Let $F = −\nabla \phi$ be a conservative force field. Suppose a particle of mass $m$ moves in this field. If $A$ and $B$ are any two points in space, prove that $$ \phi(A) + \frac{m}{2} v_A^2 = \phi(B) + \frac{m}{2} v_B^2 , $$ where $v_A = \|dr(A)/dt \|$ and $vB = \| dr(B)/dt \|$ are the magnitudes of the velocities of the particle at $A$ and $B$, respectively.
Using the fact that $F = − \nabla \phi$ is a conservative field we have $$ \int F \cdot dr = \int (-\nabla \phi) \cdot dr = \phi(A) - \phi(B) $$ We also have that $F = m \frac{d^2 r}{dt^2}$ and hence if $r = (r_1(t),r_2(t),r_3(t))$ we have R d2r dr F•dr=m • dt=(r1′′r1′ +r2′′r2′ +r3′′r3′)dt dt2 dt
I believe I can factor out $d/dt$ but I'm unsure where $1/2$ comes from...as I want $\frac{m}{2} v^2|AB$. How do I finish this?
You are on the right track
$$ -\nabla \phi = m\frac{{\rm d}^2 {\bf r}}{{\rm d}t^2} = m\frac{{\rm d}{\bf v}}{{\rm d}t} \tag{1} $$
Multiply both sides by ${\bf v}$:
$$ -\nabla \phi \cdot {\bf v} = m {\bf v} \cdot\frac{{\rm d}{\bf v}}{{\rm d}t} = \frac{1}{2}m \frac{{\rm d}}{{\rm d}t}({\bf v}\cdot {\bf v}) = \frac{1}{2}m\frac{{\rm d}v^2}{{\rm d}t} \tag{2} $$
Now integrate on both sides
$$ -\int \nabla \phi \cdot \frac{{\rm d}{\bf r}}{{\rm d}t}{\rm d}t = -\int \nabla \phi \cdot {\rm d}{\bf r} = -\int{\rm d}\phi = \frac{1}{2}m\int \frac{{\rm d}v^2}{{\rm d}t} \tag{3} $$
And from here you conclude that
$$ H = \phi + \frac{1}{2}mv^2 $$
is a constant