# Polar Curves

A **polar curve** is a shape constructed using the polar coordinate system. Polar curves are defined by points that are a variable distance from the origin (the pole) depending on the angle measured off the positive $x$-axis. Polar curves can describe familiar Cartesian shapes such as ellipses as well as some unfamiliar shapes such as cardioids and lemniscates.

$r=1-\cos{\theta}\sin{3\theta}$

Whereas Cartesian curves are useful to describe paths in terms of horizontal and vertical distances, polar curves are more useful to describe paths which are an absolute distance from a certain point. One practical use of polar curves is to describe directional microphone pickup patterns. A directional microphone will pick up different qualities of sound depending on what location the sound comes from outside of the microphone. For example, a cardioid microphone has a pickup-pattern in the shape of a cardioid.

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## Polar Coordinates System

Main Article: Polar Coordinates

See also: Convert Cartesian Coordinates to Polar and Convert Polar Coordinates to Cartesian

Each point in the polar coordinate system is given by $(r, \theta )$, where $r$ is the distance from the pole (origin) to the point, and $\theta$ is the counterclockwise angle that is made with the point, pole, and the positive $x$-axis.

To convert back and forth between polar and rectangular coordinates, we have the following formulas:

$\begin{aligned} x &= r \cos \theta \\ y &= r \sin \theta\\ r^2 &= x^2 + y^2 \\ \tan \theta^* &= \dfrac{y}{x}. \end{aligned}$

$^*$ One must consider what quadrant the point is in when computing $\theta$ from this identity.

Convert $(1, 2)$ into polar coordinates.

$x=1$ and $y=2$. The value of $r$ is computed:

$\begin{aligned} r^2 &=x^2+y^2 \\ r^2 &= 1^2+2^2 \\ r^2 &=5 \\ r &= \sqrt{5}. \end{aligned}$

The value of $\theta$ is computed:

$\begin{aligned} \tan{\theta} &= \frac{y}{x} \\ \tan{\theta} &= \frac{2}{1} \\ \theta &= \tan^{-1}(2). \end{aligned}$

The point is in the first quadrant, so the inverse tangent function should give the correct value for $\theta.$

The point in polar coordinates is $\left(\sqrt{5},\tan^{-1}(2)\right).$ $_\square$

Convert $\left( 4, \dfrac{\pi}{4} \right)$ into rectangular coordinates.

$r=4$ and $\theta=\frac{\pi}{4}.$ The value of $x$ is computed:

$\begin{aligned} x &= r\cos{\theta} \\ x &= 4\cos{\frac{\pi}{4}} \\ x &= 2\sqrt{2}. \end{aligned}$

The value of $y$ is computed:

$\begin{aligned} y &= r\sin{\theta} \\ y &= 4\sin{\frac{\pi}{4}} \\ y &= 2\sqrt{2}. \end{aligned}$

The point in rectangular coordinates is $\big(2\sqrt{2},2\sqrt{2}\big).$ $_\square$

## Polar Equations

A polar equation is any equation that describes a relation between $r$ and $\theta$, where $r$ represents the distance from the pole (origin) to a point on a curve, and $\theta$ represents the counterclockwise angle made by a point on a curve, the pole, and the positive $x$-axis.

Cartesian equations can be converted to polar equations using the same set of identities from the previous section. Likewise, polar equations can be converted to Cartesian equations using those same identities.

Convert the following equation to polar form:

$4x^2+9y^2=36.$

Use the identities $x=r\cos{\theta}$ and $y=r\sin{\theta}:$

$4r^2\cos^2{\theta}+9r^2\sin^2{\theta}=36.$

Solving for $r$ gives

$\begin{aligned} r^2\big(4\cos^2{\theta}+9\sin^2{\theta}\big) &= 36 \\\\ r^2 &= \frac{36}{4\cos^2{\theta}+9\sin^2{\theta}} \\ \\ \Rightarrow r &= \frac{\pm 6}{\sqrt{4+5\sin^2{\theta}}}. \end{aligned}$

Note that the curve is symmetric about the origin. This means that the $\pm$ symbol is not necessary to describe the full curve. Thus, the equation in polar form is

$r=\frac{6}{\sqrt{4+5\sin^2{\theta}}}.\ _\square$

One advantage of using polar equations is that certain relations that are *not* functions in Cartesian form can be expressed as functions in polar form. The problem above is an example of this. Another advantage is that certain relations are much simpler to express in polar form rather than Cartesian form.

The equation of a curve is $x^2+y^2-x=\sqrt{x^2+y^2}$.

$(\text{i})$ Find the polar equation of the curve in the form $r=f(\theta)$.

$(\text{ii})$ Sketch the curve.

$(\text{iii})$ The line $x+2y=2$ divides the region enclosed by the curve into two parts. Find the ratio of the two areas. (If the ratio is $a:b$ in its simplest form, input $a+b$ as your answer.)

There are **3** marks available for part (i), **2** marks for part (ii) and **6** marks for part (iii). In total, this question is worth **15.3%** of all available marks in the paper.

This is part of the set OCR A Level Problems.

## Transformations on Polar Curves

Certain graph transformations can be performed efficiently on polar curves. In some cases, these transformations are much easier to perform in polar form than they are to perform in Cartesian form.

Rotationsabout the pole can be performed in polar form by replacing the parameter $\theta$ with $(\theta-\phi);$ this will rotate the curve counterclockwise by $\phi$ radians.

Dilationsabout the pole can be performed in polar form by replacing the parameter $r$ with $\frac{r}{s},$ where $s$ is the scale factor.

A

reflectionabout a line$\theta=\phi$ can be performed by replacing the parameter $\theta$ with $(2\phi-\theta).$A

reflection about the polecan be performed by replacing the parameter $\theta$ with $(\theta-\pi).$

Unfortunately, **translations** are not easily performed in polar form. It is better to convert to Cartesian form if a translation is needed.

## Types of Polar Curves

There are limitless kinds of polar curves. Any equation in $r$ and $\theta$ can be drawn as a polar curve. The following examples are some of the more well-known types of polar curves.

Replacing the $\text{cosine}$ function in the equation with a $\text{sine}$ function will produce the same shape, although rotated. In most cases, this rotation will be $\frac{\pi}{2}$ radians counterclockwise.

Lines:A line has a very simple equation in polar form, provided that the line passes through the pole.

The general form equation for a line that passes through the pole is

$\theta=\alpha,$

where $\alpha$ is the angle the line makes with the positive $x$-axis. Notice that this equation contains no $r$ parameter. This implies that the line can extend indefinitely outward from the pole, but the angle it makes with positive $x$-axis remains a constant $\alpha.$

Note that any line $\theta=\alpha+\pi k$ is the

sameas the line $\theta=\alpha$ for any integer $k.$

Circles:A circle centered at the pole has a very simple equation in polar form.

The general form equation centered at the pole is

$r=a,$

where $a$ is the radius of the circle.

Cardioids:A cardioid is a heart-shaped curve that is formed by tracing the path of a point fixed on a circle as that circle rolls around another circle of the same radius.

The general form of a cardioid has equation

$r=a+a\cos{\theta},$

where $a$ is the radius of the circles described in the definition.

Limaçons:A limaçon is a more general form of a cardioid. Whereas cardioids are formed from the path traced by a point fixed

ona circle, limaçons are formed from the path traced by any point fixedtoa circle. This point could be on the circle (this would create a cardioid), or it could be any point inside or outside of the circle (this would create a different limaçon curve). As with the cardioid, the path is traced as the circle rolls around a circle of the same radius.

The general form equation of a limaçon is

$r=a+b\cos{\theta},$

where the ratio $\frac{b}{a}$ describes the shape of the limaçon:

If $\frac{b}{a}<1,$ then the point traced is inside the circle. The limaçon will have a smoothed heart shape. As $\frac{b}{a}$ approaches $0,$ the shape of the curve will approximate a circle.

If $\frac{b}{a}=1,$ then the point traced is on the circle. The limaçon will have a cardioid shape.

If $\frac{b}{a}>1,$ then the point traced is outside of the circle. The limaçon will have an inner loop. As $\frac{b}{a}$ grows larger, the inner loop will grow to the size of the larger curve, and the curve will approximate a circle.

Rose Curves:A rose curve is a sinusoidal curve graphed in polar coordinates. These kinds of curves have a flower shape, and the loops of these curves are called

petals.

The general form equation of a rose curve is

$r=a\cos(k\theta),$

where $a$ is the magnitude of each petal, and $k$ is an integer that determines how many petals there are:

If $k$ is odd, then the number of petals is $k.$

If $k$ is even, then the number of petals is $2k.$

A rose curve can also be described by the following equation:

$r=a+b\cos(k\theta).$

This equation is very similar to the limaçon equation; the only difference is the $k$ parameter. Changing the parameters $a$ and $b$ will affect both the shape of the curve and the number of petals.

Replacing the $\text{cosine}$ function with a $\text{sine}$ function will rotate the curve $\frac{\pi}{2k}$ radians counterclockwise (this is different from most other polar curves, which are rotated by $\frac{\pi}{2}$ radians).

Archimedian Spirals:An Archimedian spiral is a spiral-shaped curve that extends indefinitely outward from the pole.

The general form equation of an Archimedian spiral is

$r=a+b\theta.$

The parameter $a$ affects the initial position of the spiral, and the parameter $b$ affects the spacing of the turns of the spiral.

Lemniscates:A lemniscate is a figure-eight shaped curve. It is the locus of points where the product of the distances to two points (called foci) is a constant amount.

The general form equation of a lemniscate is

$r^2=a^2\cos(2\theta),$

where $a$ is the magnitude of one of the petals.

Replacing the $\text{cosine}$ function with a $\text{sine}$ function will rotate the curve $\frac{\pi}{4}$ radians counterclockwise (this is different from most other polar curves, which are rotated by $\frac{\pi}{2}$ radians).

Conic Sections:In the context of polar curves, a

conic sectionis the locus of points where the ratio between the distance to a point (called the focus) and the distance to a line (called the directrix) is a constant amount. This constant ratio is called theeccentricityof the conic section.

The general form equation of a conic section is

$r=\frac{l}{1+e\cos{\theta}},$

where $e$ is the eccentricity of the conic section, and $l$ is the length of the semi-latus rectum (the distance along the $y$-axis from the pole to the curve).

Using the equation above, the conic section will always have a focus at the pole. The directrix will be the line $x=\frac{1}{e}$ $(r=\frac{1}{e}\sec{\theta}$ in polar form$)$. Different values of $e$ will give different kinds of conic sections:

## Derivatives of Polar Curves

To find $\dfrac{dy}{dx}$ given that $x = r \cos \theta$ and $y = r \sin \theta$, we can find $\dfrac{dy}{d\theta}$ and $\dfrac{dx}{d\theta }$. Upon dividing the former by the latter, we get $\dfrac{dy}{dx}$.

For a polar curve,

$\dfrac{dy}{dx} = \dfrac{\dfrac{dr}{d\theta} \sin \theta + r \cos \theta}{\dfrac{dr}{d\theta} \cos \theta - r\sin \theta}.\ _\square$

**Important facts to note:**

- When $\dfrac{dy}{d\theta} = 0$, we have horizontal tangent lines.
- When $\dfrac{dx}{d\theta}= 0$, we have vertical tangent lines.
- When you find the tangent lines at the pole, let's say the slope to the tangent is $m$. Then the equation of that tangent line will be $\theta = \arctan m.$

Given $r = 1 + \cos \theta$, find the equation of all tangent lines at the pole.

Compute $\frac{dr}{d\theta}:$

$\frac{dr}{d\theta}=-\sin{\theta}.$

Then, using the theorem above, compute $\frac{dy}{dx}:$

$\begin{aligned} \frac{dy}{dx} &= \frac{-\sin^2{\theta}+r\cos{\theta}}{-\sin{\theta}\cos{\theta}-r\sin{\theta}}. \end{aligned}$

The polar curve will have a point at the pole where $r=0.$ Substituting this into the original function will give the values of $\theta$:

$\begin{aligned} 0 &= 1+\cos{\theta} \\ \cos{\theta} &= -1 \\ \theta &= \pi+2\pi k, \end{aligned}$

where $k$ is an integer. Substituting these values for $\theta$ and $r=0$ into the derivative gives

$\frac{dy}{dx}=\frac{0}{0}=(\text{undefined}).$

This implies that the curve has a cusp at $\theta=\pi+2\pi k,$ so it is not differentiable (observe that the curve is a cardioid, and a cardioid always has a cusp at the pole). Compute the limit of the derivative at $r=0$ as $\theta$ approaches $\pi:$

$\begin{aligned} \lim_{\theta\rightarrow\pi}\ \frac{-\sin^2{\theta}}{-\sin{\theta}\cos{\theta}} &= \lim_{\theta\rightarrow\pi}\ \tan{\theta} \\ &= 0. \end{aligned}$

Although the curve has a cusp at the pole, where $\theta=\pi+2\pi k,$ the cusp approaches a horizontal slope from both sides of $\theta=\pi+2\pi k$ for any integer value of $k.$ $_\square$

## Area Integrals of Polar Curves

Main Article: Polar Equations - Area

The area enclosed by a polar curve can be computed with integration.

Let $r=f(\theta)$ be the equation of a polar curve, and let $\theta=\alpha$ and $\theta=\beta$ be lines that bound an area enclosed by that polar curve. Then the area enclosed by the polar curve is

$A=\frac{1}{2} \int_{\alpha}^{\beta} \left[f(\theta)\right]^{2} \, d\theta.\ _\square$

Find the area bound by the limaçon

$r=3+2\cos{\theta}.$

Let $f(\theta)=3+2\cos{\theta}.$ First note that the period of the function is $2\pi.$ Then let the angular bounds be $\alpha=0$ and $\beta=2\pi.$ Squaring the function gives

$\left[f(\theta)\right]^2=9+12\cos{\theta}+4\cos^2{\theta}.$

The area bound by the limaçon is

$\begin{aligned} A &=\frac{1}{2} \int_{\alpha}^{\beta} \left[f(\theta)\right]^{2} \, d\theta \\ &= \frac{1}{2}\int_{0}^{2\pi} \left(9+12\cos{\theta}+4\cos^2{\theta}\right)\ d\theta \\ &= \frac{1}{2}\int_{0}^{2\pi} \left(11+12\cos{\theta}+2\cos{2\theta}\right)\ d\theta \\ &= \frac{1}{2}\Big[11\theta+12\sin{\theta}+\sin{2\theta}\Big]_{0}^{2\pi} \\ &= 11\pi.\ _\square \end{aligned}$

## Arc Length of Polar Curves

Main Article: Polar Equations - Arc Length

The length of a polar curve can be calculated with an arc length integral.

For a polar curve $r = f(\theta )$, given that the polar curve's first derivative is everywhere continuous, and the domain does not cause the polar curve to retrace itself, the arc length on $\alpha \leqslant \theta \leqslant \beta$ is given by

$\int_{\alpha}^{\beta} \sqrt{r^2 + \left(\dfrac{dr}{d\theta}\right)^2} \, d\theta.\ _\square$