why is a gradient vector field conservative?

My textbook makes a rather strange remark about gradient fields. It states that if $F=\nabla\phi$ for some function $\phi$, then F is a conservative field.

This seems like it has to be incorrect, because if we say that $\dot x =\nabla \phi(x)$ then $\phi$ has to increase as $t$ increases, and so cannot be conserved.

So why do they call $F=\nabla\phi$ a conservative field?

• Maybe I'm just not remembering something basic since it's been so many years, but why does $\dot x = \nabla \phi(x)$ mean that $\phi$ has to increase as $t$ increases? – tilper Sep 20 '16 at 13:51
• because, unless I'm mistaken, $\nabla$ gives the vector that points into the direction in which the rate of change of $\phi$ is greatest. This must either be 0 (if the function is maximized), or positive, because if $\phi$ is continuous and differentiable, then at any point at which the function can go up, by a small change of $x$, it must also go down if $x$ is changed in the other direction. – user56834 Sep 20 '16 at 13:58
• What do you understand by words "conservative field"? – Canis Lupus Sep 20 '16 at 14:02
• I think maybe you're just not understanding what the word conservative means in this context. Conservative means path-independent. That is, if $\gamma_1$ and $\gamma_2$ are two different paths starting from point $A$ and ending at point $B$ (which are both connected and differentiable and contained in the domain of $F$, yada yada), then $$\int_{\gamma_1} F\cdot dr = \int_{\gamma_2} F\cdot dr$$ – user137731 Sep 20 '16 at 14:27
• You are correct that if a path $x$ satisfies $\dot{x} = \nabla\phi(x)$, then $\phi(x)$ is nondecreasing in $t$. This can be shown by noting that $\dot{(\phi\cdot x)} = \nabla\phi(x)\cdot\dot{x} = |\nabla\phi(x)|^2$. In particular, any path satisfying $\dot{x} = \nabla\phi(x)$ cannot intersect itself unless it reaches a zero of $\nabla\phi$. The issue of whether a vector field is conservative or not depends on evaluating line integrals over paths with the same endpoints, or equivalently evaluating a line integral over a closed path. (1/2) – Joey Zou Sep 21 '16 at 2:13