A point in the plane can be described by Cartesian coordinates $(x,y)$ or by polar coordinates $(r,\theta)$.

The radial coordinate $r\ge0$ is the distance from the origin, and the angular coordinate $\theta$ gives the direction of the position vector from a chosen reference axis.

The coordinate systems are related by

$$\boxed{x=r\cos\theta},$$

$$\boxed{y=r\sin\theta}.$$

Conversely,

$$r=\sqrt{x^2+y^2},$$

while $\theta$ is chosen from the direction of the vector, taking the correct quadrant into account.

Radial and transverse unit vectors

Polar coordinates use local unit vectors whose directions depend on $\theta$.

The radial unit vector points outward from the origin:

$$\boxed{\hat{\mathbf r}=\cos\theta,\hat{\mathbf x}+\sin\theta,\hat{\mathbf y}}.$$

The transverse or azimuthal unit vector points in the direction of increasing $\theta$:

$$\boxed{\hat{\boldsymbol\theta}=-\sin\theta,\hat{\mathbf x}+\cos\theta,\hat{\mathbf y}}.$$

These vectors are perpendicular and have unit length:

$$\hat{\mathbf r}\cdot\hat{\boldsymbol\theta}=0.$$

The position vector is especially simple:

$$\boxed{\mathbf r=r\hat{\mathbf r}}.$$

The basis moves

Unlike fixed Cartesian unit vectors, $\hat{\mathbf r}$ and $\hat{\boldsymbol\theta}$ change direction when $\theta$ changes.

Differentiate $\hat{\mathbf r}$ with respect to $\theta$:

$$\frac{d\hat{\mathbf r}}{d\theta} =-\sin\theta,\hat{\mathbf x}+\cos\theta,\hat{\mathbf y} =\boxed{\hat{\boldsymbol\theta}}.$$

Similarly,

$$\frac{d\hat{\boldsymbol\theta}}{d\theta} =-\cos\theta,\hat{\mathbf x}-\sin\theta,\hat{\mathbf y} =\boxed{-\hat{\mathbf r}}.$$

If $\theta=\theta(t)$, the chain rule gives

$$\boxed{\frac{d\hat{\mathbf r}}{dt}=\dot\theta\hat{\boldsymbol\theta}},$$

$$\boxed{\frac{d\hat{\boldsymbol\theta}}{dt}=-\dot\theta\hat{\mathbf r}}.$$

These changing basis vectors are the key reason polar-coordinate velocity and acceleration contain terms that do not appear in Cartesian component formulas.

Example

A point has polar coordinates

$$r=5,\qquad \theta=\tan^{-1}\left(\frac43\right).$$

Then

$$\cos\theta=\frac35,\qquad \sin\theta=\frac45,$$

so

$$x=r\cos\theta=3,$$

$$y=r\sin\theta=4.$$

At this point,

$$\hat{\mathbf r}=\frac35\hat{\mathbf x}+\frac45\hat{\mathbf y},$$

and

$$\hat{\boldsymbol\theta}=-\frac45\hat{\mathbf x}+\frac35\hat{\mathbf y}.$$

The radial direction points from the origin toward $(3,4)$, while the transverse direction is perpendicular to it in the direction of increasing angle.

Polar coordinates are especially useful when a problem has a distinguished center, radial distance, or rotational symmetry.