- 10.2 | Plane Curves and Parametric Equations
- 10.3 | Parametric Equations and Calculus
- 10.5 | Area and Arc Length in Polar Coordinates

- first plotting a point along the \(x\)-axis of a distance from the origin equal to the distance from the origin to the point \((x,y)\), and
- second, rotating the line segment connecting the origin to this point, an angle \(\theta\).

The endpoint of this line segment, opposite the origin, is the point \((x,y)\)

When a point is identified this way, it is said to be a point in polar coordinates. For general arbitrary points in polar coordinates we use \((r,\theta)\), where the first coordinate is called the **radial coordinate** and the second coordinate is called the **angular coordinate**.

The relationship between polar and rectangular coordinates is established by the following equations

$$

x = r \cos \theta \text{ and } y = r \sin \theta.

$$

These equations allow us to rewrite a point in rectangular coordinates as a point in polar coordinates and vice-versa. More generally, these equations allow us to rewrite a rectangular equation as a polar equation and vice-versa.

To rewrite a rectangular (Cartesian) equation as a polar equation make the following substitutions and simplify as much as possible:

$$

x = r \cos \theta \text{ and } y = r\sin \theta

$$

To rewrite a polar equation as a rectangular equation look for expressions, or rewrite so that expressions of the following form appear:

$$

r \cos \theta, r \sin \theta, r^2, r, …

$$

and make the appropriate substitutions. For example, if \(r \cos \theta \) appears in a polar equation, replace it with \(x\).

$$

\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{\frac{d}{d \theta}[r \sin \theta]}{\frac{d}{d \theta}[r \cos \theta]}

= \frac{\frac{dr}{d\theta} \sin \theta+r \cos \theta}{\frac{dr}{d\theta} \cos \theta-r \sin \theta}

$$

$$

y = mx+b,

$$

provided the line is non-vertical, where \(m = \frac{dy}{dx}\Big|_{\theta = \theta_0}\) and \(b\) can be obtained by substituting the point in rectangular coordinates corresponding to \(\theta = \theta_0\). If the tangent line is vertical, then an equation for the tangent line is given by \(x = a\) where \(a\) is the \(x\)-coordinate of the point in rectangular coordinates corresponding to \(\theta = \theta_0\).

$$

r = a

$$

where \(a\) is any real number is the graph of a circle centered at the origin with radius \(a\).

If the polar equation is that of a circle centered at the origin, then the rectangular equation can be rewritten in standard form:

$$

(x-h)^2+(y-k)^2 =r^2

$$

where \((h,k)\) is the center of the circle and \(r\) is its radius.

First we rewrite the polar equation as a rectangular equation. Since \(r = \pm \sqrt{x^2+y^2}\), we have the rectangular equation

$$

\pm\sqrt{x^2+y^2} = a.

$$

Now, squaring both sides, we obtain the standard form for the equation of a circle in rectangular coordinates centered at the origin with radius \(a\):

$$

x^2+y^2 = a^2

$$

$$

r = a \cos \theta + b \sin \theta

$$

is the graph of a circle centered at \((a/2,b/2)\), when \(a \cdot b \not = 0\) and otherwise, the graph of the point \((0,0)\).

If the polar equation is that of a circle centered at \((a/2,b/2)\), then the rectangular equation can be rewritten in standard form:

$$

(x-h)^2+(y-k)^2 =r^2

$$

where \((h,k)\) is the center of the circle and \(r\) is its radius.

First we rewrite the polar equation as a rectangular equation. Since \(r = \pm \sqrt{x^2+y^2}\), \( \cos \theta = \frac{x}{r}\), and \(\sin \theta = \frac{y}{r}\) we obtain the rectangular equation as follows:

$$

\begin{aligned}

r & = a \cos \theta + b \sin \theta\\

\pm \sqrt{x^2+y^2} & = a \frac{x}{\pm \sqrt{x^2+y^2}} + b \frac{y}{\pm \sqrt{x^2+y^2}}\\

x^2+y^2 & = ax+ by

\end{aligned}

$$

Now, by completing the square on \(x\) and \(y\), we obtain the standard form of an equation of a circle:

$$

\begin{aligned}

x^2+y^2 & = ax+ by\\

x^2-ax +y^2-by & = 0\\

(x-a/2)^2 +(y-b/2)^2 & = \frac{a^2}{4} + \frac{b^2}{4}\\

\end{aligned}

$$

So, the center of the circle is at \((a/2,b/2)\) and its radius is \(r = \sqrt{\frac{a^2}{4} + \frac{b^2}{4}}\).

$$

r \sin( \theta) = a \text{ for } 0< \theta < \pi, $$ where \(a\) is a real number is the graph of the horizontal line \(y = a\).

If the polar equation is that of a non-vertical line, then the rectangular equation can be expressed in slope-intercept form:

$$

y = m x + b.

$$

By substituting \(y\) in for \(r \sin(\theta)\), we obtain the rectangular equation \(y =a\). Note that since \(\theta\) ranges from \(0\) to \(\pi\), non-inclusive \(r\) ranges from \((-\infty, \infty)\). So the line continues indefinitely in either direction.

To understand the importance of the restriction: \(0< \theta < \pi\), try graphing \(r \sin(\theta) = a \) on \([\pi/2, \pi/6]\).

$$

r = a \sin(b \theta)

$$

where \(a\) is any real number and \(b\) is a positive integer.

$$

r = a \cos(b \theta)

$$

where \(a\) is any real number and \(b\) is a positive integer.

Polar Equation | Description of Graph |
---|---|

\(r= a\) | Circle centered at the origin of radius \(a\). In the case when \(a = 0\), this is the graph of a point at the origin. |

\(\theta= a\) | Line through the origin at an angle of elevation of \(a\). In the case when \(a = 0\), the graph coincides with the \(x\)-axis. In the case when \(a = \pi/2\), the graph coincides with the \(y\)-axis. |

\(r= a \sin \theta\) | Circle centered at \((0, a/2)\) of radius \(a/2\). |

\(r= a \cos \theta\) | Circle centered at \((a/2,0)\) of radius \(a/2\). |

\(r= a \cos \theta+b \sin \theta\) | Circle centered at \((a/2,b/2)\) of radius \(\frac{1}{2}\sqrt{a^2+b^2}\). |

\(r= \theta\) | Spiral (spiraling outward, linearly). |

\(r= e^{\theta}\) | Spiral (For \(\theta \geq 0\) spiraling outward, exponentially). |

\(r= \frac{1}{\theta} \) | Spiral (For \(\theta \geq 0\) spiraling inward). |