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أدلة الدراسة > Precalculus II

Polar Coordinates

Learning Objectives

By the end of this section, you will be able to:
  • 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.
Over 12 kilometers from port, a sailboat encounters rough weather and is blown off course by a 16-knot wind. How can the sailor indicate his location to the Coast Guard? In this section, we will investigate a method of representing location that is different from a standard coordinate grid.
An illustration of a boat on the polar grid. Figure 1
 

Plotting Points Using Polar Coordinates

When we think about plotting points in the plane, we usually think of rectangular coordinates (x,y)\left(x,y\right) 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,θ)\left(r,\theta \right) 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 rr is the radius or length of the directed line segment from the pole. The angle θ\theta , measured in radians, indicates the direction of rr. We move counterclockwise from the polar axis by an angle of θ\theta , and measure a directed line segment the length of rr in the direction of θ\theta . Even though we measure θ\theta first and then rr, the polar point is written with the r-coordinate first. For example, to plot the point (2,π4)\left(2,\frac{\pi }{4}\right), we would move π4\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 Figure 2.
Polar grid with point (2, pi/4) plotted. Figure 2

Example 1: Plotting a Point on the Polar Grid

Plot the point (3,π2)\left(3,\frac{\pi }{2}\right) on the polar grid.

Solution

The angle π2\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 π2\frac{\pi }{2} direction, as shown in Figure 3.
Polar grid with point (3, pi/2) plotted. Figure 3

Try It 1

Plot the point (2,π3)\left(2,\frac{\pi }{3}\right) in the polar grid. Solution

Example 2: Plotting a Point in the Polar Coordinate System with a Negative Component

Plot the point (2,π6)\left(-2,\frac{\pi }{6}\right) on the polar grid.

Solution

We know that π6\frac{\pi }{6} is located in the first quadrant. However, r=2r=-2. We can approach plotting a point with a negative rr in two ways:
  1. Plot the point (2,π6)\left(2,\frac{\pi }{6}\right) by moving π6\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;
  2. Move π6\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 Figure 4(a). Compare this to the graph of the polar coordinate (2,π6)\left(2,\frac{\pi }{6}\right) shown in Figure 4(b).
Two polar grids. Points (2, pi/6) and (-2, pi/6) are plotted. They are reflections across the origin in Q1 and Q3. Figure 4

Try It 2

Plot the points (3,π6)\left(3,-\frac{\pi }{6}\right) and (2,9π4)\left(2,\frac{9\pi }{4}\right) on the same polar grid. Solution

 Converting Between 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,rx,y,r, and θ\theta .
cosθ=xrx=rcosθsinθ=yry=rsinθ\begin{array}{l}\begin{array}{l}\\ \cos \theta =\frac{x}{r}\to x=r\cos \theta \end{array}\hfill \\ \sin \theta =\frac{y}{r}\to y=r\sin \theta \hfill \end{array}
Dropping a perpendicular from the point in the plane to the x-axis forms a right triangle, as illustrated in Figure 5. An easy way to remember the equations above is to think of cosθ\cos \theta as the adjacent side over the hypotenuse and sinθ\sin \theta as the opposite side over the hypotenuse.
Comparison between polar coordinates and rectangular coordinates. There is a right triangle plotted on the x,y axis. The sides are a horizontal line on the x-axis of length x, a vertical line extending from thex-axis to some point in quadrant 1, and a hypotenuse r extending from the origin to that same point in quadrant 1. The vertices are at the origin (0,0), some point along the x-axis at (x,0), and that point in quadrant 1. This last point is (x,y) or (r, theta), depending which system of coordinates you use. Figure 5

A General Note: Converting from Polar Coordinates to Rectangular Coordinates

To convert polar coordinates (r,θ)\left(r,\theta \right) to rectangular coordinates (x,y)\left(x,y\right), let
cosθ=xrx=rcosθ\cos \theta =\frac{x}{r}\to x=r\cos \theta
sinθ=yry=rsinθ\sin \theta =\frac{y}{r}\to y=r\sin \theta

How To: Given polar coordinates, convert to rectangular coordinates.

  1. Given the polar coordinate (r,θ)\left(r,\theta \right), write x=rcosθx=r\cos \theta and y=rsinθy=r\sin \theta .
  2. Evaluate cosθ\cos \theta and sinθ\sin \theta .
  3. Multiply cosθ\cos \theta by rr to find the x-coordinate of the rectangular form.
  4. Multiply sinθ\sin \theta by rr to find the y-coordinate of the rectangular form.

Example 3: Writing Polar Coordinates as Rectangular Coordinates

Write the polar coordinates (3,π2)\left(3,\frac{\pi }{2}\right) as rectangular coordinates.

Solution

Use the equivalent relationships.
x=rcosθx=3cosπ2=0y=rsinθy=3sinπ2=3\begin{array}{l}\begin{array}{l}\\ x=r\cos \theta \end{array}\hfill \\ x=3\cos \frac{\pi }{2}=0\hfill \\ y=r\sin \theta \hfill \\ y=3\sin \frac{\pi }{2}=3\hfill \end{array}
The rectangular coordinates are (0,3)\left(0,3\right).
Illustration of (3, pi/2) in polar coordinates and (0,3) in rectangular coordinates - they are the same point! Figure 6

Example 4: Writing Polar Coordinates as Rectangular Coordinates

Write the polar coordinates (2,0)\left(-2,0\right) as rectangular coordinates.

Solution

See Figure 7. Writing the polar coordinates as rectangular, we have
x=rcosθx=2cos(0)=2y=rsinθy=2sin(0)=0\begin{array}{l}x=r\cos \theta \hfill \\ x=-2\cos \left(0\right)=-2\hfill \\ \hfill \\ y=r\sin \theta \hfill \\ y=-2\sin \left(0\right)=0\hfill \end{array}
The rectangular coordinates are also (2,0)\left(-2,0\right).
Illustration of (-2, 0) in polar coordinates and (-2,0) in rectangular coordinates - they are the same point! Figure 7

Try It 3

Write the polar coordinates (1,2π3)\left(-1,\frac{2\pi }{3}\right) as rectangular coordinates. Solution

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.

A General Note: Converting from Rectangular Coordinates to Polar Coordinates

Converting from rectangular coordinates to polar coordinates requires the use of one or more of the relationships illustrated in Figure 8.
cosθ=xr orx=rcosθsinθ=yr ory=rsinθr2=x2+y2tanθ=yx\begin{array}{l}\cos \theta =\frac{x}{r}\text{ or}x=r\cos \theta \hfill \\ \sin \theta =\frac{y}{r}\text{ or}y=r\sin \theta \hfill \\ {r}^{2}={x}^{2}+{y}^{2}\hfill \\ \tan \theta =\frac{y}{x}\hfill \end{array}
A right triangle with sides x, y, and r on a graph. The side x runs along the x-axis, and the point of the triangle opposite side x is the point (x, y), (r, theta). The side opposite the right angle of the triangle is labeled r. The side y is opposite the angle theta, and the vertex of angle theta is the point (0,0). Figure 8

Example 5: Writing Rectangular Coordinates as Polar Coordinates

Convert the rectangular coordinates (3,3)\left(3,3\right) to polar coordinates.

Solution

We see that the original point (3,3)\left(3,3\right) is in the first quadrant. To find θ\theta , use the formula tanθ=yx\tan \theta =\frac{y}{x}. This gives
tanθ=33tanθ=1tan1(1)=π4\begin{array}{l}\tan \theta =\frac{3}{3}\hfill \\ \tan \theta =1\hfill \\ {\tan }^{-1}\left(1\right)=\frac{\pi }{4}\hfill \end{array}
To find rr, we substitute the values for xx and yy into the formula r=x2+y2r=\sqrt{{x}^{2}+{y}^{2}}. We know that rr must be positive, as π4\frac{\pi }{4} is in the first quadrant. Thus
r=32+32r=9+9r=18=32\begin{array}{l}\begin{array}{l}\\ r=\sqrt{{3}^{2}+{3}^{2}}\end{array}\hfill \\ r=\sqrt{9+9}\hfill \\ r=\sqrt{18}=3\sqrt{2}\hfill \end{array}
So, r=32r=3\sqrt{2} and θ=π4\theta \text{=}\frac{\pi }{4}, giving us the polar point (32,π4)\left(3\sqrt{2},\frac{\pi }{4}\right).
Illustration of (3rad2, pi/4) in polar coordinates and (3,3) in rectangular coordinates - they are the same point! Figure 9

Analysis of the Solution

There are other sets of polar coordinates that will be the same as our first solution. For example, the points (32,5π4)\left(-3\sqrt{2},\frac{5\pi }{4}\right) and (32,7π4)\left(3\sqrt{2},-\frac{7\pi }{4}\right) will coincide with the original solution of (32,π4)\left(3\sqrt{2},\frac{\pi }{4}\right). The point (32,5π4)\left(-3\sqrt{2},\frac{5\pi }{4}\right) indicates a move further counterclockwise by π\pi , which is directly opposite π4\frac{\pi }{4}. The radius is expressed as 32-3\sqrt{2}. However, the angle 5π4\frac{5\pi }{4} is located in the third quadrant and, as rr is negative, we extend the directed line segment in the opposite direction, into the first quadrant. This is the same point as (32,π4)\left(3\sqrt{2},\frac{\pi }{4}\right). The point (32,7π4)\left(3\sqrt{2},-\frac{7\pi }{4}\right) is a move further clockwise by 7π4-\frac{7\pi }{4}, from π4\frac{\pi }{4}. The radius, 323\sqrt{2}, is the same.

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.

How To: Given an equation in polar form, graph it using a graphing calculator.

  1. Change the MODE to POL, representing polar form.
  2. Press the Y= button to bring up a screen allowing the input of six equations: r1,r2,...,r6{r}_{1},{r}_{2},...,{r}_{6}.
  3. Enter the polar equation, set equal to rr.
  4. Press GRAPH.

Example 6: Writing a Cartesian Equation in Polar Form

Write the Cartesian equation x2+y2=9{x}^{2}+{y}^{2}=9 in polar form.

Solution

The goal is to eliminate xx and yy from the equation and introduce rr and θ\theta . Ideally, we would write the equation rr as a function of θ\theta . To obtain the polar form, we will use the relationships between (x,y)\left(x,y\right) and (r,θ)\left(r,\theta \right). Since x=rcosθx=r\cos \theta and y=rsinθy=r\sin \theta , we can substitute and solve for rr.
 (rcosθ)2+(rsinθ)2=9 r2cos2θ+r2sin2θ=9 r2(cos2θ+sin2θ)=9 r2(1)=9Substitute cos2θ+sin2θ=1. r=±3Use the square root property.\begin{array}{ll}\text{ }{\left(r\cos \theta \right)}^{2}+{\left(r\sin \theta \right)}^{2}=9\hfill & \hfill \\ \text{ }{r}^{2}{\cos }^{2}\theta +{r}^{2}{\sin }^{2}\theta =9\hfill & \hfill \\ \text{ }{r}^{2}\left({\cos }^{2}\theta +{\sin }^{2}\theta \right)=9\hfill & \hfill \\ \text{ }{r}^{2}\left(1\right)=9 \hfill & {\text{Substitute cos}}^{2}\theta +{\sin }^{2}\theta =1.\hfill \\ \text{ }r=\pm 3\begin{array}{cccc}& & & \end{array}\hfill & \text{Use the square root property}.\hfill \end{array}
Plotting a circle of radius 3 with center at the origin in polar and rectangular coordinates. It is the same in both systems. Figure 10. (a) Cartesian form x2+y2=9{x}^{2}+{y}^{2}=9 (b) Polar form r=3r=3
Thus, x2+y2=9,r=3{x}^{2}+{y}^{2}=9,r=3, and r=3r=-3 should generate the same graph.To graph a circle in rectangular form, we must first solve for yy.
x2+y2=9 y2=9x2 y=±9x2\begin{array}{l}\begin{array}{l}\\ {x}^{2}+{y}^{2}=9\end{array}\hfill \\ \text{ }{y}^{2}=9-{x}^{2}\hfill \\ \text{ }y=\pm \sqrt{9-{x}^{2}}\hfill \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 Y1=9x2{Y}_{1}=\sqrt{9-{x}^{2}} and Y2=9x2{Y}_{2}=-\sqrt{9-{x}^{2}}. Press GRAPH.

Example 7: Rewriting a Cartesian Equation as a Polar Equation

Rewrite the Cartesian equation x2+y2=6y{x}^{2}+{y}^{2}=6y as a polar equation.

Solution

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:
r2=6yUse x2+y2=r2.r2=6rsinθSubstitutey=rsinθ. r26rsinθ=0Set equal to 0. r(r6sinθ)=0Factor and solve.r=0We reject r=0,as it only represents one point, (0,0).orr=6sinθ\begin{array}{ll}{r}^{2}=6y\hfill & \text{Use }{x}^{2}+{y}^{2}={r}^{2}.\hfill \\ {r}^{2}=6r\sin \theta \hfill & \text{Substitute}y=r\sin \theta .\hfill \\ \text{ }{r}^{2}-6r\sin \theta =0\hfill & \text{Set equal to 0}.\hfill \\ \text{ }r\left(r - 6\sin \theta \right)=0\hfill & \text{Factor and solve}.\hfill \\ r=0\hfill & \text{We reject }r=0,\text{as it only represents one point, }\left(0,0\right).\hfill \\ \text{or}r=6\sin \theta \begin{array}{cccc}& & & \end{array}\hfill & \hfill \end{array}
Therefore, the equations x2+y2=6y{x}^{2}+{y}^{2}=6y and r=6sinθr=6\sin \theta should give us the same graph.
Plots of the equations stated above - the plots are the same in both rectangular and polar coordinates. They are circles. Figure 11. (a) Cartesian form x2+y2=6y{x}^{2}+{y}^{2}=6y (b) polar form r=6sinθr=6\sin \theta
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.

Example 8: Rewriting a Cartesian Equation in Polar Form

Rewrite the Cartesian equation y=3x+2y=3x+2 as a polar equation.

Solution

We will use the relationships x=rcosθx=r\cos \theta and y=rsinθy=r\sin \theta .
 y=3x+2 rsinθ=3rcosθ+2rsinθ3rcosθ=2r(sinθ3cosθ)=2Isolate r. r=2sinθ3cosθSolve for r.\begin{array}{ll}\text{ }y=3x+2\hfill & \hfill \\ \text{ }r\sin \theta =3r\cos \theta +2\hfill & \hfill \\ r\sin \theta -3r\cos \theta =2\hfill & \hfill \\ r\left(\sin \theta -3\cos \theta \right)=2\hfill & \text{Isolate }r.\hfill \\ \text{ }r=\frac{2}{\sin \theta -3\cos \theta }\begin{array}{cccc}& & & \end{array}\hfill & \text{Solve for }r.\hfill \end{array}

Try It 4

Rewrite the Cartesian equation y2=3x2{y}^{2}=3-{x}^{2} in polar form. Solution

 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 9: Graphing a Polar Equation by Converting to a Rectangular Equation

Covert the polar equation r=2secθr=2\sec \theta to a rectangular equation, and draw its corresponding graph.

Solution

The conversion is
 r=2secθ r=2cosθrcosθ=2x=2\begin{array}{l}\text{ }r=2\sec \theta \hfill \\ \text{ }r=\frac{2}{\cos \theta }\hfill \\ r\cos \theta =2\hfill \\ x=2\hfill \end{array}
Notice that the equation r=2secθr=2\sec \theta drawn on the polar grid is clearly the same as the vertical line x=2x=2 drawn on the rectangular grid. Just as x=cx=c is the standard form for a vertical line in rectangular form, r=csecθr=c\sec \theta is the standard form for a vertical line in polar form.
Plots of the equations stated above - the plots are the same in both rectangular and polar coordinates. They are lines. Figure 12. (a) Polar grid (b) Rectangular coordinate system
A similar discussion would demonstrate that the graph of the function r=2cscθr=2\csc \theta will be the horizontal line y=2y=2. In fact, r=ccscθr=c\csc \theta is the standard form for a horizontal line in polar form, corresponding to the rectangular form y=cy=c.

Example 10: Rewriting a Polar Equation in Cartesian Form

Rewrite the polar equation r=312cosθr=\frac{3}{1 - 2\cos \theta } as a Cartesian equation.

Solution

The goal is to eliminate θ\theta and rr, and introduce xx and yy. We clear the fraction, and then use substitution. In order to replace rr with xx and yy, we must use the expression x2+y2=r2{x}^{2}+{y}^{2}={r}^{2}.
 r=312cosθr(12cosθ)=3r(12(xr))=3Use cosθ=xr to eliminate θ. r2x=3 r=3+2xIsolate r. r2=(3+2x)2Square both sides. x2+y2=(3+2x)2Use x2+y2=r2.\begin{array}{llll}\text{ }r=\frac{3}{1 - 2\cos \theta }\hfill & \hfill & \hfill & \hfill \\ r\left(1 - 2\cos \theta \right)=3\hfill & \hfill & \hfill & \hfill \\ r\left(1 - 2\left(\frac{x}{r}\right)\right)=3\hfill & \hfill & \hfill & \text{Use }\cos \theta =\frac{x}{r}\text{ to eliminate }\theta .\hfill \\ \text{ }r - 2x=3\hfill & \hfill & \hfill & \hfill \\ \text{ }r=3+2x\hfill & \hfill & \hfill & \text{Isolate }r.\hfill \\ \text{ }{r}^{2}={\left(3+2x\right)}^{2}\hfill & \hfill & \hfill & \text{Square both sides}.\hfill \\ \text{ }{x}^{2}+{y}^{2}={\left(3+2x\right)}^{2}\hfill & \hfill & \hfill & \text{Use }{x}^{2}+{y}^{2}={r}^{2}.\hfill \end{array}
The Cartesian equation is x2+y2=(3+2x)2{x}^{2}+{y}^{2}={\left(3+2x\right)}^{2}. However, to graph it, especially using a graphing calculator or computer program, we want to isolate yy.
x2+y2=(3+2x)2 y2=(3+2x)2x2 y=±(3+2x)2x2\begin{array}{l}{x}^{2}+{y}^{2}={\left(3+2x\right)}^{2}\hfill \\ \text{ }{y}^{2}={\left(3+2x\right)}^{2}-{x}^{2}\hfill \\ \text{ }y=\pm \sqrt{{\left(3+2x\right)}^{2}-{x}^{2}}\hfill \end{array}
When our entire equation has been changed from rr and θ\theta to xx and yy, we can stop, unless asked to solve for yy or simplify.
Plots of the equations stated above - the plots are the same in both rectangular and polar coordinates. They are hyperbolas. Figure 13
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.

Analysis of the Solution

In this example, the right side of the equation can be expanded and the equation simplified further, as shown above. However, the equation cannot be written as a single function in Cartesian form. We may wish to write the rectangular equation in the hyperbola’s standard form. To do this, we can start with the initial equation.
 x2+y2=(3+2x)2 x2+y2(3+2x)2=0x2+y2(9+12x+4x2)=0 x2+y2912x4x2=0 3x212x+y2=9Multiply through by 1. 3x2+12xy2=9 3(x2+4x+)y2=9Organize terms to complete the square forx. 3(x2+4x+4)y2=9+12 3(x+2)2y2=3 (x+2)2y23=1\begin{array}{ll}\text{ }{x}^{2}+{y}^{2}={\left(3+2x\right)}^{2}\hfill & \hfill \\ \text{ }{x}^{2}+{y}^{2}-{\left(3+2x\right)}^{2}=0\hfill & \hfill \\ {x}^{2}+{y}^{2}-\left(9+12x+4{x}^{2}\right)=0\hfill & \hfill \\ \text{ }{x}^{2}+{y}^{2}-9 - 12x - 4{x}^{2}=0\hfill & \hfill \\ \text{ }-3{x}^{2}-12x+{y}^{2}=9\hfill & \text{Multiply through by }-1.\hfill \\ \text{ }3{x}^{2}+12x-{y}^{2}=-9\hfill & \hfill \\ \text{ }3\left({x}^{2}+4x+\right)-{y}^{2}=-9\hfill & \text{Organize terms to complete the square for}x.\hfill \\ \text{ }3\left({x}^{2}+4x+4\right)-{y}^{2}=-9+12\hfill & \hfill \\ \text{ }3{\left(x+2\right)}^{2}-{y}^{2}=3\hfill & \hfill \\ \text{ }{\left(x+2\right)}^{2}-\frac{{y}^{2}}{3}=1\hfill & \hfill \end{array}

Try It 5

Rewrite the polar equation r=2sinθr=2\sin \theta in Cartesian form. Solution

Example 11: Rewriting a Polar Equation in Cartesian Form

Rewrite the polar equation r=sin(2θ)r=\sin \left(2\theta \right) in Cartesian form.

Solution

 r=sin(2θ)Use the double angle identity for sine. r=2sinθcosθUse cosθ=xr and sinθ=yr. r=2(xr)(yr)Simplify. r=2xyr2 Multiply both sides by r2. r3=2xy(x2+y2)3=2xyAsx2+y2=r2,r=x2+y2.\begin{array}{ll}\text{ }r=\sin \left(2\theta \right)\hfill & \text{Use the double angle identity for sine}.\hfill \\ \text{ }r=2\sin \theta \cos \theta \begin{array}{cccc}& & & \end{array}\hfill & \text{Use }\cos \theta =\frac{x}{r}\text{ and }\sin \theta =\frac{y}{r}.\hfill \\ \text{ }r=2\left(\frac{x}{r}\right)\left(\frac{y}{r}\right)\hfill & \text{Simplify}.\hfill \\ \text{ }r=\frac{2xy}{{r}^{2}}\hfill & \text{ Multiply both sides by }{r}^{2}.\hfill \\ \text{ }{r}^{3}=2xy\hfill & \hfill \\ {\left(\sqrt{{x}^{2}+{y}^{2}}\right)}^{3}=2xy\hfill & \text{As}{x}^{2}+{y}^{2}={r}^{2},r=\sqrt{{x}^{2}+{y}^{2}}.\hfill \end{array}
This equation can also be written as
(x2+y2)32=2xy or x2+y2=(2xy)23{\left({x}^{2}+{y}^{2}\right)}^{\frac{3}{2}}=2xy\text{ or }{x}^{2}+{y}^{2}={\left(2xy\right)}^{\frac{2}{3}}
 

Key Equations

Conversion formulas cosθ=xrx=rcosθsinθ=yry=rsinθr2=x2+y2tanθ=yx\begin{array}{ll}\hfill & \cos \theta =\frac{x}{r}\to x=r\cos \theta \hfill \\ \hfill & \sin \theta =\frac{y}{r}\to y=r\sin \theta \hfill \\ \hfill & {r}^{2}={x}^{2}+{y}^{2}\hfill \\ \hfill & \tan \theta =\frac{y}{x}\hfill \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,θ),θ>0\left(r,\theta \right),\theta >0, move in a counterclockwise direction from the polar axis by an angle of θ\theta , and then extend a directed line segment from the pole the length of rr in the direction of θ\theta . If θ\theta is negative, move in a clockwise direction, and extend a directed line segment the length of rr in the direction of θ\theta .
  • If rr is negative, extend the directed line segment in the opposite direction of θ\theta .
  • To convert from polar coordinates to rectangular coordinates, use the formulas x=rcosθx=r\cos \theta and y=rsinθy=r\sin \theta .
  • To convert from rectangular coordinates to polar coordinates, use one or more of the formulas: cosθ=xr,sinθ=yr,tanθ=yx\cos \theta =\frac{x}{r},\sin \theta =\frac{y}{r},\tan \theta =\frac{y}{x}, and r=x2+y2r=\sqrt{{x}^{2}+{y}^{2}}.
  • Transforming equations between polar and rectangular forms means making the appropriate substitutions based on the available formulas, together with algebraic manipulations.
  • Using the appropriate substitutions makes it possible to rewrite a polar equation as a rectangular equation, and then graph it in the rectangular plane.

Glossary

polar axis
on the polar grid, the equivalent of the positive x-axis on the rectangular grid
polar coordinates
on the polar grid, the coordinates of a point labeled (r,θ)\left(r,\theta \right), where θ\theta indicates the angle of rotation from the polar axis and rr represents the radius, or the distance of the point from the pole in the direction of θ\theta
pole
the origin of the polar grid

Section Exercises

1. How are polar coordinates different from rectangular coordinates? 2. How are the polar axes different from the x- and y-axes of the Cartesian plane? 3. Explain how polar coordinates are graphed. 4. How are the points (3,π2)\left(3,\frac{\pi }{2}\right) and (3,π2)\left(-3,\frac{\pi }{2}\right) related? 5. Explain why the points (3,π2)\left(-3,\frac{\pi }{2}\right) and (3,π2)\left(3,-\frac{\pi }{2}\right) are the same. For the following exercises, convert the given polar coordinates to Cartesian coordinates with r>0r>0 and 0θ2π0\le \theta \le 2\pi . Remember to consider the quadrant in which the given point is located when determining θ\theta for the point. 6. (7,7π6)\left(7,\frac{7\pi }{6}\right) 7. (5,π)\left(5,\pi \right) 8. (6,π4)\left(6,-\frac{\pi }{4}\right) 9. (3,π6)\left(-3,\frac{\pi }{6}\right) 10. (4,7π4)\left(4,\frac{7\pi }{4}\right) For the following exercises, convert the given Cartesian coordinates to polar coordinates with r>0,0θ<2πr>0,0\le \theta <2\pi . Remember to consider the quadrant in which the given point is located. 11. (4,2)\left(4,2\right) 12. (4,6)\left(-4,6\right) 13. (3,5)\left(3,-5\right) 14. (10,13)\left(-10,-13\right) 15. (8,8)\left(8,8\right) For the following exercises, convert the given Cartesian equation to a polar equation. 16. x=3x=3 17. y=4y=4 18. y=4x2y=4{x}^{2} 19. y=2x4y=2{x}^{4} 20. x2+y2=4y{x}^{2}+{y}^{2}=4y 21. x2+y2=3x{x}^{2}+{y}^{2}=3x 22. x2y2=x{x}^{2}-{y}^{2}=x 23. x2y2=3y{x}^{2}-{y}^{2}=3y 24. x2+y2=9{x}^{2}+{y}^{2}=9 25. x2=9y{x}^{2}=9y 26. y2=9x{y}^{2}=9x 27. 9xy=19xy=1 For the following exercises, convert the given polar equation to a Cartesian equation. Write in the standard form of a conic if possible, and identify the conic section represented. 28. r=3sinθr=3\sin \theta 29. r=4cosθr=4\cos \theta 30. r=4sinθ+7cosθr=\frac{4}{\sin \theta +7\cos \theta } 31. r=6cosθ+3sinθr=\frac{6}{\cos \theta +3\sin \theta } 32. r=2secθr=2\sec \theta 33. r=3cscθr=3\csc \theta 34. r=rcosθ+2r=\sqrt{r\cos \theta +2} 35. r2=4secθcscθ{r}^{2}=4\sec \theta \csc \theta 36. r=4r=4 37. r2=4{r}^{2}=4 38. r=14cosθ3sinθr=\frac{1}{4\cos \theta -3\sin \theta } 39. r=3cosθ5sinθr=\frac{3}{\cos \theta -5\sin \theta } For the following exercises, find the polar coordinates of the point. 40. Polar coordinate system with a point located on the third concentric circle and pi/2. 41. Polar coordinate system with a point located on the third concentric circle and midway between pi/2 and pi in the second quadrant. 42. Polar coordinate system with a point located midway between the first and second concentric circles and a third of the way between pi and 3pi/2 (closer to pi). 43. Polar coordinate system with a point located on the fifth concentric circle and pi. 44. Polar coordinate system with a point located on the fourth concentric circle and a third of the way between 3pi/2 and 2pi (closer to 3pi/2). For the following exercises, plot the points. 45. (2,π3)\left(-2,\frac{\pi }{3}\right) 46. (1,π2)\left(-1,-\frac{\pi }{2}\right) 47. (3.5,7π4)\left(3.5,\frac{7\pi }{4}\right) 48. (4,π3)\left(-4,\frac{\pi }{3}\right) 49. (5,π2)\left(5,\frac{\pi }{2}\right) 50. (4,5π4)\left(4,\frac{-5\pi }{4}\right) 51. (3,5π6)\left(3,\frac{5\pi }{6}\right) 52. (1.5,7π6)\left(-1.5,\frac{7\pi }{6}\right) 53. (2,π4)\left(-2,\frac{\pi }{4}\right) 54. (1,3π2)\left(1,\frac{3\pi }{2}\right) For the following exercises, convert the equation from rectangular to polar form and graph on the polar axis. 55. 5xy=65x-y=6 56. 2x+7y=32x+7y=-3 57. x2+(y1)2=1{x}^{2}+{\left(y - 1\right)}^{2}=1 58. (x+2)2+(y+3)2=13{\left(x+2\right)}^{2}+{\left(y+3\right)}^{2}=13 59. x=2x=2 60. x2+y2=5y{x}^{2}+{y}^{2}=5y 61. x2+y2=3x{x}^{2}+{y}^{2}=3x For the following exercises, convert the equation from polar to rectangular form and graph on the rectangular plane. 62. r=6r=6 63. r=4r=-4 64. θ=2π3\theta =-\frac{2\pi }{3} 65. θ=π4\theta =\frac{\pi }{4} 66. r=secθr=\sec \theta 67. r=10sinθr=-10\sin \theta 68. r=3cosθr=3\cos \theta 69. Use a graphing calculator to find the rectangular coordinates of (2,π5)\left(2,-\frac{\pi }{5}\right). Round to the nearest thousandth. 70. Use a graphing calculator to find the rectangular coordinates of (3,3π7)\left(-3,\frac{3\pi }{7}\right). Round to the nearest thousandth. 71. Use a graphing calculator to find the polar coordinates of (7,8)\left(-7,8\right) in degrees. Round to the nearest thousandth. 72. Use a graphing calculator to find the polar coordinates of (3,4)\left(3,-4\right) in degrees. Round to the nearest hundredth. 73. Use a graphing calculator to find the polar coordinates of (2,0)\left(-2,0\right) in radians. Round to the nearest hundredth. 74. Describe the graph of r=asecθ;a>0r=a\sec \theta ;a>0. 75. Describe the graph of r=asecθ;a<0r=a\sec \theta ;a<0. 76. Describe the graph of r=acscθ;a>0r=a\csc \theta ;a>0. 77. Describe the graph of r=acscθ;a<0r=a\csc \theta ;a<0. 78. What polar equations will give an oblique line? For the following exercises, graph the polar inequality. 79. r<4r<4 80. 0θπ40\le \theta \le \frac{\pi }{4} 81. θ=π4,r2\theta =\frac{\pi }{4},r\ge 2 82. θ=π4,r3\theta =\frac{\pi }{4},r\ge -3 83. 0θπ3,r<20\le \theta \le \frac{\pi }{3},r<2 84. π6<θπ3,3<r<2\frac{-\pi }{6}<\theta \le \frac{\pi }{3},-3<r<2

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