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Water is flowing into a pool. Its volume was recorded at five different times.

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I want to estimate the instantaneous rate of change at the 20 minute mark

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So this is volume, and this is time.

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Water is flowing in, so volume is a continuous function. Would this be continuous?

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Yes. The volume would not instantly jump; it changes steadily.

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We do not know what the function looks like.

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It could look something like this, but we do not know. It would probably be a smooth

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function like that.

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We're trying to find the instantaneous rate of change which is the slope of what?

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At one point, it is the tangent line.

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Right at the 20 minute mark.

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We're trying to guess the slope of that line right here.

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Why can't we do what we did on Friday?

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Why can't we just pick values really, really close and find the average rate of change?

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It's asking for the instantaneous rate of change.

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It's definitely asking for that.

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But last class, we did this, like the average rate of change.

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For example, we used the average rate of change [f(2.99) − f(3)] / (2.99 − 3).

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Then we kept using numbers

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that get closer and closer to the target.

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Remember this?

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OK.

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So why can't we do that for this problem?

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And what do we not have here?

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We have no equation

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To use that method, we would need to be given

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a function formula such as f(x) = …

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But we do not know what the formula is.

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So here we can't do this method where you keep plugging in numbers closer and closer

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to that number.

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Does this make sense?

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Okay.

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So we need some other way to estimate it.

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So we find the average rate of change between these two, and that would give us the slope

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of that line.

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And then do the same thing between 20 and 25, which would give us the slope of that

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line. Then what should we do with the two slopes? Average them.

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One possible—but less reliable—idea is to draw the tangent line we expect and then

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use the known point plus a second estimated point. That would still be a guess,

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so it is better to use the table data. That is what we will

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do. First, find the average rate of change on the interval [15, 20].

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AROC on [15, 20] = (3,500 − 2,500) / (20 − 15).

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That is 200.

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Is that 200?

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Okay.

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What are the units?

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The volume values are measured in gallons, and the time values are

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measured in minutes, so the units are gallons per minute.

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The slope of the green line is 200 gallons per minute.

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Is that 300?

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The slope of the orange line is 300 gallons per minute.

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Okay. We found the first slope, 200 gallons per minute, and then

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the orange slope, 300 gallons per minute. Our best estimate

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is the average of those two slopes.

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We are estimating the instantaneous rate of change at one single point in time—20 minutes—not over an interval

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of time. At that exact moment, our best estimate is (200 + 300) / 2 = 250 gallons per minute.

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This method is not very complicated.

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So when are we going to do this method?

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Use this method when only a table or graph of values is given.

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Use it when no function formula is available. If a function or equation is given, use

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the method from Friday: find average rates over increasingly small intervals. Does that make sense?

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Okay