Polar Coordinates

This isn't strictly a calculus topic but it comes up a lot with calculus so I'm sticking it here. I'm not really sure what "part" of math coordinate systems belong to; they are used all over the place and have geometric and algebraic interpretations.

Definition

Polar coordinates are a 2d orthogonal coordinate system where coordinates are given by two components, radius and angle, i.e. $P=(r,\theta ).$

The radius is the distance from the origin to $P$ and the angle is the angle formed with a line measured counter-clockwise from the positive $x$-axis to a line passing through $P$.

Converting Between Rectangular and Polar

$$x=r\cos \theta ,y=r\sin \theta$$

$$r=\sqrt{x^2+y^2},\theta =\text{atan2}(y,x)$$

Derivatives

If we parameterize such that $x=x(t)$ and $y=y(t)$ and want to find $x^{\prime }$ and $y^{\prime }$ in polar coordinates, we have:

$$\begin{array}{cc}\frac{dx}{dt}=x(t{)}^{\prime }=(r\cos \theta {)}^{\prime }=r^{\prime }\cos \theta -{\theta }^{\prime }r\sin \theta ,\frac{dy}{dt}=y(t{)}^{\prime }=(r\sin \theta {)}^{\prime }=r^{\prime }\sin \theta +{\theta }^{\prime }r\cos \theta & \text{(a)}\end{array}$$

thus:

$$\begin{array}{cc}\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{dy}{dx}=\frac{r^{\prime }sin\theta +{\theta }^{\prime }r\cos \theta }{r^{\prime }\cos \theta -{\theta }^{\prime }r\sin \theta }& \text{(b)}\end{array}$$

Doing some tedious algebra from here, we can arrive at definitions of $r^{\prime }$ and ${\theta }^{\prime }$ in terms of $x,y,x^{\prime },y^{\prime }\text{and}r$, but it's easier if we start with the following, letting $x=x(t),y=y(t),r=r(t),\theta =\theta (t)$:

$$\begin{array}{cc}{r}^2=x^2+y^2,\tan \theta =\frac{y}{x}& \text{(c)}\end{array}$$

Differentiating the first equation in (c) with respect to $t$ we get:

$$\begin{array}{cc}2r{r}^{\prime }=2x{x}^{\prime }+2y{y}^{\prime },r{r}^{\prime }=x{x}^{\prime }+y{y}^{\prime },r^{\prime }=\frac{x{x}^{\prime }+y{y}^{\prime }}{r}& \text{(d)}\end{array}$$

Rewriting the second equaton in (c) and then differentiating with respect to $t$ we get:

$$\begin{array}{cc}\theta ={\tan }^{-1}\frac{y}{x},{\theta }^{\prime }=\frac{1}{1+(\frac{y}{x}{)}^2}\cdot \frac{y^{\prime }x-y{x}^{\prime }}{x^2}=\frac{y^{\prime }x-y{x}^{\prime }}{x^2+y^2}=\frac{y^{\prime }x-y{x}^{\prime }}{r^2}& \text{(e)}\end{array}$$

Summarizing (d) and (e) we have:

$$\begin{array}{cc}\frac{dr}{dt}=r^{\prime }=\frac{x{x}^{\prime }+y{y}^{\prime }}{r},\frac{d\theta }{dt}={\theta }^{\prime }=\frac{y^{\prime }x-y{x}^{\prime }}{r^2}& \text{(f)}\end{array}$$

Therefore, the derivative of $r$ with respect to $\theta$ is:

$$\frac{\frac{dr}{dt}}{\frac{d\theta }{dt}}=\frac{dr}{d\theta }=\frac{x{x}^{\prime }+y{y}^{\prime }}{r}\cdot \frac{r^2}{y^{\prime }x-y{x}^{\prime }}=\frac{r(x{x}^{\prime }+y{y}^{\prime })}{y^{\prime }x-y{x}^{\prime }}$$

Radial and Transverse Components of Motion

Given a particle at point $B(r,\theta )$ moving in a plane, we can define the radial and transverse components of the particle's motion.

Radial: The particle's motion in the direction away from the origin, that is, in the direction that purely increases the $r$ coordinate of the particle's position.

Transverse: The particle's motion in the direction tangential to its radius, that is, in the direction that purely increases the $\theta$ coordinate of the particle's position.

Let's call the radial direction $R$ and the transverse direction $T$. Also recognize that the radial and transverse directions are orthogonal - moving in the radial direction doesn't change the transverse component and vice versa.

Let's assume the particle at $B(r_1,{\theta }_1)$ is moving in the direction of point $L(r_2,{\theta }_2)$, at velocity $v_0$. We can then say that the velocity vector for the particle is given as the vector $\mathbf{v}=⟨v_0,\theta -\rho ⟩$.

Then, the component of $\mathbf{v}$ in the $R$ direction is the projection of $\mathbf{v}$ onto $R$, whose magnitude is given by $\Vert \Vert {\mathbf{v}}_{\mathbf{R}}\Vert \Vert =v_0\cos (\theta -\rho )$, and the component of $\mathbf{v}$ in the $T$ direction is the projection of $\mathbf{v}$ onto $T$, whose magnitude is given by $\Vert \Vert {\mathbf{v}}_{\mathbf{T}}\Vert \Vert =v_0\sin (\theta -\rho )$. The following image shows this geometrically:

Here, you can see that we can sort of think of the $R$ direction as the $x$-axis and the $T$ direction as the $y$-axis when thinking about how to find the amount of motion in the $R$ and $T$ directions. It's important to remember, though, that for a moving particle, its position is changing constantly, and so are the $R$ and $T$ directions, so they don't make good fixed references.

We must also point out that ${\mathbf{v}}_{\mathbf{R}}$ is moving in the negative $R$ direction here and so is reducing the radius. In vector component form it would be given as $⟨-v_0\cos (\theta -\rho ),\theta ⟩$ and the derivative of the radius with respect to time would be given by $\frac{dr}{dt}=-v_0\cos (\theta -\rho )$.

Also, while $\Vert \Vert {\mathbf{v}}_{\mathbf{T}}\Vert \Vert =v_0\sin (\theta -\rho )$, this doesn't tell us the change in $\theta$ with respect to time, it only tells us the amount of linear movement in the tangential direction. The change in angle also depends on the radius - the further from the origin we are, the less the angle will change. The derivative of the angle $\theta$ with respect to time is then given by $\frac{d\theta }{dt}=\frac{v_0\sin (\theta -\rho )}{r}$. Note: The division by $r$ here needs better justification. It's intuitive that the bigger $r$, the less the change in angle, but it's not clear why it should be a linear factor.

Other Notes

I found the notes in this document on converting a system of rectangular differential equations to polar coordinates particularly helpful.

polar integrals