 ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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# Average Rate of Change - Problem 1

Norm Prokup ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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Calculating the average rate of change is much like calculating average velocity (in fact, finding average velocity is simply a special case of calculating the average rate of change). In order to determine the average rate of change, divide the difference between the quantities by the difference between the times at which those quantities are measured. Like average velocity, average rate of change can be represented by finding the slope between two points on a graph. Remember that rate of change is the change of some quantity per unit of time.

Let’s take a look at another average change problem. In this problem we have a barrel of maple syrup tapped at t equals zero. The total amount f(t) in gallons, of maple syrup poured at time t, is given by this table. So t is measured in minutes and these are increments of two minutes. And here are my measurements for the number of gallons that have leaked out of the barrel.

Part A says compute the average rates of change of f(t) on these two intervals; first between 0 and 4 and then between 4 and 12. So let’s do between 4 and 12. Let’s do between 0 and 4 first.

Average rate of change is f(4) minus f(0) over f(0); this is the change in the quantity f, over the change in time. f(4) I can get from the table, it’s 54. f(0) is 0. In the numerator I get 54. Remember this is going to be in gallons. In the denominator I get 4, and that’s going to be in minutes. This is the same as 27 over 2 or 13.5 gallons per minute.

What about during this time interval between 4 and 12? Well that’s going to be f(12) minus f(4) over 12 minus 4. So again, taking a look at the table, f(12) is 100, f(4) is 54, so I have 100 minus 54 over 12 minus 4. 100 minus 54 is 46, and that’s in gallons. 12 minus 4 is 8 and that’s in minutes. So I have to simplify this 46 divide by 8. Well 40 divide by 8 is 5, 6 divided by 8 is .75, so this is 5.75 gallons per minute.

Now part B asks me to represent these average rates of change on this graph. So I have a graph of the number of gallons that have leaked out of the barrel, over time, and I got a couple point plotted. These are the two important points here. It turns out that average rate of change can be represented by the slope of a secant line.

For example the average rate of change between t equals 0 and t equals 4 is the slope of the secant line. Now that average rate of change was 13.5 gallons per minute. So the slope will be 13.5 gallons per minute. How does that work? Well slope remember is rise over run. The rise here is the difference in y values; 54 minus 0, that’s 54 gallons. The run is 4 minus 0 or 4 minutes. We calculated that before and that is 13.5 gallons per minute. The average rate of change from 4 to 12, would be represented by the slope of this secant line. And that slope was 5.72 gallons per minute.

And you can see that it’s less, it’s not as steep as this slope here. Of course this comes from the rise which is 100 minus 54, 46 gallons, and 12 minus 4, 8 minutes. So this slope is, and remember it’s the slope of the secant line that gives you the average rate of change of a function. So this is a really important way to understand average rate of change. It’s the slope of a secant line between two points.