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Polar Coordinates Functions

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Have you ever seen a nautilus shell? While this curve may seem difficult to express in terms of rectangular coordinates, we can very naturally express it in terms of **polar coordinates**. This curve is an example of a **polar function**, a type of function that is often used to describe curves that are geometrically defined in terms of distance and angles.

This article will introduce functions in polar coordinates and how to convert functions from rectangular to polar coordinates. If you are seeking information on finding the derivatives of polar functions, see the article Derivatives of Polar Curves.

A **function in polar coordinates** or a **polar function** is a function of the form

\[ r = f(\theta), \]

where \(r\) and \(\theta\) satisfy that

\[ \begin{align} x &= r \cos \theta, \mbox{ and } \\ y &= r \sin \theta . \end{align}\]

Where a function in rectangular coordinates takes an \(x\) value and returns a \(y\) value, a polar function takes an angle and returns a distance from the origin.

For more information on polar coordinates, what they are, and how to work with them, see the article Polar Coordinates.

Polar functions differ from polar curves in that a function can have only one output for any input, while a curve may have multiple outputs for any given input. For example, the equation

\[ r^2 = \cos (2\theta )\]

defines a polar curve, but does not define a polar function. In particular, when

\[ \theta = \frac{\pi}{6}, \]

both

\[ r = \sqrt{\frac{1}{2}} \mbox{ and } r = -\sqrt{\frac{1}{2}} \]

satisfy that

\[ \begin{align} r^2 &= \cos\left( 2\frac{\pi}{6} \right) \\ &= \cos\left( \frac{\pi}{3} \right) \\ &= \frac{1}{2}. \end{align}\]

For more information on polar curves that may not be polar functions, see the article Polar Curves.

It can be quite useful to transform functions written in rectangular coordinates to polar functions, and vice versa. Some functions are more naturally written in one coordinate system than the other, so converting between coordinate systems can simplify calculations like taking derivatives and finding integrals immensely. In general, functions with

\[ x^2 + y^2 \]

terms are likely to have nice polar forms.

Some important identities to know when working with functions in polar coordinates are the following:

\( y = r\sin \theta\)

\(x = r\cos \theta \)

\(x^2 + y^2 = r^2 \)

\(\tan \theta = \frac{x}{y} \)

If you choose to use the relation

\[ \theta = \tan^{-1}\left( \frac{x}{y} \right), \]

remember to be careful with which quadrant your points are in, since for any \(x\) and \(y\) there are many \(\theta\) values satisfying

\[\tan \theta = \frac{x}{y}. \]

For more information on selecting the correct angle, see the article Polar Coordinates.

If you are converting a function or equation written in rectangular coordinates to a function or equation in polar coordinates, there are two steps you can follow:

*Step 1:* Replace every instance of \(x\) with \(r \cos \theta \), every instance of \(y\) with \(r \sin\theta \), and every instance of \(x^2 + y^2\) with \( r^2\).

*Step 2:* Simplify the expression. The identity \[ \cos^2\theta + \sin^2\theta = 1 \] can be especially useful in this step.

Let's look at some examples.

Rewrite the equation

\[ x^2 + y^2 = e^{xy} \]

in polar form.

Answer:

*Step 1:* Doing the replacement gives you

\[ r^2 = e^{(r\cos \theta )(r \sin\theta)}.\]

*Step 2:* Simplify the expression. First, you can rewrite the expression as:

\[ r^2 = e^{r^2\cos \theta\sin\theta}.\]

The question now is if this expression is simplified enough. Since there are no clear substitutions that would make it look any simpler to evaluate for a particular \(\theta \) you can call this fully simplified.

So the conversion to polar form is

\[ r^2 = e^{r^2\cos \theta\sin\theta}.\]

Sometimes you might not be sure if you are done simplifying.

Write the function \( y = x^3\) in polar form.

Answer:

*Step 1:* Doing the replacement, you get:

\[ r\sin\theta = r^3 \cos^3\theta .\]

*Step 2:* Then simplifying gives you

\[ r^2 = \frac{\sin\theta}{\cos^3\theta}.\]

You might be tempted to take the square root of both sides. But remember, if you do that you actually get two solutions,

\[ \begin{align} r &=\sqrt{\frac{\sin\theta}{\cos^3\theta} },\quad \mbox{ and } \\ r&= - \sqrt{\frac{\sin\theta}{\cos^3\theta} } .\end{align}\]

So you can only take the square root if both of the answers are exactly the same! In this case it is better to say that the polar curve is

\[ r\sin\theta = r^3 \cos^3\theta .\]

Converting a function written in polar coordinates to a function written in rectangular coordinates can be a bit more tricky, but following the steps below often gives a good start.

*Step 1: * Replace every instance of \(\theta\) with expressions in terms of \(x\), \(y\), and \(r\). The relations

\[ \begin{align} \cos\theta &= \frac{x}{r} \\ \sin\theta &= \frac{y}{r}\\ \tan\theta &= \frac{y}{x} \end{align}\]

are frequently useful here, as are the double and half-angle trig identities.

*Step 2:* Rearrange the equation and replace every instance of \(r\) with expressions in terms of \(x\) and \(y\), using the relation

\[x^2 + y^2 = r^2 \]

and squaring \(r\) terms as necessary.

*Step 3:* Simplify the equation.

Let's look at some examples.

Rewrite the equation \(r = 3\sin\theta \) in rectangular form.

Answer:

*Step 1:* The first step is to rewrite it. Using the identity \(x = r\sin\theta \), which is equivalent to the expression

\[ \sin\theta = \frac{x}{r}, \]

you can write

\[ r = 3\frac{x}{r} .\]

*Step 2:* Multiplying both sides by \(r\) gives you

\[r^2 = 3x.\]

Then you can use the identity \(x^2 + y^2 = r^2 \) to write

\[ x^2 + y^2 = 3x.\]

*Step 3:* This is already simplified, so you are done!

Let's look at an example where you actually need to simplify a little.

Rewrite the equation

\[ r = \frac{5}{2+3\sin\theta } \]

in rectangular form.

Answer:

*Step 1: *Rewrite all the \(\theta\) terms. In this case, you only need to use

\[ \sin\theta = \frac{y}{r} \]

to get

\[ r = \frac{5}{2+3\frac{y}{r} }. \]

*Step 2:* Next, rearrange the equation, multiplying both sides by the denominator to get the \(r\) terms on the same side of the equation

\[ r\left(2+3\frac{y}{r} \right) = 5,\]

or

\[2r + 3y = 5.\]

You can solve this for \(r\) to get

\[ 2r = 5-3y.\]

You really wish that was an \(r^2\) on the left side because then you would be almost done. But it isn't, so you need to square both sides of the equation to get

\[ \begin{align} 4r^2 &= (5-3y)^2 \\ &= 25 - 30y + 9y^2. \end{align}\]

Now you can replace the \(r^2\) with \(x^2 + y^2 \), giving you

\[4(x^2 + y^2) = 25 - 30y + 9y^2 .\]

Step 3: All that is left to do is simplify it! Generally for this kind of equation you want all of the variables on one side, and all of the constants on the other side. So you get

\[ 4x^2 + 4y^2 +30y - 9y^2 = 25,\]

or

\[ 4x^2 -5y^2 + 30y = 25.\]

Many familiar curves can be written as functions in polar coordinates. Two such families of curves are circles and lines. Other important polar curves, many of which are polar functions, are discussed in the article Polar Curves.

For each of the types of curves below, let's look at their polar equations using a combination of geometric and algebraic reasoning. The idea is that circles and lines both have nice geometric definitions that simplify the algebra in certain cases.

In general, before carrying out the steps outlined above for converting between polar and rectangular forms, it can be useful to consider whether the equation you are working with has any geometric properties that can simplify computations.

There are four types of circles that we can write particularly nicely in polar coordinates. These are:

circles centered at the origin

circles of radius a centered at the point \( (0,a)\)

circles of radius a centered at the point \( (a,0)\)

circles of radius \(\sqrt{a^2+b^2}\) centered at the point \( (a, b)\)

The generic polar formula for a circle with radius a centered at \( (b, c)\) is quite complicated and often not particularly useful, but these four particular cases show up often and are important to understand.

To derive the function for a circle of radius a centered at the origin, we first recall the geometric definition of a circle.

A circle with center \(O\) and radius \(a\) must satisfy the condition that, given any point \(P\) on the circle, the line segment \(OP\) has length \(a\).

In this case, the point \(O\) happens to be the origin, so this means that any point on the circle must be distance \(a\) from the origin.

You can write this geometric condition in terms of polar coordinates.

Given a point \( (r,\theta )\), it is on the circle if \(r=a\), since the center of the circle is at the origin and the \(r\) coordinate describes distance from the origin. There are no conditions on the \(\theta\) coordinate, since a circle centered at the origin can have points at any angle. Thus, the function describing the circle of radius \(a\) centered at the origin in polar coordinates is

\[r=a.\]

This describes a circle because it takes an angle \(\theta\) and always returns \(a\) as the corresponding distance from the origin.

Notice that you did not have to do any algebraic work in this case; the equation fell right out of the geometric definition of a circle.

It is important to note that the graph of the circle does not describe a function in rectangular coordinates, as it does not satisfy the vertical line test. However, it is a function when written in polar coordinates. This is one of the advantages of working with polar coordinates: many curves that **do not** describe functions in rectangular coordinates **do** define functions in polar coordinates.

The function describing a circle of radius \(a\) through the point \( (a,0) \) in polar coordinates is

\[ r^2 = 2a\cos\theta .\]

You can prove this by starting with the equation for a circle in rectangular coordinates, then converting the equation to polar coordinates.

First, in rectangular coordinates, the equation for a circle of radius a through the point \( (a,0)\) is

\[ (x-a)^2 + y^2 = a^2.\]

Doing the same replacing as in the previous examples, you get that

\[ (r\cos\theta - a)^2 + (r\sin\theta )^2 = a^2.\]

Now it is a matter of simplifying with algebra to get:

\[ r^2 \cos^2 \theta - 2ar\cos\theta + a^2 + r^2\sin^2\theta = a^2,\]

or

\[r^2 = 2ar\cos\theta ,\]

which is exactly what you were hoping for.

The function describing a circle of radius \(a\) through the point \( (0,a)\) in polar coordinates is

\[ r = 2a \sin\theta .\]

Just as before, you can prove this by starting with the equation for a circle in rectangular coordinates, then converting the equation to polar coordinates.

This is the most general example. You can take the equation for the function describing a circle of radius \( \sqrt{a^2 + b^2}\) through the point \( (a,b)\) in polar coordinates and substitute in the origin, or a point on the horizontal axis, or a point on the vertical axis, to get any of the three equations listed above.

The equation for the function describing a circle of radius \( \sqrt{a^2 + b^2}\) through the point \( (a,b)\) is

\[ r = 2a\cos\theta + 2b\sin\theta .\]

You can prove this by starting with the equation for a circle in rectangular coordinates, then converting the equation to polar coordinates, just like in the examples above.

Lines can also be written as polar functions. There are three different polar equations for lines that we generally work with, depending on the information we are given:

Lines through the origin with slope \( \tan a\)

Lines with the point \( (a, b)\) closest to the origin

Lines with slope \(m\) through the point \( (a, b)\)

The line through the origin with slope \( \tan a\) is defined by

\[ \theta = a.\]

The vertical line through the origin has equation

\[ \theta = \frac{\pi}{2}.\]

Both of these equations come straight from the definition of the tangent. Since \(\theta = a\) is the set of all points \( (r, \theta )\) that are at angle \(a\) from the positive \(x\)-axis, it gives a line passing through the origin.

The slope of a line is its rise over its run. This is the same as the tangent the angle makes with respect to the origin.

Thus, the line through the origin with slope \( \tan a\) has the equation \(\theta = a \).

The polar equation for the line with point \( (a, b)\) closest to the origin is

\[ r = a\sec (\theta - b).\]

Here, instead of assuming that \((a, b)\) is in Cartesian coordinates (as you have until this point), you need to assume that the point \( (a, b)\) is in polar coordinates.

To see why this might be the case, take a look at the picture below. In this image, the point labelled \(A = (r, \theta) \) is an arbitrary point on the line.

Since the point \(P = (a,b) \) is that point on the line closest to the origin, the line segment \(OP\) is perpendicular to the line. So, the triangle \(\Delta OAP\) is a right triangle. Since \(\Delta OAP\) is a right triangle,

\[ \cos (\theta - b) = \frac{a}{r}.\]

You can rearrange this to get that

\[ r = \frac{a}{\cos (\theta - b)} = a\sec (\theta - b),\]

which is exactly what you wanted.

The line with slope \(m\) through a point \( (0,b)\) is given by

\[ r = \frac{b}{\sin \theta - m \cos\theta }.\]

Here, you need to assume that \((a, b)\) is in rectangular coordinates.

First, recall that the equation for a line with slope \(m\) through a point \( (0,b) \) is \(y = mx+b\). Doing the same replacement as in the previous examples you get

\[ r\sin\theta - b = mr\cos\theta .\]

Then you just need to simplify to get that

\[ r = \frac{b}{\sin \theta - m \cos\theta }.\]

Spherical coordinates are generalizations of polar coordinates to three dimensions. For more information on spherical coordinates, see the article Polar Coordinates. You can use the equations found in that article to convert functions in spherical coordinates to functions in rectangular coordinates, and vice-versa.

- A function in polar coordinates is a function that takes in an angle theta and returns a radius \(r\).
- Not all polar equations or polar curves are polar functions.
- Circles and lines are two important types of functions that can be written in polar form.
- To convert a polar function to a function in rectangular coordinates (or vice-versa), use the equations for converting between polar and rectangular coordinates.

Examples of functions in polar coordinates are Archimedean spirals, rose curves, and cardioids.

More about Polar Coordinates Functions

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