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- [Voiceover] All right, part c. What time t, where zero is less than or equal to t is less than or equal to eight, is the amount of water in the pipe at a minimum? Justify your answer. All right, well, let's define a function w that represents the amount of water in the pipe at anytime, t, and then we could figure out how to figure out the minimum value for w over that interval. So let's just say w of t. So that's the amount of water in the pipe at any given time, t. It's going to be equal to, well, we start with 30 cubic feet of water in the pipe that t equals zero that tells us that right up here. So we start with that much. And then we are going to add or take away some amount. So then it's going to be plus. We're going to sum up from time zero to time t. Remember this is the function of t. Well, let's sum up our net inflow times little small changes in time. We will sum up all those small changes in time from zero to t and that will give us our aggregate inflow or aggregate net inflow. So what is our net inflow? Well, you have the function R which says the rate which the water is entering and your function d which is the rate which water is exiting. Instead of using t as our variable, I'm going to use x since t is already one of our variable, is one of our boundaries of integration. We can use any variable we want. It's really just a bit of a placeholder variable for the integration. So our net inflow is R, I'll say R of x minus D of x and then dx. Once again, this is our net rate of inflow. This is our inflow minus our outflow so our net rate of inflow times small changes, we could view this as time although it's x, and then we sum up from zero to time is equal to t. So this is how much water we are going to have in the pipe at any time t. So let's now think about at what point does w hit a minimum in the central. And there's three possibilities where w could hit a minimum point. It could hit a minimum at right at the beginning at each at w of zero. It could hit it at the end of our interval at w of eight at t equals eight or it could be some place in between where w is at a local maximum point. In which case, the derivative of w will be equal to zero. So let's first evaluate w at the end points. So w of zero. Oh, we know that. That's the boundary of a zero to zero that they gave it to us in the problem is the initial state is we have 30 cubic feet of water in the pipe already. And now, what is w of eight? Well, this is going to be 30 plus the definite integral from zero to eight of R of x minus D of x, minus D of x, dx. And lucky for us, we're allowed to use a calculator so let's use a calculator to evaluate this. And so, let me get my calculator out. Let me go see the definition of R of x, D of x right over here so I can type them in. And what I'm gonna do and there's a bunch of ways of more advance you get with a calculator that might save you some time in front of the AP test. I'm actually gonna define a function that is R of x minus D of x. So everywhere I see t, I'm gonna substitute it. I'm gonna use x instead. And then I could use that later on to integrate it or to solve it in some way. So let's do that. So y1 is gonna be 20 times sine of x squared divided by 35, close parenthesis, minus open parenthesis. You have negative 0.04 times t to the third power or sorry, so I'll write times x, x to the third power plus 0.4 times x squared plus 0.96 times x. And then I can close these parentheses. So does that make sense? 20 times sine of x squared divided by 35 minus negative 0.04 x to the third plus 0.4 x squared plus 0.96 times x. All right, there you go. I have defined y1. And now I can use that to evaluate different things. So let me now go back here. And so let me evaluate what this is going to be. This is going to be 30 plus the definitive role of the math functions. You scroll down a little bit. You have, this is the function for definite integral, function integral. The function is y1 and I can go to vars. I'm gonna get my y vars. It's gonna be a function variable. So y1 is what I select. I could have typed it in but this will hopefully save time and I can reuse y1. And my variable integration is x. I'm going from zero to eight. From my lower bound is zero. My upper bound is eight. Let's see, did I type everything in? All right. And I'm gonna munch a little bit. And there you have it. It's approximately 48.544 cubic feet of water at t equals eight. So let me write that, 48.544. So this is approximately 48.544 cubic feet. And now let's see if there's any point in between where w hits a local maximum point or a local minimum point I should say. I think I said a local maximum earlier on. I should say a local minimum point. And so if it's at a local or maximum point really, the derivative of w is going to be zero. So let's see, at what t do we get a zero derivative? So w prime of t. Well, the derivative or a constant with respect to t is zero and the derivative of this with respect to t and this comes from the fundamental theorem of calculus, this is going to be R of t minus D of t. Well, once again, this is a function. We have t as the upper bound. And so we have whatever here that we are integrating but it is now going to be a function of t. And so in order for this to be equal to zero, we have to figure out when does R of t minus D of t equals zero. And lucky for us, we've already typed in R of t minus D of t. We defined that as y sub one on our calculator. So let's go back here. And now I can use the Solver. So go to math and then let's see if I scroll up. Let's see or this is, I get to the Solver. I could have scrolled down as well. It's right below definite integral. So I go to the Solver. And I say the equation zero equals, and I could just go to y1. So go to my y variables, function. I select y1. So in my equation is equal to y1. And I press Enter. And now I could put an initial guess for what x value is going to solve that equation. Usually a t value but I'm using x as a variable here. And now I do alpha and I quickly solve here. So a little blue you see right above the Enter. And it gets to, well, zero is one of them. Let's see if I can get another one. So let's see if I can get, let's see if I started two, alpha, solve, what it munch on that a little bit. Okay, so this is actually within the interval. So approximately 3.272. So t is approximately 3.272. That's where we have a local minimum point but now we have to evaluate w there. So we have to evaluate w at 3.272 to figure out if it's truly lower than w zero. W of eight is higher than w of zero. So, how do we do that? Well, luckily, we can go back. So second quit. And that should be stored into the variable x. Yep, it's right there. So then we can say, well, let's calculate. We wanna calculate the function evaluated when t is 3.272. So our function is 30 plus the definite integral. Go to the math. So definite integral. Once again, we have y1. So let me go to my y variables, function, y1. We already defined that as R of x minus D of x. Our variable of integration is x. Our lower bound is zero and our upper bound is 3.272 we've restored in the variable x. So we can actually, it's a little confusing. This is saying what's our variable integration. This is the actual variable. This is actual value for upper bound. And so let's munch on it a little bit and we get 27.965. So this is approximately equal to 27.965. So now we're ready to answer. At what time t is the amount of water in the pipe at a minimum? We can see that at time 3.272, we have less water in the pipe than either right when we started or right at the end of our interval at t equals eight. So, at what time t? We say t is equal to 3.272. We could write w of 3.272 is less than w of zero which is less than w of eight. So this indeed just by knowing the derivative of the zero, it could be a minimum or maximum. But the fact that it's lower than both of the end points, this tells us that, hey, this is a minimum, minimum point right over there or we could say that t equals 3.272 is where we hit out minimum value.
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