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Current time:0:00Total duration:8:33

- [Voiceover] The rate at which rainwater flows into a drainpipe is modeled by the function R, where R of t is equal
to 20sin of t squared 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 gonna go between 0 and 8. The pipe is partially blocked, allowing water to drain out the other end of the pipe at rate modeled by D of t. It's equal to -0.04t to the third power plus 0.4t squared plus
0.96t 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 0. Alright, Part A. How many cubic feet of rainwater flow into the pipe during the 8 hour time interval 0 is less than or equal to t is less than or equal to 8? Alright, so we know the rate, the rate that things flow
into the rainwater pipe. Let me draw a little rainwater pipe here just so that we can
visualize what's going on. So 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. And they even tell us that
there is 30 cubic feet of water right in the beginning. But these are the rates of
entry and the rates of exiting. So they're asking 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 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 wanna do is we wanna sum up these amounts over very small changes in time to go from time is equal to 0, 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 times D of t,
this is how much flows, what volume flows in over
a very small interval, dt, and then we're gonna 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 20sin of t squared 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 wanna do definite integrals so I can click math right over here, move down. So this function, fn integral, this is a integral of a function, or a function integral right over here, so we press Enter. And the way that you do it is you first define the function, then you put 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 gonna write 20sin of and just cuz it's easier
for me to input x than t, I'm gonna use x, but if you just do this as sin of x squared over 35 dx you're gonna get the same value so you're going to get x squared divided by 35. Close that parentheses. So that is my function there. Actually, I don't know if
it's going to understand. Let me put the times 2nd, insert, times just to make sure it understands that. Ok, 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've put a t here and integrated it with respect to t, we would get the same value. Comma, my lower bound is 0. And my upper bound is 8. And then close the parentheses and let the calculator
munch on it a little bit. And we get 76.570 so this is approximately Seventy-six point five, seven, zero. Now let's tackle the next part. Is the amount of water in the
pipe increasing or decreasing at time t is equal to 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
of things 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, then things are flowing into the pipe, well then the amount of
water would be decreasing. Let me be clear, so amount, if R of t greater than, actually let me write it this way, if R of 3, t equals 3 cuz t is given in hour. t is measured in hours. If R of 3 is greater than D of 3, then D of 3, If R of 3 is greater than D of 3 that means water's flowing in at a higher rate than leaving. So that means that water in pipe, let me right then, then water in pipe Increasing. Increasing. And then if it's the other way around, if D of 3 is greater than R of 3, then water in pipe decreasing, then you're draining faster
than you're putting into it. Then water in pipe decreasing. decreasing. So we just have to evaluate
these functions at 3. So let's see R. Actually I can do it right over here. So let me make a little line here. R of 3 is equal to, well let me get my calculator out. This is going to be, whoops, not that calculator, Let me get this calculator out. 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 times sin of 3 squared is 9, divided by 35, and it gives us, this is equal to approximately 5.09. So this is approximately 5.09 and D of 3 is going to be approximately, let me get the calculator back out. So it is, We have -0.04 times 3 to the third power, so times 27, plus 0.4 times 9, times 9, t squared. plus 0.96 times t, times 3. And this gives us 5.4. So this is equal to 5.4. So D of 3 is greater than R of 3, so water decreasing. We're draining faster than
we're getting water into it so water is decreasing.

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