Calculus II Lesson 16: Polar Coordinates

  1. Warm Up
    1. Upcoming
    2. Advising
  2. Polar Coordinates
    1. Playing with desmos
    2. Physics analogy
    3. Converting
    4. Exercise
  3. Graphs
    1. Special Curves
  4. Calculus
  5. Playing with desmos again

Warm Up

Hand in:

  1. Section 7.2, #104 Find the area of the region bounded by the curve $x(t) = 2\cot(t)$, $y(t) = 2\sin^2(t)$, over $0 \leq t \leq \pi$.
  2. 7.2 #114: Find the length of one arch of the cycloid $x(t) = 4(t - \sin(t))$, $y(t) = 4(1 - \cos(t))$.

Graph for #2:

For #2, first you need to figure out one “one arch” corresponds to in terms of the values of $t$. Looking at the graph, it starts when $y = 0$. One arch is the next time that $y = 0$. So set $y(t) = 0$ and solve for $t$ first.

Upcoming

Advising

Next week is advising week. Make an appointment with your advisor.

Polar Coordinates

The idea behind polar coordinates it hat we describe points on the plane in an alternate way. Instead of using $(x, y)$, which describe horizontal and vertical distances from the origin, we use $(r, \theta)$, which describe:

As we said last time, this is useful if our motivating geometry is “circular” or “rotational”.

Playing with desmos

Graph of $y = x^2 + a$ on Desmos

The goal is to be able to ask similar questions about polar graphs: what if we graph $r$ as a function of $\theta$? For example, in the above link we can graph $r = \cos(3\theta)$ and its transformations.

Physics analogy

Recall: what is a vector (in physics)?

A quantity which has magnitude and direction. For example, a 2 pound force horizontally pushing an object to the right is not the same as a 2 pound force pushing an object down. Often times, 2D vectors are described as points on the plane: $(x, y)$ coordinates. The magnitude would be their distance from the origin, and the direction would just be whatever direction they are pointing in. But if we wanted to, we could just use polar to describe those ideas directly: the magnitude is $r$, direction is $\theta$.

Converting

Question: How do we convert from $(x, y)$ coordinates to $(r, \theta)$ (and back)?

Example: express the point $(1, \sqrt{3})$ in polar coordinates.

Take a look at the following figure from the textbook:

Figure 7.27: a point on the plane in polar / rectangular coordinates

Notice that we can use what we know about triangles to convert. So to go from $(x, y)$ to $(r, \theta)$, we use the following formulas:

To go the other way, notice that $\frac{x}{r} = \cos(\theta)$ and $\frac{y}{r} = \sin(\theta)$, so we have the following formulas:

Now some definitions. In polar, we don’t really have an $x$-axis, exactly, so what exactly is $\theta$ measuring? We define a single point, the pole (which corresponds to the origin in rectangular coordinates), and from that pole, we draw a ray (half of a line, so it only extends in one direction) going immediately to the right (forever). That ray is referred to as the polar axis. This corresponds to the positive $x$-axis.

The “gridlines” for a polar graph would look a little different: for the $r$-values, you’d have concentric circles, and for $\theta$, you have rays at each angle.

Figure 7.28: The polar coordinate system. A series of concentric circles is drawn with spokes indicating different values between 0 and 2π in increments of π/12. The first quadrant starts with 0 where the x axis would be, then the next spoke is marked π/12, then π/6, π/4, π/3, 5π/12, π/2, and so on into the second, third, and fourth quadrants. The polar axis is noted near the former x axis line.

Exercise

Plot the following points expressed in polar, and express them in $(x, y)$ coordinates:

  1. $(1, \frac{\pi}{2})$
  2. $(2, -\frac{\pi}{4})$
  3. $(3, \frac{3\pi}{4})$

(After going over the exercise): Notice that $(2, -\frac{\pi}{4}) = (2, \frac{7\pi}{4}) = \ldots$ The same point can be represented in infinitely many ways (in polar). Why is this?

Click for the answer

It's the same reason that sin and cos are periodic functions: every $2\pi$, the angle $\theta$ gets to the same point on the plane.

Graphs

Previously saw an example of a polar graphs. In general, we want to graph functions of the form $r = f(\theta)$.

Let’s do an exmaple: $r = 2\cos(\theta)$. Like we did with parametric equations, we start by making a table:

\[\begin{array}{c|c} \theta & r = 2\cos(\theta) \\ \hline 0 & 2 \\ \frac{\pi}{6} & \sqrt{3} \approx 1.73 \\ \frac{\pi}{4} & \sqrt{2} \approx 1.41 \\ \frac{\pi}{3} & 1 \\ \frac{\pi}{2} & 0 \\ \frac{2\pi}{3} & -1 \\ \pi & -2 \end{array}\]

So what does the graph look like?

Special Curves

Section 7.3 of the textbook has a good list of examples of types of polar curves. It’s worth going through some of those. Here are just a couple:

Calculus

Any polar curve $r = f(\theta)$ can be converted to paramteric equations:

So actually $\frac{dy}{dx}$ can be found as before: $\frac{y^\prime(\theta)}{x^\prime(\theta)}$. You just need to use the product rule for each of $y$ and $x$. For example, let’s find the equation of the line tangent to the rose $r = 2\sin(2\theta)$ at $\theta = \frac{\pi}{4}$.

First let’s find the point. When $\theta = \pi/4$:

Now let’s find the derivatives:

So the slope is $\sqrt{2} / -\sqrt{2}$ or $-1$.

Now that we have the point and the slope, we have our equation: $y - \sqrt{2} = -(x - \sqrt{2})$. Graph:

Playing with desmos again

Question: How might we “rotate” $y = x^2$ by $45^\circ$? ($\frac{\pi}{4}$)

Solution? Switch to polar! $y$ becomes $r\sin(\theta)$, and $x$ becomes $r\cos(\theta)$. Since $y = x^2$, we have: