Polar Graphs

studied byStudied by 19 people
5.0(2)
Get a hint
Hint

r = a sin(θ)

1 / 19

flashcard set

Earn XP

Description and Tags

all coordinates are in polar form unless noted, take all graphs with a grain of salt because the graphs may ask for ask horizontal or vertical shift that these equations do not consider

Pre-Calculus

20 Terms

1

r = a sin(θ)

circle with diameter a, passing through the origin and (a, π/2), b is

graph pictured is r = sin(θ)

<p>circle with diameter a, passing through the origin and (a, π/2), b is</p><p>graph pictured is r = sin(θ)</p>
New cards
2

r = a cos(θ)

circle with diameter a, passing through the origin and (a, 0˚)

graph pictured is r = cos(θ)

<p>circle with diameter a, passing through the origin and (a, 0˚)</p><p>graph pictured is r = cos(θ)</p>
New cards
3

r = a sin(nθ), n is odd

rose curve with n as the number of petals around the origin and a as the amplitude of the petals. graph starts at the origin, graph is rotated 90˚ from cosine rose curve, negative graph will reflect around the polar axis

graph pictured is r = cos(5θ)

<p>rose curve with n as the number of petals around the origin and a as the amplitude of the petals. graph starts at the origin, graph is rotated 90˚ from cosine rose curve, negative graph will reflect around the polar axis</p><p>graph pictured is r = cos(5θ)</p>
New cards
4

r = a sin(nθ), n is even

rose curve with 2n as the number of petals around the origin and a as the amplitude of the petals. graph starts at the origin, graph is rotated 90˚ from cosine rose curve

graph shows r = sin(4θ)

<p>rose curve with 2n as the number of petals around the origin and a as the amplitude of the petals. graph starts at the origin, graph is rotated 90˚ from cosine rose curve</p><p>graph shows r = sin(4θ)</p>
New cards
5

r = a cos(nθ), n is odd

rose curve with n as the number of petals around the origin and a as the amplitude of the petals. graph starts at the “peak“ of a petal, graph is rotated 90˚ from sine rose curve

graph pictured is r = cos(5θ)

<p>rose curve with n as the number of petals around the origin and a as the amplitude of the petals. graph starts at the “peak“ of a petal, graph is rotated 90˚ from sine rose curve</p><p>graph pictured is r = cos(5θ)</p>
New cards
6

r = a cos(nθ), n is even number

rose curve with 2n as the number of petals around the origin and a as the amplitude of the petals. graph starts at the “peak“ of a petal, graph is rotated 90˚ from sine rose curve

graph shows r = cos(4θ)

<p>rose curve with 2n as the number of petals around the origin and a as the amplitude of the petals. graph starts at the “peak“ of a petal, graph is rotated 90˚ from sine rose curve</p><p>graph shows r = cos(4θ)</p>
New cards
7

r^2 = a^2 sin(2θ)

lemniscate with both petals symmetric on the π/4 ray and a as the amplitude of the petals.

graph shows r^2 = sin(2θ)

<p>lemniscate with both petals symmetric on the π/4 ray and a as the amplitude of the petals.</p><p>graph shows r^2 = sin(2θ)</p>
New cards
8

r^2 = a^2 cos(2θ)

lemniscate with both petals symmetric on the polar axis and a as the amplitude of the petals.

graph shows r^2 = cos(2θ)

<p>lemniscate with both petals symmetric on the polar axis and a as the amplitude of the petals.</p><p>graph shows r^2 = cos(2θ)</p>
New cards
9

r = a ± b sin(θ), a/b < 1

looped limaçon symmetric about the π/2 ray, the amplitude of the “main” part of the limaçon is as tall as the positive part of the y = a ± b sin(x) graph, and the small loop is as tall as the negative part of the y = a ± b sin(x) graph, flipping sign of sin() will flip about the polar axis

graph shows r = 1 + 2 sin(θ)

<p>looped limaçon symmetric about the π/2 ray, the amplitude of the “main” part of the limaçon is as tall as the positive part of the y = a ± b sin(x) graph, and the small loop is as tall as the negative part of the y = a ± b sin(x) graph, flipping sign of sin() will flip about the polar axis</p><p>graph shows r = 1 + 2 sin(θ)</p>
New cards
10

r = a ± b cos(θ), a/b < 1

looped limaçon symmetric about the polar axis, the amplitude of the “main” part of the limaçon is as tall as the positive part of the y = a ± b cos(x) graph, and the small loop is as tall as the negative part of the y = a ± b cos(x) graph, flipping sign of cos() will flip about the π/2 ray

graph shows r = 1 + 2 cos(θ)

<p>looped limaçon symmetric about the polar axis, the amplitude of the “main” part of the limaçon is as tall as the positive part of the y = a ± b cos(x) graph, and the small loop is as tall as the negative part of the y = a ± b cos(x) graph, flipping sign of cos() will flip about the π/2 ray</p><p>graph shows r = 1 + 2 cos(θ)</p>
New cards
11

r = a ± b cos(θ), a/b = 1

cardioid/limaçon symmetric about the polar axis, the amplitude of the cardioid is 2a, point at the origin, flipping sign of cos() will flip about the π/2 ray

graph shows r = 1 + cos(θ)

<p>cardioid/limaçon symmetric about the polar axis, the amplitude of the cardioid is 2a, point at the origin, flipping sign of cos() will flip about the π/2 ray</p><p>graph shows r = 1 + cos(θ)</p>
New cards
12

r = a ± b sin(θ), a/b = 1

cardioid/limaçon symmetric about the π/2 ray, the amplitude of the cardioid is 2a, point at the origin, flipping sign of sin() will flip about the polar axis

graph shows r = 1 + sin(θ)

<p>cardioid/limaçon symmetric about the π/2 ray, the amplitude of the cardioid is 2a, point at the origin, flipping sign of sin() will flip about the polar axis</p><p>graph shows r = 1 + sin(θ)</p>
New cards
13

r = aθ

spiral of archimedes, starts at the origin and proceeds counter-clockwise if restricted to positive values and clockwise if restricted to negative values

graph shows r = θ/2, θ > 0

<p>spiral of archimedes, starts at the origin and proceeds counter-clockwise if restricted to positive values and clockwise if restricted to negative values</p><p>graph shows r = θ/2, θ &gt; 0</p>
New cards
14

r = a ± b sin(θ), 1 < a/b < 2

dimpled limaçon, symmetric about the π/2 ray, the amplitude of the limaçon is “peak“ of graph, dimple at the closest value to r = 0, flipping sign of sin() will flip about the polar axis

graph is r = 1 + 2/3 sin(θ)

<p>dimpled limaçon, symmetric about the π/2 ray, the amplitude of the limaçon is “peak“ of graph, dimple at the closest value to r = 0, flipping sign of sin() will flip about the polar axis</p><p>graph is r = 1 + 2/3 sin(θ)</p>
New cards
15

r = a ± b cos(θ), 1 < a/b < 2

dimpled limaçon, symmetric about the polar axis, the amplitude of the cardioid is 2a, dimple at the closest value to r = 0, flipping sign of sin() will flip about the π/2 ray

graph is r = 1 + 2/3 cos(θ)

<p>dimpled limaçon, symmetric about the polar axis, the amplitude of the cardioid is 2a, dimple at the closest value to r = 0, flipping sign of sin() will flip about the π/2 ray</p><p>graph is r = 1 + 2/3 cos(θ)</p>
New cards
16

r = a ± b sin(θ), a/b ≥ 2

convex limaçon, symmetric about the π/2 ray, the amplitude of the limaçon is “peak“ of graph, flat part at the closest value to r = 0, flipping sign of sin() will flip about the polar axis

graph is r = 1 + 1/2 sin(θ)

<p>convex limaçon, symmetric about the π/2 ray, the amplitude of the limaçon is “peak“ of graph, flat part at the closest value to r = 0, flipping sign of sin() will flip about the polar axis</p><p>graph is r = 1 + 1/2 sin(θ)</p>
New cards
17

r = a ± b sin(θ), a/b ≥ 2

convex limaçon, symmetric about the polar axis, the amplitude of the limaçon is “peak“ of graph, flat part at the closest value to r = 0, flipping sign of sin() will flip about the π/2 ray

graph is r = 1 + 1/2 cos(θ)

<p>convex limaçon, symmetric about the polar axis, the amplitude of the limaçon is “peak“ of graph, flat part at the closest value to r = 0, flipping sign of sin() will flip about the π/2 ray</p><p>graph is r = 1 + 1/2 cos(θ)</p>
New cards
18

r = a + b sin(nθ) OR r = a + b cos(nθ), n is odd positive integer > 1, b > a

rose curve with petals inside of the main petals, amplitude of both petals determined by value of b - a (small petal) and b + a (big petal), 2n is the number of petals

graph shows r = 1/2 + sin(3θ)

<p>rose curve with petals inside of the main petals, amplitude of both petals determined by value of b - a (small petal) and b + a (big petal), 2n is the number of petals</p><p>graph shows r = 1/2 + sin(3θ)</p>
New cards
19

r = a + b sin(nθ) OR r = a + b cos(nθ), n is even positive integer > 1, b > a

rose curve with petals in between the main petals, amplitude of both petals determined by value of b - a (small petal) and b + a (big petal), 2n is the number of petals

graph shows r = 1/2 + cos(4θ)

<p>rose curve with petals in between the main petals, amplitude of both petals determined by value of b - a (small petal) and b + a (big petal), 2n is the number of petals</p><p>graph shows r = 1/2 + cos(4θ)</p>
New cards
20

r = b/(sinθ - m cosθ)

straight line for the equation y = mx + b, where m and b are equivalent numbers

graph shows r = 1/(sin θ - 2 cos θ), or y = 2x + 1

<p>straight line for the equation y = mx + b, where m and b are equivalent numbers</p><p>graph shows r = 1/(sin θ - 2 cos θ), or y = 2x + 1</p>
New cards

Explore top notes

note Note
studied byStudied by 5 people
... ago
5.0(1)
note Note
studied byStudied by 14 people
... ago
5.0(1)
note Note
studied byStudied by 79 people
... ago
5.0(4)
note Note
studied byStudied by 2 people
... ago
4.0(1)
note Note
studied byStudied by 73 people
... ago
5.0(1)
note Note
studied byStudied by 27 people
... ago
4.5(2)
note Note
studied byStudied by 9 people
... ago
5.0(1)
note Note
studied byStudied by 32 people
... ago
4.5(2)

Explore top flashcards

flashcards Flashcard (335)
studied byStudied by 33 people
... ago
5.0(1)
flashcards Flashcard (115)
studied byStudied by 14 people
... ago
5.0(1)
flashcards Flashcard (27)
studied byStudied by 6 people
... ago
5.0(1)
flashcards Flashcard (44)
studied byStudied by 8 people
... ago
5.0(1)
flashcards Flashcard (94)
studied byStudied by 3 people
... ago
5.0(1)
flashcards Flashcard (75)
studied byStudied by 307 people
... ago
4.5(2)
flashcards Flashcard (172)
studied byStudied by 2 people
... ago
5.0(1)
flashcards Flashcard (632)
studied byStudied by 70 people
... ago
5.0(1)
robot