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Suppose that for the tank in Exercise 23 the pump breaks down after $ 4.7 \times 10^5 J $ of work has been done. What is the depth of the water remaining in the tank?

The water remaining is about 2 meters deep.

Applications of Integration

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Missouri State University

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University of Nottingham

Okay. So we're using the work we already did for question 23. Um which it all boiled down to 8000 G. And we are size has why squared -Y cubed over three from three, which is the top of the container. But now I need to know how deep the water was, where it stopped, how far down it went right the lower limit which I'm just going to call each. So I'm going to plug all this in and I know that that equals work, right? And I know that work is 4.7 times 10 to the 5th power jules. So I'm gonna plug all this in um and see what I get. So okay, 8000 times gravity. Oh well We get whenever plug in three. So we get 45 Over 2 -9. Just so this fraction subtraction works out more easily. I'm going to call 9:18 over to instead just getting a common denominator minus 5/2. 8 squared minus h cubed over three. Okay, And that equals 4.7 Times 10 to the 5th power. Okay, so this right here, it becomes 27 over to just kind of kind of simplify as we go now. So you're going to get uh negative 5/2 H squared- H cubed over three Equals 4.7 times 10. 5th power over 8000 G from that out uh -27/2. Okay. Um and then if I just want to write this a little bit more neatly, I will get H cubed over three plus five halves eight sq. So I'm I'm taking a negative out of this side of bringing it over here. Uh huh. For no reason other than preference, I think it makes it a little bit nicer if our side with our independent variable just have a ton of negatives in it -4.7 Times zero Times 10 To the 5th power over 8000. So there's a little icon in this textbook that you really want to notice that basically says that we're not solving this algebraic lee, which makes this way easier. Uh, we're going to solve it graphically. So you want to graph this, replacing your ages with X's ideally. Um I used gizmos and you find where it crosses the, basically finding the intercepts of the graph. Well actually I should add one thing, which is that the equation we plug in needs to be sold for zero. So I suppose what we're actually plugging in is h cubed over three plus five halves, H squared -27/2 Plus 4.7 times 10, uh, 10 to the 5th power over 8000 sheets. So I might have done a little bit more than I needed to their, uh, I forgot that we know we want to just solve for zero. Certainly do that. I have that up here. This is what our intercepts going to be. 2.092. So we'll say about two m, that's as deep as it went.

Texas State University

Applications of Integration