math project...shoot me

Discussion in 'General Discussions' started by lazypando, Feb 8, 2009.

  1. lazypando

    lazypando Level IV

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    So I've been working on this with triple integrals to find the volumes and center of masses and then applying Archimedes' Principle (a body floats in a fluid at the level at which the weight of the displaced fluid equals the weight of the body) but it's taking way too long. >__>
    Anyone have a easier way to solve this?
     
  2. Zer0

    Zer0 Level IV

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    For part 1, I would assume that it IS possible and obtain a contradiction.
    Just find the mass of the cone and the paraboloid (0.5*pi*r^2 * h) with p as a variable
    Then solve for the equation of the paraboloid such that it satisfies Archimedes principle and you should obtain a contradiction somewhere along the way

    Part 2 shouldn't be much harder since all you're adding to the equation is a torus. Just apply Archimedes principle again and solve for a special case that works (keep the paraboloid constant and solve for the equation of the torus such that it satisfies the requirements).

    I don't think there's any need for triple integrals or anything of that sort. Just use some volume formulae.
     
  3. lazypando

    lazypando Level IV

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    eh???
    For part 1, you can't actually find the mass since you don't know density and if you try to solve for density, you'll get something in terms of D. But you don't know D either.

    yeah...I do know what Archimedes' Priciple is since I did post a definition of it....what more, Archimedes discovered the principle while in a bathtub
     
  4. Zer0

    Zer0 Level IV

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    Right, you don't solve for p nor D (in fact its impossible since you're told that there are no such parameters to satisfy those constraints)
    You find the mass in terms of p and D, so M = V*p where V is the total volume
    Then you write the equation for mass of the displaced fluid (should also be in terms of p and D) and set it equal to M

    You should find that the equation has no solutions

    I didn't work out all the math, but this should theoretically work :p
     
  5. lazypando

    lazypando Level IV

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    .......................?
    volume of solid = [(9D+24)/2]*pi
    volume of displaced liquid = [(9D+21)/2]*pi

    so then
    mass of solid = p*[(9D+24)/2]*pi
    volume of displaced liquid = p*[(9D+21)/2]*pi but the density of water is 1 so volume = [(9D+21)/2]*pi

    then with archimedes' principle
    p*[(9D+24)/2]*pi = [(9D+21)/2]*pi
    which gives you
    p = (9D+21)/(9D+24)

    which tell you that unless your D is negative (which is impossible), the thing's going to float
    so then...that mean's I've just went around in a mental circle? o_O
     
  6. Ice Nine

    Ice Nine Level IV

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    This problem is really confusing. I think what you have done is correct, and so I think the problem with the design is the location of the centroid. I don't understand what your teacher means by the center of mass of the displaced liquid. If that is just (2+D)/2 below the surface of the water, then you can solve and show that in order for the centroid of your object to be lower than this, D would have to be negative...I don't know if this is what your teacher means though...

    EDIT: I think D would have to be between -2 and 0

    EDIT EDIT: I must be doing something wrong, because I don't know how the addition of the torus would help...

    EDIT EDIT EDIT: The addition of the torus makes sense because it will keep it afloat while allowing you to increase D (lowering the center of mass), but not from the math that I've done

    EDIT EDIT EDIT EDIT: Okay, last one...I've concluded that the "center of mass of the displaced liquid" is not (2+D)/2, rather it needs to have some dependence on p. I still don't understand what it means though, because the displaced volume of water cannot have a centroid without defining the shape of said volume. In order to define the shape of the volume you have to define the container that the liquid is in, since liquids cannot assume shapes on their own...this is very confusing to me...BUT I'M DONE WITH IT NOW BECAUSE IT'S NOT MY PROBLEM :p
     
  7. Zer0

    Zer0 Level IV

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    Um, pando, I don't think that's the right equation for the volume of the solid. Sum the volume of the cone and the volume of the paraboloid in terms of D, then multiply by p.
     
  8. lazypando

    lazypando Level IV

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    It is. :yup:
    The volume of the solid is the volume of the cone which is (1/3)pi(3^2)(D+4) plus the volume of the paraboloid which is basically the area bounded by the 1st quadrant above the line y = Dx^2/9 and below the line y=D rotated about an axis. But I am certain that those volumes are correct cause I've checked those answers with a few other people already.

    haha yeah...that's basically how I proved that it wasn't possible
    I have about half of it done now