How do I derive the formula for radial acceleration when there is no uniform circular motion? [duplicate]

My lecturer states that $$a_r=\dfrac{v_t^2}{r}=\omega^2r$$ where $$v_t$$ is tangential velocity, he also wrote that this is derived the same way that radial acceleration is derived in uniform circular motion.

I know the derivation for uniform circular motion but I simply can't see how the derivation can be the same for non uniform motion. (the derivation I know is the graphical one where you manipulate the infinitesimal velocity and angular position.)

• Differentiate $\vec{r}=r\hat{r}$ twice with respect to time, taking care to consider the derivative of $\hat{r}$. Youâ€™ll find both the radial and the tangential acceleration. – G. Smith Oct 3 '19 at 23:18
• – G. Smith Oct 4 '19 at 0:18

For a non-uniform circular motion your graphical demonstration remains valid. Simply, $$\frac{d\theta}{dt}=\omega$$ is no longer constant in time. You could write $$\omega(t)$$ to make it explicitly time-dependent.