Polar Equations and Graphs

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    When trying to figure out the graphs of polar equations we

    can convert them to rectangular equations particularly if

    we recognize the graph in rectangular coordinates.

    7!r We could square both sides

    492 !r

    Now use our conversion:

    222yxr !

    4922 ! yx

    We recognize this as a circle

    with center at (0, 0) and a

    radius of 7.

    On polar graph paper it will centered at the origin and out 7

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    Let's try another:

    3

    TU ! Take the tangent of both sides

    !

    3tantan TU

    Now use our conversion:

    3!x

    y

    We recognize this as a line with slope square root of 3.

    3!

    x

    y!Utan

    Multiply both

    sides byx

    xy 3!

    To graph on a polar plot

    we'd go to whereand make a line. 3

    TU !

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    Let's try another: 5sin !r

    Now use our conversion:

    We recognize this as a

    horizontal line 5 units belowthe origin (or on a polar plot

    below the pole)

    Usinry !

    5!y

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    Sometimes converting to rectangular equations doesn't

    help us figure out what the graph would look like or it is

    not necessary.

    Usin1!rThe only way we know how toconvert is if there is an rin front of

    the sin U term so we'll multiply both

    sides by r.Usin2 rrr !

    Now use our conversions:

    222yxr ! 22 yxr ! Usinry !

    yyxyx ! 2222 I still don't

    know what the

    graph looks

    like!In these cases we'll plot points, choosing

    a U from the polar form and finding a

    corresponding rvalue.

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    Before we do this, if we knew something about the

    symmetry of the graph we may not have to find as many

    points. TESTS FOR SYMMETRY

    Symmetry with

    Respect to the Polar

    Axis (xaxis)

    Replace U by - U

    and if you get

    original equation

    back

    Symmetry with

    Respect to the Line

    U = T/2 (yaxis)

    Replace U by T - U

    and if you get

    original equation

    back

    (r,U)

    (r, -U)

    Symmetry with

    Respect to the Pole

    (Origin)

    Replace rby - r

    and if you get

    original equation

    back

    (r,U)(r,T-U)

    (r,U)

    (- r,U)

    These tests are sufficient but not necessary so if test fails you don't know anything.

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    Usin1!rLet's test for symmetry

    Polar Axis: U! sin1r Usin1!rFAILS

    Line U = T/2: UT! sin1r

    UTUT sincoscossin1 !r

    Use the difference formula

    0 -1

    Usin1!r

    This IS the original equation!

    Pole: Usin1! r

    Not the original equation.

    So this graph is symmetric

    with respect to the line U = T/2

    (yaxis). We will only need tochoose U's on the right side of

    the graph then and we can use

    symmetry to get the other half.

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    U Usin1!r

    2

    T 211 !

    3

    T 87.1

    2

    31 }

    Let's let each unit be 1/4.

    6

    T

    5.1211 !

    0 101 !

    6

    T

    2

    1

    2

    11 !

    3

    T

    13.02

    31 }

    2

    T

    011 !Let's plot the symmetric points

    This type of graph is called a

    cardioid.

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    Equations of carioids would look like one of the following:

    r= a(1 + cosU) r= a(1 + sinU)

    r= a(1 - cosU) r= a(1 - sinU)

    wherea > 0

    All graphs of cardioids pass through the pole.

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    Let's test for symmetry

    Polar Axis: U! cos23r

    YES!

    Line U = T/2: UT! cos23r

    UTUT sinsincoscos23 !r

    Use the difference formula

    -1 0

    Ucos23!r

    Not the original equation

    Pole: Ucos23! r

    Not the original equation.

    So this graph is symmetric

    with respect to the polar axis

    (xaxis). We will only need to

    choose U's on the top half of

    the graph then and we can use

    symmetry to get the other half.

    Let's try another: Ucos23!r

    Ucos23!r

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    Equations of limacons without inner loops would look like

    one of the following:

    r= a +b cosU r= a +b sinU

    r= a - b cosU r= a - b sinU

    wherea > 0, b > 0, and a >b

    These graphs DO NOT pass through the pole.

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    Let's test for symmetry

    Polar Axis: U! cos21r

    YES!

    Line U = T/2: UT ! cos21r

    UTUT sinsincoscos21 !r

    Use the difference formula

    -1 0

    Ucos21!r

    Not the original equation

    Pole: Ucos21! r

    Not the original equation.

    So this graph is symmetric

    with respect to the polar axis

    (xaxis). We will only need to

    choose U's on the top half of

    the graph then and we can use

    symmetry to get the other half.

    Let's try another: Ucos21!r

    Ucos21!r

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    U

    0 3121 !

    6

    T

    73.22

    321 }

    Let's let each unit be 1/2.

    3

    T

    2

    2

    121 !

    2

    T 1021 !

    3

    2T0

    2

    121 !

    6

    5T

    73.02

    321 }

    T 1121 !

    Let's plot the symmetric points

    This type of graph is called a

    limacon with an innerloop.Ucos21!r

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    Equations of limacons with inner loops would look like

    one of the following:

    r= a +b cosU r= a +b sinU

    r= a - b cosU r= a - b sinU

    wherea > 0, b > 0, and a < b

    These graphs will pass through the pole twice.

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    Let's test for symmetry

    Polar Axis: U! 2cos2rYES!

    Line U = T/2: UT! 2cos2r

    UU 2cos22cos2 !!rYES!

    Pole:

    Since graph is symmetric

    to both the polar axis and

    the line U =T/2 it will also

    be with respect to the pole.

    So this graph is symmetric

    with respect to the pole, the

    polar axis and the line U =T/2.

    We will only need to choose

    U's between 0 and T/2.

    Let's try another: U2cos2!r

    U2cos2!r

    UT 22cos2 !

    cos is periodic so can drop the 2T

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    U

    0 212 !

    6

    T1

    2

    12 !

    Let's let each unit be 1/2.

    4

    T

    002 !

    3

    T1

    2

    12 !

    2

    T

    212 !

    Let's plot the symmetric points

    This type of graph is called a rose

    with 4 petals. U2cos2!r

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    Equations of rose curves would look like one of the

    following:

    r= a cos(nU) r= a sin(nU)

    Wheren even has 2n petals and

    n odd has n petals

    (n {

    0 ors

    1)

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    Let's test for symmetry

    Polar Axis: U! 2sin42rFAILS

    Line U = T/2: UT! 2sin42r

    UU 2sin22sin42 !!r FAILS

    Pole:

    So this graph is symmetric

    with respect to the pole.

    Let's try another: U2sin42 !r

    U2sin42 !r

    UT 22sin4 !

    sin is periodic so can drop the 2T

    U2sin42

    ! r

    U2sin42 !r

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    U

    0 004 !

    6

    T

    322

    34 !

    Let's let each unit be 1/4.

    4

    T

    4

    14 !

    3

    T32

    2

    34 !

    2

    T

    004 !

    This type of graph is called a

    lemniscate U2sin42 !r r

    0

    9.1s

    2s

    9.1s

    0

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    Equations of lemniscates would look like one of the

    following:

    r2 = a2 cos(2U) r2 = a2 sin(2U)

    These graphs will pass through the pole and are

    propeller shaped.

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    Table 7 in your book on page 732

    summarizes all of the polar graphs.

    You can graph these on your calculator. You'll need tochange to polar mode and also you must be in radians.

    If you are in polar function mode when you hit your

    button to enter a graph you should see r1 instead ofy1.

    Your variable button should now put in U on TI-83's and

    it should be a menu choice in 85's & 86's.

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    Have fun plotting pretty pictures!

    Limacon With InnerLoop

    made with TI Calculator

    Rose with 7 petals made

    with graphing program on

    computer

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    Acknowledgement

    I wish to thank Shawna Haider from Salt Lake Community College, Utah

    USA for her hard work in creating this PowerPoint.

    www.slcc.edu

    Shawna has kindly given permission for this resource to be downloadedfrom www.mathxtc.com and for it to be modified to suit the Western

    Australian Mathematics Curriculum.

    Stephen Corcoran

    Head of Mathematics

    St Stephens School Carramar

    www.ststephens.wa.edu.au