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Problem-solving strategy: plotting a curve in polar coordinates

  1. Create a table with two columns. The first column is for θ , and the second column is for r .
  2. Create a list of values for θ .
  3. Calculate the corresponding r values for each θ .
  4. Plot each ordered pair ( r , θ ) on the coordinate axes.
  5. Connect the points and look for a pattern.

Watch this video for more information on sketching polar curves.

Graphing a function in polar coordinates

Graph the curve defined by the function r = 4 sin θ . Identify the curve and rewrite the equation in rectangular coordinates.

Because the function is a multiple of a sine function, it is periodic with period 2 π , so use values for θ between 0 and 2 π . The result of steps 1–3 appear in the following table. [link] shows the graph based on this table.

θ r = 4 sin θ θ r = 4 sin θ
0 0 π 0
π 6 2 7 π 6 −2
π 4 2 2 2.8 5 π 4 −2 2 −2.8
π 3 2 3 3.4 4 π 3 −2 3 −3.4
π 2 4 3 π 2 4
2 π 3 2 3 3.4 5 π 3 −2 3 −3.4
3 π 4 2 2 2.8 7 π 4 −2 2 −2.8
5 π 6 2 11 π 6 −2
2 π 0
On the polar coordinate plane, a circle is drawn with radius 2. It touches the origin, (2 times the square root of 2, π/4), (4, π/2), and (2 times the square root of 2, 3π/4).
The graph of the function r = 4 sin θ is a circle.

This is the graph of a circle. The equation r = 4 sin θ can be converted into rectangular coordinates by first multiplying both sides by r . This gives the equation r 2 = 4 r sin θ . Next use the facts that r 2 = x 2 + y 2 and y = r sin θ . This gives x 2 + y 2 = 4 y . To put this equation into standard form, subtract 4 y from both sides of the equation and complete the square:

x 2 + y 2 4 y = 0 x 2 + ( y 2 4 y ) = 0 x 2 + ( y 2 4 y + 4 ) = 0 + 4 x 2 + ( y 2 ) 2 = 4.

This is the equation of a circle with radius 2 and center ( 0 , 2 ) in the rectangular coordinate system.

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Create a graph of the curve defined by the function r = 4 + 4 cos θ .


The graph of r = 4 + 4 cosθ is given. It vaguely looks look a heart tipped on its side with a rounded bottom instead of a pointed one. Specifically, the graph starts at the origin, moves into the second quadrant and increases to a rounded circle-like figure. The graph is symmetric about the x axis, so it continues its rounded circle-like figure, goes into the third quadrant, and comes to a point at the origin.
The name of this shape is a cardioid, which we will study further later in this section.

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The graph in [link] was that of a circle. The equation of the circle can be transformed into rectangular coordinates using the coordinate transformation formulas in [link] . [link] gives some more examples of functions for transforming from polar to rectangular coordinates.

Transforming polar equations to rectangular coordinates

Rewrite each of the following equations in rectangular coordinates and identify the graph.

  1. θ = π 3
  2. r = 3
  3. r = 6 cos θ 8 sin θ
  1. Take the tangent of both sides. This gives tan θ = tan ( π / 3 ) = 3 . Since tan θ = y / x we can replace the left-hand side of this equation by y / x . This gives y / x = 3 , which can be rewritten as y = x 3 . This is the equation of a straight line passing through the origin with slope 3 . In general, any polar equation of the form θ = K represents a straight line through the pole with slope equal to tan K .
  2. First, square both sides of the equation. This gives r 2 = 9 . Next replace r 2 with x 2 + y 2 . This gives the equation x 2 + y 2 = 9 , which is the equation of a circle centered at the origin with radius 3. In general, any polar equation of the form r = k where k is a positive constant represents a circle of radius k centered at the origin. ( Note : when squaring both sides of an equation it is possible to introduce new points unintentionally. This should always be taken into consideration. However, in this case we do not introduce new points. For example, ( −3 , π 3 ) is the same point as ( 3 , 4 π 3 ) . )
  3. Multiply both sides of the equation by r . This leads to r 2 = 6 r cos θ 8 r sin θ . Next use the formulas
    r 2 = x 2 + y 2 , x = r cos θ , y = r sin θ .

    This gives
    r 2 = 6 ( r cos θ ) 8 ( r sin θ ) x 2 + y 2 = 6 x 8 y .

    To put this equation into standard form, first move the variables from the right-hand side of the equation to the left-hand side, then complete the square.
    x 2 + y 2 = 6 x 8 y x 2 6 x + y 2 + 8 y = 0 ( x 2 6 x ) + ( y 2 + 8 y ) = 0 ( x 2 6 x + 9 ) + ( y 2 + 8 y + 16 ) = 9 + 16 ( x 3 ) 2 + ( y + 4 ) 2 = 25.

    This is the equation of a circle with center at ( 3 , −4 ) and radius 5. Notice that the circle passes through the origin since the center is 5 units away.
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Source:  OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2
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