1. H.Melikyan/12001 Graphs of Polar Equations Dr.Hayk Melikyan Departmen of Mathematics and CS melikyan@nccu.edu

2. H.Melikyan/12002 Recall that a polar equation is an equation whose variables are r and è. The graph of a polar equation is the set of all points whose polar coordinates satisfy the equation. We use polar grids like the one shown to graph polar equations. The grid consists of circles with centers at the pole. This polar grid shows five such circles. A polar grid also shows lines passing through the pole, In this grid, each fine represents an angle for which we know the exact values of the trigonometric functions.  0 6 ˝ 4 3  2  3 22 4 3  6 5  6 7  4 5  3 4  3 5  4 7  6 11  2 3  42 Using Polar Grids to Graph Polar Equations  

3. H.Melikyan/12003 ˝ 0 6  4 ˝ 3  2  3 2  4 3  6 5  6 7  4 5  3 4  3 5  4 7  6 11  2 3  42 SolutionWe construct a partial table of coordinates using multiples of   6. Then we plot the points and join them with a smooth curve, as shown. (-4,  ) 4 cos ˝  = 4(-1) = -4 ˝ (-3.5,5  / 6 ) 4 cos5  / 6 = 4(-  3/2)=-2  3=-3.5 5/65/6 (-2, 2  / 3 ) 4 cos2  / 3 = 4(- 1/2) = -2 2  / 3 (0,  / 2 ) 4 cos  / 2 = 4 0 = 0  / 2 (2,  / 3 ) 4 cos  / 3 = 4 1/2 = 4  / 3 (3.5,  / 6 ) 4 cos  / 6 = 4  3/2=2  3=3.5  / 6 (4, 0)4 cos 0 = 4 1 = 40 (r,  )r  4 cos  (0,  / 2)(2,  / 3)(3.5,  / 6) (4, 0) or (-4,  ) (-3.5, 5  / 6) (-2, 2  / 3)  Text Example  Graph the polar equation r  4 cos  with  in radians.

4. H.Melikyan/12004 The graphs of r = a cos  and r = a sin  Are circles. r = a cos  r = a sin  The graphs of r = a cos  and r = a sin  Are circles. r = a cos  r = a sin  ˝0  / 2 3  / 2 ˝0 /2/2 a a Circles in Polar Coordinates

5. H.Melikyan/12005 SolutionWe apply each of the tests for symmetry. Polar Axis: Replace  by  in r  1  cos  : r  1  cos (  ) Replace  by  in r  1  cos . r  1  cos  The cosine function is even: cos (  )  cos . Because the polar equation does not change when  is replaced by , the graph is symmetric with respect to the polar axis. Text Example  Check for symmetry and then graph the polar equation: r  1  cos .

6. H.Melikyan/12006 The Line     2 : Replace (r,  ) by (  r,  ) in r  1  cos  :  r  1  cos(  ) Replace r by  r and  by –  in  r  1  cos(  ).  r  1 – cos  cos(  )  cos . r  cos   1 Multiply both sides by  1. Because the polar equation r  1  cos  changes to r  cos   1 when (r,  ) is replaced by (  r,  ), the equation fails this symmetry test. The graph may of may not be symmetric with respect to the line     2. Solution The Pole: Replace r by  r in r  1  cos  :  r  1 – cos  Replace r by –r. r  cos   1 Multiply both sides by  1. Because the polar equation r  1  cos  changes to r  cos   1 when r is replaced by  r, the equation fails this symmetry test. The graph may or may not be symmetric with respect to the pole. Text Example cont.

7. H.Melikyan/12007 Now we are ready to graph r  1  cos . Because the period of the cosine function is 2r, we need not consider values of  beyond 2 . Recall that we discovered the graph of the equation r  1  cos  has symmetry with respect to the polar axis. Because the graph has symmetry, we may be able to obtain a complete graph without plotting points generated by values of  from 0 to 2 . Let's start by finding the values of r for values of  from 0 to . Solution 21.871.501.000.500.130r  5/65/6 2/32/3  2 /3/3 /6/6 0  The values for r and  are in the table. Examine the graph. Keep in mind that the graph must be symmetric with respect to the polar axis.  0 6  4  3 ˝ 2  3 2  4 3  6 5  6 7  4 5  3 4  3 5  4 7  6 11  2 3  21 Text Example cont. 

8. H.Melikyan/12008 Thus, if we reflect the graph from the last slide about the polar axis, we will obtain a complete graph of r  1  cos , shown below. Solution  0 6  4  3  2  3 2  4 3  6 5  6 7  4 5  3 4  3 5  4 7  6 11  2 3  12 Text Example cont.

9. H.Melikyan/12009 The graphs of r  a  b sin , r  a  b sin , r  a  b cos , r  a  b cos , a > 0, b > 0 are called limacons. The ratio a  b determines a limacon's shape. Inner loop if a  b < 1 Heart shaped if a  b  1 Dimpled with no inner No dimple and no inner and called cardiods loop if 1< a  b < 2 loop if a  b  2. The graphs of r  a  b sin , r  a  b sin , r  a  b cos , r  a  b cos , a > 0, b > 0 are called limacons. The ratio a  b determines a limacon's shape. Inner loop if a  b < 1 Heart shaped if a  b  1 Dimpled with no inner No dimple and no inner and called cardiods loop if 1< a  b < 2 loop if a  b  2.  0 2  2 33  0 2  2 33  0 2  2 33  0 2  2 33 Limacons

10. H.Melikyan/120010 Example v Graph the polar equation y= 2+3cos 

11. H.Melikyan/120011 Example v Graph the polar equation y= 2+3cos  Solution:

12. H.Melikyan/120012 The graphs of r  a sin n  and r  a cos n , a does not equal 0, are called rose curves. If n is even, the rose has 2n petals. If n is odd, the rose has n petals. r  a sin 2  r  a cos 3  r  a cos 4  r  a sin 5  Rose curve Rose curve with 4 petalswith 3 petalswith 8 petalswith 5 petals The graphs of r  a sin n  and r  a cos n , a does not equal 0, are called rose curves. If n is even, the rose has 2n petals. If n is odd, the rose has n petals. r  a sin 2  r  a cos 3  r  a cos 4  r  a sin 5  Rose curve Rose curve with 4 petalswith 3 petalswith 8 petalswith 5 petals n = 2 a  0 2  2 33  0 2  2 33 n = 5 a  0 2  2 33 n = 4 a  0 2  2 33 n = 3 a a Rose Curves

13. H.Melikyan/120013 Example v Graph the polar equation y=3sin2 

14. H.Melikyan/120014 Example v Graph the polar equation y=3sin2  Solution:

15. H.Melikyan/120015 Lemniscate: r 2 = a 2 cos 2  Lemniscates  The graphs of r 2 = a 2 sin 2  and r 2 = a 2 cos 2  are called lemniscates