Derive the expression for centripetal acceleration.

Consider the position vectors and velocity vectors shift through the some angle $\theta$ in a small invertal of time $\Delta t$ as shown in figure

(ii) In uniform circular motion.
$r=\left|\overrightarrow{r_1}\right|=\left|\overrightarrow{r_2}\right|\text{and}v=\left|\overrightarrow{v_1}\right|=\left|\overrightarrow{v_2}\right|$
(iii) From figure, the geometrical relationship between the magntitude of position and velocity vectors is given by.
$\frac{\Delta r}{r}=-\frac{\Delta v}{v}=\theta$
(iv) Here the negative sign implies that $\theta v$ points radially inward, towards the center of the circle.
$\Delta v=-v\left(\frac{\Delta r}{r}\right)$
(v) Dividing both sides by $\Delta t,$ we get,
$a=\frac{\Delta v}{\Delta t}=-\frac{v}{r}\left(\frac{\Delta r}{\Delta t}\right)$
(vi) Applying the limit $\Delta t\to 0,$ We get,
$\frac{dv}{dt}=-\frac{v}{r}\left(\frac{dr}{dt}\right)$
(vii) Since $a_c=\frac{dv}{dt}$ and $v=\frac{dr}{dt}$, we can write,
$a_c=-\frac{v^2}{r}=\frac{v^2}{r}$
where $a_c$ is the centripetal acceleration.