# Work done by Three Dimensional Inverse Square Field

1. Apr 10, 2016

### RyanTAsher

1. The problem statement, all variables and given/known data

Find the work done by the three-dimensional inverse-square field

$F(r) = \frac {1} {||r||^3} r$

on a particle that moves along the line segment from P(6, 2, 3) to Q(4,2,4)

2. Relevant equations

$\int_C F \bullet dr = \int_a^b f(h(t), g(t)) \sqrt {(\frac {dx} {dt})^2 + (\frac {dy} {dt})^2} = \int_a^b f(h(t),g(t)) ||\vec r'(t)|| dt$

3. The attempt at a solution

I cannot figure out where to start. I've been generally confused by line integrals the entire section. The only relationship I can see is with the 3rd equation I provided in the relevant equations where we have a magnitude $||\vec r'(t)||$. But then what would be the function F? Is it the $\frac {1} {||r||^3} r$? But that is a function of F(r), so I'm not sure how this changes the dynamics of the problem...? The most I could thing of would be along the lines of

$\int_C F \bullet dr = \int_a^b f(h(t),g(t)) ||\vec r'(t)|| dt$

$\int_C F \bullet dr = \int_a^b \frac {r} {||r||^3} ||\vec r'(t)|| dt$

but now we have a function of r AND t inside this integral, but the integral is with respect to t, so could I just pull the function of r out?

$\int_C F \bullet dr = \frac {r} {||r||^3} \int_a^b ||\vec r'(t)|| dt$

but then what is the function for r(t), I don't understand parameterization very well, and I'm not sure if I'm supposed to make some kind of parametric substitution?

2. Apr 10, 2016

### BvU

What did you do with the dot product ? In your first equation you pretend $\vec F$ and $d\vec r$ are collinear -- they are not !

And your $F(r) = \frac {1} {||r||^3} r$ should be $\vec F(\vec r) = \frac {1} {||r||^3} \vec r$ or
$\vec F(\vec r) = \frac {1} {\vec r\cdot\vec r} \hat r$

3. Apr 10, 2016

### RyanTAsher

I'm not sure I follow. Since F is a function of r in this case, unlike it is usually a function re-parameterized, I'm not sure how to deal with F as a function of r, in general.

4. Apr 10, 2016

### BvU

A force is a vector. Your inverse-square forces point away from the origin. So the force vector at position $(x,y,z)$ has a magnitude that only depends on $|\vec r| = \sqrt{x^2+y^2+z^2}$ but a direction that depends on each of x, y and z.

You path is from $(6,2,3)$ to $(4,2,4)$. I could draw the path as a line segment in the plane $y = 2$, but that doesn't help much. Your job (as I understand it *) is to find a way to express $\vec F\cdot d\vec s$ ( I use $\vec s$ for the path direction, to distinguish from the position vector $\vec r$) and integrate.

*) there are other more indirect ways to calculate this work, but they depend on establishing that $\vec F$ is conservative and finding the corresponding potential field; my impression is that you are supposed to actually do the integral in this exercise.