Derivatives of Unit Vectors in Polar Coordinates

From ProofWiki
Jump to navigation Jump to search

Theorem

Consider a particle $p$ moving in the plane.

Let the position of $p$ be given in polar coordinates as $\polar {r, \theta}$.

Let:

$\mathbf u_r$ be the unit vector in the direction of the radial coordinate of $p$
$\mathbf u_\theta$ be the unit vector in the direction of the angular coordinate of $p$


Then the derivative of $\mathbf u_r$ and $\mathbf u_\theta$ with respect to $\theta$ can be expressed as:

\(\ds \dfrac {\d \mathbf u_r} {\d \theta}\) \(=\) \(\ds \mathbf u_\theta\)
\(\ds \dfrac {\d \mathbf u_\theta} {\d \theta}\) \(=\) \(\ds -\mathbf u_r\)


Proof

By definition of sine and cosine:

\(\text {(1)}: \quad\) \(\ds \mathbf u_r\) \(=\) \(\ds \mathbf i \cos \theta + \mathbf j \sin \theta\)
\(\text {(2)}: \quad\) \(\ds \mathbf u_\theta\) \(=\) \(\ds -\mathbf i \sin \theta + \mathbf j \cos \theta\)

where $\mathbf i$ and $\mathbf j$ are the unit vectors in the $x$-axis and $y$-axis respectively.


DerivativesOfUnitPolarVectors.png


Differentiating $(1)$ and $(2)$ with respect to $\theta$ gives:


\(\ds \dfrac {\d \mathbf u_r} {\d \theta}\) \(=\) \(\ds -\mathbf i \sin \theta + \mathbf j \cos \theta\)
\(\ds \) \(=\) \(\ds \mathbf u_\theta\)
\(\ds \dfrac {\d \mathbf u_\theta} {\d \theta}\) \(=\) \(\ds -\mathbf i \cos \theta - \mathbf j \sin \theta\)
\(\ds \) \(=\) \(\ds -\mathbf u_r\)

$\blacksquare$


Sources