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Vector field

  1. May 16, 2007 #1
    1. The problem statement, all variables and given/known data

    A particle is attracted toward the origin by a force proportional to the cube of it's distance from the origin. (...)

    What would be this Force equal to (in xy plane)?

    3. The attempt at a solution

    So distance is Sqrt[x^2+y^2]... and from here I don't know what to do...

    F=( ? i , ? j)
  2. jcsd
  3. May 16, 2007 #2


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    The force should also point in the direction of the origin. So consider F=k*(x*i,y*j). The magnitude of this force is directly proportional to distance from the origin. How would you change it to get the right proportionality (BTW - did you mean 'inversely' proportional to cube of the distance?).
  4. May 16, 2007 #3
    F=k*(x^3*i,y^3*j), k<0 ?

    Is this good then?

    // I did not mean inversely.
  5. May 16, 2007 #4


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    That i) does not point towards the origin anymore and ii) is NOT proportional to distance cubed. It's proportional to sqrt(x^6+y^6). Distance cubed is sqrt(x^2+y^2)^3. Not at all the same thing. If (x,y) is already proportional to distance, why not just multiply it by distance squared?
  6. May 16, 2007 #5


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    Dick SAID you must F=k*(x*i,y*j). Here k is "proportional to the cube of it's distance from the origin". Yes, the "distance from the origin is [itex]\sqrt{x^2+ y^2}= (x^2+ y^2)^{1/2}[/itex]. What is the cube of that? And since the particle is "attracted", the vector must be directed toward the origin. (xi, yj) is directed away from the origin.
  7. May 16, 2007 #6
    TO DICK.

    F=(x*i, y*j)
    DIV F = d/dx(x) + d/dy(y) = 2 which means it goes outward and not to the origin...

    // EDIT

    F=-(x^2 + y^2)^3/2 * (x*i, y*j)
    Last edited: May 16, 2007
  8. May 16, 2007 #7


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    You can point it in the right direction by putting a -k in front. But we're getting there. Your latest effort points in the right direction, but (x,y) is proportional to distance and (x^2+y^2)^(3/2) is proportional to distance cubed. So the product is proportional to distance to the fourth. You just need one minor adjustment.
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