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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 1d

When the pipe overflows.

## Want to join the conversation?

- Why is it =50 instead of >50?

The question says it can hold 50 cubic feet BEFORE overflowing, so why isn't it >50 ?(1 vote) - Would W(w)=50 be considered to be a functional?(0 votes)

## Video transcript

- [Voiceover] Part d. The pipe can hold 50 cubic feet
of water before overflowing. For t greater than eight,
water continue to flow into and out of the
pipe at the gives rates until the pipe begins to overflow. Write, but do not solve, an equation involving one or more integrals that gives the time w when the
pipe will begin to overflow. All right, so the pipe
is going to overflow. So we're going to figure
out that gives the time w and the pipe begins to overflow. So the pipe will begin to overflow when it crosses 50 cubic feet of water or you could say right when
you hit 50 cubic feet of water then it will begin to overflow. So we could figure out at
what time does the pipe have 50 cubic feet of water in it. And so we could just say,
well, w of this time. So I use uppercase W as my function for how much water is in the pipe. So capital W of lowercase w is going to be equal to 50. And so you would just solve for the w. And it said write but do
not solve an equation. Well, just to make this
a little bit clear, if uppercase W of
lowercase w is going to be 30 plus the integral from zero to w. Actually, now since I don't
have t as one of my vars, I could just say R of t minus D of t, td. So let me just do that. R of t minus D of t, dt. So this is the amount of total water in the pipe at time w. Well, this is going to be equal to 50. So we have just written
an equation involving one or more integrals
that gives the time w with the pipe will begin to overflow. So if you could solve for w, that's the time that the
pipe begins to overflow. We are assuming that it doesn't just right get to 50 and then
somehow come back down that it will cross 50 at this time around. So you could test a little
bit more if you want. You could try that slightly larger w or you could see that the rate, that you have more flowing in than flowing out of that time which this R of w is gonna
be greater than D of w so it means you're only
gonna be increasing so you get across over right at that time. So if you want the w
though, you'd solve this. Now another option, it's okay, we know w is going to be greater than eight. So you could say, okay,
how much water do we have right at time equals eight? We figure that out at the last problem. And so you could say we,
at time equals eight, we have that much, 48.544, and this is approximation
that's pretty close. Plus the amount of water
we accumulate between time, eight and time w of R of t minus D of t, dt is equal to 50. Either one of these would
get you to the same place. So w at right, when do we
hit 50 cubic feet of water? And then if you want to test it further, you can make sure that your
rate is increasing right at that or you have a net positive
inflow of water at that point which tells you that you're
just about to start overflowing.