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

All right, part C. At what time ( t ) where ( 0 \leq t \leq 8 ) 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 any time ( t ), and then we can figure out how to determine the minimum value for ( w ) over that interval. So let's just say ( w(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 at ( t = 0 ); they tell us that right up here. So we start with that much, and then we are going to add or take away some amount.

Then it's going to be plus; we're going to sum up from time 0 to time ( t ). Remember, this is a function of ( t ). Well, let's sum up our net inflow times small changes in time; we'll sum up all those small changes in time from zero to ( t ), and that'll give us our aggregate inflow, our aggregate net inflow.

So what is our net inflow? Well, you have the function ( r ), which says the rate at which the water is entering, and your function ( d ), which is the rate at which water is exiting. Instead of using ( t ) as our variable here, I'm going to use ( x ) since ( t ) is already one of our variable boundaries of integration, and 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(x) - d(x) , dx ). Once again, this is our net rate of inflow; this is our inflow minus our outflow. Our net rate of inflow times small changes; you could view this as time, although it's ( x ), and then we sum up from zero to time, which is equal to ( t ). So this is how much water we are going to have in the pipe at any time ( t ).

Now, let's think about at what point does ( w ) hit a minimum in this interval. There are three possibilities where ( w ) could hit a minimum point. It could have hit a minimum right at the beginning at ( w(0) ); it could hit it at the end of our interval at ( w(8) ) at ( t = 8 ); or it could be someplace 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 endpoints. ( w(0) ): well, we know that the bounds are going to be zero to zero; they even gave that to us in the problem. The initial state is we have 30 cubic feet of water in the pipe already. Now, what is ( w(8) )? Well, this is going to be ( 30 + ) the definite integral from 0 to 8 of ( r(x) - d(x) , dx ).

Lucky for us, we're allowed to use a calculator, so let's use a calculator to evaluate this. Let me get my calculator out, and actually, let me go see the definitions of ( r(x) ) and ( d(x) ) right over here so I can type them in. What I'm going to do, and there are a bunch of ways the more advanced you get with a calculator that might save you some time if you're on the AP test, I'm going to actually define a function that is ( r(x) - d(x) ).

So everywhere I see a ( t ), I'm going to substitute it; I'm going to use an ( x ) instead. 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 going to be ( 20 \times \sin\left(\frac{x^2}{35}\right) - \left(-0.04 \times x^3 + 0.4 \times x^2 + 0.96 \times x\right) ). Does that make sense? ( 20 \times \sin\left(\frac{x^2}{35}\right) - (-0.04 x^3 + 0.4 x^2 + 0.96 x) ).

All right, there you go; I have defined ( y1 ), and now I can use that to evaluate different things. Let me now go back here, and so let me evaluate what this is going to be. This is going to be ( 30 + ) the definite integral. I'll go to the math functions; you scroll down a little bit; you have this is the function for the definite integral function.

The function is ( y1 ), and I can go to vars, go to my ( y ) vars; it's going to 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 of integration is ( x ), and I'm going from ( 0 ) to ( 8 ), from my lower bound to ( 0 ) and my upper bound ( 8 ).

Let's see, did I type everything in all right? Then let it munch a little bit, and there you have it; it's approximately ( 48.544 ) cubic feet of water at ( t = 8 ). So let me write that: ( 48.544 ).

So this is approximately ( 48.544 ) cubic feet. 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.

So if it's at a local minimum or maximum point, really the derivative of ( w ) is going to be zero. Let's see at what ( t ) do we get a zero derivative. So ( w' (t) ), well, the derivative of a constant with respect to ( t ) is zero, and the derivative of this with respect to ( t ), and this comes right out of the fundamental theorem of calculus, this is going to be ( r(t) - d(t) ).

Once again, this is a function; we have ( t ) as the upper bound, and so we have whatever here that we were integrating, but it's now going to be a function of ( t ). In order for this to be equal to 0, we need to figure out when does ( r(t) - d(t) = 0 ).

Lucky for us, we've already typed in ( r(t) - d(t) ); we've defined that as ( y1 ) on our calculator. So let's go back here, and now I can use the solver. Go to math, and then let's see if I scroll up. This is the solver; I get to the solver by scrolling down as well; it's right below the definite integral.

So I go to the solver, and I say the equation ( 0 = ) and now I could just go to ( y1 ), so go to my ( y ) variables function; I select ( y1 ), and I press enter. Now I put an initial guess for what ( x ) value is going to solve that equation. It's really a ( t ) value, but I'm using ( x ) as a variable here, and now I do alpha, and I click solve.

Here, it’s a little in blue; you see it right above the enter, and it gets to, well, zero is one of them. Let's see if I can get another one. Let's see if I can get see if I start at ( 2 ), alpha, solve, let it munch on that a little bit.

Okay, so this is actually within the interval. So, ( 3 , \text{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(3.272) ) to figure out if it is truly lower than either ( w(0) ) or ( w(8) ) is even higher than ( w(0) ).

So how do we do that? Well, luckily we can go back; so ( \text{second, quit} ), and that should be stored under the variable ( x ); yep, it's right there. So, then we can say, well, let's calculate; we want to calculate the function evaluated when ( t = 3.272 ).

So our function is ( 30 + ) the definite integral; go to the math, so definite integral, and once again, we have ( y1 ). So let me go to my ( y ) variables; function ( y1 ), we already defined that as ( r(x) - d(x) ).

Our variable of integration is ( x ), our lower bound is ( 0 ), and our upper bound is ( 3.272 ), which we stored in the variable ( x ).

So we can actually, it's a little confusing; this is saying what's our variable of integration? This is the actual variable; this is the actual value for our 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 = 8 ).

So, at what time ( t )? We say ( t = 3.272 ). We could write ( w(3.272) < w(0) < w(8) ). So this indeed, just by knowing that the derivative there is zero, could be a minimum or a maximum. But the fact that it's lower than both of the endpoints tells us that, hey, this is a minimum point, right over there.

Or we could say that ( t = 3.272 ) is where we hit our minimum value.