# I Real Analysis with Physics

1. Jan 28, 2017

### aliens123

Suppose I wanted to prove the work-kinetic energy theorem. This means that I want to show that
$\frac{1}{2}m( \vec {v}^2_f - \vec{v}^2_i)=\int_{x_1}^{x_2} \vec{F} \cdot dx$.

So, I go ahead and start on the right side:

$\int_{x_1}^{x_2} (m \frac{d\vec{v}}{dt}) \cdot dx = m \int_{x_1}^{x_2} (\frac{d\vec{x}}{dt}) \cdot dv=m \int_{x_1}^{x_2} \vec{v} \cdot dv=\frac{1}{2}m( \vec {v}^2_f - \vec{v}^2_i)$.

And I say that I am done. But my question is, how do we rigorously argue that the following step is valid?:
$\int_{x_1}^{x_2} (m \frac{d\vec{v}}{dt}) \cdot dx = m \int_{x_1}^{x_2} (\frac{d\vec{x}}{dt}) \cdot dv$

In other words, if we were in a real analysis class, what would allow us to switch the $d\vec{v}$ with the $d\vec{x}$, using just the formal definition of an integral? Intuitively if we think of these as representing infinitesimally small amounts which are multiplied, then obviously the multiplication is commutative. But this is not very satisfying. What role does the $d\vec{x}$ actually play in the integral?

2. Jan 29, 2017

### BvU

Doesn't this look like a partial integration to an analysis expert ?
 never mind, just woke up.

3. Jan 29, 2017

### PeroK

The first thing is to define what is meant by a line integral along a curve. First you have to parameterise the curve and in this case using time $t$ is the best option. By definition:

$\int_{C} \vec{F} \cdot \vec{dr} = \int_{t_1}^{t_2} \vec{F}(\vec{r(t)}) \cdot \vec{r'(t)} dt$

Where $\vec{r(t)}$ is a parameterisation of the curve $C$.

In this case we have:

$\vec{F}(\vec{r(t)}) \cdot \vec{r'(t)} = m \vec{r''(t)}\cdot \vec{r'(t)} = m(\frac12) \frac{d}{dt}(\vec{r'(t)} \cdot \vec{r'(t)}) = m(\frac12) \frac{d}{dt}(v^2(t))$

Hence:

$\int_{C} \vec{F} \cdot \vec{dr} = \frac12 m \int_{t_1}^{t_2} \frac{d}{dt}(v^2(t)) dt = \frac12 m (v^2(t_2) - v^2(t_1))$

The last integral is just an ordinary integral wrt $t$ and we can apply the fundamental theorem of calculus.