Finding the potential energy of a conservative force

Question

My solution to show whether each of the forces is conservative is to check whether $\textrm{curl}\, \vec {F}$ is zero. It is $0$ if conservative and not $0$ if not conservative, which I have no problem with. But how can one find the potential energies of $\vec{F}$ from that relationship? I don't really have a clue, and any help would be highly appreciated.

Answer 1

Compute the line integral $\int_{\mathbf{a}}^{\mathbf{x}} \mathbf{F} \cdot d\mathbf{r}$, where $\mathbf{a}$ is an arbitrary point. If $F$ is conservative, this is well-defined and has the right derivative (gradient), by the Fundamental Theorem of Calculus for line integrals, and hence it is $U$, up to a constant.

For example, the first one is conservative, and the integral can be computed as $$ U(X,Y,Z) = \int_{0}^{(X,Y,Z)} x \, dx + 2y \, dy + 3z \, dz = \frac{1}{2}(X^2-0) + (Y^2-0) + \frac{3}{2}(Z^2-0). $$ The second one is also conservative, and $x\frac{dy}{dt} + y\frac{dx}{dt} = \frac{d}{dt}(xy), $ so the potential is $U(X,Y,Z)=XY$.

Answer 2

As an example, if ${\bf F} = (y,x,0)$, then ${\bf F} = \nabla f$ where $f_x(x,y,z) = y$, $f_y(x,y,z) = x$, and $f_z(x,y,z) = 0$. Now, an antiderivative of $f_x(x,y,z)$ with respect to $x$ is $xy$, an antiderivative of $f_y(x,y,z)$ with respect to $y$ is $xy$, and an antiderivative of $f_z(x,y,z)$ with respect to $z$ is $0$. Adding together all unique terms of those antiderivatives gives us that $f(x,y,z) = xy + C$ where $C = 0$ in this case since $f(0,0,0) = 0$ by assumption. (Here I used $f$ instead of $U$; sorry).

Answer 3

If $\bf F$ is conservative, it has some potential $U$, that is, a function for which ${\bf F} = -\nabla U$. Writing this equation in components gives thesystem $$ \left\{ \begin{array}{rcl} F_x &=& -\frac{\partial U}{\partial x}\\ F_y &=& -\frac{\partial U}{\partial y}\\ F_z &=& -\frac{\partial U}{\partial z} \end{array} \right. $$ of partial differential equations. Finding a potential amounts to solving this system.

So, for example, for the field $\bf F$ in (a), the system is $$-\frac{\partial U}{\partial x} = x, \qquad -\frac{\partial U}{\partial y} = 2 y, \qquad -\frac{\partial U}{\partial z} = 3 z .$$ Integrating the first equation gives that $U$ must have the form $$U(x, y, z) = -\frac{1}{2} x^2 + A(y, z)$$ for some $A(y, z)$, and likewise integrating the second and third equations give respectively that $U$ can also be written as $$U(x, y, z) = -y^2 + B(x, z) = -\frac{3}{2} z^2 + C(x, y) .$$ Putting these forms together gives that $f$ has the form $$U(x, y, z) = -\frac{1}{2} x^2 - y^2 - \frac{3}{2} z^2 + K$$ for some constant $K$. We can check that all such functions are solutions, by the way, simply by computing $-\nabla U$ and checking that it coincides with the given $\bf F$.