1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    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
    5h=10b
    b=5/10 h
    b=1/2h

    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

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Horizontal Eavestrough Problem
  1. Horizontal Asymptotes (Replies: 2)

  2. Horizontal asymtote (Replies: 26)

Loading...