Why does $F$ being conservative guarantee that $\operatorname{curl} F = 0$

Question

Clairaut's Theorem guarantees that

If $f$ is a function of three variables that has continuous second-order partial derivatives, then $\operatorname{curl} f = 0$. This can be confirmed by evaluating $\nabla \times \nabla f$.

Now here's where I'm lost: apparently, a consequence of this is that

If F is conservative, then $\operatorname{curl}$ F $ = 0$

where F is a vector field.

Apparently the above follows trivially from the initial statement guaranteed by Clairaut's Theorem, but I don't see how/why it follows. Why does $F$ being conservative guarantee that $\operatorname{curl} F = 0$ ?

Answer 1

You have it the wrong way around. Zero curl doesn't imply conservative without some assumptions on the actual space in question (like if the field is defined and differentiable on all of $\Bbb R^3$). But that's another question entirely.

What they're saying is that conservative implies zero curl. Their second statement follows from the first in that if $\mathbf F$ is a conservative field, then there is some $f$ such that $\nabla f = \mathbf F$. Thus $\nabla \times \mathbf F = \nabla\times \nabla f = 0$.

Answer 2

As an addendum to Arthur's answer, the reason your text is saying that the conclusion follows from Clairaut's theorem is that if $F$ is conservative, $F = \nabla f$ for some scalar function $f$. To see this, pick an arbitrary point $x_0$ and define $$f(x) = \int_0^1 \langle F[\gamma(t)], \gamma'(t)\rangle \,dt$$ where $\gamma:[0,1]\to\mathbb{R}^3$ is an arbitrary smooth curve from $x_0$ to $x$. The definition of a conservative vector field is exactly the condition that this integral is independent of $\gamma$ and that therefore $f$ is well-defined and smooth. Now for any vector $v$, pick a path $\gamma$ that ends with $\gamma'(1) = v$, and you have $$\langle\nabla f ,v\rangle = \frac{d}{dh} \int_0^{1+h}\langle F[\gamma(t)], \gamma'(t)\rangle \,dt = \langle F(x), v\rangle.$$