Relation between angular acceleration and linear acceleration

Derivation of equation of the relation between linear acceleration and angular acceleration:

Deduce the equation from the General form

Deduce the equation from the Differential form

We know that angular acceleration is $\alpha=\frac{\Delta \omega}{\Delta t} \qquad (1)$ $\alpha=\frac{\Delta (v/r)}{\Delta t} \qquad \left( \because \omega=\frac{v}{r} \right)$ $\alpha=\frac{1}{r} \frac{\Delta v}{\Delta t}$ If $\Delta t \rightarrow 0$ then the above equation can be written as $\alpha=\frac{1}{r} \: \underset{\Delta t \rightarrow 0}{Lim} \frac{\Delta v}{\Delta t}$ Where $\underset{\Delta t \rightarrow 0}{Lim} \: \frac{\Delta v}{\Delta t}$ = Instantaneous Acceleration $(a)$ $\alpha=\frac{a}{r}$

We know that angular acceleration is $\alpha=\frac{d \omega}{dt} \qquad (1)$ $\alpha=\frac{d (v/r)}{dt} \qquad \left( \because \omega=\frac{v}{r} \right)$ $\alpha=\frac{1}{r} \frac{dv}{dt} \qquad (2)$ Where $\frac{dv}{dt}$ = Instantaneous Acceleration $(a)$ Now equation $(2)$ can be written as $\alpha=\frac{a}{r}$