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Polar Coordinates
Plot points using polar coordinates.
Plotting Points Using Polar Coordinates
When we think about plotting points in the plane, we usually think of rectangular coordinates \((x,y)\) in the Cartesian coordinate plane. However, there are other ways of writing a coordinate pair and other types of grid systems. In this section, we introduce to polar coordinates, which are points labeled \((r,\theta )\) and plotted on a polar grid. The polar grid is represented as a series of concentric circles radiating out from the pole, or the origin of the coordinate plane.
The polar grid is scaled as the unit circle with the positive x-axis now viewed as the polar axis and the origin as the pole. The first coordinate \(r\) is the radius or length of the directed line segment from the pole. The angle \(\theta ,\) measured in radians, indicates the direction of \(r.\) We move counterclockwise from the polar axis by an angle of \(\theta ,\) and measure a directed line segment the length of \(r\) in the direction of \(\theta .\) Even though we measure \(\theta\) first and then \(r,\) the polar point is written with the r-coordinate first. For example, to plot the point \((2,\frac{\pi }{4}),\) we would move \(\frac{\pi }{4}\) units in the counterclockwise direction and then a length of 2 from the pole. This point is plotted on the grid in .
Example
Try it.
Plot the point \((3,\frac{\pi }{2})\) on the polar grid.
Solution
The angle \(\frac{\pi }{2}\) is found by sweeping in a counterclockwise direction 90° from the polar axis. The point is located at a length of 3 units from the pole in the \(\frac{\pi }{2}\) direction, as shown in .
Example
Try it.
Plot the point \((-2,\ \frac{\pi }{6})\) on the polar grid.
Solution
We know that \(\frac{\pi }{6}\) is located in the first quadrant. However, \(r=-2.\) We can approach plotting a point with a negative \(r\) in two ways:
- Plot the point \((2,\frac{\pi }{6})\) by moving \(\frac{\pi }{6}\) in the counterclockwise direction and extending a directed line segment 2 units into the first quadrant. Then retrace the directed line segment back through the pole, and continue 2 units into the third quadrant;
- Move \(\frac{\pi }{6}\) in the counterclockwise direction, and draw the directed line segment from the pole 2 units in the negative direction, into the third quadrant.
See (a). Compare this to the graph of the polar coordinate \((2,\frac{\pi }{6})\) shown in (b).
Converting from Polar Coordinates to Rectangular Coordinates
When given a set of polar coordinates, we may need to convert them to rectangular coordinates. To do so, we can recall the relationships that exist among the variables \(x,\ y,\ r,\) and \(\theta .\)
\[\begin{array}{l}\begin{array}{l} \\ \cos \ \theta =\frac{x}{r}\to x=r\cos \ \theta \end{array} \\ \sin \ \theta =\frac{y}{r}\to y=r\sin \ \theta \end{array}\]Dropping a perpendicular from the point in the plane to the x-axis forms a right triangle, as illustrated in . An easy way to remember the equations above is to think of \(\cos \ \theta\) as the adjacent side over the hypotenuse and \(\sin \ \theta\) as the opposite side over the hypotenuse.
Example
Try it.
Write the polar coordinates \((3,\frac{\pi }{2})\) as rectangular coordinates.
Solution
Use the equivalent relationships.
\[\begin{array}{l}x=r\cos \ \theta \\ x=3\cos \ \frac{\pi }{2}=0 \\ y=r\sin \ \theta \\ y=3\sin \ \frac{\pi }{2}=3\end{array}\]The rectangular coordinates are \((0,3).\) See .
Example
Try it.
Write the polar coordinates \((-2,0)\) as rectangular coordinates.
Solution
See . Writing the polar coordinates as rectangular, we have
\[\begin{array}{l}x=r\cos \ \theta \\ x=-2\cos (0)=-2 \\ \\ y=r\sin \ \theta \\ y=-2\sin (0)=0\end{array}\]The rectangular coordinates are also \((-2,0).\)
Converting from Rectangular Coordinates to Polar Coordinates
To convert rectangular coordinates to polar coordinates, we will use two other familiar relationships. With this conversion, however, we need to be aware that a set of rectangular coordinates will yield more than one polar point.
Example
Try it.
Convert the rectangular coordinates \((3,3)\) to polar coordinates.
Solution
We see that the original point \((3,3)\) is in the first quadrant. To find \(\theta ,\) use the formula \(\tan \ \theta =\frac{y}{x}.\) This gives
\[\begin{array}{l}\tan \ \theta =\frac{3}{3} \\ \tan \ \theta =1 \\ \theta ={\tan }^{-1}(1) \\ \theta =\frac{\pi }{4}\end{array}\]To find \(r,\) we substitute the values for \(x\) and \(y\) into the formula \(r=\sqrt{{x}^{2}+{y}^{2}}.\) We know that \(r\) must be positive, as \(\frac{\pi }{4}\) is in the first quadrant. Thus
\[\begin{array}{l}r=\sqrt{{3}^{2}+{3}^{2}} \\ r=\sqrt{9+9} \\ r=\sqrt{18}=3\sqrt{2}\end{array}\]So, \(r=3\sqrt{2}\\) and \(\theta \text{=}\frac{\pi }{4},\) giving us the polar point \((3\sqrt{2},\frac{\pi }{4}).\) See .
Transforming Equations between Polar and Rectangular Forms
We can now convert coordinates between polar and rectangular form. Converting equations can be more difficult, but it can be beneficial to be able to convert between the two forms. Since there are a number of polar equations that cannot be expressed clearly in Cartesian form, and vice versa, we can use the same procedures we used to convert points between the coordinate systems. We can then use a graphing calculator to graph either the rectangular form or the polar form of the equation.
Example
Try it.
Write the Cartesian equation \({x}^{2}+{y}^{2}=9\) in polar form.
Solution
The goal is to eliminate \(x\) and \(y\) from the equation and introduce \(r\) and \(\theta .\) Ideally, we would write the equation \(r\) as a function of \(\theta .\) To obtain the polar form, we will use the relationships between \((x,y)\) and \((r,\theta ).\) Since \(x=r\cos \ \theta\) and \(y=r\sin \ \theta ,\) we can substitute and solve for \(r.\)
\[\begin{array}{lllll}\ {(r\cos \ \theta )}^{2}+{(r\sin \ \theta )}^{2}=9 & \\ {r}^{2}{\cos }^{2}\theta +{r}^{2}{\sin }^{2}\theta =9 & \\ {r}^{2}({\cos }^{2}\theta +{\sin }^{2}\theta )=9 & \\ {r}^{2}(1)=9 & {\text{Substitute cos}}^{2}\theta +{\sin }^{2}\theta =1. \\ r=\pm 3\begin{array}{llll} & & & \end{array} & \text{Use the square root property}.\end{array}\]Thus, \({x}^{2}+{y}^{2}=9,r=3,\) and \(r=-3\) should generate the same graph. See .
To graph a circle in rectangular form, we must first solve for \(y.\)
\[\begin{array}{l}\begin{array}{l} \\ {x}^{2}+{y}^{2}=9\end{array} \\ {y}^{2}=9-{x}^{2} \\ y=\pm \sqrt{9-{x}^{2}}\end{array}\]Note that this is two separate functions, since a circle fails the vertical line test. Therefore, we need to enter the positive and negative square roots into the calculator separately, as two equations in the form \({Y}_{1}=\sqrt{9-{x}^{2}}\) and \({Y}_{2}=-\sqrt{9-{x}^{2}}.\) Press GRAPH.
Example
Try it.
Rewrite the Cartesian equation \(y=3x+2\) as a polar equation.
Solution
We will use the relationships \(x=r\cos \ \theta\) and \(y=r\sin \ \theta .\)
\[\begin{array}{lllll}\ y=3x+2 & \\ r\sin \ \theta =3r\cos \ \theta +2 & \\ r\sin \ \theta -3r\cos \ \theta =2 & \\ r(\sin \ \theta -3\cos \ \theta )=2 & \text{Isolate }r. \\ r=\frac{2}{\sin \ \theta -3\cos \ \theta }\begin{array}{llll} & & & \end{array} & \text{Solve for }r.\end{array}\]Condensed — the full section is in OpenStax Algebra and Trigonometry 2e.
Identify and Graph Polar Equations by Converting to Rectangular Equations
We have learned how to convert rectangular coordinates to polar coordinates, and we have seen that the points are indeed the same. We have also transformed polar equations to rectangular equations and vice versa. Now we will demonstrate that their graphs, while drawn on different grids, are identical.
Example
Try it.
Covert the polar equation \(r=2\text{sec}\ \theta\) to a rectangular equation, and draw its corresponding graph.
Solution
The conversion is
\[\begin{array}{lll}r & = & 2\text{sec}\ \theta \\ r & = & \frac{2}{\cos \ \theta } \\ r\ \cos \ \theta & = & 2 \\ x & = & 2\end{array}\]Notice that the equation \(r=2\text{sec}\ \theta\) drawn on the polar grid is clearly the same as the vertical line \(x=2\) drawn on the rectangular grid (see ). Just as \(x=c\) is the standard form for a vertical line in rectangular form, \(r=c\text{sec}\ \theta\) is the standard form for a vertical line in polar form.
A similar discussion would demonstrate that the graph of the function \(r=2\text{csc}\ \theta\) will be the horizontal line \(y=2.\) In fact, \(r=c\text{csc}\ \theta\) is the standard form for a horizontal line in polar form, corresponding to the rectangular form \(y=c.\)
Example
Try it.
Rewrite the polar equation \(r=\frac{3}{1-2\cos \ \theta }\) as a Cartesian equation.
Solution
The goal is to eliminate \(\theta\) and \(r,\) and introduce \(x\) and \(y.\) We clear the fraction, and then use substitution. In order to replace \(r\) with \(x\) and \(y,\) we must use the expression \({x}^{2}+{y}^{2}={r}^{2}.\)
\[\begin{array}{llll}r & = & \frac{3}{1-2\ \cos \ \theta } & \\ r(1-2(\frac{x}{r})) & = & 3 & \\ r-2x & = & 3 & \text{Use}\ \cos \ \theta =\frac{x}{r}\ to\ \text{eliminate}\ \theta \\ r-2x & = & 3 & \\ r & = & 3+2x & \text{Isolate}\ r \\ {r}^{2} & = & {(3+2x)}^{2} & \text{Square}\ \text{both}\ \text{sides} \\ {x}^{2}+{y}^{2} & = & {(3+2x)}^{2} & \text{Use}\ {x}^{2}+{y}^{2}={r}^{2}\end{array}\]The Cartesian equation is \({x}^{2}+{y}^{2}={(3+2x)}^{2}.\) However, to graph it, especially using a graphing calculator or computer program, we want to isolate \(y.\)
\[\begin{array}{l}{x}^{2}+{y}^{2}={(3+2x)}^{2} \\ {y}^{2}={(3+2x)}^{2}-{x}^{2} \\ y=\pm \sqrt{{(3+2x)}^{2}-{x}^{2}}\end{array}\]When our entire equation has been changed from \(r\) and \(\theta\) to \(x\) and \(y,\) we can stop, unless asked to solve for \(y\) or simplify. See .
The “hour-glass” shape of the graph is called a hyperbola. Hyperbolas have many interesting geometric features and applications, which we will investigate further in Analytic Geometry.
Condensed — the full section is in OpenStax Algebra and Trigonometry 2e.
Key Equations
| Conversion formulas | \(\begin{array}{ll} & \cos \ \theta =\frac{x}{r}\to x=r\cos \ \theta \\ & \sin \ \theta =\frac{y}{r}\to y=r\sin \ \theta \\ & {r}^{2}={x}^{2}+{y}^{2} \\ & \tan \ \theta =\frac{y}{x}\end{array}\) |
Key Concepts
- The polar grid is represented as a series of concentric circles radiating out from the pole, or origin.
- To plot a point in the form \((r,\theta ),\ \theta >0,\) move in a counterclockwise direction from the polar axis by an angle of \(\theta ,\) and then extend a directed line segment from the pole the length of \(r\) in the direction of \(\theta .\) If \(\theta\) is negative, move in a clockwise direction, and extend a directed line segment the length of \(r\) in the direction of \(\theta .\) See .
- If \(r\) is negative, extend the directed line segment in the opposite direction of \(\theta .\) See .
- To convert from polar coordinates to rectangular coordinates, use the formulas \(x=r\cos \ \theta\) and \(y=r\sin \ \theta .\) See and .
- To convert from rectangular coordinates to polar coordinates, use one or more of the formulas: \(\cos \ \theta =\frac{x}{r},\sin \ \theta =\frac{y}{r},\tan \ \theta =\frac{y}{x},\) and \(r=\sqrt{{x}^{2}+{y}^{2}}.\) See .
- Transforming equations between polar and rectangular forms means making the appropriate substitutions based on the available formulas, together with algebraic manipulations. See , , and .
- Using the appropriate substitutions makes it possible to rewrite a polar equation as a rectangular equation, and then graph it in the rectangular plane. See , , and .
Plotting Points Using Polar Coordinates
When we think about plotting points in the plane, we usually think of rectangular coordinates \((x,y)\) in the Cartesian coordinate plane. However, there are other ways of writing a coordinate pair and other types of grid systems. In this section, we introduce to polar coordinates, which are points labeled \((r,\theta )\) and plotted on a polar grid. The polar grid is represented as a series of concentric circles radiating out from the pole, or the origin of the coordinate plane.
The polar grid is scaled as the unit circle with the positive x-axis now viewed as the polar axis and the origin as the pole. The first coordinate \(r\) is the radius or length of the directed line segment from the pole. The angle \(\theta ,\) measured in radians, indicates the direction of \(r.\) We move counterclockwise from the polar axis by an angle of \(\theta ,\) and measure a directed line segment the length of \(r\) in the direction of \(\theta .\) Even though we measure \(\theta\) first and then \(r,\) the polar point is written with the r-coordinate first. For example, to plot the point \((2,\frac{\pi }{4}),\) we would move \(\frac{\pi }{4}\) units in the counterclockwise direction and then a length of 2 from the pole. This point is plotted on the grid in .
Example
Try it.
Plot the point \((3,\frac{\pi }{2})\) on the polar grid.
Solution
The angle \(\frac{\pi }{2}\) is found by sweeping in a counterclockwise direction 90° from the polar axis. The point is located at a length of 3 units from the pole in the \(\frac{\pi }{2}\) direction, as shown in .
Example
Try it.
Plot the point \((-2,\ \frac{\pi }{6})\) on the polar grid.
Solution
We know that \(\frac{\pi }{6}\) is located in the first quadrant. However, \(r=-2.\) We can approach plotting a point with a negative \(r\) in two ways:
- Plot the point \((2,\frac{\pi }{6})\) by moving \(\frac{\pi }{6}\) in the counterclockwise direction and extending a directed line segment 2 units into the first quadrant. Then retrace the directed line segment back through the pole, and continue 2 units into the third quadrant;
- Move \(\frac{\pi }{6}\) in the counterclockwise direction, and draw the directed line segment from the pole 2 units in the negative direction, into the third quadrant.
See (a). Compare this to the graph of the polar coordinate \((2,\frac{\pi }{6})\) shown in (b).
Converting from Polar Coordinates to Rectangular Coordinates
When given a set of polar coordinates, we may need to convert them to rectangular coordinates. To do so, we can recall the relationships that exist among the variables \(x,\ y,\ r,\) and \(\theta .\)
\[\begin{array}{l}\begin{array}{l} \\ \cos \ \theta =\frac{x}{r}\to x=r\cos \ \theta \end{array} \\ \sin \ \theta =\frac{y}{r}\to y=r\sin \ \theta \end{array}\]Dropping a perpendicular from the point in the plane to the x-axis forms a right triangle, as illustrated in . An easy way to remember the equations above is to think of \(\cos \ \theta\) as the adjacent side over the hypotenuse and \(\sin \ \theta\) as the opposite side over the hypotenuse.
Example
Try it.
Write the polar coordinates \((3,\frac{\pi }{2})\) as rectangular coordinates.
Solution
Use the equivalent relationships.
\[\begin{array}{l}x=r\cos \ \theta \\ x=3\cos \ \frac{\pi }{2}=0 \\ y=r\sin \ \theta \\ y=3\sin \ \frac{\pi }{2}=3\end{array}\]The rectangular coordinates are \((0,3).\) See .
Example
Try it.
Write the polar coordinates \((-2,0)\) as rectangular coordinates.
Solution
See . Writing the polar coordinates as rectangular, we have
\[\begin{array}{l}x=r\cos \ \theta \\ x=-2\cos (0)=-2 \\ \\ y=r\sin \ \theta \\ y=-2\sin (0)=0\end{array}\]The rectangular coordinates are also \((-2,0).\)
Converting from Rectangular Coordinates to Polar Coordinates
To convert rectangular coordinates to polar coordinates, we will use two other familiar relationships. With this conversion, however, we need to be aware that a set of rectangular coordinates will yield more than one polar point.
Example
Try it.
Convert the rectangular coordinates \((3,3)\) to polar coordinates.
Solution
We see that the original point \((3,3)\) is in the first quadrant. To find \(\theta ,\) use the formula \(\tan \ \theta =\frac{y}{x}.\) This gives
\[\begin{array}{l}\tan \ \theta =\frac{3}{3} \\ \tan \ \theta =1 \\ \theta ={\tan }^{-1}(1) \\ \theta =\frac{\pi }{4}\end{array}\]To find \(r,\) we substitute the values for \(x\) and \(y\) into the formula \(r=\sqrt{{x}^{2}+{y}^{2}}.\) We know that \(r\) must be positive, as \(\frac{\pi }{4}\) is in the first quadrant. Thus
\[\begin{array}{l}r=\sqrt{{3}^{2}+{3}^{2}} \\ r=\sqrt{9+9} \\ r=\sqrt{18}=3\sqrt{2}\end{array}\]So, \(r=3\sqrt{2}\\) and \(\theta \text{=}\frac{\pi }{4},\) giving us the polar point \((3\sqrt{2},\frac{\pi }{4}).\) See .
Transforming Equations between Polar and Rectangular Forms
We can now convert coordinates between polar and rectangular form. Converting equations can be more difficult, but it can be beneficial to be able to convert between the two forms. Since there are a number of polar equations that cannot be expressed clearly in Cartesian form, and vice versa, we can use the same procedures we used to convert points between the coordinate systems. We can then use a graphing calculator to graph either the rectangular form or the polar form of the equation.
Example
Try it.
Write the Cartesian equation \({x}^{2}+{y}^{2}=9\) in polar form.
Solution
The goal is to eliminate \(x\) and \(y\) from the equation and introduce \(r\) and \(\theta .\) Ideally, we would write the equation \(r\) as a function of \(\theta .\) To obtain the polar form, we will use the relationships between \((x,y)\) and \((r,\theta ).\) Since \(x=r\cos \ \theta\) and \(y=r\sin \ \theta ,\) we can substitute and solve for \(r.\)
\[\begin{array}{lllll}\ {(r\cos \ \theta )}^{2}+{(r\sin \ \theta )}^{2}=9 & \\ {r}^{2}{\cos }^{2}\theta +{r}^{2}{\sin }^{2}\theta =9 & \\ {r}^{2}({\cos }^{2}\theta +{\sin }^{2}\theta )=9 & \\ {r}^{2}(1)=9 & {\text{Substitute cos}}^{2}\theta +{\sin }^{2}\theta =1. \\ r=\pm 3\begin{array}{llll} & & & \end{array} & \text{Use the square root property}.\end{array}\]Thus, \({x}^{2}+{y}^{2}=9,r=3,\) and \(r=-3\) should generate the same graph. See .
To graph a circle in rectangular form, we must first solve for \(y.\)
\[\begin{array}{l}\begin{array}{l} \\ {x}^{2}+{y}^{2}=9\end{array} \\ {y}^{2}=9-{x}^{2} \\ y=\pm \sqrt{9-{x}^{2}}\end{array}\]Note that this is two separate functions, since a circle fails the vertical line test. Therefore, we need to enter the positive and negative square roots into the calculator separately, as two equations in the form \({Y}_{1}=\sqrt{9-{x}^{2}}\) and \({Y}_{2}=-\sqrt{9-{x}^{2}}.\) Press GRAPH.
Example
Try it.
Rewrite the Cartesian equation \(y=3x+2\) as a polar equation.
Solution
We will use the relationships \(x=r\cos \ \theta\) and \(y=r\sin \ \theta .\)
\[\begin{array}{lllll}\ y=3x+2 & \\ r\sin \ \theta =3r\cos \ \theta +2 & \\ r\sin \ \theta -3r\cos \ \theta =2 & \\ r(\sin \ \theta -3\cos \ \theta )=2 & \text{Isolate }r. \\ r=\frac{2}{\sin \ \theta -3\cos \ \theta }\begin{array}{llll} & & & \end{array} & \text{Solve for }r.\end{array}\]Condensed — the full section is in OpenStax Precalculus 2e.
Identify and Graph Polar Equations by Converting to Rectangular Equations
We have learned how to convert rectangular coordinates to polar coordinates, and we have seen that the points are indeed the same. We have also transformed polar equations to rectangular equations and vice versa. Now we will demonstrate that their graphs, while drawn on different grids, are identical.
Example
Try it.
Covert the polar equation \(r=2\text{sec}\ \theta\) to a rectangular equation, and draw its corresponding graph.
Solution
The conversion is
\[\begin{array}{lll}r & = & 2\text{sec}\ \theta \\ r & = & \frac{2}{\cos \ \theta } \\ r\ \cos \ \theta & = & 2 \\ x & = & 2\end{array}\]Notice that the equation \(r=2\text{sec}\ \theta\) drawn on the polar grid is clearly the same as the vertical line \(x=2\) drawn on the rectangular grid (see ). Just as \(x=c\) is the standard form for a vertical line in rectangular form, \(r=c\text{sec}\ \theta\) is the standard form for a vertical line in polar form.
A similar discussion would demonstrate that the graph of the function \(r=2\text{csc}\ \theta\) will be the horizontal line \(y=2.\) In fact, \(r=c\text{csc}\ \theta\) is the standard form for a horizontal line in polar form, corresponding to the rectangular form \(y=c.\)
Example
Try it.
Rewrite the polar equation \(r=\frac{3}{1-2\cos \ \theta }\) as a Cartesian equation.
Solution
The goal is to eliminate \(\theta\) and \(r,\) and introduce \(x\) and \(y.\) We clear the fraction, and then use substitution. In order to replace \(r\) with \(x\) and \(y,\) we must use the expression \({x}^{2}+{y}^{2}={r}^{2}.\)
\[\begin{array}{llll}r & = & \frac{3}{1-2\ \cos \ \theta } & \\ r(1-2(\frac{x}{r})) & = & 3 & \\ r-2x & = & 3 & \text{Use}\ \cos \ \theta =\frac{x}{r}\ to\ \text{eliminate}\ \theta \\ r-2x & = & 3 & \\ r & = & 3+2x & \text{Isolate}\ r \\ {r}^{2} & = & {(3+2x)}^{2} & \text{Square}\ \text{both}\ \text{sides} \\ {x}^{2}+{y}^{2} & = & {(3+2x)}^{2} & \text{Use}\ {x}^{2}+{y}^{2}={r}^{2}\end{array}\]The Cartesian equation is \({x}^{2}+{y}^{2}={(3+2x)}^{2}.\) However, to graph it, especially using a graphing calculator or computer program, we want to isolate \(y.\)
\[\begin{array}{l}{x}^{2}+{y}^{2}={(3+2x)}^{2} \\ {y}^{2}={(3+2x)}^{2}-{x}^{2} \\ y=\pm \sqrt{{(3+2x)}^{2}-{x}^{2}}\end{array}\]When our entire equation has been changed from \(r\) and \(\theta\) to \(x\) and \(y,\) we can stop, unless asked to solve for \(y\) or simplify. See .
The “hour-glass” shape of the graph is called a hyperbola. Hyperbolas have many interesting geometric features and applications, which we will investigate further in Analytic Geometry.
Condensed — the full section is in OpenStax Precalculus 2e.
Key Equations
| Conversion formulas | \(\begin{array}{ll} & \cos \ \theta =\frac{x}{r}\to x=r\cos \ \theta \\ & \sin \ \theta =\frac{y}{r}\to y=r\sin \ \theta \\ & {r}^{2}={x}^{2}+{y}^{2} \\ & \tan \ \theta =\frac{y}{x}\end{array}\) |
Key Concepts
- The polar grid is represented as a series of concentric circles radiating out from the pole, or origin.
- To plot a point in the form \((r,\theta ),\ \theta >0,\) move in a counterclockwise direction from the polar axis by an angle of \(\theta ,\) and then extend a directed line segment from the pole the length of \(r\) in the direction of \(\theta .\) If \(\theta\) is negative, move in a clockwise direction, and extend a directed line segment the length of \(r\) in the direction of \(\theta .\) See .
- If \(r\) is negative, extend the directed line segment in the opposite direction of \(\theta .\) See .
- To convert from polar coordinates to rectangular coordinates, use the formulas \(x=r\cos \ \theta\) and \(y=r\sin \ \theta .\) See and .
- To convert from rectangular coordinates to polar coordinates, use one or more of the formulas: \(\cos \ \theta =\frac{x}{r},\sin \ \theta =\frac{y}{r},\tan \ \theta =\frac{y}{x},\) and \(r=\sqrt{{x}^{2}+{y}^{2}}.\) See .
- Transforming equations between polar and rectangular forms means making the appropriate substitutions based on the available formulas, together with algebraic manipulations. See , , and .
- Using the appropriate substitutions makes it possible to rewrite a polar equation as a rectangular equation, and then graph it in the rectangular plane. See , , and .
Practice (40)
Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.
-
Plot the point \((3,\frac{\pi }{2})\) on the polar grid.
Bonisa impendulo
The angle \(\frac{\pi }{2}\) is found by sweeping in a counterclockwise direction 90° from the polar axis. The point is located at a length of 3 units from the pole in the \(\frac{\pi }{2}\) direction, as shown in .
-
Plot the point \((2,\ \frac{\pi }{3})\) in the polar grid.
Bonisa impendulo
-
Plot the point \((-2,\ \frac{\pi }{6})\) on the polar grid.
Bonisa impendulo
We know that \(\frac{\pi }{6}\) is located in the first quadrant. However, \(r=-2.\) We can approach plotting a point with a negative \(r\) in two ways:
- Plot the point \((2,\frac{\pi }{6})\) by moving \(\frac{\pi }{6}\) in the counterclockwise direction and extending a directed line segment 2 units into the first quadrant. Then retrace the directed line segment back through the pole, and continue 2 units into the third quadrant;
- Move \(\frac{\pi }{6}\) in the counterclockwise direction, and draw the directed line segment from the pole 2 units in the negative direction, into the third quadrant.
See (a). Compare this to the graph of the polar coordinate \((2,\frac{\pi }{6})\) shown in (b).
-
Plot the points \((3,-\frac{\pi }{6})\) and \((2,\frac{9\pi }{4})\) on the same polar grid.
-
Write the polar coordinates \((3,\frac{\pi }{2})\) as rectangular coordinates.
Bonisa impendulo
Use the equivalent relationships.
\[\begin{array}{l}x=r\cos \ \theta \\ x=3\cos \ \frac{\pi }{2}=0 \\ y=r\sin \ \theta \\ y=3\sin \ \frac{\pi }{2}=3\end{array}\]The rectangular coordinates are \((0,3).\) See .
-
Write the polar coordinates \((-2,0)\) as rectangular coordinates.
Bonisa impendulo
See . Writing the polar coordinates as rectangular, we have
\[\begin{array}{l}x=r\cos \ \theta \\ x=-2\cos (0)=-2 \\ \\ y=r\sin \ \theta \\ y=-2\sin (0)=0\end{array}\]The rectangular coordinates are also \((-2,0).\)
-
Write the polar coordinates \((-1,\frac{2\pi }{3})\) as rectangular coordinates.
Bonisa impendulo
\((x,y)=(\frac{1}{2},-\frac{\sqrt{3}}{2})\)
-
Convert the rectangular coordinates \((3,3)\) to polar coordinates.
Bonisa impendulo
We see that the original point \((3,3)\) is in the first quadrant. To find \(\theta ,\) use the formula \(\tan \ \theta =\frac{y}{x}.\) This gives
\[\begin{array}{l}\tan \ \theta =\frac{3}{3} \\ \tan \ \theta =1 \\ \theta ={\tan }^{-1}(1) \\ \theta =\frac{\pi }{4}\end{array}\]To find \(r,\) we substitute the values for \(x\) and \(y\) into the formula \(r=\sqrt{{x}^{2}+{y}^{2}}.\) We know that \(r\) must be positive, as \(\frac{\pi }{4}\) is in the first quadrant. Thus
\[\begin{array}{l}r=\sqrt{{3}^{2}+{3}^{2}} \\ r=\sqrt{9+9} \\ r=\sqrt{18}=3\sqrt{2}\end{array}\]So, \(r=3\sqrt{2}\\) and \(\theta \text{=}\frac{\pi }{4},\) giving us the polar point \((3\sqrt{2},\frac{\pi }{4}).\) See .
-
Write the Cartesian equation \({x}^{2}+{y}^{2}=9\) in polar form.
Bonisa impendulo
The goal is to eliminate \(x\) and \(y\) from the equation and introduce \(r\) and \(\theta .\) Ideally, we would write the equation \(r\) as a function of \(\theta .\) To obtain the polar form, we will use the relationships between \((x,y)\) and \((r,\theta ).\) Since \(x=r\cos \ \theta\) and \(y=r\sin \ \theta ,\) we can substitute and solve for \(r.\)
\[\begin{array}{lllll}\ {(r\cos \ \theta )}^{2}+{(r\sin \ \theta )}^{2}=9 & \\ {r}^{2}{\cos }^{2}\theta +{r}^{2}{\sin }^{2}\theta =9 & \\ {r}^{2}({\cos }^{2}\theta +{\sin }^{2}\theta )=9 & \\ {r}^{2}(1)=9 & {\text{Substitute cos}}^{2}\theta +{\sin }^{2}\theta =1. \\ r=\pm 3\begin{array}{llll} & & & \end{array} & \text{Use the square root property}.\end{array}\]Thus, \({x}^{2}+{y}^{2}=9,r=3,\) and \(r=-3\) should generate the same graph. See .
To graph a circle in rectangular form, we must first solve for \(y.\)
\[\begin{array}{l}\begin{array}{l} \\ {x}^{2}+{y}^{2}=9\end{array} \\ {y}^{2}=9-{x}^{2} \\ y=\pm \sqrt{9-{x}^{2}}\end{array}\]Note that this is two separate functions, since a circle fails the vertical line test. Therefore, we need to enter the positive and negative square roots into the calculator separately, as two equations in the form \({Y}_{1}=\sqrt{9-{x}^{2}}\) and \({Y}_{2}=-\sqrt{9-{x}^{2}}.\) Press GRAPH.
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Rewrite the Cartesian equation \({x}^{2}+{y}^{2}=6y\) as a polar equation.
Bonisa impendulo
This equation appears similar to the previous example, but it requires different steps to convert the equation.
We can still follow the same procedures we have already learned and make the following substitutions:
\[\begin{array}{lllll}{r}^{2}=6y & \text{Use }{x}^{2}+{y}^{2}={r}^{2}. \\ {r}^{2}=6r\sin \ \theta & \text{Substitute}\ y=r\sin \ \theta . \\ {r}^{2}-6r\sin \ \theta =0 & \text{Set equal to 0}. \\ r(r-6\sin \ \theta )=0 & \text{Factor and solve}. \\ r=0 & \text{We reject }r=0,\ \text{as it only represents one point, }(0,0). \\ r=6\sin \ \theta \begin{array}{llll} & & & \end{array} & \end{array}\]Therefore, the equations \({x}^{2}+{y}^{2}=6y\) and \(r=6\sin \ \theta\) should give us the same graph. See .
The Cartesian or rectangular equation is plotted on the rectangular grid, and the polar equation is plotted on the polar grid. Clearly, the graphs are identical.
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Rewrite the Cartesian equation \(y=3x+2\) as a polar equation.
Bonisa impendulo
We will use the relationships \(x=r\cos \ \theta\) and \(y=r\sin \ \theta .\)
\[\begin{array}{lllll}\ y=3x+2 & \\ r\sin \ \theta =3r\cos \ \theta +2 & \\ r\sin \ \theta -3r\cos \ \theta =2 & \\ r(\sin \ \theta -3\cos \ \theta )=2 & \text{Isolate }r. \\ r=\frac{2}{\sin \ \theta -3\cos \ \theta }\begin{array}{llll} & & & \end{array} & \text{Solve for }r.\end{array}\] -
Rewrite the Cartesian equation \({y}^{2}=3-{x}^{2}\) in polar form.
Bonisa impendulo
\(r=\sqrt{3}\)
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Covert the polar equation \(r=2\text{sec}\ \theta\) to a rectangular equation, and draw its corresponding graph.
Bonisa impendulo
The conversion is
\[\begin{array}{lll}r & = & 2\text{sec}\ \theta \\ r & = & \frac{2}{\cos \ \theta } \\ r\ \cos \ \theta & = & 2 \\ x & = & 2\end{array}\]Notice that the equation \(r=2\text{sec}\ \theta\) drawn on the polar grid is clearly the same as the vertical line \(x=2\) drawn on the rectangular grid (see ). Just as \(x=c\) is the standard form for a vertical line in rectangular form, \(r=c\text{sec}\ \theta\) is the standard form for a vertical line in polar form.
A similar discussion would demonstrate that the graph of the function \(r=2\text{csc}\ \theta\) will be the horizontal line \(y=2.\) In fact, \(r=c\text{csc}\ \theta\) is the standard form for a horizontal line in polar form, corresponding to the rectangular form \(y=c.\)
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Rewrite the polar equation \(r=\frac{3}{1-2\cos \ \theta }\) as a Cartesian equation.
Bonisa impendulo
The goal is to eliminate \(\theta\) and \(r,\) and introduce \(x\) and \(y.\) We clear the fraction, and then use substitution. In order to replace \(r\) with \(x\) and \(y,\) we must use the expression \({x}^{2}+{y}^{2}={r}^{2}.\)
\[\begin{array}{llll}r & = & \frac{3}{1-2\ \cos \ \theta } & \\ r(1-2(\frac{x}{r})) & = & 3 & \\ r-2x & = & 3 & \text{Use}\ \cos \ \theta =\frac{x}{r}\ to\ \text{eliminate}\ \theta \\ r-2x & = & 3 & \\ r & = & 3+2x & \text{Isolate}\ r \\ {r}^{2} & = & {(3+2x)}^{2} & \text{Square}\ \text{both}\ \text{sides} \\ {x}^{2}+{y}^{2} & = & {(3+2x)}^{2} & \text{Use}\ {x}^{2}+{y}^{2}={r}^{2}\end{array}\]The Cartesian equation is \({x}^{2}+{y}^{2}={(3+2x)}^{2}.\) However, to graph it, especially using a graphing calculator or computer program, we want to isolate \(y.\)
\[\begin{array}{l}{x}^{2}+{y}^{2}={(3+2x)}^{2} \\ {y}^{2}={(3+2x)}^{2}-{x}^{2} \\ y=\pm \sqrt{{(3+2x)}^{2}-{x}^{2}}\end{array}\]When our entire equation has been changed from \(r\) and \(\theta\) to \(x\) and \(y,\) we can stop, unless asked to solve for \(y\) or simplify. See .
The “hour-glass” shape of the graph is called a hyperbola. Hyperbolas have many interesting geometric features and applications, which we will investigate further in Analytic Geometry.
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Rewrite the polar equation \(r=2\sin \ \theta\) in Cartesian form.
Bonisa impendulo
\({x}^{2}+{y}^{2}=2y\) or, in the standard form for a circle, \({x}^{2}+{(y-1)}^{2}=1\)
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Rewrite the polar equation \(r=\sin (2\theta )\) in Cartesian form.
Bonisa impendulo
\[\begin{array}{lllll}\ r=\sin (2\theta ) & \text{Use the double angle identity for sine}. \\ r=2\sin \ \theta \cos \ \theta \begin{array}{llll} & & & \end{array} & \text{Use }\cos \ \theta =\frac{x}{r}\text{ and }\sin \ \theta =\frac{y}{r}. \\ r=2(\frac{x}{r})(\frac{y}{r}) & \text{Simplify}. \\ r=\frac{2xy}{{r}^{2}} & \text{ Multiply both sides by }{r}^{2}. \\ {r}^{3}=2xy & \\ {(\sqrt{{x}^{2}+{y}^{2}})}^{3}=2xy & \text{As}\ {x}^{2}+{y}^{2}={r}^{2},r=\sqrt{{x}^{2}+{y}^{2}}.\end{array}\]This equation can also be written as
\[{({x}^{2}+{y}^{2})}^{\frac{3}{2}}=2xy\ \text{or}\ {x}^{2}+{y}^{2}={(2xy)}^{\frac{2}{3}}\] -
How are polar coordinates different from rectangular coordinates?
Bonisa impendulo
For polar coordinates, the point in the plane depends on the angle from the positive x-axis and distance from the origin, while in Cartesian coordinates, the point represents the horizontal and vertical distances from the origin. For each point in the coordinate plane, there is one representation, but for each point in the polar plane, there are infinite representations.
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How are the polar axes different from the x- and y-axes of the Cartesian plane?
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Explain how polar coordinates are graphed.
Bonisa impendulo
Determine \(\theta\) for the point, then move \(r\) units from the pole to plot the point. If \(r\) is negative, move \(r\) units from the pole in the opposite direction but along the same angle. The point is a distance of \(r\) away from the origin at an angle of \(\theta\) from the polar axis.
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How are the points \((3,\frac{\pi }{2})\) and \((-3,\frac{\pi }{2})\) related?
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Explain why the points \((-3,\frac{\pi }{2})\) and \((3,-\frac{\pi }{2})\) are the same.
Bonisa impendulo
The point \((-3,\frac{\pi }{2})\) has a positive angle but a negative radius and is plotted by moving to an angle of \(\frac{\pi }{2}\) and then moving 3 units in the negative direction. This places the point 3 units down the negative y-axis. The point \((3,-\frac{\pi }{2})\) has a negative angle and a positive radius and is plotted by first moving to an angle of \(-\frac{\pi }{2}\) and then moving 3 units down, which is the positive direction for a negative angle. The point is also 3 units down the negative y-axis.
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\((7,\frac{7\pi }{6})\)
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\((5,\pi )\)
Bonisa impendulo
\((-5,0)\)
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\((6,-\frac{\pi }{4})\)
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\((-3,\frac{\pi }{6})\)
Bonisa impendulo
\((-\frac{3\sqrt{3}}{2},-\frac{3}{2})\)
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\((4,\frac{7\pi }{4})\)
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\((-4,6)\)
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\((3,-5)\)
Bonisa impendulo
\((\sqrt{34},5.253)\)
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\((-10,-13)\)
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\(y=4{x}^{2}\)
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\(y=2{x}^{4}\)
Bonisa impendulo
\(r=\sqrt[3]{\frac{sin\theta }{2co{s}^{4}\theta }}\)
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\({x}^{2}+{y}^{2}=4y\)
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\({x}^{2}+{y}^{2}=3x\)
Bonisa impendulo
\(r=3\cos \theta\)
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\({x}^{2}-{y}^{2}=x\)
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\({x}^{2}-{y}^{2}=3y\)
Bonisa impendulo
\(r=\frac{3\sin \theta }{\cos (2\theta )}\)
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\({x}^{2}+{y}^{2}=9\)
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\({x}^{2}=9y\)
Bonisa impendulo
\(r=\frac{9\sin \theta }{{\cos }^{2}\theta }\)
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\({y}^{2}=9x\)
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\(r=3\sin \ \theta\)
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\(r=4\cos \ \theta\)
Bonisa impendulo
\({x}^{2}+{y}^{2}=4x\) or \(\frac{{(x-2)}^{2}}{4}+\frac{{y}^{2}}{4}=1;\) circle
Symbols used here
The non-negative number whose square (n-th power) is x.
Ratio of a circle's circumference to its diameter, 3.14159…
The usual name for an angle.
Ratios of sides in a right triangle; coordinates on the unit circle.
Both signs at once: x = 3 ± 2 means 5 and 1.
1/360 of a full turn. 180° = π radians.
The angle whose sine is the given value (and likewise arccos, arctan).
How to: Polar Coordinates
- Plot points using polar coordinates.
- Convert from polar coordinates to rectangular coordinates.
- Convert from rectangular coordinates to polar coordinates.
- Transform equations between polar and rectangular forms.
- Identify and graph polar equations by converting to rectangular equations.
- Plot the point
- Move
- Given the polar coordinate
Questions people ask
Why radians instead of degrees?
A radian is the angle whose arc equals the radius, so it is a pure ratio rather than an arbitrary 1/360 of a turn. With radians the derivative of sin x is exactly cos x; with degrees a factor of π/180 appears everywhere.
Why does sin x = 1/2 have infinitely many solutions?
Sine repeats every full turn, and within one turn it reaches 1/2 twice (at π/6 and 5π/6). Add any whole number of turns to either and it is still true.
How do I remember the exact values?
Two triangles: the 45-45-90 with sides 1, 1, √2 and the 30-60-90 with sides 1, √3, 2. Every value for 30°, 45° and 60° is a ratio of those sides; symmetry gives the rest of the circle.
Zama ngokwakho
Parts of this page are adapted from OpenStax Algebra and Trigonometry 2e (CC BY-NC-SA 4.0), OpenStax Precalculus 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.
IiNkqubo Trigonometry
The unit circleTrigonometric equationsTrigonometric identitiesDegrees and radiansRight-triangle trigonometry (SOH-CAH-TOA)Law of sines and law of cosinesGraphs of sine, cosine and tangentInverse trigonometric functions