Find work done by the force field F in moving the particle from $(-1, 1)$ to $(3, 2)$

This sounds good till we are given that $\textbf{F} = \dfrac{2x}{y}\textbf{ i }- \dfrac{x^2}{y^2}\textbf{ j }$

Can someone explain how to understand this problem, does the word conservative field have a meaning if the field is not continuous everywhere

  • $\begingroup$ It is continuous for example when $y>0$. And both of the given points satisfy this condition. So yes, conservative field can be defined on some subset of the entire space. $\endgroup$ – Quang Hoang Aug 25 '14 at 13:33
  • $\begingroup$ but how do we find the workdone, isnt the workdone path dependent, from where cut the x-axis $\endgroup$ – Holy cow Aug 25 '14 at 13:40
  • $\begingroup$ It is path independent whenever the path lies in the domain of the field. For an example, think about the water flow inside of a container. $\endgroup$ – Quang Hoang Aug 25 '14 at 13:42
  • 1
    $\begingroup$ @QuangHoang, you mean path-independent, of course. $\endgroup$ – Travis Aug 25 '14 at 13:45

The vector field is continuous on its domain, which is $\mathbb{R}^2 - \{y = 0\}$. It's nevertheless quite reasonable to ask how not being defined on the $x$-axis affects a given problem. For integrations of functions along curves, it just means that the paths we consider must trace out curves contained entirely within the domain; in our situation, this just means we're making a statement about the work done by $\mathbf{F}$ along some curve from $(-1, 1)$ to $(3, 2)$ that doesn't cross the $x$-axis (and of course, conservative means that the work doesn't depend on which such path the particle takes).

Vector fields not being defined everywhere are quite common in practice, maybe most importantly as the force fields generated by point-sources where the force equation obeys an inverse-square law.


You can check that there exists a potential function, $U$, such that $\mathbf{F} = -\nabla U $, since $\nabla \wedge \mathbf{F} = 0$ (then, $\mathbf{F}$ is known as a force field). You can also verify that $U$ is given by:

$$ U(x,y) =- \frac{x^2}{y},$$

which is continuous in its domain, which is the same as the domain of $\mathbf{F}$.

Therefore, you can simply compute the work done by the (conservative) force $\mathbf{F}$ as the difference of potential between the two points $A(-1,1)$ and $B(3,2)$, that is to say:

$$\color{blue}{W_{A\to B} = U(A)-U(B)}$$

Hope this helps.


Some thoughts:

An example of such a force which is not defined somewhere would be the force created by a punctual mass $M$ actuating on a mass $m$, which is given by:

$$ \mathbf{F} = - G \frac{Mm}{|\mathbf{r}|^3} \mathbf{r}.$$

The gravitational potential, $\phi = G M m/|\mathbf{r}|^2$ is not defined at $(x,y,z) = (0,0,0)$ but we can still compute the work done by gravity when actuating along a certain path. We call this potential energy.


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