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Isosceles Triangle related rates problem

  1. Mar 6, 2010 #1
    1. The problem statement, all variables and given/known data
    A trough is 9 ft long and its ends have the shape of isosceles triangles that are 5 ft across at the top and have a height of 1 ft. If the trough is being filled with water at a rate of 14 ft3/min, how fast is the water level rising when the water is 8 inches deep?

    Variables:
    b=5 ft
    h=1 ft
    l=9 ft
    2. Relevant equations
    v=(1/2)bhl
    dv/dt=14
    dh/dt=?
    when h=2/3 ft
    3. The attempt at a solution
    I tried doing it the straightforward way:
    v=(1/2)bhl
    dv/dt=(1/2)(5)(2/3)(dh/dt)(9)
    14=(1/2)(5)(2/3)(dh/dt)(9)
    dh/dt=14/15

    This is wrong. Then I tried using similar triangles to get the new base to go along with the height of 8 in. For that I got 10/3 ft. So,
    v=(1/2)bhl
    dv/dt=(1/2)(10/3)(2/3)(dh/dt)(9)
    14=(1/2)(10/3)(2/3)(dh/dt)(9)
    dh/dt= 7/5

    This is also incorrect. I really don't know why. Any help would be greatly appreciated!
     
  2. jcsd
  3. Mar 6, 2010 #2

    tiny-tim

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    Welcome to PF!

    Hi hks118! Welcome to PF! :smile:

    You need to write b as a function of h before you differentiate. :wink:
     
  4. Mar 6, 2010 #3
    Re: Welcome to PF!

    Thanks for the reply!

    So b=5h correct? but if I plug in my h value, I get 10/3, which was wrong.:confused:
     
  5. Mar 7, 2010 #4

    tiny-tim

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    (just got up :zzz: …)

    What was your equation for dv/dt?
     
  6. Mar 7, 2010 #5
    Well, v=1/2bhl so I figured dv/dt=1/2bh(dh/dt)l. Or do I need db/dt and dl/dt as well?
     
  7. Mar 7, 2010 #6

    tiny-tim

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    That wouldn't be right even if b was constant. :confused:

    Try again. :smile:
     
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