Proof Of Conservation Of Energy Yesterday I was trying to prove The law of conservation of energy, But in a different approach:
Using the concept of vector field  here is how it went:
I assumed a force field $\mathbf{F}=f(x,y,z){\mathbf{ \hat{i}}}+g(x,y,z)\mathbf{\hat{j}}+h(x,y,z){\mathbf{\hat{k}}}$
I also assumed that the  $\mathbf{∇ \times F=0}$ (meaning that the field is conservative).
Then I assumed a 3-D space in which there is a particle with velocity $v$.
Acc. to me the equation should be:
$E=(1/2)mv^2+∫F.ds$
Then taking the derivative on both sides with respect to time:
$dE/dx=mvv'+d/dx(∫F.ds)$
We need to prove that $dE/dt=0$
But I went very deep into the maths and finally messed up everything . So, could anyone please explain the proof in my way?
 A: We can treat conservation of the sum of potential energy and kinetic energy as a theorem of newtonian mechanics.

The following derivation starts with the case where a single force is acting, and there is a single degree of freedom. Towards the end the generalization to multiple degrees of freedom is discussed.
As to the form in which I give the derivation:
I believe the form I use is particularly well suited for generalizing from the concept of mechanical energy to other forms of energy.

Let $F=ma$ be granted as axiom.
We define potential energy as the negative of force acting over distance.
To emphasize how tight the interconnections are I give the relevant equations/expressions below as an uninterrupted sequence.
Scoll down to the group (1.1) to (1.8)
(1.4) states the integral of force acting over distance.
To represent the integral $ \int_{s_0}^s F \ ds $ in a diagram: put the acceleration along the vertical axis, and the position coordinate along the horizontal axis.
In the case of uniform acceleration: then the integral $ \int_{s_0}^s F \ ds $ corresponds to a rectangular area.
We can create an animation, showing how that area increases as a function of time
Remaining with uniform acceleration for the time being:
As we know: the area of a rectangle remains the same when we go to double the height, and half the width. More generally: multiply the height with the factor 'a', and divide the width with the factor 'a': the area remains the same.
In the line tagged (1.5) the operations that are performed execute that rescaling. The vertical coordinate is multiplied with the factor 't' for the time elapsed, and the horizontal coordinate is divided by the factor 't'. The result: $ \int_{s_0}^s a \ ds = \int_{v_0}^v  v \ dv $
Represented in a diagram: the result of the coordinated rescaling is that both along the vertical and the horizontal axis velocity (as a function of time) is plotted.
The corresponding animation: now the area (increasing as a function of time) is the area of a triangle.

Significantly, the above reasoning generalizes to non-uniform acceleration.
In the course of (1.5):
-the differential $ds$ is substituted with $dt$, with corresponding change of limits
-the differential $dt$ is substituted with $dv$, with corresponding change of limits

$ ds = v \ dt  \qquad \qquad (1.1) $
$ a \ dt = dv  \qquad \qquad (1.2) $

$ F = ma \qquad \qquad (1.3) $
$ \int_{s_0}^s F \ ds = \int_{s_0}^s ma \ ds \qquad \qquad (1.4) $
$ \int_{s_0}^s a \ ds = \int_{t_0}^t  a \ v  \ dt
= \int_{t_0}^t  v \ a \ dt = \int_{v_0}^v  v \ dv
= \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2 \qquad (1.5) $
$ \int_{s_0}^s F \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2 \qquad \qquad (1.6) $
$ \Delta E_p = -\int_{s_0}^s F \ ds \qquad \qquad (1.7) $
$ \Delta E_k = -\Delta E_p \quad \Leftrightarrow \quad \Delta E_k + \Delta E_p = 0 \qquad \qquad (1.8) $

(1.7) gives the definition of potential energy.
As we know, that definition is subject to a condition. Potential energy is well defined only if the value of that integral is independent of how the object moves from the start point to the end point. As we know: a force with that particular property is referred to as a 'conservative force'.

Transformation between representations
We have that we can represent mechanics taking place in terms of force-and-acceleration, or in terms of potential energy and kinetic energy.
The relation between those two representations is the operation of integration.
Take $F=ma$, integrate both sides with respect to the spatial coordinate, and the result is the Work-Energy theorem: (1.6)

Second derivative and squaring
Position, velocity and acceleration. There is symmetry in that group of three physical attributes. The work-energy theorem (1.6) can be seen as arising from symmetry properties.

Multiple degrees of freedom
As we know, this integration generalizes to multiple degrees of freedom.
-Potential energy: evaluate the integral for each degree of freedom
-Kinetic energy: since the expression for kinetic energy is quadratic we get the benefit of Pythagoras' theorem. If there are multiple degrees of freedom then the total kinetic energy is the sum of the component kinetic energies.

Generalization
The pattern of the derivation, the steps from (1.4) to (1.6), is applicable for other physics phenomena. In the process the concept of Energy is generalized to areas beyond Mechanics.
Example: the electrodynamics of an LC circuit.
As we know: electric oscillation in an LC circuit is analogous to mechanical oscillation. The simplest case of mechanical oscillation is when the restoring force is according to Hooke's law.
$V$ voltage (electromotive force)
$I$ current in the circuit (charge through the circuit per unit of time)
$L$ Inductance (counterpart of inertia)
$C$ Capacitance
In the case of an LC circuit: the simplest case is:
-A capacitance such that the electromotive force increases linear with the amount of accumulated charge
-An inductance such that the relation between electromotive force and time derivative of is linear.
Richard Fitzpatrick gives for the total energy in the case of electric oscillation in an LC circuit:
$$ E = \tfrac{1}{2} CV^2  + \tfrac{1}{2}LI^2   \tag{2.1}  $$
The expression for the energy of the current is proportional to the square of the current because change of current is the second time derivative of charge through the circuit.
A: If $\mathbf F$ is conservative, then $\mathbf F = -\nabla U$ for some scalar field $U$. From there, $E = \frac{1}{2}mv^2 + U$, and
$$\frac{dE}{dt} = m\mathbf v \cdot \mathbf a + \nabla U \cdot \mathbf v = \mathbf v \cdot \big(m\mathbf a + \nabla U\big) = \mathbf v \cdot \big(m\mathbf a - \mathbf F\big)$$
According to Newton's 2nd law, the right hand side vanishes, so $dE/dt = 0$.

I'm guessing that you're getting tripped up on the chain rule.  It may help to write everything somewhat more explicitly:
$$E(\mathbf x,\mathbf v) = \frac{1}{2}m|\mathbf v|^2 + U(\mathbf x)$$
$$\frac{dE}{dt} = \frac{\partial E}{\partial \mathbf v} \cdot \frac{d\mathbf v}{dt} + \frac{\partial E}{\partial \mathbf x} \cdot \frac{d\mathbf x}{dt} = m\mathbf v \cdot \mathbf a + \nabla U \cdot \mathbf v  = \ldots$$
