why are conservative vector fields curl-free?

Question

The book told me that, if a vector field $\vec{F} = Mi + Nj$ is conservative, then

$$ M_y = N_x $$

But why is this true?

Answer 1

$$M=f_x$$ $$N=f_y$$ $$M_y=\cdots$$ $$N_x=\cdots$$

Answer 2

The important idea is that if $f$ is of class $C^2$ (meaning it is at least twice differentiable, and those derivatives are continuous), then mixed partials are equal (which is called Clairaut's theorem), and therefore a quick calculation shows that the curl of a gradient is zero.

For what it's worth, you should be aware that the requirement that $M_y=N_x$ is only a necessary and not a sufficient condition for $F$ to be a conservative vector field. There are topological obstructions which prevent curl-free vector fields from being conservative, a fact which marks the beginning of de Rham cohomology.