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4.2: Polar-Cartesian Transformations

Difficulty Level: At Grade Created by: CK-12

Learning Objectives

• Convert polar points to rectangular form
• Convert rectangular points to polar form
• Convert some graphs from rectangular to polar

Introduction

You have learned that the polar coordinate system offers a way to plot points in a plane. Also, you have extensive experience using the rectangular (x - y) coordinate system. Sometimes it is helpful to transform a point or a whole graph from one coordinate system to the other.

In this section you will learn how to convert polar points to rectangular, and vice versa. You will also see some surprising ways to write conic sections in polar form.

Polar to Rectangular

Sometimes a problem will be given as a coordinate in polar form but rectangular form may be needed.

To transform the polar point (4,3π4)\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 the point on polar axis. Drop a line from the point to the polar axis. The distance from the pole to where the line 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:

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

Example: 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 (4.342, -1)

Rectangular to Polar

Going from rectangular coordinates to polar coordinates is also possible, but it takes a bit more work. Suppose we want t 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 45o 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 you use 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 θref. Find the actual angle by analyzing which quadrant the angle must be given the signs of x and y.

Example: 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*}

Now, 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 4th 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*}.

Conic Section Transformations

Circle:

x2 + y2 = k2 is the equation of a circle with a radius of k in rectangular coordinates. 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 = k2

or by squaring the terms:

r2 cos2 θ + r2 sin2 θ = k2

or by factoring the r2 from both terms on the left

r2 (cos2 θ + sin2 θ) = k2

and recalling the identity cos2 θ + sin2 θ = 1

r2 = k2

r = ±k is an equation for a circle in polar units

When r is equal to a constant, the polar graph is a circle. This was confirmed previously when introduced to polar graphing.

In the graph above of r = 2, at every angle the distance to the pole is 2 units. That is the meaning of r = 2. (Note that circles in the polar coordinate system are analogous to horizontal lines in the rectangular coordinate system).

Optional: Parabola, Hyperbola and Ellipse in Polar form

We could convert the traditional equations for parabolas, hyperbolas, and ellipses into polar form using the same conversions

x = r cos θ and y = r sin θ

But, there is a much more elegant way to graph parabola, hyperbolas and ellipses in polar form. Below we introduce this method.

First, for these conic sections, recalling the definition of eccentricity will be useful.

\begin{align*}\frac{PF}{PD} = \mbox{e} \ \mbox{(eccentricity)}\end{align*}.

e is the ratio of the distance between a point on the conic and its focus (PF) and the distance between the point on the conic and the conic’s directrix (PD)

For convenience the focus is placed at the pole (0, 0)

The distance between the point on the conic and conic’s focus (on the pole) is equal to r, and as seen in the diagram, the distance between the point on the conic and the conic’s directrix, PD is equal to p + FA.

\begin{align*}e = \frac{r}{p + FA}\end{align*}

Also in the diagram, note that \begin{align*}\frac{FA}{r} = \mbox{cos} \ \theta\end{align*}

\begin{align*}e = \frac{r}{p + r \ \mbox{cos} \ \theta}\end{align*}

solving for r:

ep + er cos θ = r

or

r - er cos θ = ep

factoring for r: r(1 - e cos θ) = ep

and finally we get Equation A:

\begin{align*}r = \frac{ep}{1-e \ \mbox{cos} \ \theta}\end{align*}

In this equation the denominator contains 1 - e cos θ term

Examine the graph above: the conic opens to the right. This equation is valid for any conic that is opening to the right or if the major axis is the x-axis.

The following shows the other forms of conics in polar form:

\begin{align*}r = \frac{ep}{1-e \ \mbox{sin} \ \theta}\end{align*}

opens to up, and

\begin{align*}r = \frac{ep}{1+e \ \mbox{sin} \ \theta}\end{align*}

opens down.

Or, if these are the equations of an ellipse, then the major axis is parallel to the x-axis

\begin{align*}r = \frac{ep}{1+e \ \mbox{cos} \ \theta}\end{align*} opens to the left or vertically or major axis is the y-axis

If the polar equation is given with a constant in the numerator and with a sum of a constant and a term that contains either sin θ or cos θ in the denominator, the polar equation is a conic.

To graph such a polar equation these are steps to follow:

1. First determine which conic will match the given polar equation. To do this: Be sure the constant in the denominator is Equation A is equal to 1. If it is not, then divide the three terms in Equation A by this constant. Once the constant is ONE, the coefficient of the trig term is equal to e. If e > 1, the conic section is a hyperbola. If e = 1, the conic is a parabola and if e < 1, then the conic section is an ellipse.

When e = 0, the conic is a circle.

2. The numerator is the product of e and p (the distance between the focus and the directrix). Substituting the value of e, from step (1), p can now be found.

3. Examine the four general conic polar equations to see how the conic will open. This is determined by the sign of the trig term and whether the trig term is sin or cos.

Example: Graph the polar equation: \begin{align*}r = \frac{9}{3 -3 \ \mbox{sin} \ \theta}\end{align*}

1. Divide all terms by 3 to put into the form \begin{align*}1 -e \ \mbox{sin} \ \theta\end{align*} or \begin{align*}r = \frac{3}{1 - \ \mbox{sin} \ \theta}\end{align*}

2. This has -e sin θ in denominator- therefore it opens up, e = 1 the conic is a parabola

3. ep = 3 and e = 1, p = 3

Example: Graph the polar equation: \begin{align*}r = \frac{3}{1 + 2 \ \mbox{cos} \ \theta}\end{align*}

1. Since the constant in the numerator is already 1, no need to divide all terms.

2. e = 2, the coefficient of the cosine term. Therefore the conic is a hyperbola ep = 3, p = 1.5

3. This has a +2cosθ in denominator, so the conic opens left or horizontally.

Example: Graph the polar equation: \begin{align*}r = \frac{5}{2 + 3 \ \mbox{sin} \ \theta}\end{align*}

1. divide all terms by 2: \begin{align*}r = \frac{5/2}{1 + {(3/2)} \ \mbox{sin} \ \theta}\end{align*}

2. e = 3/2, hyperbola., ep = 5/2, p = 5/3

3. 1 + 3/2sinθ in denominator- opens vertically

Example: Graph \begin{align*}r = \frac{2}{2 - \mbox{cos} \ \theta}\end{align*}

1. divide all terms by 2: \begin{align*}r = \frac{1}{1- \dfrac{1}{2} \ \mbox{cos} \ \theta}\end{align*}

2. \begin{align*}e = \frac{1}{2}\end{align*}, the conic an ellipse, ep = 1 or p = 2

3. the denominator contains \begin{align*}1 - \frac{1}{2} \ \mbox{cos} \ \theta\end{align*}, major axis is horizontal

Applications, Technological Tools

Technology note: Some graphing calculators can convert polar points to rectangular form and rectangular coordinates to polar form.

On the TI-83/84: go to [ANGLE](or [2nd]function) [APPS]. Scroll down to 5 or “R-Pr(“ and press [Enter]. Next, enter the rectangular coordinates and close the parenthesis. Press [Enter], the “r” value appears. Scroll down to 6R-Pθ, and the polar angle appears in decimal radian form.

Also under the [ANGLE] menu, commands 7 and 8 allow transformation from polar form to rectangular form.

Lesson Summary

There are at least two coordinate systems- rectangular and polar. Each can be used to graph a variety of equations. Additionally, once a point or an equation is graphed in one system, these points and equations can be transformed into the other system.

Points to Consider

Sometimes an equation written in rectangular form, it can reveal characteristics that are hidden, unless transformed into polar. The same can be true for polar equations when transformed into rectangular form. The key here is to be able to transform one type into the other with ease, so that all characteristics of the equation can be known.

Review Questions

1. Convert the following: Convert the following rectangular coordinates to polar coordinates a. \begin{align*}(3, 3\sqrt{3})\end{align*} b. \begin{align*}(-2, 2)\end{align*} Convert the following polar coordinates to rectangular coordinates: c. \begin{align*}\left (4, \frac{2\pi}{3} \right )\end{align*} d. \begin{align*}\left (-1, \frac{5\pi}{6} \right )\end{align*}
2. Identify the conic section that each polar equation represents. Use the three steps to graph each polar equation: a. \begin{align*}r = \frac{1}{1 + \mbox{sin} \ \theta}\end{align*} b. \begin{align*}r = \frac{1}{1 - \mbox{cos} \ \theta}\end{align*} c. \begin{align*}r = \frac{6}{3 - 2\ \mbox{cos} \ \theta}\end{align*} d. \begin{align*}r = \frac{12}{5 + 3\ \mbox{cos} \ \theta}\end{align*} e. \begin{align*}r = \frac{8}{2 - 4\ \mbox{cos} \ \theta}\end{align*}

1. a. \begin{align*}(6,66^\circ)\end{align*} b. \begin{align*}(2\sqrt{2}, 225^\circ)\end{align*} c. \begin{align*}(-2, 2\sqrt{3})\end{align*} d. \begin{align*}\left ( \frac{\sqrt{3}}{2}, - \frac{1}{2} \right )\end{align*}
2. a. parabola e = 1, + sin θ → opens down b. parabola e = 1, - cos θ → opens to the right c. ellipse \begin{align*}e = \frac{2}{3} < 1\end{align*} d. ellipse \begin{align*}e = \frac{3}{5} < 1\end{align*} e. hyperbola, e = 2 > 1

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