what path reduces magnetic field strength I'm having trouble solving this particular problem in Colley's Vector Calculus.  I believe my trouble lies in not being able to set up a differential equation.  Here is the problem;
Igor, the inchworm, is crawling along graph paper in a magnetic field.  The intensitiy of the field at the point $(x,y)$ is given by $M(x,y)=3x^2+y^2+5000$.  If Igor is at the point $(8,6)$, describe the curve along which he should travel if he wishes to reduce the field as rapidly as possible.
So, the section this is in is Directional Derivatives and the Gradient, and so this is a gradient problem.  Since we are minimizing magnetic field strength, we are looking for $-\nabla{M(x,y)}$ and we can easily find this at any point.  In this case, $(8,6)$.  But how do we go about doing this starting at a point and following the path for the entire magnetic field?  
EDIT:  If it isn't necessary, please don't give me the answer.  Guidance would be much more appreciated, since I want to understand what is happening here.
 A: Since Igor always moves in the direction of fastest decrease of $M(x, y)$, he follows a path $(x(s), y(s))$ such that $\frac{d}{ds}(x(s), y(s)) = -\nabla M(x, y)$, where $s$ is the distance or arc-length along $(x(s), y(s))$.  Now $\nabla M(x,y)= (6x, 2y)$, so Igor's path satisfies the system of ordinary differential equations $x'(s) = -6x$, $y'(s) = -2y$.  How convenient!  The variables are already seperated for Igor and ourselves!  It is easy to see, by solving this simple, decoupled system,  that $x(s) = x_0e^{-6s}$ and $y(s) = y_0e^{-2s}$, where the initial point is $(x_0, y_0)$.  So Igor's path is $(8e^{-6s}, 6e^{-2s})$.  Note that this path, as does any path satisfying $\frac{d}{ds}(x(s), y(s)) = -\nabla M(x, y)$, heads right into the point $(0, 0)$, where $M(x, y)$ clearly attains its global minimum value of $5000$.   And, there I suppose, Igor gets stuck! (Actually, he never quite gets there! ;))  Don't know exactly what the graph paper has to do with this, though . . .   ;)
A: Your system of differential equations is
$$
\begin{cases}
x'(t) = - 6 x(t)\\
y'(t) = - 2 y(t)
\end{cases}
$$
which can be easily seen is solved by $x(t) = e^{-6t}$, $y=e^{-2t}$.
