The significance of the magnitude of a vector derivative

circlesclassical-mechanicsgeometrytrigonometryvectors

My Mathematics BSc course notes on circular motion use

\begin{align}
\overrightarrow{e_r}
&:=\overrightarrow{i}\cos\theta+\overrightarrow{j}\sin\theta,\tag{1}\\
\overrightarrow{e_\theta}
&:=\frac{d}{d\theta}(\overrightarrow{e_r})\tag{2.1}\\
&=-\overrightarrow{i}\sin\theta+\overrightarrow{j}\cos\theta.\tag{2.2}
\end{align}

Now, $\overrightarrow{e_r}$ can be drawn as a unit vector parallel to the radius; and $\overrightarrow{e_\theta}$ can be drawn as a unit vector perpendicular to the radius, in particular as a tangent to the circle.

My understanding of scalar derivatives is that the derivative is the gradient of the tangent, and the tangent can take any finite or infinite length.

Here, the derivative is the tangent itself, so what is the significance of its length?

Best Answer

Here, the derivative represents the velocity of a body undergoing circular motion. Since this velocity is always a tangent to the location of the body, it indicates that the velocity is always tangential to the circle on which the body moves.

The length of this vector gives you the magnitude of the velocity. In this case, the length is 1, but if you are moving on an ellipse, or a larger circle, the equation becomes $$e_r = a\cos \theta i + b\sin \theta j$$ $$\frac {de_r}{d\theta} = -a\sin\theta i + b\cos\theta j$$

here, the magnitude plays a more important role as you can see clearly that the velocity is not the same at different points of motion if $a \ne b$.

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