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Math Help - Differential Equations

  1. #1
    Member classicstrings's Avatar
    Mar 2006

    Differential Equations

    A water trough, 6 metres long, has uniform cross section in the shape of a semicircle with horizontal diameter 2 metres. It contains water to a depth of x metres. THe centre of the circle is at O and the triangle POQ is isosceles with angle POQ = @.

    I've managed to get parts. But am stuck..

    a. Find the area of the segment that is filled with water.
    A = 0.5(@-sin@)

    b. Find the volume of water in trough.
    V = 6*A = 3(@-sin@)

    c. If x metres is the depth of the water in the trough, find
    i. x in terms of @
    ii. dx/d@
    iii. dx/dt in terms of @ and d@/dt

    d. Find dV/dt.
    dV/dt = 3(1-cos@)*(d@/dt)

    e. If d@/dt = 0.05 rad/min, when @ = pie/3, find
    i. Rate in m^3 per min at which volume of water is increasing.
    ii. Rate in metres per min at which water level is rising.
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  2. #2
    Eater of Worlds
    galactus's Avatar
    Jul 2006
    Chaneysville, PA
    May I ask, why would someone in a calculus or DE class spell \pi as 'pie'?.

    This is a math forum, not a baking forum

    The depth of the water in the trough is the 'middle ordinate' of the segment.

    It is given by x=R(1-cos(\frac{\theta}{2}))=1-cos(\frac{\theta}{2})
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