1.11 - Polar Coordinates

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Introduction

Polar coordinates $(r, \theta)$ are often used for problems involving circular or rotational symmetry.

Polar Coordinates Definition:

In two dimensions, a point $P$ can be represented using polar coordinates $(r, \theta)$, where:

$r$ (the radial coordinate): the distance from the origin to the point $P$
$\theta$ (the angular coordinate): the angle measured counterclockwise from the positive $x$-axis to the line connecting the origin to $P$

By convention:

$r \geq 0$ (distance is always non-negative)
$\theta$ is typically measured in radians, with $\theta \in [0, 2\pi)$ or $\theta \in (-\pi, \pi]$
The origin $(0, 0)$ in Cartesian coordinates corresponds to $r = 0$, where $\theta$ is undefined

Interactive Polar Coordinate Plot

Use the interactive plot below to explore polar coordinates. Adjust the sliders to change $r$ and $\theta$, and see how the point moves in both the polar plot (showing $r$ and $\theta$) and the corresponding Cartesian coordinates. You can also click on the plot to set a point directly. Check the "Show unit vectors" checkbox to display the unit vectors $\hat{r}$ and $\hat{\theta}$ at the point.

Polar Coordinate Visualization

r: 3.0

θ (deg): 45°

Show unit vectors:

Polar: (r=3.0, θ=45°) | Cartesian: (x=2.12, y=2.12)

Transformation Between Polar and Cartesian Coordinates

Converting between polar and Cartesian coordinates is straightforward using basic trigonometry. Given a point with polar coordinates $(r, \theta)$, we can find its Cartesian coordinates $(x, y)$ and vice versa using trigonometry.

Polar to Cartesian Transformation:

\begin{aligned} x &= r\cos\theta \\ y &= r\sin\theta \end{aligned}

These equations follow directly from the definitions of sine and cosine in a right triangle, where $r$ is the hypotenuse and $x$ and $y$ are the adjacent and opposite sides, respectively.

Cartesian to Polar Transformation:

\begin{aligned} r &= \sqrt{x^2 + y^2} \\ \theta &= \arctan\left(\frac{y}{x}\right) \end{aligned}

However, care must be taken when determining $\theta$:

The formula $\theta = \arctan(y/x)$ gives correct results only when $x > 0$ (quadrants I and IV)
When $x < 0$ and $y \geq 0$ (quadrant III), we need $\theta = \arctan(y/x) + \pi$
When $x < 0$ and $y < 0$ (quadrant II), we need $\theta = \arctan(y/x) - \pi$

In practice, many programming languages and calculators provide a function $\text{atan2}(y, x)$ that automatically handles all these cases and returns the correct angle in the range $(-\pi, \pi]$.

Example 1: Converting from Polar to Cartesian

Convert the polar coordinates $(r, \theta) = (5, \pi/3)$ to Cartesian coordinates.

Solution:

Using the transformation equations:

\begin{aligned} x &= r\cos\theta = 5\cos(\pi/3) = 5 \cdot \frac{1}{2} = \frac{5}{2} \\ y &= r\sin\theta = 5\sin(\pi/3) = 5 \cdot \frac{\sqrt{3}}{2} = \frac{5\sqrt{3}}{2} \end{aligned}

Therefore, the Cartesian coordinates are $\left(\frac{5}{2}, \frac{5\sqrt{3}}{2}\right)$.

Example 2: Converting from Cartesian to Polar (Simple Case)

Convert the Cartesian coordinates $(x, y) = (3, 4)$ to polar coordinates.

Solution:

\begin{aligned} r &= \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5 \\ \theta &= \arctan\left(\frac{4}{3}\right) \approx 0.927 \text{ radians} \approx 53.13° \end{aligned}

Since $x > 0$ and $y > 0$, the point is in the first quadrant, and the simple arctangent formula works correctly.

Example 3: Converting from Cartesian to Polar

Convert the Cartesian coordinates $(x, y) = (-3, 4)$ to polar coordinates.

Solution:

First, calculate $r$:

r = \sqrt{x^2 + y^2} = \sqrt{(-3)^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5

For $\theta$, since $x < 0$ and $y > 0$, the point is in the second quadrant. We use:

\theta = \arctan\left(\frac{y}{x}\right) + \pi = \arctan\left(\frac{4}{-3}\right) + \pi = \arctan\left(-\frac{4}{3}\right) + \pi

Using a calculator, $\arctan(-4/3) \approx -0.927$ radians, so:

\theta \approx -0.927 + \pi \approx 2.214 \text{ radians}

Therefore, the polar coordinates are approximately $(5, 2.214)$ radians, or $(5, 126.87°)$ in degrees.

Unit Vectors in Polar Coordinates

In Cartesian coordinates, we use the constant unit vectors $\hat{x}$ and $\hat{y}$ that point in the positive $x$ and $y$ directions, respectively. In polar coordinates, it is natural to define unit vectors that point in the radial and angular directions. However, unlike Cartesian unit vectors, these polar unit vectors depend on the angle $\theta$ and change direction as we move around the plane.

To derive the polar unit vectors, we start with the position vector in Cartesian coordinates and replace the Caresian coordinates $(x, y)$ with the polar coordinates $(r, \theta)$ using the transformation equations:

\mathbf{r} = x\hat{x} + y\hat{y} = r\cos\theta\hat{x} + r\sin\theta\hat{y}

The unit vector $\hat{r}$ (radial unit vector) points in the direction of increasing $r$, which is radially outward from the origin. We can find this by taking the partial derivative of the position vector $\mathbf{r}$ with respect to the polar coordinate$r$ and normalizing:

\frac{\partial \mathbf{r}}{\partial r} = \cos\theta\hat{x} + \sin\theta\hat{y}

The magnitude of this vector is:

\left|\frac{\partial \mathbf{r}}{\partial r}\right| = \sqrt{\cos^2\theta + \sin^2\theta} = 1

Since it already has unit magnitude, the radial unit vector is:

Radial Unit Vector:

\hat{r} = \cos\theta\hat{x} + \sin\theta\hat{y}

The unit vector $\hat{\theta}$ (angular unit vector) points in the direction of increasing $\theta$, which is tangent to the circle of constant $r$ and perpendicular to $\hat{r}$. We can find this by taking the partial derivative of $\mathbf{r}$ with respect to $\theta$ and normalizing:

\frac{\partial \mathbf{r}}{\partial \theta} = -r\sin\theta\hat{x} + r\cos\theta\hat{y}

The magnitude of this vector is:

\left|\frac{\partial \mathbf{r}}{\partial \theta}\right| = r\sqrt{\sin^2\theta + \cos^2\theta} = r

Dividing by the magnitude $r$ gives us the angular unit vector:

Angular Unit Vector:

\hat{\theta} = -\sin\theta\hat{x} + \cos\theta\hat{y}

We can verify that $\hat{r}$ and $\hat{\theta}$ are orthogonal (perpendicular) by computing their dot product:

\begin{aligned} \hat{r} \cdot \hat{\theta} &= (\cos\theta\hat{x} + \sin\theta\hat{y}) \cdot (-\sin\theta\hat{x} + \cos\theta\hat{y}) \\ &= \cos\theta(-\sin\theta)(\hat{x} \cdot \hat{x}) + \cos\theta(\cos\theta)(\hat{x} \cdot \hat{y}) + \sin\theta(-\sin\theta)(\hat{y} \cdot \hat{x}) + \sin\theta(\cos\theta)(\hat{y} \cdot \hat{y}) \\ &= -\cos\theta\sin\theta + 0 + 0 + \sin\theta\cos\theta = 0 \end{aligned}

In this derivation we used the orthonormality of the Cartesian unit vectors: $\hat{x} \cdot \hat{x} = 1$, $\hat{y} \cdot \hat{y} = 1$, and $\hat{x} \cdot \hat{y} = 0$. We find that the dot product is zero, confirming that $\hat{r}$ and $\hat{\theta}$ are orthogonal.

Key Properties of Polar Unit Vectors:

$\hat{r}$ points radially outward from the origin
$\hat{\theta}$ points counterclockwise, tangent to circles of constant $r$
Both unit vectors depend on $\theta$ and change direction as we move around the plane
$\hat{r} \cdot \hat{\theta} = 0$ (they are orthogonal)
$\hat{r} \times \hat{\theta} = \hat{z}$ (where $\hat{z}$ is the unit vector perpendicular to the plane, pointing out of the page in a right-handed coordinate system)

Projecting the Position Vector onto the Polar Coordinate Basis

The position vector is very special in polar coordinates. Because it, by definition, points from the origin to a point in the plane, it always points in the radial direction. Thus, it can be expressed soley in terms of the radial polar unit vector:

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

In contrast, the Cartesian position vector requires both $x$ and $y$ unit vectors, unless the point happens to lie along either the $x$ or $y$ axis:

\mathbf{r} = x\hat{x} + y\hat{y}

While the polar form looks simpler, we must remember that $\hat{r}$ itself depends on $\theta$. As we will see in the next lecture, this $\theta$ dependence has both advantages (in rotationally symmetric systems) and disadvantages (more complicated time derivative calculations).

Projecting Arbitrary Vectors onto the Polar Coordinate Basis

Many vectors that we care aboutdo not originate at the origin. Imagine, for example, a particle at polar coordinates $(r,\theta)$. If we wish to find the polar components of the particle's velocity or acceleration, or the net force acting on it, etc., we need to take into account the fact that the orientation of the polar basis vectors depend on where the particle is in the plane, specifically, what the polar coordinate $\theta$ is.

To find the polar components of the vector, we project the vector onto the polar basis vectors (see Chapter 1.2 Scalar Product). The radial component of the vector is given by: $$v_r = \mathbf{v} \cdot \hat{r}$$ and the angular component of the vector is given by: $$v_\theta = \mathbf{v} \cdot \hat{\theta}$$ where $\hat{r}$ and $\hat{\theta}$ are the polar unit vectors at the particle's position.

In practice, if we know the Cartesian components of our vector, we need to write the polar unit vectors in terms of Cartesian unit vectors in order to evaluate the dot product: $$\begin{aligned}v_r &= \mathbf{v} \cdot \hat{r}\\ &= (v_x\hat{x} + v_y\hat{y}) \cdot (\cos\theta\hat{x} + \sin\theta\hat{y})\\ &= v_x\cos\theta + v_y\sin\theta \end{aligned}$$ and similarly for $v_\theta$: $$\begin{aligned}v_\theta &= \mathbf{v} \cdot \hat{\theta}\\ &= (v_x\hat{x} + v_y\hat{y}) \cdot (-\sin\theta\hat{x} + \cos\theta\hat{y})\\ &= -v_x\sin\theta + v_y\cos\theta. \end{aligned}$$

Example 4: Expressing a Vector in Terms of Polar Unit Vectors

Suppose an object at Cartesian coordinates $2\hat{x}+2\hat{y}$ has a velocity vector $\mathbf{v} = 3\hat{x} + 4\hat{y}$ (m/s) in Cartesian coordinates. Express this velocity vector in terms of the polar unit vectors $\hat{r}$ and $\hat{\theta}$.

Solution:

We want to calculate $$ \begin{aligned}v_r &= \mathbf{v} \cdot \hat{r} \\ v_\theta &= \mathbf{v} \cdot \hat{\theta} . \end{aligned}$$ Because the object is at position $2\hat{x}+2\hat{y}$, its angular position is $\theta = \pi/4$. At $\theta = \pi/4$, the polar unit vectors are:

\begin{aligned} \hat{r} &= \cos(\pi/4)\hat{x} + \sin(\pi/4)\hat{y} = \frac{1}{\sqrt{2}}\hat{x} + \frac{1}{\sqrt{2}}\hat{y} \\ \hat{\theta} &= -\sin(\pi/4)\hat{x} + \cos(\pi/4)\hat{y} = -\frac{1}{\sqrt{2}}\hat{x} + \frac{1}{\sqrt{2}}\hat{y} \end{aligned}

We want to find $v_r$ and $v_\theta$ such that $\mathbf{v} = v_r\hat{r} + v_\theta\hat{\theta}$. In other words, we want to project our velocity vector onto the basis vectors of the polar coordinate system. We can find these components using the dot product:

\begin{aligned} v_r &= \mathbf{v} \cdot \hat{r} = (3\hat{x} + 4\hat{y}) \cdot \left(\frac{1}{\sqrt{2}}\hat{x} + \frac{1}{\sqrt{2}}\hat{y}\right) = \frac{3}{\sqrt{2}} + \frac{4}{\sqrt{2}} = \frac{7}{\sqrt{2}} \\ v_\theta &= \mathbf{v} \cdot \hat{\theta} = (3\hat{x} + 4\hat{y}) \cdot \left(-\frac{1}{\sqrt{2}}\hat{x} + \frac{1}{\sqrt{2}}\hat{y}\right) = -\frac{3}{\sqrt{2}} + \frac{4}{\sqrt{2}} = \frac{1}{\sqrt{2}} \end{aligned}

Therefore, at position $2\hat{x}+2\hat{y}$, the velocity vector can be expressed as:

\mathbf{v} = \frac{7}{\sqrt{2}}\hat{r} + \frac{1}{\sqrt{2}}\hat{\theta} = \frac{1}{\sqrt{2}}(7\hat{r} + \hat{\theta})

Example 5: Finding Components of a Vector at Different Angles

A velocity vector is given by $\mathbf{v} = 5\hat{x}$ (pointing purely in the $x$-direction). Find the velocity components in polar coordinates $(v_r,v_\theta)$ $v_r = \mathbf{v} \cdot \hat{r}$ and angular component $v_\theta = \mathbf{v} \cdot \hat{\theta}$ at $\theta = 0$, $\theta = \pi/2$, and $\theta = \pi$.

Solution:

At $\theta = 0$: $\hat{r} = \hat{x}$ and $\hat{\theta} = \hat{y}$

\begin{aligned} v_r &= (5\hat{x}) \cdot \hat{x} = 5 \\ v_\theta &= (5\hat{x}) \cdot \hat{y} = 0 \end{aligned}

At $\theta = \pi/2$: $\hat{r} = \hat{y}$ and $\hat{\theta} = -\hat{x}$

\begin{aligned} v_r &= (5\hat{x}) \cdot \hat{y} = 0 \\ v_\theta &= (5\hat{x}) \cdot (-\hat{x}) = -5 \end{aligned}

At $\theta = \pi$: $\hat{r} = -\hat{x}$ and $\hat{\theta} = -\hat{y}$

\begin{aligned} v_r &= (5\hat{x}) \cdot (-\hat{x}) = -5 \\ v_\theta &= (5\hat{x}) \cdot (-\hat{y}) = 0 \end{aligned}

This example illustrates that the same vector $\mathbf{v}$ has different components in the polar basis depending on the angle $\theta$. This is because the polar unit vectors $\hat{r}$ and $\hat{\theta}$ themselves depend on $\theta$.