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Simple calculus question

  1. Nov 1, 2007 #1
    1. The problem statement, all variables and given/known data
    A reservoir of square cross-section has sides sloping at an angle of 45 degrees with the vertical. The side of the bottom is p feet in length, and water flows in the reservoir at the rate of c cubic feet per minute. Find an expression for the rate at which the surface of the water is rising at the instant its depth is h feet. Calculate this rate when p = 17, h = 4 and c = 35.


    2. Relevant equations
    Not stated


    3. The attempt at a solution
    I don't know how to begin, I'm sorry. :frown:
     
  2. jcsd
  3. Nov 1, 2007 #2

    Dick

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    Start by trying to find an expression for the volume V as a function of h and p.
     
  4. Nov 1, 2007 #3
    What shape is the reservoir?
     
  5. Nov 1, 2007 #4

    Dick

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    Did you read the problem? It's a truncated pyramid with square cross section.
     
  6. Nov 1, 2007 #5
    Hmm. I didn't know that. How did you come up with that?
     
  7. Nov 1, 2007 #6

    Dick

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    I read the first sentence of the problem you posted.
     
  8. Nov 1, 2007 #7
    Did you even read the problem? You asked the question and it seems like you haven't even read it. The first sentence describes the shape of the resevoir.
     
  9. Nov 1, 2007 #8
    Of course I read it but I didn't know what that was. Sorry.
     
  10. Nov 1, 2007 #9
    Anyway, the volume of a truncated pyramid is

    [tex]V = \frac{H}{3}(A + a + \sqrt{Aa})[/tex]

    right?
     
    Last edited: Nov 1, 2007
  11. Nov 1, 2007 #10

    Dick

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    Right, if the A and a are the areas of the top and bottom.
     
  12. Nov 1, 2007 #11
    What should I do next?
     
  13. Nov 1, 2007 #12

    Dick

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    Apply that formula to your problem. If the reservoir is filled to height h, what are the areas of the top and bottom?
     
  14. Nov 1, 2007 #13
    So the area of A = p2 and a = (p+2h)2.
     
  15. Nov 1, 2007 #14

    Dick

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    Now you are cooking. What's the formula then for the volume V in terms of p and h?
     
  16. Nov 1, 2007 #15
    Then I'll just substitute those in the formula for the volume:

    [tex]\frac{H}{3}[p^2 + p(p + 2h) + (p + 2h)^2][/tex]
     
  17. Nov 1, 2007 #16

    Dick

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    H=h, right? Now to make life easier later, I would expand that out. Now you have V as a function of p and h. What next? Please don't say "I don't know".
     
  18. Nov 1, 2007 #17
    Yeah, I should have written it as 1/3h instead of H/3 actually.

    Thus,

    [tex]p^2h + 2ph^2 + \frac{4}{3}h^3[/tex]
     
  19. Nov 1, 2007 #18

    Dick

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    Better and better. What next?
     
  20. Nov 1, 2007 #19
    I'm stuck. :(
     
  21. Nov 1, 2007 #20
    Maybe we should take time in consideration?
     
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