# How to understand the rearrangement of differentials in the derivation of kinetic energy using calculus?

When the formula of kinetic energy is derived using calculus, the differentials in the integral are rearranged like below. Could you point to how to understand this rearrangement of differentials ? o_O I could not understand what is going on behind the scene even after reading posts about separating x and y here and here...

$$\begin{array}{l}{\Delta K=m \int \frac{d \mathbf{v}}{d t} \cdot d \mathbf{r}} \\ {\Delta K=m \int \frac{d \mathbf{r}}{d t} \cdot d \mathbf{v}} \\ {\Delta K=m \int \mathbf{v} \cdot d \mathbf{v}}\end{array}$$

• For your further interest, kinetic energy can be related as: $$\mathbf{v}=\frac{\partial E}{\partial \mathbf{p}}$$ or alternatively, $$\Delta E=\int \mathbf{v} \cdot d\mathbf{p}$$ See also another answer of mine here. – Ng Chung Tak Nov 26 '19 at 2:59

Funny, we did this same exact lesson today in my AP physics class. I didn't think I'd be applying it so soon.

Anyways, the key is understanding the chain rule, namely (in general): $$\frac{dF}{dx}=\frac{dF}{du}\frac{du}{dx}$$

That's what's going on here. We have:

$$W=m\int_{x_0}^{x_f}{\frac{dv}{dt}{dx}}$$

Which can be rewritten as: $$W=m\int_{x_0}^{x_f}{\frac{dv}{dx}\frac{dx}{dt}dx}$$

Since $$\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}$$

Recall now that $$v=\frac{dx}{dt}$$, this becomes: $$W=m\int_{x_0}^{x_f}{v\frac{dv}{dx}{dx}}$$

$$dx$$ cancels leaving you with $$W=m\int_{v_0}^{v_f}{vdv}$$

Perhaps it is easier to convince if you don't use differentials: \begin{align}ma~dr\approx \dfrac{m\Delta v\Delta r}{\Delta t} = m\dfrac{\Delta r}{\Delta t}\Delta v \end{align} $$\Delta v, \Delta r, \Delta t$$ are real numbers so regular arithmetic can be performed on them. In the limit you replace $$\Delta$$ by $$d$$.