2015 AP Calculus AB/BC 1ab | AP Calculus AB solved exams | AP Calculus AB | Khan Academy

The rate at which rainwater flows into a drain pipe is modeled by the function R, where R of T is equal to 20 sin of T^2 over 35 cubic feet per hour. T is measured in hours, and 0 is less than or equal to T which is less than or equal to 8. So T is going to go between zero and eight. The pipe is partially blocked, allowing water to drain out the other end of the pipe at a rate modeled by D of T, which is equal to 0.04 T^3 plus 0.4 T^2 plus 0.96 T, cubic feet per hour for the same interval. Right over here, there are 30 cubic feet of water in the pipe at time T equals zero.

All right, part A: how many cubic feet of rainwater flow into the pipe during the eight-hour time interval, 0 is less than or equal to T is less than or equal to 8? All right, so we know the rate, the rate that the things flow into the rainwater pipe. In fact, we could let me draw a little rainwater pipe here just so we can visualize what's going on.

So if this is, if that is the pipe right over there, things are flowing in at a rate of R of T, and things are flowing out at a rate of D of T. They even tell us that there's 30 cubic feet of water right in the beginning. But these are the rates of entering and the rates of exiting. So how much water? They're asking how much, how many cubic feet of water flow into, so enter into the pipe during the 8-hour time interval.

So if you have your rate, this is the rate at which things are flowing into it. They give it in cubic feet per hour. If you multiply it times some change in time, even an infinitesimally small change in time, so DT, this is the amount that flows in over that very small change in time. And so what we want to do is we want to sum up these very small amounts over very small changes in time to go from time is equal to zero all the way to time is equal to 8.

So this expression right over here, this is going to give us how many cubic feet of water flow into the pipe. Once again, what am I doing? R of T * DT. This is how much flows, what volume flows in over very small interval DT. Then, we're going to sum it up from T equals 0 to T equals 8. That's the power of the definite integral.

And so, this is going to be equal to the integral from 0 to 8 of 20 sin of T^2 over 35 DT. And lucky for us, we can use calculators in this section of the AP exam. So let's bring out a graphing calculator where we can evaluate definite integrals. And so let's see, we want to do definite integrals, so I can click math right over here, move down.

So this function FN integral, this is an integral of our function or function integral right over here. So let me press enter. And the way that you do it is you first define the function, then you press a comma, then you say what variable is the variable that you're integrating with respect to, and then you put the bounds of integration.

So I'm going to write 20 sin of, and just because it's easier for me to input X than T, I'm going to use x, but if you just use this as sin of x^2 over 35 DX, you're going to get the same value. So you're going to get x^2 divided by 35, divided by 35; close that parenthesis. So that is my function. There, actually I don't know if it's going to understand. Let me, I don't know if it's going to understand, so let me put the times, so second insert times just to make sure it understands that.

Okay, so that's my function and then let me throw a comma here, make it clear that I'm integrating with respect to X. I could have put a T here and integrated with respect to T; we'd get the same value, comma, my lower bound is zero, and my upper bound is eight. Close the parenthesis and then let the calculator munch on it a little bit, and we get 76.570. So this is approximately 76.570.

Now let's tackle the next part: Is the amount of water in the pipe increasing or decreasing at time T equals 3 hours? Give a reason for your answer. Well, what would make it increasing? Well, if the rate at which things are going in is larger than the rate at which things are going out, then the amount of water would be increasing. But if it's the other way around, if we're draining faster at T equals 3 than things are flowing into the pipe, then the amount of water would be decreasing.

Let me be clear. So, amount: if R of T is greater than D of T, actually let me write it this way: if R of 3, when T equals 3 (because T is given in hours), if R of 3 is greater than D of 3. Whoops! Then D of 3. If R of 3 is greater than D of 3, that means water is flowing in at a higher rate than leaving. So that means that water in the pipe, right, then the water in the pipe is increasing.

And then, if it's the other way around, if D of 3 is greater than R of 3, then water in the pipe is decreasing. Then you're draining faster than you're putting into it, then water in the pipe is decreasing. So we just have to evaluate these functions at three. So let's see, R of 3 is equal to, well let me get my calculator out.

This is going to be, oops not that calculator; let me get this calculator out. So, and I'm assuming that things are in radians here. So I already put my calculator in radian mode. So it's going to be 20 * sin of (3^2) over 35, and it gives us this is equal to approximately 5.09. So this is approximately 5.09. And D of 3, D of 3 is going to be approximately, so look at the calculator back out.

So it is, we have 0.04 times 3^3, so times 27 plus 0.4 * 3^2 plus 0.96 * 3. And this gives us 5.4. So this is equal to, so this is equal to 5.4. So D of 3 is greater than R of 3, so water is increasing. Sorry, water is decreasing. We're draining faster than we're getting water into it, so water is decreasing.