Polar Coordinates¶
To describe a point $P$ in the plane requires two pieces of information. In the Cartesian coordinates that we're used to, these two pieces of information are the horizontal and vertical distances from the origin. In polar coordinates, these two pieces of information are the linear distance from the origin and the angle of rotation from the positive $x$-axis. We describe the situation in detail.
The following video lecture is by a YouTuber named Chris Odden. It's much better than anything I can create. Enjoy!
Video lecture
The Polar Transformation(s)¶
Suppose the polar coordinates of a point are known, $P(r,\theta)$. Then the rectangular coordinates of $P$ are given by
$$
T(r,\theta) = \begin{cases} x = r\cos\theta, \\ y = r\sin\theta. \end{cases}
$$
If the rectangular coordinates of a point are known, $P(x,y)$, and if $x>0$, then the polar coordinates of $P$ are given by
$$
T^{-1}(x,y) = \begin{cases} r = \sqrt{x^2 + y^2}, \\ \theta = \tan^{-1}\left(\tfrac{y}{x}\right). \end{cases}
$$
1. Exercise¶
Explain why the restriction $x>0$ is required in the definition of $T^{-1}$. How should the definition change for $x < 0$? What about for $x = 0$, $y \neq 0$?
Answer
As Chris Odden mentioned in the YouTube video lecture above, the range of arc-tangent is $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$. These values of $\theta$ correspond only to the right-hand side of the $xy$-plane; that is, the points for which $x > 0$.
The video lecture also tells us how to handle the case when $x < 0$. In this case we should define $\theta = \pi + \tan^{-1}\left(\tfrac{y}{x}\right)$.
At this point, the only points that are missing are the ones for which $x=0$. For these we define
$$
\theta = \begin{cases} \frac{\pi}{2} & \text{if}\ x=0\ \text{and}\ y > 0, \\[0.5 ex]
-\frac{\pi}{2} & \text{if}\ x=0\ \text{and}\ y < 0.\end{cases}
$$
We are still missing one point from the plane: the origin $(0,0)$ in Cartesian coordinates. This point is called the pole in polar coordinates. It is uniquely described by the property that $r = 0$. (So, it only takes one piece of information to describe the pole!)
2. Example¶
Describe the graphs of the equations $\theta = k$ and $r = k$ in the polar plane, where $k$ represents a positive constant. Sketch the graphs for a few values of $k$ for each equation.
Check your answer
The graphs of $\theta = k$ are rays that eminate from the origin in the direction of the angle $\theta$.
The graphs of $r=k$ are concentric circles of radius $k$.
These graphs can be combined on a single plane to form a polar grid.
See the polar grid
Polar Curves¶
A polar curve is the set of all points in the polar plane that satisfy an equation $F(r,\theta) = 0$, where $F$ is a function of $r$ and $\theta$. We will be most interested in the case when it's possible to solve this equation for $r$. In such cases we obtain a formula for a polar function, $r = f(\theta)$ or $r = \rho(\theta)$.
We investigate some fundamental examples below.
You may find it useful to use a polar grid, instead of trying to plot the polar graph on a Cartesian grid. I've included a polar grid below.
Blank polar grid
3. Example¶
Plot the cardioid, $r = 1 - \sin\theta$.
Video solution
4. Example¶
Plot the four-leaved rose, $r = \cos(2\theta)$.
Video solution
5. Example¶
Describe the graph of the polar function $r = \cos\theta$.
Hint
Multiply the entire equation by $r$, then use the polar transformation above to write the equation in rectangular coordinates.
Solution
Following the advice of the hint, multiplying the entire equation by $r$ yields the new equation $r^2 = r\cos\theta$.
Now subbing in $r^2 = x^2 + y^2$ and $r\cos\theta = x$, the equation becomes
$$
x^2 + y^2 = x.
$$
Subtracting $x$ from both sides and completing the square yields
$$
\left(x-\tfrac{1}{2}\right)^2 + y^2 = \tfrac{1}{4}.
$$
So the graph of this polar equation is the circle centered at $\left(\tfrac{1}{2},0\right)$ with radius $\frac{1}{2}$.
Think through how this graph is traced out and determine the smallest domain for which the circle is traced out exactly once.
Tangent Lines¶
Suppose $r = \rho(\theta)$ is a polar function that gives rise to a polar curve. We wish to compute the slope of the tangent line to this curve at any point on the curve. To do so, we'll take advantage of the ideas studied in Section 10.2. But first we must find a parametrization of the polar curve.
Recall that the polar transformation is
$$
\begin{cases}
x = r\cos\theta, \\
y = r\sin\theta.
\end{cases}
$$
Since $r = \rho(\theta)$ is a function of $\theta$, then these equation define $x$ and $y$ as functions of the single variable, $\theta$.
$$
\begin{cases}
x = \rho(\theta)\cos\theta, \\
y = \rho(\theta)\sin\theta.
\end{cases}
$$
Therefore we may always regard a polar curve corresponding to a polar function as curve in the $xy$-plane that is parametrized by the angle $\theta$.
6. Exercise¶
Use the formula for $\tfrac{dy}{dx}$ for a parametrized curve that we derived in Section 10.2 to find a formula the slope of the tangent line to a polar curve.
Answer
$$\frac{dy}{dx} = \frac{\dot{\rho}(\theta)\sin\theta + \rho(\theta)\cos\theta}{\dot{\rho}(\theta)\cos\theta - \rho(\theta)\sin\theta}$$
7. Exercise¶
Compute $\tfrac{dy}{dx}$ in terms of $\theta$ for the polar function $r = \cos\theta$.
Answer
$$\frac{dy}{dx} = -\dfrac{-\sin^2\theta + \cos^2\theta}{-2\cos\theta\sin\theta} = -\cot(2\theta)$$
Discussion¶
Questions? You can ask them below.