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## AP®︎ Calculus AB (2017 edition)

### Course: AP®︎ Calculus AB (2017 edition) > Unit 12

Lesson 2: AP Calculus AB 2015 free response- 2015 AP Calculus AB/BC 1ab
- 2015 AP Calculus AB/BC 1c
- 2015 AP Calculus AB/BC 1d
- 2015 AP Calculus AB 2a
- 2015 AP Calculus AB 2b
- 2015 AP Calculus 2c
- 2015 AP Calculus AB/BC 3a
- 2015 AP Calculus AB/BC 3b
- 2015 AP Calculus AB/BC 3cd
- 2015 AP Calculus AB/BC 4ab
- 2015 AP Calculus AB/BC 4cd
- 2015 AP Calculus AB 5a
- 2015 AP Calculus AB 5b
- 2015 AP Calculus AB 5c
- 2015 AP Calculus AB 5d
- 2015 AP Calculus AB 6a
- 2015 AP Calculus AB 6b
- 2015 AP Calculus AB 6c

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# 2015 AP Calculus AB/BC 1c

Minimum water in the pipe.

## Want to join the conversation?

- around7:40how come the time interval where the amount of water is at a minimum happens to be where the two lines intersect? ,is it because at that time the area between the curves (representing the amount of water inside) is at a minimum?(6 votes)
- The points of minimum and maximum are the points where the derivative of the function is equal to 0 (0 raise over any run, or just a flat horizontal line). For more detailed description, I would advise to review "Optimization" in "Differential calculus" section on Khan Academy.(2 votes)

- When you are finding where there is a local minimum (about8:00), how do you know that the value t=3.272 is a local minimum and not a local maximum? When the derivative of a function is 0 the function can have either a max or min. Wouldn't you have to check to make sure it is indeed a min and not a max?(6 votes)
- I feel like that is a point of interest, because I found that a little bit strange as well. One thing to point out is that one item to look at would be in part b, where we found that at t = 3, the graph was decreasing. While this isn't to say that this is always
*100 %*accurate, you can look at the trend and see that when**t = 3.272**, the**derivative = 0**and the graph was decreasing moments before it became 0, you can reason that the point at t will most likely be equal to the minimum point on the graph.(2 votes)

- Wouldn't it be easier to just create a new function:

N(t) = R(t) - D(t)

Graph N(t), find where N(t) equals zero (it equals 0 at t=0 and t=3.272). Then find which zero it changes from negative to positive at (t=3.272), then call that the answer? And reason that the endpoints t=0 and t=8 can be disregarded because the graph of N(t), which is the rate of change of the water, goes down is negative before hitting t=3.272 and then continues on positive till t=8?(3 votes) - 1.How did you get 3.272?

2. How can you find w(3.272) with a TI-84 Plus calculator? The one used in this video has different features.(3 votes) - Would it be possible to evaluate R(t) - D(t) = 0, as well as the integral of (sin(t^2/35)dt) without a calculator?(2 votes)
- I'm pretty sure it would be possible. But I don't know how.(0 votes)

- At10:04, and during the entire rest of the problem, wouldn't it be easier simply to find the intercept between the two functions R(t), D(t)? Is it only a coincidence that it worked for this problem.(2 votes)
- I think your question needs a bit more elaboration. Wouldn't it be easier to find what exactly? There are 3 parts to this question and all are slightly different in what they're asking.(0 votes)

- Around1:20, Sal uses R(x) - D(x) in the integral rather than R(t)-D(t). Can someone please explain?(1 vote)
- As he explains, he as used the variable t for the upper bondary of the integral (W is expressed as a function of time so t is a common choice here), so another variable name has to be chosen for the integration. He's using x, for it's the most conventional choice, but any other letter could do the job.(2 votes)

- At10:30ish Khan says it could be either a maximum or a minimum but because its lower than both the endpoints its clearly a minimum, what if the limits of integration on the BC exam are like 0 to 45 then should we find all the points where the f'(x) graph reaches 0 and try all of them or is there a different method anyone knows about?(1 vote)
- Would it be acceptable to show graphically that before the intersection between R(t) and D(t) more water is being drained since D(t)>R(t) while from that point on more water is entering the pipe than being drained since R(t)>D(t), making the time of intersection the time when water is at the lowest level?(1 vote)
- at7:44, how to understand "started at 2, alpha"? I know Sal was trying to find another x that satisfies the equation.(1 vote)

## Video transcript

- [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.