Physics equations/Geometrical discussion of uniform circular motion

uniform circular motion (here the Latin d was used instead of the Greek Δ

Using the figure we define the distance traveled by a particle during a brief time interval, $\mathrm{\Delta }t$, and the (vector) change in velocity:

1     $\mathrm{\Delta }\ell =|{\overrightarrow{r}}_2-{\overrightarrow{r}}_1|$, and $\mathrm{\Delta }v=|{\overrightarrow{v}}_2-{\overrightarrow{v}}_1|$

2     $\mathrm{\Delta }\ell =v\mathrm{\Delta }t$ (rate times time equals distance).

3     $\mathrm{\Delta }\overrightarrow{v}=\overrightarrow{a}\mathrm{\Delta }t$ (definition of acceleration).

4     $\mathrm{\Delta }v=a\mathrm{\Delta }t$ (taking the absolute value of both sides).

5     $\frac{\mathrm{\Delta }v}{v}=\frac{\mathrm{\Delta }\ell }{r}$ (by similar triangles). Substituting (2) and (4) yields:

6     $\frac{a\mathrm{\Delta }t}{v}=\frac{v\mathrm{\Delta }t}{r}$, which leads to $\frac{a}{v}=\frac{v}{r}$, and therefore:

7     $a=\frac{v^2}{r}$