2015 AP Calculus AB/BC 3a | AP Calculus AB solved exams | AP Calculus AB | Khan Academy

Johanna jogs along a straight path for (0 \leq t \leq 40). Johanna's velocity is given by a differentiable function (v). Selected values of (v(t)), where (t) is measured in minutes and (v(t)) is measured in meters per minute, are given in the table above we see that right over there.

Use the data in the table to estimate the value of (v(16)) and (v'(16)). And we see here they don't even give us (v(16)). So how do we think about (v'(16))? Well, just remind ourselves this is the rate of change of (v) with respect to time when time is equal to 16.

So we can estimate it, and that's the key word here: estimate. We can estimate (v'(16)) by thinking about what is our change in velocity over our change in time around 16. When we look at it over here, they don't give us (v(16)) but they give us (v(12)) and they give us (v(20)).

So let's figure out our rate of change between (t = 12) and (t = 20). Our change in velocity is going to be (v(20) - v(12)) and then our change in time is going to be (20 - 12).

So this is going to be equal to (v(20)) is 240, (v(12)) we see right there is 200. So this is going to be (40) over (8), which is equal to (5). And we would be done.

Oh, for good measure, it's good to put the units there. So this is our rate; this is how fast the velocity is changing with respect to time. The units are going to be meters per minute per minute, so we could write this as meters per minute squared. And we're done.

Now, if you want to get a little bit more of a visual understanding of this—and what I'm about to do, you would not actually have to do on the actual exam—let's just try to graph these points here. Let me give myself some space to do it.

So if we were to try to graph it, I'll just do a very rough graph here. Let's say this is (y = v(t)), and we see that (v(t)) goes as low as -220 and goes as high as 240. So let me give, I want to draw the horizontal axis someplace around here. (t) is positive, so it looks something like that.

Let's make this 200, and let's make that 300. And then this would be 200 and 100, and that would be negative 100, and that would be negative 200, and then that would be negative 300. We see on the (t)-axis our highest value is 40; we go between 0 and 40.

These obviously aren't at the same scale, but this is going to be 0 if we put 40 here, and then if we put 20 in between, so this would be 10. Let me do a little bit to the right; that would be 10, and then that would be 30.

What points do they give us? They give us—I'll do these in orange—(0, 0), so that is right over there. They give us when time is 12, our velocity is 200, so when the time is 12, which is right over there, our velocity is going to be 200, so that's that point.

When our time is 20, our velocity is going to be 240. So when our time is 20, our velocity is 240, which is going to be right over there. Then when our time is 24, our velocity is -220. So she switched directions; so 24 is going to be roughly over here; it's going to be -220, so -220 might be right over there.

Finally, when time is 40, her velocity is 150, positive 150. So at 40, it's positive 150. These are just sample points from her velocity function.

But what we wanted to do is we wanted to find, in this problem, we want to say, "Okay, when (t = 16), what is the rate of change?" We don't know much about—we don't know what (v(16)) is, but what we could do is, and this is essentially what we did in this problem, we could say, "Alright, well we can approximate what the function might do by roughly drawing a line here."

So if you draw a line there and you say, "Alright, well (v(16)) or (v'(16)), I should say, (v'(16)) is going to be approximately the slope of this line." That's going to be our best job, based on the data that they've given us, of estimating the value of (v'(16)).

So we literally just did (\Delta v) over (\Delta t) to get the slope of this line, which is our best approximation for the derivative when (t = 16).