# 10.4 | Polar Coordinates and Polar Graphs

The traditional way to identify points in the Cartesian plane is by means of rectangular coordinates $$(x,y)$$, where $$x$$ represents the distance traveled right or left (for negative values of $$x$$) and $$y$$ represents the distance traveled up or down (for negative values of $$y$$), with respect to the origin. The same point can be reached by

• 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 $$x$$.

If $$r$$ is a function of $$\theta$$ then
$$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{\frac{d}{d \theta}[r \sin \theta]}{\frac{d}{d \theta}[r \cos \theta]} = \frac{\frac{dr}{d\theta} \sin \theta+r \cos \theta}{\frac{dr}{d\theta} \cos \theta-r \sin \theta}$$
An equation of the tangent line to the graph of the polar equation $$r = g(\theta)$$ at $$\theta = \theta_0$$ is given by the rectangular equation
$$y = mx+b,$$
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 $$\theta = \theta_0$$.
Show that the graph of the polar equation
$$r = a$$
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 $$a$$:
$$x^2+y^2 = a^2$$

Show that the graph of the polar equation
$$r = a \cos \theta + b \sin \theta$$
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 $$r = \sqrt{\frac{a^2}{4} + \frac{b^2}{4}}$$.

Show that the graph of the polar equation
$$r \sin( \theta) = a \text{ for } 0< \theta < \pi,$$ 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 $$[\pi/2, \pi/6]$$.

Sketch the graph of the polar equation
$$r = a \sin(b \theta)$$
where $$a$$ is any real number and $$b$$ is a positive integer.

Sketch the graph of the polar equation
$$r = a \cos(b \theta)$$
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).