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Homework Help: Horizontal Eavestrough Problem

  1. Apr 5, 2009 #1
    1. The problem statement, all variables and given/known data
    A horizontal eavestrough 3m long has a triangular cross section 10 cm across the top and 10 cm deep. During a rainstorm the water in the trough is rising at the rate of 1 cm/min when the depth is 5 cm.

    a) How fast is the volume of water in the trough increasing?
    b) After the rain stopped, the water drained out of the trough at the rate of 0.06 m3/min. How fast is the surface of the water falling when the depth is 1 cm?

    3. The attempt at a solution
    V =[tex]\frac{1}{2}[/tex] bhl

    Solving in terms of b:
    b 5
    -- = --
    h 10
    Last edited: Apr 5, 2009
  2. jcsd
  3. Apr 5, 2009 #2


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    Gold Member

    I'm not entirely sure what you're doing here. You're given dl and dh at a certain time and you're asked to determine dV.
  4. Apr 5, 2009 #3
    I know this formula that relates Volume with base, height and lenght of the water trough. The formula is V=1/2 bhl

    From similiar triangles:
    5/10 = b/h
    b=5/10 h

    Then substituting 1/2h for b into the formula we get:
    V=1/2(1/2 h)hl
    V=1/2(1/2 h2 l)
    V=1/4 h2 l
    Now differentiating h we get:
    dV/dt = 2(1/4)h dh/dt l
    dV/dt = 1/2h dh/dt l
    Now submitting in all the values we get:
    dV/dt = 1/2(5 cm)(1 cm/min)(300 cm)
    dV/dt = 750 cm3/min

    However the answer is 1500 cm3/min. Now my question is WHAT HAPPENED!??!?! i did everything correct from everything ive learned but i still dont get the correct answer.... :( please help me understand thank you.

    as for b) ill wait until i get the first part correct.
  5. Apr 5, 2009 #4


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    When the trough is full b=10, and h=10. When the trough is half full b=5 and h=5, right? The correct relation is b=h, not b=(1/2)h. I would rethink the logic that lead you to that.
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