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# A swimming pool is $20 ft$ wide, $40 ft$ long, $3 ft$ deep at the shallow end, and $9 ft$ deep at its deepest point. A cross-section is shown in the figure. If the pool is being filled at a rate of $0.8 ft^3/min,$ how fast is the water level rising when the depth at the deepest point is $5 ft?$

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Derivatives

Differentiation

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##### Catherine R.

Missouri State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

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### Video Transcript

So for this problem, we want to keep our whole objective in mind. So we have a swimming pool, and this is going to be a view of it where this is the water, right? So we know that it is 3 ft deep. Um, it's going to be 40 ft long, and then we'll have going to be a the rectangular prism. So we'll draw that real quick, and we see that it's going to be 20 ft wide. So based on this, um, we also see that 3 ft deep and a shallow end, But then it's 9 ft deep at the deep end, and we have a cross section shut in the figure. So if we know the pool is being filled at a rate of, um, 8 ft cubed per minute. So right there, we want to stop. We want to recognize this is a rate. This is a change in volume because its feet cubed per minute. So this is a DVD t a change in volume with respect to time, and that's going to equal 0.8 feet cube permanent. And when we use it in our calculations, we're going to use 0.8 As long as the units all work out mhm, then we want to determine how fast the water level is rising. Some right here. We're giving our change in volume with respect to time. And we want to know a change in height. We want to know the water level. So the water level is this right here or this right here. This is just the height. It has nothing to do with the volume. Such a d h d t. This is what we want. So what it's up to us to do is to determine the relationship between the volume and the Hank and depending on the cross section that will have a different relationship. So what we do is we find the relationship between the volume and the height, and that will be some equation. Then we differentiate that relationship. And once we differentiate the relationship, we will have an equation that we can get in terms of D HDTV that will be equal to something in terms of d v d t. But we already know DVD T. So once we plug this in, that is how we are going to get the GT which is the water level rising at a certain point now, one other step we're gonna have to take is we need to know what's happening exactly at 5 ft. So we need to figure out the time at which it reaches that 5 ft. So in order to do that, we have to know the height 15 ft. Once we have our relationship between volume and this, we plug in our 5 ft here. That gives us the volume here at 5 ft and then that goes into our equation, which will then give us the actual rate of change.

California Baptist University

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Derivatives

Differentiation

##### Catherine R.

Missouri State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

Lectures

Join Bootcamp