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the rate at which rainwater flows into a drain pipe is modeled by the function R where R of T is equal to 20 sine of T squared over 35 cubic feet per hour T is measured in hours and zero is less than or equal to T which is less than or equal to 8 so T is going to go between 0 & 8 the pipe is partially blocked allowing water to drain out the other end of the pipe at a rate modeled by D of T it's equal to negative 0.04 T to the third power plus 0.47 and 60 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 rain water flow into the pipe during the eight hour time interval zero is less than or equal to T is less than or equal to 8 all right so we know the rate the rate that though that the that 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 and they even tell us that there's 30 cubic feet of water right in the beginning but these are the rates of entry 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 these amounts over very small changes in time to go from time is equal to zero all the way to time is equal to eight 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 how what volume flows in over a very small interval DT and then we're to sum it up from T equals zero to T equals eight that's the power of the definite integral and so this is going to be equal to the integral from zero to eight of twenty sine 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 want to do definite integral so that I can click math right over here move down so this function FN integral this is a this is a integral of our function or a function integral right over here somebody press ENTER and the way that you do it is you first define the function then you press 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 going to write 20 sign of and just because it's easier for me to input X than T I'm going to use X but if you just do this a sine of x squared over 35 DX you're getting at the same value so you're going to get x squared 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'm just going to understand let me put the times 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 zero but my lower bound is zero and my upper is 8 and close the parentheses and then let the calculator munch on it a little bit and we get seventy six point five seven Oh so this is approximately 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 three hours give a reason for your answer well what would make it increasing well if if the rate at which things are going in is larger than the rate of things going out then R then R the amount of water would be increasing but if it's the other way around if we're draining faster at T equals three then things are being then things are flowing into the pipe well then R then the amount of water would be decreasing let me be clear so amount so if if R of T greater than actually write it this way if R of three T equals three because T is given in our T is measured in hours if R 3 is greater than d of three whoops then D D of three if R three squared in the D three that means water is flowing in at a higher rate than leaving so that means that water and pipe right then then water in pipe increasing increasing and then if it's the other way around if D of three is greater than R of three then water and pipe decreasing and your draining faster than your putting into it then water in pipe decreasing decreasing so we just have to evaluate these functions at three so let's see are actually I can do it right over here so then we come along the line here so R of three is equal to well let me get my I calculator out this is going to be oops that calculator let me get this calculator out so and I'm assuming that things are in that things are in radians here so I already put my calculator in Radian mode so it's going to be twenty times sine of whoops 20 times sine of 3 squared is 9 divided by 35 and it gives us this is equal to five point approximately five point zero nine so this is approximately five point zero nine and to do three DF 3 is going to be approximately so the calculator back out so it is we have negative 0.04 times times 3 to the third power so times 27 plus plus 0.4 times 9 times 90 squared plus plus zero point zero oh no it's just zero zero point nine six nine six times T times three and this gives us five point four so this is equal to so this is equal to five point four so D of three is greater than R of three so water increasing sorry water decreasing we're draining faster than we're punk we're getting water into it so water water is decreasing

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