Essential Knowledge 2.1A1

Essential Knowledge 2.1A1 Students will know that the difference quotients

and express the average rate of change of a function over an interval.

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Introduction

If you wanted to know how fast a car was going, how would you calculate it? Maybe you'd say "I'd get a radar gun."

But I asked how you would calculate it. That radar gun isn't magical, it's performing a calculation. So what is it doing? We'll come back to this later. Let's start with something a little simpler.

Constant Speed

Traffic counters are tubes record when a car drives over them. They are normally used to provide data about how busy a road is, so city planners can decide if more lanes need to be added. But they could also be used to find the speed of a car.

We want to measure how fast the blue car goes, so we put two traffic counters on the road. The traffic counters are 100 feet apart. Now we have everything we need to determine the speed of a car.

Example 1

If the car hit the first counter at 2:36:41.4 pm (41.4 seconds after 2:36 pm) and hit the second counter at 2:36:43.1 pm, how fast was the car traveling?

Well, to find the rate of anything, you can use the "difference quotient". The difference quotient calculates the change in one variable compared to a change in another.

The mathematical symbol for change (when subtracting) is Δ. Spelled "delta". So let's slowly transform this equation into the one we'll use for calculus.

For speed, we care about comparing distance traveled to time elapsed, so let's plug those in

Change in distance just means final distance minus starting distance. Change in time means final time minus starting time.

We'll stop here for now. Let's finally plug in our numbers and solve this problem.

This is about 40.1 mph.

A graph of distance traveled vs. time elapsed is shown below.

Example 2

Let's do another example, but this time go all the way to the calculus form of a difference quotient.

If the car hit the first counter at 6:13:27.5 pm (27.5 seconds after 6:13 pm) and hit the second counter at 6:13:28.6 pm, how fast was the car traveling?

This time let's do the calculus thing and think about the math as a function. So, as time increases, the distance the car travels increases. This means we could say that distance is a function of time. In other words, d = f(t). Now our difference quotient would look like this.

Now, we'll just call time 2 "x" and time 1 "a" and we'll have one of the general forms of the difference quotient.

Now if we can just get rid of that t₂ in the numerator, we would would be done. Well, lets see here. t₂ is just whatever time it was when the car hit the first counter plus the amount of time that passed before it hit the second counter. In other words, t₂ = t₁ + Δt. Now we've got it.

We'll call time 1 "a" again, but this time we'll call Δt "h". And we get the following general equation

is the other general form of the difference quotient. Both have their uses.

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the "difference quotient" used in calculus, a→x form.

the "difference quotient" used in calculus, h→0 form.

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Finally getting back to our example, we can solve using either equation. Let'd do both.

For the second form of the difference quotient, we would need to solve for Δt first.

Δt = h = 6:13:28.6 pm - 6:13:27.5 pm = 1.1 sec

!NB I basically think of the equations as distance and time whenever I get confused. Just think of

• x as t or t₂

• a as t₁

• f as a position function

• f(x) as d or d₂

• f(a) as d₁

• h as Δt

and that should help you remember what all the parts really mean.

Non-Constant Speed

We didn't need calculus for the car example above, because we assumed the car was cruising along at a constant speed as it crossed the traffic counters. Let's look at a situation where speed changes as time goes on.

If I drop a ball off of a 1000 foot tall building, how fast will the ball be traveling at any point in time? For example, how fast will the ball be traveling after falling for one second? Two seconds? Five seconds? Any number of seconds?

So you do an experiment. You can't measure speed, but you can measure position. You buy camera and an extremely long tape measure. You extend the tape measure from the top of the building to the ground, then you film the ball traveling down along the tape measure. You make a table of the distance the ball has traveled after each second, and the results are as follows. Remember the building is 1000 feet tall.

Let's try to figure out the speed when t = 2 sec. Well, we have our formula

which for distance vs time is the same thing as

But a critical question plagues us: which times do I use if I only care about t = 2? t = 2 and t = 1?

(negative because ball is going down)

Why not t = 2 and t = 3?

Yikes! Those are really different. I could use t = 1 and t = 3, but neither of those even involve the time we care about. In truth, there is no way to solve this. We would need a bit more information, and we would need calculus. For now, all we could do is estimate.

If you were forced to answer this question without any more information, your best answer would probably be "somewhere between 48 and 80 feet per second."