# Vector field

Logik

## Homework Statement

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)?

## 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)

Homework Helper
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?).

Logik
F=k*(x^3*i,y^3*j), k<0 ?

Is this good then?

// I did not mean inversely.

Homework Helper
F=k*(x^3*i,y^3*j), k<0 ?

Is this good then?

// I did not mean inversely.

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?

Homework Helper
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 $\sqrt{x^2+ y^2}= (x^2+ y^2)^{1/2}$. 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.

Logik
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)

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