Solving F(r) and Finding Potential Function in r

[ElDavidas]

I'm going over an exam question and it reads as follows:

"Determine if the force F(r) is conservative (justify your answer) and, if it is, find a potential function in the case

F(r) = \frac {1} {||r||} (-xi + yj)

where r = xi + yj"

I know that F = -grad V where V is the potential function. In order to find the potential function V from the above, do you integrate grad V for each unit vector i and j?

Another thing is, I'm not very sure how to manipulate the r. Would you use the quotient rule in order to find \frac {dV} {dydx} as an example?

If somebody managed to answer the question and display their answer showing how they did it, that would be fantastic!

 

[HallsofIvy]

||r|| is, of course \sqrt{x^2+ y^2} or (x²+ y²)^1/2.

F(x,y)= -x(x²+ y²)^-1/2i+ y(x²+ y²)^-1/2j

If we write that as f(x,y)i+ g(x,y)j then \frac{\partial f}{\partial y}= (-x)(-\frac{1}{2})(x^2+ y^2)^{-\frac{3}{2}}(2y)= xy(x^2+y^2)^{\frac{-3}{2}}.

 

[ElDavidas]

\frac{\partial f}{\partial y}= (-x)(-\frac{1}{2})(x^2+ y^2)^{-\frac{3}{2}}(2y)= xy(x^2+y^2)^{\frac{-3}{2}}.

So is this the potential function of V?

 

[Benny]

This looks like a fairly well fudged problem.

If the field is conservative then it is the gradient of a scalar function. If you were to 'show' that it is conservative by finding such a scalar function then aren't you assuming that the field is conservative in the first place? It's probably better to show that curl F = 0 (I haven't done the computation for this particular field).

As for finding the scalar function (assuming that the field if converative) I would start by writing the field as:

<br /> \mathop F\limits^ \to = - \frac{x}{{\sqrt {x^2 + y^2 } }}\mathop i\limits^ \to + \frac{y}{{\sqrt {x^2 + y^2 } }}\mathop j\limits^ \to <br />

<br /> \nabla \phi \left( {x,y} \right) = \mathop F\limits^ \to <br />

To determine phi, there are various procedures. Here is the one I use.

<br /> \phi \left( {x,y} \right) = \int {\frac{{\partial \phi }}{{\partial x}}} dx<br />

<br /> = \int {\left( { - \frac{x}{{\sqrt {x^2 + y^2 } }}} \right)} dx<br /> by definition since \nabla \phi \left( {x,y} \right) = \frac{{\partial \phi }}{{\partial x}}\mathop i\limits^ \to + \frac{{\partial \phi }}{{\partial y}}\mathop j\limits^ \to = \mathop F\limits^ \to

<br /> = - \sqrt {x^2 + y^2 } + A\left( y \right) + c<br />

Now differentiate the above expression for phi with respect to y and you'll get some expression. But the expression you get is also equal to something else, it should be easy to deduce just what I'm referring to. Consider the components of F and the relation between phi and F, I highlighted it earlier in my post.

On second thought, I'm not so sure that grad(anything) = F. You could try to compute to curl to check if I'm wrong (I wouldn't be suprised it's 12:55 in the morning over here :zzz: ). But just from looking at the equation for F, it seems to lack some kind of necessary 'symmetry'. Nevermind, it's just me. Probably best to wait for some other responses. I hope that was of some help.