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Homework Help: Exciting riverbed problem

  1. Aug 5, 2009 #1
    1. The problem statement, all variables and given/known data
    The cross-section of a channel is parabolic. It is 3 metres wide at the top and 2 metres deep. Find the depth of water, to the nearest cm, when the channel is half full.

    2. Relevant equations

    3. The attempt at a solution
    I found the function for the graph to be [tex]y=x^2-3x[/tex]. Then I figured that I could just plug (0,0) and (3,0) into the integrand, [tex]\frac{\1}{3}x^3-1.5x^2[/tex], and then divide by two to get the answer. I'm pretty sure that the problem lies with the method, not my math. I have a feeling that's not the above method does not solve the problem, but I'm not sure why that is so.

    For reference, my method gets 225cm while the book has 126cm.
    Last edited by a moderator: Aug 5, 2009
  2. jcsd
  3. Aug 5, 2009 #2

    Gib Z

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    Well, I would have personally interpreted the graph to be such that the parabola passed through the points (-1.5, 2) (0,0) and (1.5, 2). With those points we can find the equation of the parabola, sub in y=1 and solve for x.

    Your equations has a width of 3, but its depth is not 2 as the description states.
  4. Aug 5, 2009 #3
    Ok thanks, now I'm getting 137cm, and its probably because my function is a bit out of the points. Also, Is my latex correct, I'm supposed to see the [tex] stuff right?
  5. Aug 5, 2009 #4

    Gib Z

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    To start tex code, use [ tex ] but to end it its [ /tex ], without the spaces of course.
  6. Aug 5, 2009 #5


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    I have corrected the LaTex. You just need "[/tex]" at the end, not "[ tex ]".

    The simplest way to do this would be to take the origin of your coordinate system to be at the vertex so that the equation is just [itex]y= ax^2[/itex]. That is, the parabola passes through (0,0), (3/2, 2), and (-3/2, 2) and you determine a by [itex]2= a(3/2)^2[/itex].

    From that, the "full" volume is [itex]\int_{-3/2}^{3/2}2- ax^2 dx[/itex] and the depth half full, h, is given by [itex]a\int_{-3/2}^{3/2} h- ax^2 dx= (1/2)\int_{-3/2}^{3/2}2- ax^2 dx[/itex].
  7. Aug 19, 2009 #6
    I agree with Halls of Ivy that you should probably look at the way you're integrating.
    Once you've gotten the proper equation for a parabola, I see two methods. .
    A) Make a rectangle from the x axis up to your line y = 2, ending at the points (+1.5, 2) and (-1.5, 2). What is the area of this box?
    How does the area given by the definite integral relate to this?
    B) More tricky method: Take the inverse of the parabola, and integrate directly the area of water, and use that.
    Last edited: Aug 19, 2009
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