homework and exercises - Prove the Total Mechanical Energy of the System is Conserved via Differential Equations

Consider the dynamics of a particle P shown: Particle in 3D space with Radius r Newton's second law states that: $$\frac{d}{dt}(m\dot r) = \mathbf F$$

where, $\boldsymbol{r}$ is the position vector of the particle, $\boldsymbol{\dot r}$ is its velocity vector and $m$ is its mass. $\mathbf F$ is the total external force (vector) acting on P. Prove the following result regarding the existent of mechanical energy integral of motion of the system:

Theorem: For the dynamical system, if $F = -\nabla V (\boldsymbol{r}) $ i.e the force, $\mathbf F$, is the negative gradient of a scalar function $V(\mathbf r)$ of only the position vector, then the total mechanical energy of the system is conserved. Moreover, the function $V(\mathbf r)$ is nothing but the potential energy of the system.

Hints:

The above theorem states that if $F = -\nabla V (\boldsymbol{r}) $ then the mechanical energy is an integral of motion.

The gradient of a scalar function, $V$ is written as:

$\nabla V = \partial V / \partial \boldsymbol{r} $

Also, the gradient of a scalar quantity is a vector. Do not worry about writing it in terms of the components

Recall the mechanical energy is the sum of KE and PE. Mechanical energy is given by the standard form:

$$T=1/2 ∗ m\boldsymbol{\dot r} \cdot \boldsymbol{\dot r} $$

where $\boldsymbol{\dot r} \cdot \boldsymbol{\dot r}$ represents the dot product of $\boldsymbol{\dot r} $, which is the velocity vector, with itself (i.e. square of the speed: $\boldsymbol{\dot r} \cdot \boldsymbol{\dot r} = v^2$ )

Also, let $G = G(\mathbf r, t)$ be a scalar function of a vector, $r$, and a scalar, $t$. Then, the total derivative of $G$ is related to its partial derivatives by:

$$dG = \frac{\partial G}{ \boldsymbol{\partial r}}\cdot d\mathbf r + \frac{\partial G}{\partial t}\cdot dt$$

with $\frac{\partial G}{ \boldsymbol{\partial r}} $ identified as the gradient of G, which is a vector quantity, and $\frac{\partial G}{ \boldsymbol{\partial r}}\cdot d\mathbf r $ as the dot product between the gradient of $G$ and $d\boldsymbol{r}$

Do not assume that $V(\boldsymbol{r})$ is the potential energy as this should be a natural conclusion of the analysis.

To begin, take the time derivative of the kinetic energy. Recall that if $c = \boldsymbol{a} \cdot \boldsymbol{b} $ is the dot product of two vectors, then the time derivative of c is: $$\dot c = \boldsymbol{\dot a} \cdot \boldsymbol{b} + \boldsymbol{a} \cdot \boldsymbol{\dot b}$$

ASSUME THE PARTICLE'S MASS, $m$ is constant

Edit: Attempted solution: $$\frac {d}{dt}(m \dot r) = F $$ $$\frac {d}{dr}(m \dot r)\frac {dr}{dt} = F$$ $$d(mv)v = Fdr$$ $$\int mvdv = \int Fdr$$ $$\frac {1}{2}mv^2 + \int-Fdr = C $$

Seems to prove the above theorem, however this kind of seems to ignore the fact that the variables are vectors. How can a similar approach be applied to the vectors? And still derive the potential energy at the end?

Basically, I'm asking how the potential energy can be an outcome of $$\frac {d}{dt}(\frac{1}{2} m \dot r \cdot \dot r) $$ and $$\frac{d}{dt}(m\dot r) = \mathbf F$$