Cardioids and Rose Curves
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Cardioids and Rose Curves. After Exploration (p. 574). THEOREM: Let a be a (+) real number, then. r = 2a sin circle: radius a; center (0, a). r = -2a sin circle: radius a; center (0, -a). r = 2a cos circle: radius a; center (a, 0).

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Cardioids and Rose Curves

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Cardioids and rose curves

Cardioids and Rose Curves


Cardioids and rose curves

After Exploration (p. 574)

THEOREM:Let a be a (+) real number, then

  • r = 2a sincircle: radius a; center (0, a)

  • r = -2a sincircle: radius a; center (0, -a)

  • r = 2a coscircle: radius a; center (a, 0)

  • r = -2a coscircle: radius a; center (-a, 0)


Cardioids and rose curves

  • Symmetry

I.

II.

III.

(r, )

(r, )

(r, )

(r,   )

(-r, )

(r, -)

  • symmetric WRT the polar axis (and x-axis)

  • symmetric WRT  = /2 (and y-axis)

  • symmetric WRT the pole (and origin)

  • replace  w/ -. If an equivalent equation results, the graph is symmetric WRT the polar axis

  • replace  w/   . If an equivalent equation results, the graph is symmetric WRT  = /2

  • replace r w/ -r. If an equivalent equation results, the graph is symmetric WRT the pole


Cardioids and rose curves

  • Cardioid ~ A heart-shaped curve

Forms:

r = a(1 + cos)

r = a(1  cos)

where a > 0, the graph of a cardioid passes through the pole.

r = a(1 + sin)

r = a(1  sin)


Cardioids and rose curves

Example:

Sketch r = 3(1 + sin)

3(1 + sin)

0 3

30 4.5

60 5.6

90 6

120 5.6

150 4.5

180 3

210 1.5

240 .4

270 0

** Each circle represents 2

300 .4

330 1.5


Cardioids and rose curves

  • The Rose Curve

Forms:

r = a sin n

r = a cos n

if n is odd, there are n leaves.

if n is even, there are 2n leaves.


Cardioids and rose curves

Example:

Sketch r = sin2

“4 leaf Rose”

sin 2

0 0

30 .866

45 1

60 .866

90 0

Falls on (0,0)

120 -.866

135 -1

150 -.866

Falls on (0,0)

180 0

210 .866

225 1

240 .866

Falls on (0,0)

270 0

300 -.866

315 -1

330 -.866

Falls on (0,0)

360 0

** Each circle represents .5


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