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Posers and Puzzles

Posers and Puzzles

  1. 16 Feb '13 16:52
    Having another go at:
    http://www.redhotpawn.com/board/showthread.php?threadid=142353
    which never really got sorted

    the question is, if you have a paper circle, and you cut a slice out of it (from the middle), and you form both parts into paper cones, then which slice gives the largest difference in volume of the two cones.

    If the angle cut out of the circle is A (which we define as a 2pi_fract ranging from 0 to 1_ then the angle left in the other circle is 1-A

    and since distance round circle is Angle*radius, let us define the circle radius to be 1 so:

    diameter of one cone = A*2pi, diameter of other cone = 2pi - A*2pi

    since D = 2.pi.R, then the radius of the first cone is A and the radius of the second cone is 1 - A

    For the equation of conic volume we also need height, the long side of both circles is 1, and long_side^2 = height^2 + radius^2, so:

    height of cone 1 = sqrt(1-A^2)
    height of cone 2 = sqrt(1-(1-A)^2)

    and finally the equation for conic volume is V = 1/3*pi*r^2*h

    so volume of first cone = 1/3*pi*A^2* sqrt(1-A^2)
    volume of second cone = 1/3pi*( 1 - A)^2* sqrt(1-(1-A)^2)

    and the difference in the two volumes is:
    1/3*pi * (A^2*sqrt(1-A^2) - ( 1 - A)^2* sqrt(1-(1-A)^2))

    we will find the turning points of the difference by finding where the derivative of this difference is zero, so we need to differentiate it, I will work on that next but I'd appreciate it if someone could check the above is correct!

    Part of the differentiation is obviously going to involve variable substitution as the two part of the equation are both of the form X^2*sqrt(1-X^2)
  2. 16 Feb '13 19:27
    ok, first let us take the challenge of differentiating X^2*sqrt(1-X^2)

    x^2*sqrt(1-x^2)

    d/dx {X^2*sqrt(1-x^2)}

    the product rule for differentiation is:

    d/dx (uv) = v du/dx + u(dv/dx)

    so u = X^2 and V = sqrt(1-x^2)

    du/dx = sqrt(1-x^2).2x + x^2.d/dx{sqrt(1-x^2)}

    now d/dx{sqrt(1-x^2)} has to be done by the chain rule
    let u = 1-x^2 then d/dx{sqrtt(u)} = du/dx * d/du{sqrt(u)}

    so d/dx{sqrt(1-x^2)} = -2x*1/(2.sqrt(1-x^2) = -x/sqrt(1-x^2)

    and therefore

    d/dx {X^2*sqrt(1-x^2)} =
    2x.sqrt(1-x^2)-x^3/sqrt(1-x^2) =
    (2x - 2x^3 - x^3)/sqrt(1-x^2) =
    (2x - x^3)/sqrt(1-x^2)
  3. 16 Feb '13 22:24
    ok, so now we are trying to find:

    d/dx {1/3*pi * (A^2*sqrt(1-A^2) - ( 1 - A)^2* sqrt(1-(1-A)^2)) }

    and we know

    d/dx {X^2*sqrt(1-x^2)} = (2x - x^3)/sqrt(1-x^2)

    the (A^2*sqrt(1-A^2) is the differential we have already worked out, so we are left with

    -d/dA{( 1 - A)^2* sqrt(1-(1-A)^2)}

    let u = 1-A

    then du/dA = -1 and -d/dA{( 1 - A)^2* sqrt(1-(1-A)^2)} =
    d/du{u^2*sqrt(1-u^2)
    = (2u - u^3)/sqrt(1-u^2) {from our previous working}

    I'll have to check that carefully so that is all for now
  4. Subscriber joe shmo On Vacation
    Strange Egg
    17 Feb '13 03:04
    Just trying to throw something out there ( and I didn't work this out), but how about changing the complex algebra, to a trigonometric function by trigonometric substitution, then doing the differentiation with the dummy variable?

    ie,

    sin(B) = A^2/(Sqrt(1-A^2)) ect...

    V = 1/3*pi*(sin(B)-...)
  5. Subscriber joe shmo On Vacation
    Strange Egg
    17 Feb '13 03:28
    never mind... i'm sleeping(that's multiplication, not division)!
  6. 22 Feb '13 22:52 / 3 edits
    Originally posted by iamatiger
    ok, so now we are trying to find:

    d/dx {1/3*pi * (A^2*sqrt(1-A^2) - ( 1 - A)^2* sqrt(1-(1-A)^2)) }

    and we know

    d/dx {X^2*sqrt(1-x^2)} = (2x - x^3)/sqrt(1-x^2)

    the (A^2*sqrt(1-A^2) is the differential we have already worked out, so we are left with

    -d/dA{( 1 - A)^2* sqrt(1-(1-A)^2)}

    let u = 1-A

    then du/dA = -1 and -d/dA{( 1 - A)^2* sqr ...[text shortened]... -u^2) {from our previous working}

    I'll have to check that carefully so that is all for now
    oops, noticed an error:

    d/dx {X^2*sqrt(1-x^2)} = (2x - x^3)/sqrt(1-x^2) was wrong, it should read

    d/dx {X^2*sqrt(1-x^2)} = (2x - 3x^3)/sqrt(1-x^2)

    so now with u=1-a
    d/dA{( 1 - A)^2* sqrt(1-(1-A)^2)} = (2u - 3u^3)/sqrt(1-u^2)

    =(1-A)(2 - 3(1-A)^2)/sqrt(2A-A^2)

    so
    d/dx {1/3*pi * (A^2*sqrt(1-A^2) - ( 1 - A)^2* sqrt(1-(1-A)^2)) }

    =pi/3 *( A(2 - 3A^2)/sqrt(1-A^2) + (1-A)(2 - 3(1-A)^2)/sqrt(2A-A^2) )

    so, to find the maximum and minimum of the required function we now have to solve:

    A(2 - 3A^2)/sqrt(1-A^2) + (1-A)(2 - 3(1-A)^2)/sqrt(1 - (1-A)^2) = 0

    this is very tricky!
  7. 23 Feb '13 20:42 / 3 edits
    I'm fairly sure the equation has derivable direct solution (we can turn it to an order 9 polynomial but that is no better), so I have had to resort to numeric methods which give A = 0.852022572512388 or A = 1 - 0.852022572512388, which is a 306.728 degree slice