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Urgently Needed: Related Rates Swimming Pool Problem

  1. Jun 18, 2008 #1
    1. The problem statement, all variables and given/known data
    A swimming pool is 24 m long by 8 m wide, 1 m deep at the shallow end and 3 m deep at the deep end, the bottom being an inclined plane. If water is pumped into the empty pool at a rate of 2m^3/min, then how fast is the water level rising at the moment when the water is 1 m deep at the end of the pool.


    2. Relevant equations

    dh/dt = 1/As x dV/dt

    As: area of exposed surface


    3. The attempt at a solution

    Basically I used ratios to get the length of the pool when the height is 1 m deep.

    1/3 = x/24
    So x = 8 m.

    Then As = 8*8 = 64 m^2

    dh/dt = 2/64
    = 1/32 m/min

    I am not sure what I am doing wrong. I have my diagram attached. Please tell me what I am doing wrong. Also, if you have the answer please tell me because I need this urgently. I can use the answer and then work it back to get everything right.

    Thanks.
     

    Attached Files:

  2. jcsd
  3. Jun 18, 2008 #2

    Dick

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    The surface area 'As' is a function of time as well, it's not just a constant you can factor out.
     
  4. Jun 18, 2008 #3
    So would that I mean I have to use some other formula rather than the As one. Because my teacher told me that formula is good for any questions which have exposed area.
     
  5. Jun 18, 2008 #4

    Dick

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    Volume(t)=(1/2)*As(t)*h(t) since the volume is 1/2 of a rectangular solid. So you have to use the product rule to get dV/dt. Your teacher may have exaggerated the usefulness of "dh/dt = 1/As x dV/dt". That's only good if As is a constant.
     
    Last edited: Jun 18, 2008
  6. Jun 19, 2008 #5
    Um, I am sorry I cannot understand. Perhaps if you could explain without As because it's confusing to me.
    Please. Thanks.
     
  7. Jun 19, 2008 #6

    Dick

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    Good idea. Write a formula for the volume of the pool without using As. V=(1/2)*h*(8m)*(h*24/3). h*24/3 is the length of the filled part. I got that from you 1/3=x/24 by replacing 1 with h. Now how is dV/dt related to dh/dt? Notice As=(8m)*(h*24/3).
     
  8. Jun 29, 2008 #7
    sorry for the late reply.
    I am extremely confused.

    How did you get the formula for the volume of pool? Also, I understand that as the height increases, the volume increases too.

    Exposed surface will be a rectangle, so area would be length by width.
    I understand the width is 8 m and the length is (24h)/3.

    But if we solve the volume formula you gave, we end up with the same conclusion that dh/dt is 1/32 m/min.

    Please do explain. I really need to know how this works. Thanks.
     
  9. Jun 30, 2008 #8

    Dick

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    I got the formula by multiplying the area of the triangular cross section of the pool by the width. Yes, you get the same answer of 1/32 m/min. Because that's the correct answer. There is also nothing wrong with dV/dt=(1/As)*dh/dt. You seemed to be so convinced you were wrong that you convinced me. Why do you think 1/32 m/min is wrong?
     
  10. Jun 30, 2008 #9
    LOL.The reason why I think it's wrong is because my teacher did not give me marks for the answer 1/32 m/min. That is why I think there must be some twist to it.

    And when I showed my calculations for h/3 = x/24 the teacher also put a "x" mark there.


    Maybe then I did it right, I am not sure.
     
  11. Jun 30, 2008 #10

    Dick

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    Well, I think you did it right.
     
  12. Jul 1, 2008 #11
    Here is a solution that I present only because you say the (1/32) solution didn’t go well with your instructor. My math, being what it is, I would treat it with suspicion.

    Let b be the length of the surface of the water in the 24 meter direction.
    Let h be the depth of the water at the deep end of the pool.
    Let v be the volume of water in the pool at time t.

    b = 12 h
    v = (1/2) h (12 h) 8 = 48 h^2
    dv/dt = 96 h dh/dt
    dh/dt = 1/(96 h) dv/dt

    At h = 1 and dv/dt = 2
    dh/dt = 1/(96×1)×2 = 1/48
     
  13. Jul 1, 2008 #12

    Dick

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    That would be fine, except I question the b=12h. At h=0, b=0. At h=3, b=24. So b=8h.
     
  14. Jul 1, 2008 #13
    The bottom of the pool terminates at the shallow end at h=2, not h=3. At h=2, b=24.
     
  15. Jul 1, 2008 #14

    Dick

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    Oooooooops. You've got it. I was looking at the attached JPG and reading the little tick mark near the 3m label as the depth of the slanted section. Silly me. Sorry. Thanks!
     
  16. Jul 1, 2008 #15
    Thanks jimvoit. I finally get it. Your answer was very detailed and it helped me understand this question. The answer 1/48 seems better than the one I got, also the fact that it is 2m and not 3m for the ratios was a small mistake that I had made, but your post was really helpful.
    You are the best.
    Thanks again.
     
  17. Jul 1, 2008 #16
    I’m glad we were able to help out. If there is a lesson here, it is that things can go wrong in at least 2 ways…First if the translation from the description of the physical situation to the math is faulted, and second if the math is faulted. The first way is a real headache because perfectly good math can lead to the wrong answer.
     
  18. Jul 1, 2008 #17

    Dick

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    You can also go wrong by focusing on the big picture and forgetting to check the little details. As we did. Thanks again.
     
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