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Math Help - more challenging related rates cone problem

  1. #1
    Eater of Worlds
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    more challenging related rates cone problem

    Everyone has seen the old, cliche related rates problem where a cone is being filled with water. How fast is the height of the water changing at the instant the water is h units deep...yada, yada?.

    In these problems, the apex of the cone is always pointed down.

    What if the cone were laying on its side?. Makes it a little tougher, huh?.

    Suppose we have a conical tank with radius 10 ft and height 24 feet. Water is flowing in at 20 ft^3 per minute. How fast is the depth increasing when the water is 6 feet deep?. Only the cone is on its side, not apex down.
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    Bar0n janvdl's Avatar
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    Quote Originally Posted by galactus View Post
    Everyone has seen the old, cliche related rates problem where a cone is being filled with water. How fast is the height of the water changing at the instant the water is h units deep...yada, yada?.

    In these problems, the apex of the cone is always pointed down.

    What if the cone were laying on its side?. Makes it a little tougher, huh?.

    Suppose we have a conical tank with radius 10 ft and height 24 feet. Water is flowing in at 20 ft^3 per minute. How fast is the depth increasing when the water is 6 feet deep?. Only the cone is on its side, not apex down.
    I'd love to see the solution to this one...
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    I, too, J. I was hoping some smarty pants would tackle it. I may ponder it today. Maybe relate it in terms of an angle to the x-axis?. Here is a haphazard diagram.
    Last edited by galactus; November 24th 2008 at 05:38 AM.
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    Bar0n janvdl's Avatar
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    What if we just swap b and h around? So if the base was 2 and the height 3, why not make the base 3 and the height 2? I think an approach like this might be the easiest, but keep in mind that the numbers are just for explanatory purposes. We may have to find the height and base using trig i think.
    Last edited by janvdl; October 29th 2007 at 11:33 PM. Reason: Mistake
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    As the water rises up the cone, we have a horizontal plane intersecting a cone. That makes a parabola. We need to find an equation that relates that.

    Does anyone know enough about conic sections to find the equation of this parabola?. I must admit I am at a loss as to how to. It appears we have a line which makes up a slant of the cone and a plane which slices the cone horizontally and creates parabolas. Perhaps we have to find the angle to relate them?.

    The equation of a cone is z^{2}=ax^{2}+ay^{2}

    The equation of a plane is z=ay+c

    If you set these equal you get y=\frac{a}{2c}x^{2}-\frac{c}{2a}

    How to find a and c?.

    I have asked others about this problem but this is all the further I have gotten.
    Last edited by galactus; October 31st 2007 at 03:42 PM.
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    Well, kids, I may have a solution to this ugly related rates problem.

    When we slice a cone with a plane, we get a parabola.

    The rate of change of volume is equal to the cross sectional area at that instant times the rate of change of the height.

    \frac{dV}{dt}=A(t)\cdot\frac{dh}{dt}

    \frac{dh}{dt}=\frac{\frac{dV}{dt}}{A(t)}

    That means, \frac{dh}{dt}=\frac{20}{\text{area of parabolic sector formed by plane at h=6}}

    So, I set about using a little trig and geometry to find various angles and lengths.

    Per the diagram:

    law of cosines to find angle BAC => 10=\sqrt{26^{2}+26^{2}-2(26)(26)cos(BAC)}=45.24 \;\ degrees

    6cot(45.24)=5.95

    angle ACB = 67.38

    6cot(67.38)=2.50

    26-(2.5+5.95)=17.55

    Now, look down the 'throat' of the cone. It's a circle with a plane cut through it. This makes a parabola.

    The width of the parabola where it enters the circle making up the base of the cone is 18. Which I found by using geometry:

    The small triangle with base GC has hypoteneuse 6.5

    So, the middle ordinate is given by:

    6.5=10(1-cos(x/2)), so we can solve for the angle x subtended by the chord formed by the plane slicing through the cone.

    x=2cos^{-1}(7/20)=139.025 \;\ degrees

    chord at 6.5 feet is given by:

    20sin(139.025/2)=18.735

    The length of it is 17.55.

    So, the area of the parabola is \frac{2}{3}(17.55)(18.735)=219.20

    Now, all we have to do is divide it into 20.

    \boxed{\frac{20}{219.20}=0.0912 \;\ ft/min \;\ or \;\ 1.095 \;\ in/min}

    This sounds very reasonable. I am confident it is correct.
    Last edited by galactus; November 24th 2008 at 05:38 AM.
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