer-interval average: \(12.01\,\text{m/s}\) versus \(13\,\text{m/s}\).

a) \(13\,\text{m/s}\) b) Approximately \(12.01\,\text{m/s}\) c) The estimated instantaneous velocity is slightly smaller than the average velocity over \([1,3]\).

- Treat radius as the input and volume as the output. - Compute output change before dividing by input change. - The units should be volume units per radius unit.

1. The change in volume is \(\frac{32\pi}{3}-\frac{4\pi}{3}=\frac{28\pi}{3}\,\text{in}^3\). 2. The change in radius is \(2-1=1\,\text{in}\). 3. The average rate is \(\frac{28\pi}{3}\,\text{in}^3/\text{in}\).

\(\frac{28\pi}{3}\,\text{in}^3/\text{in}\)

- Represent each average rate as a secant slope. - Equal average rates correspond to equal secant slopes. - Keep the input differences in the same order as the output differences.

1. The slope of \(\overline{AB}\) is \(\frac{10-2}{5-1}=2\). 2. The slope of \(\overline{CD}\) is \(\frac{y-3}{2-(-2)}=\frac{y-3}{4}\). 3. Set the average rates equal: \(\frac{y-3}{4}=2\). 4. Solving gives \(y=11\).

\(y=11\)