# How to create a path defined by the gradient vector?

## Recommended Posts

Suppose I have a function $f(x,y)$, along with a gradient at some arbitrary point $(x_0, y_0)$

The surface is smooth and $C^1$, there thus also exists some gradient vector:

$<\frac{df}{dx}(x_0, y_0) , \frac{df}{dy}(x_0, y_0)>$ I want to "follow" this vector to the next point $(x_1, y_1)$, here there is another gradient vector $<\frac{df}{dx}(x_1, y_1) , \frac{df}{dy}(x_1, y_1)>$, I repeat this process for up to $(x_n, y_n)$.

How do I write an expression for the curve? It seems like I would have to integrate along a vector somehow?

Edited by rakuenso
##### Share on other sites

You need to solve equations like

$\frac{d x^{A}}{dt} = X^{A}(x(t))$

in a given chart for the curves. You need some initial condition $x^{A}(0) = x^{A}_{0}$ and then by the existence and uniqueness theorem of ODEs you are guaranteed a solution (at lest locally).

It maybe possible that the integral curve is only defined for some subset of $\mathbb{R}$. You will have to be careful that your parameter $t$ does not exceed the interval.

##### Share on other sites

Are $x^A$ my variables? like $x=x^1,y=x^2$ etc. And what erxactly is $X^A$?

How do you arbitrarily parametrize $x^A$ wrt to t? Or is this $x^A(t)$ exactly what I'm trying to solve?

I also have difficulties seeing where the integration along a "vector" part comes in.

For two dimensions, so far I've written the $n$ th point $(x_n, y_n)$ as the following, over some interval tiny interval $t$ (in the case of the unit vector it is equal to the magnitude of the vector)":

$x_n = x_0 + \frac{1}{t} [\frac{\partial f}{\partial x}(x_0,y_0) + \frac{\partial f}{\partial x}(x_1,y_1) + ... + \frac{\partial f}{\partial x}(x_n,y_n) ]$

$y_n = y_0 + \frac{1}{t} [\frac{\partial f}{\partial y}(x_0,y_0) + \frac{\partial f}{\partial y}(x_1,y_1) + ... + \frac{\partial f}{\partial y}(x_n,y_n) ]$

note that $(x_1,y_1), (x_2,y_2) .. etc$ can be found using the above definition

The path then would be comprised of the points $(x_0,y_0), (x_1,y_1) ... (x_n, y_n)$, which when taken over the $\displaystyle\lim_{n\to\infty}\displaystyle\lim_{t\to\infty}$ becomes continuous and never ending. Surely, given some function $f(x,y)$ this path can be traced out using numerical methods, but a method of finding an analytic solution seems much more difficult.

Edited by rakuenso
##### Share on other sites

By curve you are looking for $x^{A}(t)$.

I think you need to do some reading on vector fields, integral curves and flows. All fundamental things in differential geometry.

##### Share on other sites

eh? I've done calc II and we went over quite a bit of vector calculus. But all the stuff we did dealt with flows, and integrals of dot products (which is just a scalar), and stuff like stoke's theorem. here it seems like I'm integrating an actual vector as opposed to a dot product. So the problem is analagous to finding the integral curves for a vector field (which here is the gradient) using ODEs?

nm I think I get what you mean now, if my vector field is $< \frac{\partial V}{\partial x}, \frac{\partial V}{\partial y} >$, say $V=\frac{1}{|x-y|}$ then I would form the autonomous system, as it is independent of time:

$\frac{dx}{dt} = \frac{\partial V}{\partial x} = \frac{-1}{|x-y|(x-y)}$

$\frac{dy}{dt} = \frac{\partial V}{\partial y} = \frac{1}{|x-y|(x-y)}$

so I would have to look for an integral curve with components (x(t), y(t)) that obeys the above equations

Edited by rakuenso
##### Share on other sites

Yes. You will need some initial condition, which I think you have stated in your original post. You can appeal to known theorems of ODEs to help you if needed.

## Create an account

Register a new account