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# 6.3: Converting Between Systems

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Convert rectangular coordinates to polar coordinates.
• Convert equations given in rectangular form to equations in polar form and vice versa.

## Polar to Rectangular

Just as \begin{align*}x\end{align*} and \begin{align*}y\end{align*} are usually used to designate the rectangular coordinates of a point, \begin{align*}r\end{align*} and \begin{align*}\theta\end{align*} are usually used to designate the polar coordinates of the point. \begin{align*}r\end{align*} is the distance of the point to the origin. \begin{align*}\theta\end{align*} is the angle that the line from the origin to the point makes with the positive \begin{align*}x-\end{align*}axis. The diagram below shows both polar and Cartesian coordinates applied to a point \begin{align*}P\end{align*}. By applying trigonometry, we can obtain equations that will show the relationship between polar coordinates \begin{align*}(r, \theta)\end{align*} and the rectangular coordinates \begin{align*}(x, y)\end{align*}

The point \begin{align*}P\end{align*} has the polar coordinates \begin{align*}(r, \theta)\end{align*} and the rectangular coordinates \begin{align*}(x, y)\end{align*}.

Therefore

\begin{align*}x& = r \cos \theta && r^2 = x^2+y^2\\ y&= r \sin \theta && \tan \theta = \frac{y}{x}\end{align*}

These equations, also known as coordinate conversion equations, will enable you to convert from polar to rectangular form.

Example 1: Given the following polar coordinates, find the corresponding rectangular coordinates of the points: \begin{align*}W(4,-200^\circ),H \left (4, \frac{\pi}{3} \right )\end{align*}

Solution:

a) For \begin{align*} W(4,-200^\circ), r = 4 \end{align*} and \begin{align*}\theta = -200^\circ\end{align*}

\begin{align*}x & = r \cos \theta && y = r \sin \theta\\ x &= 4 \cos (-200^\circ) && y = 4 \sin(-200^\circ)\\ x &= 4(-.9396) && y = 4(.3420)\\ x & \approx - 3.76 && y \approx 1.37\end{align*}

The rectangular coordinates of \begin{align*}W\end{align*} are approximately \begin{align*}(-3.76, 1.37)\end{align*}.

b) For \begin{align*}H \left ( 4,\frac{\pi}{3} \right ), r = 4\end{align*} and \begin{align*}\theta = \frac{\pi}{3}\end{align*}

\begin{align*}x &= r \cos \theta && y = r \sin \theta\\ x &= 4 \cos \frac{\pi}{3} && y = 4 \sin \frac{\pi}{3}\\ x &= 4 \left ( \frac{1}{2} \right ) && y = 4 \left ( \frac{\sqrt{3}}{2} \right )\\ x &= 2 && y = 2 \sqrt{3}\end{align*}

The rectangular coordinates of \begin{align*}H\end{align*} are \begin{align*}(2, 2 \sqrt{3})\end{align*} or approximately \begin{align*}(2, 3.46)\end{align*}.

In addition to writing polar coordinates in rectangular form, the coordinate conversion equations can also be used to write polar equations in rectangular form.

Example 2: Write the polar equation \begin{align*}r = 4 \cos \theta \end{align*} in rectangular form.

Solution:

\begin{align*}r &= 4 \cos \theta\\ r^2 &= 4r \cos \theta && Multiply \ both \ sides \ by \ r.\\ x^2 + y^2 &= 4x && r^2 = x^2 + y^2 \ and \ x = r \cos \theta\end{align*}

The equation is now in rectangular form. The \begin{align*}r^2\end{align*} and \begin{align*}\theta\end{align*} have been replaced. However, the equation, as it appears, does not model any shape with which we are familiar. Therefore, we must continue with the conversion.

\begin{align*}x^2 - 4x + y^2 &= 0\\ x^2 - 4x + 4 + y^2 &= 4 && Complete \ the \ square \ for \ x^2 - 4x.\\ (x - 2)^2 + y^2 &= 4 && Factor \ x^2 - 4x + 4.\end{align*}

The rectangular form of the polar equation represents a circle with its centre at (2, 0) and a radius of 2 units.

This is the graph represented by the polar equation \begin{align*}r = 4 \cos \theta\end{align*} for \begin{align*}0 \le \theta \le 2 \pi\end{align*} or the rectangular form \begin{align*}(x - 2)^2 + y^2 = 4.\end{align*}

Example 3: Write the polar equation \begin{align*}r = 3 \csc \theta\end{align*} in rectangular form.

Solution:

\begin{align*}r &= 3 \csc \theta\\ \frac{r}{\csc \theta} &= 3 && divide \ by \csc \theta\\ r \cdot \frac{1}{\csc \theta} &= 3\\ r \sin \theta &= 3 && \sin \theta = \frac{1}{\csc \theta}\\ y &= 3 && y = r \sin \theta\end{align*}

## Rectangular to Polar

When converting rectangular coordinates to polar coordinates, we must remember that there are many possible polar coordinates. We will agree that when converting from rectangular coordinates to polar coordinates, one set of polar coordinates will be sufficient for each set of rectangular coordinates. Most graphing calculators are programmed to complete the conversions and they too provide one set of coordinates for each conversion. The conversion of rectangular coordinates to polar coordinates is done using the Pythagorean Theorem and the Arctangent function. The Arctangent function only calculates angles in the first and fourth quadrants so \begin{align*}\pi\end{align*} radians must be added to the value of \begin{align*}\theta\end{align*} for all points with rectangular coordinates in the second and third quadrants.

In addition to these formulas, \begin{align*}r = \sqrt{x^2 + y^2}\end{align*} is also used in converting rectangular coordinates to polar form.

Example 4: Convert the following rectangular coordinates to polar form.

\begin{align*}P (3, -5)\end{align*} and \begin{align*}Q (-9, -12)\end{align*}

Solution: For \begin{align*}P (3, -5) \ x = 3\end{align*} and \begin{align*}y = -5\end{align*}. The point is located in the fourth quadrant and \begin{align*}x > 0\end{align*}.

\begin{align*}r &= \sqrt{x^2 + y^2} && \theta = Arc \ \tan \frac{y}{x}\\ r &= \sqrt{(3)^2 + (-5)^2} && \theta = \tan^{-1} \left ( -\frac{5}{3} \right )\\ r &= \sqrt{34} && \theta \approx - 1.03\\ r &\approx 5.83\end{align*}

The polar coordinates of \begin{align*}P (3, -5)\end{align*} are \begin{align*}P (5.83, -1.03)\end{align*}.

For \begin{align*}Q (-9, -12) \ x = -9\end{align*} and \begin{align*}y = -5\end{align*}. The point is located in the third quadrant and \begin{align*}x < 0\end{align*}.

\begin{align*}r & = \sqrt{x^2 + y^2} && \theta = Arc \tan \frac{y}{x} + \pi \\ r & = \sqrt{(-9)^2 + (-12)^2} && \theta = \tan^{-1} \left (\frac{-12}{-9} \right ) + \pi \\ r & = \sqrt{225} && \theta \approx 4.07 \\ r & = 15\end{align*}

The polar coordinates of \begin{align*}Q (-9, -12)\end{align*} are \begin{align*}Q (15, 4.07) \end{align*}

## Converting Equations

To write a rectangular equation in polar form, the conversion equations of \begin{align*}x = r \cos \theta\end{align*} and \begin{align*}y = r \sin \theta\end{align*} are used.

Example 5: Write the rectangular equation \begin{align*}x^2 + y^2 = 2x\end{align*} in polar form.

Solution: Remember \begin{align*}r = \sqrt{x^2 + y^2}, r^2 = x^2 + y^2\end{align*} and \begin{align*}x = r \cos \theta\end{align*}.

\begin{align*}x^2 + y^2 &= 2x\\ r^2 &= 2(r \cos \theta) && Pythagorean \ Theorem \ and \ x = r \cos \theta\\ r^2 &= 2r \cos \theta\\ r &= 2 \cos \theta && Divide \ each \ side \ by \ r\end{align*}

Example 6: Write the rectangular equation \begin{align*}(x - 2)^2 + y^2 = 4\end{align*} in polar form.

Solution: Remember \begin{align*}x = r \cos \theta\end{align*} and \begin{align*}y = r \sin \theta\end{align*}.

\begin{align*}&(x - 2)^2 + y^2 = 4\\ &(r \cos \theta - 2)^2 + (r \sin \theta)^2 = 4 && x = r \cos \theta \ and \ y = r \sin \theta\\ &r^2 \cos^2 \theta - 4r \cos \theta + 4 + r^2 \sin^2 \theta = 4 && expand \ the \ terms\\ &r^2 \cos^2 \theta - 4r \cos \theta + r^2 \sin^2 \theta = 0 && subtract \ 4 \ from \ each \ side\\ &r^2 \cos^2 \theta + r^2 \sin^2 \theta = 4r \cos \theta && isolate \ the \ squared \ terms\\ &r^2 (\cos^2 \theta + \sin^2 \theta) = 4r \cos \theta && factor \ r^2 - a \ common \ factor\\ &r^2 = 4r \cos \theta && Pythagorean \ Identity\\ &r = 4 \cos \theta && Divide \ each \ side \ by \ r\end{align*}

If the graph of the polar equation is the same as the graph of the rectangular equation, then the conversion has been determined correctly.

\begin{align*}(x-2)^2+y^2=4\end{align*}

The rectangular equation \begin{align*}(x - 2)^2 + y^2 = 4\end{align*} represents a circle with center (2, 0) and a radius of 2 units. The polar equation \begin{align*}r = 4 \cos \theta\end{align*} is a circle with center (2, 0) and a radius of 2 units.

## Converting Using the Graphing Calculator

You have learned how to convert back and forth between polar coordinates and rectangular coordinates by using the various formulae presented in this lesson. The TI graphing calculator allows you to use the angle function to convert coordinates quickly from one form to the other. The calculator will provide you with only one pair of polar coordinates for each pair of rectangular coordinates.

Example 7: Express the rectangular coordinates of \begin{align*}A (-3, 7)\end{align*} as polar coordinates.

Polar coordinates are expressed in the form \begin{align*}(r,\theta)\end{align*}. An angle can be measured in either degrees or radians, and the calculator will express the result in the form selected in the \begin{align*}\fbox{MODE}\end{align*} menu of the calculator.

Press \begin{align*}\fbox{MODE}\end{align*} and cursor down to Radian Degree. Highlight Degree. Press \begin{align*} \begin{array} {|c|} \hline 2^{\text{nd}}\\ \hline \end{array} \end{align*} \begin{align*}\fbox{mode}\end{align*} to return to home screen. To access the angle menu of the calculator press \begin{align*} \begin{array} {|c|} \hline 2^{\text{nd}}\\ \hline \end{array} \end{align*} \begin{align*}\fbox{APPS}\end{align*} and this screen will appear:

Cursor down to 5 and press \begin{align*}\fbox{ENTER}\end{align*}. The following screen will appear . Press -3, 7) \begin{align*}\fbox{ENTER}\end{align*} and the value of \begin{align*}r\end{align*} will appear . Access the angle menu again by pressing \begin{align*} \begin{array} {|c|} \hline 2^{\text{nd}}\\ \hline \end{array} \end{align*} \begin{align*}\fbox{APPS}\end{align*}. When the angle menu screen appears, cursor down to 6 and pres \begin{align*}\fbox{ENTER}\end{align*} or press 6 on the calculator. The screen will appear. Press -3, 7) \begin{align*}\fbox{ENTER}\end{align*} and the value of \begin{align*}\theta\end{align*} will appear.

This procedure can be repeated to determine the rectangular coordinates in radians. Before starting, press \begin{align*}\fbox {MODE}\end{align*} and cursor down to Radian Degree and highlight Radian.

Example 8: Express the polar coordinates of \begin{align*}(300, 70^\circ)\end{align*} in rectangular form.

The angle \begin{align*}\theta\end{align*} is given in degrees so the mode menu of the calculator should also be set in degree. Therefore, press \begin{align*}\fbox{MODE}\end{align*} and cursor down to Radian Degree and highlight degree. Press \begin{align*} \begin{array} {|c|} \hline 2^{\text{nd}}\\ \hline \end{array} \end{align*} \begin{align*}\fbox{mode}\end{align*} to return to home screen. To access the angle menu of the calculator press \begin{align*} \begin{array} {|c|} \hline 2^{\text{nd}}\\ \hline \end{array} \end{align*} \begin{align*}\fbox{APPS}\end{align*} and this screen will appear:

Cursor down to 7 and press \begin{align*}\fbox{ENTER}\end{align*} or press 7 on the calculator. The following screen will screen will appear: Press 300, 70) and the value of \begin{align*}x\end{align*} will appear Access the angle menu again by pressing \begin{align*} \begin{array} {|c|} \hline 2^{\text{nd}}\\ \hline \end{array} \end{align*} \begin{align*}\fbox{APPS}\end{align*}. When the angle menu screen appears, cursor down to 8 and pres \begin{align*}\fbox{ENTER}\end{align*} or press 8 on the calculator. The screen will appear. Press 300,70) \begin{align*}\fbox{ENTER}\end{align*} and the value of \begin{align*}y\end{align*} will appear .

## Points to Consider

• When we convert coordinates from polar form to rectangular form, the process is very straightforward. However, when converting a coordinate from rectangular form to polar form there are some choices to make. For example the point 0,1 could translate to \begin{align*}(1,2 \pi)\end{align*} or to \begin{align*}(1,- 4 \pi)\end{align*}, and so on.
• Are there any advantages to using polar coordinates instead of rectangular coordinates? List any situations in which this is the case. What types of curves are easier to draw with polar coordinates?
• List situations in which rectangular coordinates are preferable.

## Review Questions

1. For the following polar coordinates that are shown on the graph, determine the rectangular coordinates for each point.
2. Write the following polar equations in rectangular form.
1. \begin{align*}r = 6 \cos \theta\end{align*}
2. \begin{align*}r \sin \theta = -3\end{align*}
3. \begin{align*}r = 2 \sin \theta\end{align*}
4. \begin{align*}r \sin^2 \theta =3 \cos \theta\end{align*}
3. Write the following rectangular points in polar form.
1. \begin{align*}A(-2, 5)\end{align*} using radians
2. \begin{align*}B(5, -4)\end{align*} using radians
3. \begin{align*}C(1, 9)\end{align*} using degrees
4. \begin{align*}D(-12, -5)\end{align*} using degrees
4. Write the rectangular equations in polar form.
1. \begin{align*}(x - 4)^2 + (y - 3)^2 = 25\end{align*}
2. \begin{align*}3x - 2y = 1\end{align*}
3. \begin{align*}x^2 + y^2 - 4x + 2y = 0\end{align*}
4. \begin{align*}x^3 = 4y^2\end{align*}

1. For \begin{align*}A,r = -4\end{align*} and \begin{align*}\theta = \frac{5 \pi}{4}\end{align*} \begin{align*}x &= r \cos \theta && y = r \sin \theta\\ x &= -4 \cos \frac{5 \pi}{4} && y = -4 \sin \frac{5 \pi}{4} \\ x &= -4 \left (-\frac{\sqrt{2}}{2} \right ) && y = -4 \left (- \frac{\sqrt{2}}{2} \right ) \\ x &= 2 \sqrt{2} && y = 2 \sqrt{2}\end{align*} For \begin{align*}B, r = -3\end{align*} and \begin{align*}\theta = 135^\circ\end{align*} \begin{align*}x &= r \cos \theta && y = r \sin \theta\\ x &= -3 \cos 135^\circ && y = -3 \sin 135^\circ \\ x &= -3 - \frac{\sqrt{2}}{2} && y = -3 \frac{\sqrt{2}}{2} \\ x &= \frac{3 \sqrt{2}}{2} && y = \frac{-3 \sqrt{2}}{2}\end{align*} For \begin{align*}C\end{align*}, \begin{align*}r = 5\end{align*} and \begin{align*}\theta = \left (\frac{2 \pi}{3} \right )\end{align*} \begin{align*}x & = r \cos \theta && y = r \sin \theta \\ x & = 5 \cos \frac{2\pi}{3} && y = 5 \sin \frac{2\pi}{3} \\ x & = 5 \left (- \frac{1}{2} \right ) && y = 5 \left (\frac{\sqrt{3}}{2} \right ) \\ x & = - 2.5 && y = \frac{5 \sqrt{3}}{2}\end{align*}
1. \begin{align*}r & = 6 \cos \theta \\ r^2 & = 6r \cos \theta \\ x^2 + y^2 & = 6x \\ x^2 - 6x + y^2 & = 0 \\ x^2 - 6x + 9 + y^2 & = 9 \\ (x - 3)^2 + y^2 & = 9\end{align*}
2. \begin{align*}r \sin \theta & = -3 \\ y & = -3\end{align*}
3. \begin{align*}r & = 2 \sin \theta \\ r^2 & = 2 r \sin \theta \\ x^2 + y^2 & = 2 y \\ y^2 - 2y & = - x^2 \\ y^2 - 2y + 1 & = -x^2 + 1 \\ (y - 1)^2 & = -x^2 +1 \\ x^2 + (y - 1)^2 & = 1\end{align*}
4. \begin{align*}r \sin^2 \theta & = 3 \cos \theta \\ r^2 \sin^2 \theta & = 3 r \cos \theta \\ y^2 & = 3x\end{align*}
2. a. For \begin{align*}A (-2, 5) x = -2\end{align*} and \begin{align*}y = 5\end{align*}. The point is located in the second quadrant and \begin{align*}x < 0\end{align*}. \begin{align*}r = \sqrt{(-2)^2 + (5)^2} = \sqrt{29} \approx 5.39, \ \theta = Arc\ \tan \frac{5}{-2} + \pi = 1.95.\end{align*} The polar coordinates for the rectangular coordinates \begin{align*}A(-2,5)\end{align*} are \begin{align*}A(5.39,1.95)\end{align*} b. For \begin{align*}B (5,-4) x = 5\end{align*} and \begin{align*}y = -4\end{align*}. The point is located in the fourth quadrant and \begin{align*}x > 0\end{align*}. \begin{align*}r = \sqrt{(5)^2 + (-4)^2} = \sqrt{41} \approx 6.4, \ \theta = \tan^{-1} \left (\frac{-4}{5} \right ) \approx -0.67\end{align*} The polar coordinates for the rectangular coordinates \begin{align*}B(5, -4)\end{align*} are \begin{align*}A(6.40, -0.67)\end{align*} c. \begin{align*}C(1, 9)\end{align*} is located in the first quadrant. \begin{align*}r = \sqrt{1^2 + 9^2} = \sqrt{82} \approx 9.06, \ \theta = \tan^{-1} \frac{9}{1} \approx 83.66^\circ.\end{align*} d. \begin{align*}D(-12, -5)\end{align*} is located in the third quadrant and \begin{align*}x < 0\end{align*}. \begin{align*}r = \sqrt{(-12)^2 + (-5)^2} = \sqrt{169} = 13, \ \theta = \tan^{-1} \frac{5}{12} + 180^\circ \approx 202.6^\circ.\end{align*}
1. \begin{align*}(x - 4)^2 + (y - 3)^2 & = 25 \\ x^2 - 8x + 16 + y^2 - 6y + 9 & = 25 \\ x^2 - 8x + y^2 - 6y + 25 & = 25 \\ x^2 - 8x + y^2 - 6y & = 0 \\ x^2 + y^2 - 8x - 6y & = 0 && \text{From graphing}\ r - 8 \cos \theta - 6 \sin \theta = 0,\ \text{we see that the} \\ r^2 - 8(r \cos \theta) - 6(r \sin \theta) & = 0 && \text{additional solutions are}\ 0\ \text{and}\ 8. \\ r^2 - 8r \cos \theta - 6r \sin \theta & = 0 \\ r(r - 8 \cos \theta - 6 \sin \theta) & = 0 \\ r = 0\ \text{or}\ r - 8 \cos \theta - 6 \sin \theta & = 0 \\ r = 0\ \text{or}\ r & = 8 \cos \theta + 6 \sin \theta\end{align*}
2. \begin{align*}3x - 2y & = 1 \\ 3r \cos \theta - 2r \sin \theta & = 1 \\ r (3 \cos \theta - 2 \sin \theta) & = 1 \\ r & = \frac{1}{3 \cos \theta - 2 \sin \theta}\end{align*}
3. \begin{align*}x^2 + y^2 - 4x + 2y & = 0 \\ r^2 \cos^2 \theta + r^2 \sin^2 \theta - 4 r \cos \theta + 2 r \sin \theta & = 0 \\ r^2 (\sin^2 \theta + \cos^2 \theta) - 4 r \cos \theta + 2 r \sin \theta & = 0 \\ r (r - 4 \cos \theta + 2 \sin \theta) & = 0 \\ r = 0\ \text{or}\ r - 4 \cos \theta + 2 \sin \theta & = 0 \\ r = 0\ \text{or}\ r & = 4 \cos \theta - 2 \sin \theta\end{align*}
4. \begin{align*}x^3 & = 4 y^2 \\ (r \cos \theta)^3 & = 4 (r \sin \theta)^2 \\ r^3 \cos^3 \theta & = 4r^2 \sin^2 \theta \\ \frac{4r^2 \sin^2 \theta}{r^3 \cos^3 \theta} & = 1 \\ \frac{4 \tan^2 \theta \sec \theta}{r} & = 1 \\ 4 \tan^2 \theta \sec \theta & = r\end{align*}

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