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The velocity of circular motion in polar coordinates is like this; $$\vec v(t) = \frac{d}{dt}\vec r(t) = \frac{dR}{dt} \hat u_R (t)\ +\ R \frac{d\hat u_R}{dt} $$ where $R$ is the radius, and $\hat u_R (t)$ is unit vector parallel to the radius.

Why the differential is different? I want to know how to split $\frac {d}{dt}\vec r(t)$ to $$\frac{d}{dt}\vec r(t) = \frac{dR}{dt} \hat u_R (t)\ +\ R \frac{d\hat u_R}{dt} $$

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I hope I understood you right. Write $\vec{r}(t)=R(t)\hat{u}_R(t)$ and apply the product formula for differentials. This explaines the equation. The 'problem' with your notation is that you omit the $t$-dependency of $R$ and $\hat{u}_R$. A more complete expression would be: \begin{align} \frac{d}{dt} \vec{r}(t) = \frac{d}{dt}(R(t) \cdot \hat{u}_R(t)) = \frac{d}{dt}(R(t)) \cdot \hat{u}_R(t) + R(t) \cdot \frac{d}{dt}(\hat{u}_R(t)) \end{align}

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  • $\begingroup$ Oh! is $R$ also depend on $t$? Then, is it okay to understand the equation as the Product Rule? $\endgroup$ – S. Yoo Apr 11 at 11:33
  • $\begingroup$ Yes. If you picture $r(t)=R\cdot \hat{u}_R(t)$, you will see that $R$ has to be $t$-dependend. Otherwise you could only move around with constant distance $R$ to the origin. $\endgroup$ – Caroline Apr 11 at 11:36

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