# Flow line in conservative vector field

## Homework Statement

Recall that a flow line, c(t), of a vector field F has c'(t)=F(c(t)) at all times t. Show all work below.
a.) Let c(t) be the flow line of a particle moving in a conservative force field F=-grad(f), where f:R^3->R, f(x,y,z) >=0 for all (x,y,z), represents the potential energy at each point in space. Prove that the particle will always move towards a point with lower potential energy. What is the limit of F(c(t)) as t goes to infinity?

c'(t)=F(c(t))

## The Attempt at a Solution

We are trying to show that f(c(t)) is a decreasing function of t. So far, I have that

But I am not sure if this really helps at all. Is this the right direction to go in, or am I completely off base?

STEMucator
Homework Helper
Recall that vectors in a vector field are tangent to the flow lines. Since ##\vec c(t)## is a flow line, you know that ##\vec F(\vec c(t)) = \vec c'(t)##. That is, the vector field of the flow line is equal to the tangent to the flow line for all times ##t##.

You know ##\vec F## is conservative, and so ##\vec F = - \vec{\nabla f} = -(f_x \hat i + f_y \hat j + f_z \hat k)## for some potential function ##f(x, y, z) \geq 0##. The gradient of a function gives the direction of fastest increase. So minus the gradient gives the direction of fastest decrease.

Hence ##\vec F(\vec c(t)) = - \vec{\nabla f}(\vec c(t)) = \vec c'(t)##. Notice the tangent to the flow line is equal to the negative of the gradient of the potential function (which is decreasing). So it must be the case ##\vec c'(t)## is also decreasing. So as the particle moves along the flow line ##\vec c(t)##, its potential energy is decreasing.

Now what about ##\displaystyle \lim_{t \rightarrow \infty} \vec F(\vec c(t))##?