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Instantaneous Rate of Change - Problem 3
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To estimate the instantaneous rate of change of an object, calculate the average rate of change over smaller and smaller time intervals. When is data is given in a table, the information for smaller time intervals may not be given. So, in order to estimate the instantaneous rate of change, find the average rate of change between two subsequent intervals and average those values (for example, if given information for t=2,4,6, and you want to find the instantaneous rate of change at t=4, calculate the average rate of change from t=2 to t=4, and t=4 to t=6, and calculate the average of those values). Recall that the average rate of change is the change in some quantity divided by the change in time. When the rate of change is positive, that means that quantity is increasing. When the rate of change is negative, the quantity is decreasing.

Let’s take a look at another example. I have a problem, the population of vampires in Pennsylvania is given by the table and I have time, the actual year, 1500, 1600, 1700, so these are the beginning of each century and then the population of vampires, 512 in 1500, 311 in 1600 and so forth. Part a asks me to find the average rate of the population form 1800 to 1900. This is a straight forward change in population over change in time calculation.

It’s p of 1900 minus p (1800) over 1900 minus 1800 p (1900), 69, p (1800) is 114. So this is 69 minus 114 and of course 1900 minus 1800 is 100. 69 minus 144 is -45, over 100 and that’s -.45. -.45 vampires per year, those are the units. What I’m working up towards here, is the idea of instantaneous rate of change, and how you deal with instantaneous rate of change, when you’ve got a table. So that’s what we’re going to do in part b.

It says approximate the rate that the vampire population was decreasing in 1900. The way I’m going to do this is I’m going to work with this average rate of change. That’s the average over this interval, from 1800 to 1900. And then I’m going to find the average rate of change from 1900 to 2000 and average the 2 average rates of change.

Instantaneous rate of change is the limit of average rate of change, as the time increment goes to zero. Here, I can’t make the time increment get any smaller than 100 years. So I have to work with what I have in the data, and that’s 100 years apart. So I’m going to average these two average rates of change.

I already have the first value. Between 1800 and 1900, my average rate of change was -.45 vampires per year. And then between 1900 and 2000, I need to calculate that. That’s going to be p (2000) minus p (1900) over 2000 minus 1900. Let’s take a look at the table to get those values. P (2000) is 42. P (1900) is 69 so this is going to be 42 minus 69 over 100. That’s -27 over 100. That’s -.27 vampires per year.

Now remember, the idea here is to take these two average rates of change and average them. This will be our approximation for the rate of decrease in 1900, but it’s as good as we can do. The rate of decrease is approximately the average of these two values. -.45 plus -.27 all over 2. That happens to be -.36 vampires per year.

When you’re looking at a table of data like this, you can see that the values are decreasing. We’re getting in finding our average rate of decrease in 1900 is we’re estimating how fast the vampire’s population’s decreasing at that time, minus .36 vampires per year.

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