You will see during this lesson that points can be converted from rectangular form to polar form with a little algebra and trigonometry.

Can the equation of a shape be converted also? How about a circle, for instance?

### Polar and Cartesian Transformation

#### Polar Form to Rectangular Form

Sometimes a problem will be given with coordinates in polar form but rectangular form may be needed.

To transform the polar point \begin{align*}\left (4, \frac{3\pi}{4} \right )\end{align*} into rectangular coordinates: first identify (r, θ).

r = 4 and \begin{align*}\theta = \frac{3\pi}{4}\end{align*}.

Second, draw a vertical line from the point to the polar axis (the horizontal axis). The distance from the pole to where the line you just drew intersects the polar axis is the x value, and the length of the line segment from the point to the polar axis is the y value.

These distances can be calculated using trigonometry:

*x* = *r* cos θ and *y* = *r* sin θ

\begin{align*}x = 4 \ \mbox{cos} \ \frac{3\pi}{4}\end{align*} and \begin{align*}y = 4 \ \mbox{sin} \ \frac{3\pi}{4}\end{align*} or \begin{align*}x = -2 \sqrt{2} \ \ y = 2\sqrt{2}\end{align*}

\begin{align*}\left (4, \frac{3\pi}{4} \right )\end{align*} in polar coordinates is equivalent to \begin{align*}(-2\sqrt{2}, \ 2\sqrt{2})\end{align*} in rectangular coordinates.

#### Rectangular Form to Polar Form

Going from rectangular coordinates to polar coordinates is also possible, but it takes a bit more work. Suppose we want to find the polar coordinates of the rectangular point (2, 2). To begin doing this operation, the distance that the point (2, 2) is from the origin (the radius, *r*) can be found by

\begin{align*}r = \sqrt{x^2 + y^2}\end{align*}

\begin{align*}r = \sqrt{2^2 + 2^2}\end{align*}

\begin{align*}r = \sqrt{8} = 2\sqrt{2}\end{align*}

The angle that the line segment between the point and the origin can be found by

\begin{align*}\mbox{tan} \ \theta = \frac{y}{x}\end{align*}

\begin{align*}\mbox{tan} \ \theta = \frac{2}{2}\end{align*}

\begin{align*}\mbox{tan} \ \theta = 1\end{align*}

\begin{align*}\theta = \mbox{tan}^{-1} 1\end{align*}

\begin{align*}\theta = \frac{\pi}{4}\end{align*}

Since this point is in the first quadrant (both the *x* and *y* coordinate are positive) the angle must be 45^{o} or \begin{align*}\frac{\pi}{4}\end{align*} radians. It is also possible that when tan *θ* = 1 the angle can be in the third quadrant, or \begin{align*}\frac{5\pi}{4}\end{align*} radians. But this angle will not satisfy the conditions of the problem, since a third quadrant angle must have both *x* and *y* negative.

Note: when using \begin{align*}\mbox{tan} \ \theta = \frac{y}{x}\end{align*} to find the measure of θ you should consider, at first, the quotient \begin{align*}\mbox{tan} \ \theta = \left | \frac{y}{x} \right |\end{align*} and find the first quadrant angle that satisfies this condition. This angle will be called the reference angle, denoted *θ** _{ref}*. Find the actual angle by analyzing which quadrant the angle must be given the signs of

*x*and

*y*.

### Examples

#### Example 1

Earlier, you were asked if the equation of a circle could be converted from rectangular form to polar form.

Equation of a circle: *x*^{2} + *y*^{2} = *k*^{2} is the equation of a circle with a radius of *k* in rectangular coordinates.

The equation of a circle is extremely simple in polar form. In fact, a circle on a polar graph is analogous to a horizontal line on a rectangular graph!

You can transform this equation to polar form by substituting the polar values for *x*, *y*. Recall *x* = *r* cos *θ* and *y* = *r* sin *θ*.

(*r* cos *θ*)^{2} + (*r* sin *θ*)^{2} = *k*^{2},

square the terms: *r*^{2} cos^{2} *θ* + *r*^{2} sin^{2} *θ* = *k*^{2},

factor the *r*^{2} from both terms on the left: *r*^{2} (cos^{2} *θ* + sin^{2} *θ*) = *k*^{2}

recall the identity: cos^{2} *θ* + sin^{2} *θ* = 1

*r*^{2} = *k*^{2}

Therefore: \begin{align*}r = \pm k\end{align*} is an equation for a circle in polar units.

When r is equal to a constant, the polar graph is a circle.

#### Example 2

Transform the polar coordinates \begin{align*}\left (2, \frac{11\pi}{6} \right )\end{align*} to rectangular form.

\begin{align*}r = 2\end{align*} and \begin{align*}\theta = \frac{11\pi}{6}\end{align*}

\begin{align*}x = r \ \mbox{cos} \ \theta\end{align*} and \begin{align*}y = r \ \mbox{sin} \ \theta\end{align*}

\begin{align*}x = 2 \ \mbox{cos} \ \frac{11\pi}{6}\end{align*} and \begin{align*}y = 2 \ \mbox{sin} \ \frac{11\pi}{6}\end{align*} or \begin{align*}x = 3\sqrt{2} \ \ y = -1\end{align*}

\begin{align*}\left (2, \frac{11\pi}{6} \right )\end{align*} is equivalent to \begin{align*}(3\sqrt{2}, -1)\end{align*} or in decimal form, approximately \begin{align*}(4.342, -1)\end{align*}.

#### Example 3

Find the polar coordinates for \begin{align*}(3, -3\sqrt{3})\end{align*}.

\begin{align*}x = 3\end{align*} and \begin{align*}y = -3\sqrt{3}\end{align*}

Draw a right triangle in standard form. Find the distance the point is from the origin and the angle the line segment that represents this distance makes with the +x axis:

\begin{align*}r = \sqrt{3^2 + (-3\sqrt{3})^2}\end{align*}

\begin{align*}= \sqrt{9 + 27}\end{align*}

\begin{align*}= \sqrt{36}\end{align*}

\begin{align*}= 6\end{align*}

And for the angle,

\begin{align*}\mbox{tan} \ \theta_{ref} = \left | \frac{(-3\sqrt{3})}{3} \right |\end{align*}

\begin{align*}\mbox{tan} \ \theta_{ref} = \sqrt{3}\end{align*}

\begin{align*}\theta_{ref} = \mbox{tan}^{-1} \sqrt{3}\end{align*}

\begin{align*}\theta_{ref} = \frac{\pi}{3}\end{align*}

So, \begin{align*}\theta_{ref} = \frac{\pi}{3}\end{align*} and we can look at the signs of *x* and *y* -- (+, -) -- to see that \begin{align*}\theta = \frac{5\pi}{3}\end{align*} since it is a 4^{th} quadrant angle.

The rectangular point \begin{align*}(3, -3\sqrt{3})\end{align*} is equivalent to the polar point \begin{align*}\left (6, \frac{5\pi}{3} \right )\end{align*}.

Recall that when solving for *θ*, we used

\begin{align*}\mbox{tan} \ \theta = \left | \frac{(-3\sqrt{3})}{3} \right |\end{align*} or \begin{align*}\mbox{tan} \ \theta = \sqrt{3}\end{align*}

We found

\begin{align*}\theta = \frac{5\pi}{3}\end{align*}. BUT, θ could also be \begin{align*}\theta = \frac{2\pi}{3}\end{align*}. You must examine the signs of each coordinate to see that the angle must be in the fourth quadrant in rectangular units or between \begin{align*}\frac{3\pi}{2}\end{align*} and 2π in polar units. Of the two possible angles for θ, only \begin{align*}\frac{5\pi}{3}\end{align*} is valid. Note that when you use tan^{-1} on a calculator you will always get an answer in the range \begin{align*}- \frac{\pi}{2} \le \theta \le \frac{\pi}{2}\end{align*}.

#### Example 4

Convert the following rectangular coordinates to polar coordinates.

- \begin{align*}(3, 3\sqrt{3})\end{align*}

\begin{align*}(6,66^\circ)\end{align*}

- \begin{align*}(-2, 2)\end{align*}

\begin{align*}(2\sqrt{2}, 225^\circ)\end{align*}

Convert the following polar coordinates to rectangular coordinates.

- \begin{align*}\left (4, \frac{2\pi}{3} \right )\end{align*}

\begin{align*}(-2, 2\sqrt{3})\end{align*}

- \begin{align*}\left (-1, \frac{5\pi}{6} \right )\end{align*}

\begin{align*}\left ( \frac{\sqrt{3}}{2}, - \frac{1}{2} \right )\end{align*}

#### Example 5

Express the equation in rectangular form: \begin{align*}r = 6 cos \theta\end{align*}.

\begin{align*}r^2 = 6 r cos \theta\end{align*} : multiply both sides by \begin{align*}r\end{align*}

\begin{align*}x^2 + y^2 = 6x\end{align*} : Using \begin{align*}x^2 + y^2 = r^2\end{align*} and \begin{align*}x = r cos \theta\end{align*}

\begin{align*}\therefore x^2 + y^2 = 6x\end{align*} is the equation in rectangular form.

#### Example 6

Express the equation in rectangular form: \begin{align*}r = 6\end{align*}.

This one is easy:

\begin{align*}r = 6\end{align*} is the polar form of the equation for a circle

\begin{align*}r^2 = 6^2\end{align*} : square both sides

\begin{align*}x^2 + y^2 = 36\end{align*} : Using \begin{align*}x^2 + y^2 = r^2\end{align*} and simplifying

\begin{align*}\therefore x^2 + y^2 = 36\end{align*} is the equation in rectangular form.

### Review

- How is the point with polar coordinates \begin{align*}(5, \pi)\end{align*} represented in rectangular coordinates?

Plot each point below in polar coordinates (r, θ). Then write the rectangular coordinates (x, y) for the point.

- \begin{align*}(3, 60^o)\end{align*}
- \begin{align*}\left( -10, \frac{\pi}{3} \right)\end{align*}
- \begin{align*}\left( 15, {\pi} \right)\end{align*}

The rectangular coordinates (x, y) are given. For each question: a) Find two pairs of polar coordinates (r, θ), one with r > 0 and the other with r < 0. b) Express θ in radians, and round to the nearest hundredth.

- \begin{align*}(5, -5)\end{align*}
- \begin{align*}(0, 10)\end{align*}
- \begin{align*}(-8, 6)\end{align*}

Transform each polar equation to an equation using rectangular coordinates. Identify the graph, and give a rough sketch or description of the sketch.

- \begin{align*}\theta =\frac{\pi}{10}\end{align*}
- \begin{align*}r = 8\end{align*}
- \begin{align*}r sin \theta = 7\end{align*}
- \begin{align*}r cos \theta = -3\end{align*}

Transform each rectangular equation to an equation using polar coordinates. Identify the graph, and give a rough sketch or description of the sketch.

- \begin{align*}x^2 + y^2 - 2x = 0\end{align*}
- \begin{align*}y = \sqrt{3} x\end{align*}
- \begin{align*}y = -5\end{align*}
- \begin{align*}xy = 15\end{align*}

### Review (Answers)

To see the Review answers, open this PDF file and look for section 4.2.