11.4 – Graphing in Polar Coordinates

11.4 – Graphing in Polar CoordinatesPolar Symmetries

(𝒓 ,𝜽 )=(𝒓 ,− 𝜽 ) (𝒓 ,𝜽 )=(𝒓 ,𝝅−𝜽 ) (𝒓 ,𝜽 )=(𝒓 ,𝝅+𝜽 )𝒙−𝒂𝒙𝒊𝒔 𝒔𝒚𝒎𝒎𝒆𝒕𝒓𝒚 𝒚 −𝒂𝒙𝒊𝒔 𝒔𝒚𝒎𝒎𝒆𝒕𝒓𝒚 𝒐𝒓𝒊𝒈𝒊𝒏 𝒔𝒚𝒎𝒎𝒆𝒕𝒓𝒚



𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒔𝒚𝒎𝒎𝒆𝒕𝒓𝒊𝒆𝒔 𝒓=𝒄𝒐𝒔 𝜽−𝟒𝒓 (𝝅𝟑 )=−𝟑 .𝟓𝒓 (− 𝝅𝟑 )=−𝟑 .𝟓𝒙−𝒂𝒙𝒊𝒔 𝒔𝒚𝒎 .

𝒓 (𝝅𝟑 )=−𝟑 .𝟓𝒓 (𝝅− 𝝅𝟑 )=−𝟒 .𝟓𝒏𝒐𝒚 −𝒂𝒙𝒊𝒔 𝒔𝒚𝒎 .

𝒓 ( 𝝅𝟑 )=−𝟑 .𝟓𝒓 (𝝅+𝝅𝟑 )=−𝟒 .𝟓𝒏𝒐𝒐𝒓𝒊𝒈𝒊𝒏𝒔𝒚𝒎 .



𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒔𝒚𝒎𝒎𝒆𝒕𝒓𝒊𝒆𝒔 𝒓=−𝟔𝒔𝒊𝒏𝜽

𝒓 (𝝅𝟔 )=−𝟏𝟐𝒓 (− 𝝅𝟔 )=𝟏𝟐𝒏𝒐𝒙 −𝒂𝒙𝒊𝒔 𝒔𝒚𝒎 .

𝒓 (𝝅𝟔 )=−𝟏𝟐𝒓 (𝝅− 𝝅𝟔 )=−𝟏𝟐𝒚 −𝒂𝒙𝒊𝒔 𝒔𝒚𝒎 .

𝒓 (𝝅𝟔 )=−𝟏𝟐𝒓 (𝝅+𝝅𝟔 )=𝟏𝟐𝒏𝒐𝒐𝒓𝒊𝒈𝒊𝒏𝒔𝒚𝒎 .

11.4 – Graphing in Polar CoordinatesSlope of a Tangent for

𝒙=𝒓𝒄𝒐𝒔 𝜽𝒎=

𝒅𝒚𝒅𝒙

𝒚=𝒓𝒔𝒊𝒏𝜽𝒅𝒚𝒅𝒙 =

𝒅𝒚𝒅 𝜽𝒅𝒙𝒅 𝜽

𝒙= 𝒇 (𝜽 )𝒄𝒐𝒔𝜽 𝒚=𝒇 (𝜽 )𝒔𝒊𝒏𝜽

𝒅𝒚𝒅𝜽= 𝒇 ′ (𝜽 ) 𝒔𝒊𝒏𝜽+ 𝒇 (𝜽 )𝒄𝒐𝒔 𝜽 𝒅𝒙

𝒅𝜽= 𝒇 ′ (𝜽 )𝒄𝒐𝒔 𝜽− 𝒇 (𝜽 ) 𝒔𝒊𝒏𝜽

𝒅𝒚𝒅𝒙 =

𝒅𝒚𝒅 𝜽𝒅𝒙𝒅 𝜽

=𝒇 ′ (𝜽 )𝒔𝒊𝒏𝜽+ 𝒇 (𝜽 )𝒄𝒐𝒔𝜽 𝒇 ′ (𝜽 )𝒄𝒐𝒔 𝜽− 𝒇 (𝜽 ) 𝒔𝒊𝒏𝜽

𝒅𝒚𝒅𝒙 =

𝒓 ′ 𝒔𝒊𝒏𝜽+𝒓 𝒄𝒐𝒔 𝜽 𝒓 ′𝒄𝒐𝒔𝜽−𝒓 𝒔𝒊𝒏𝜽

𝒅𝒚𝒅𝒙 =

𝒇 ′ (𝜽 ) 𝒔𝒊𝒏𝜽+ 𝒇 (𝜽 )𝒄𝒐𝒔 𝜽 𝒇 ′ (𝜽 )𝒄𝒐𝒔 𝜽− 𝒇 (𝜽 )𝒔𝒊𝒏𝜽


𝒅𝒚𝒅𝒙 =


When passes through the pole, and the slope of the tangent is:𝒅𝒚𝒅𝒙 =


=𝒇 ′ (𝜽 )𝒔𝒊𝒏𝜽𝒇 ′ (𝜽 )𝒄𝒐𝒔 𝜽

= 𝒔𝒊𝒏𝜽𝒄𝒐𝒔𝜽=𝒕𝒂𝒏𝜽

𝒅𝒚𝒅𝒙 =



𝒅𝒚𝒅𝒙 =


Given the polar equation, find the polar coordinates and the slope of the tangent at

𝒅𝒚𝒅𝒙 =


𝒓=𝒔𝒊𝒏(𝟐𝜽 )

𝒓=𝒔𝒊𝒏(𝟐(𝝅𝟒 ))𝒓=𝟏(𝟏 , 𝝅𝟒 )

𝒅𝒚𝒅𝒙 =

(𝟎)𝒔𝒊𝒏(𝝅𝟒 )+(𝟏)𝒄𝒐𝒔 (𝝅𝟒 ) (𝟎)𝒄𝒐𝒔 ( 𝝅𝟒 )−(𝟏)𝒔𝒊𝒏( 𝝅𝟒 )

𝒓 ′=𝟐𝒄𝒐𝒔 (𝟐𝜽)

𝒓 ′=𝟐𝒄𝒐𝒔 (𝟐(𝝅𝟒 ))𝒓 ′=𝟎

𝒅𝒚𝒅𝒙 =

√𝟐𝟐

− √𝟐𝟐

=−𝟏

At

11.4 – Graphing in Polar CoordinatesPolar Equations

Examples:

𝒓=𝒂𝜽=𝜽𝟎

𝒓=𝟑

Basic Equations:

𝜽=𝝅𝟔

A line through the pole at the angle .

A circle with its center at the pole.

http://www.researchgate.net/post/How_to_use_bilinear_interpolation_in_grids_circle_with_angles_and_grids


Example:𝜽=𝟎

𝒓=𝟒 𝒔𝒊𝒏𝜽

Plotting Polar Curves

𝜽=𝝅𝟔

𝜽=𝝅𝟑

𝒓=𝟎 𝒓=𝟐

𝒓=𝟑 .𝟒𝟔

(𝒓 ,𝜽)

𝜽=𝝅𝟐𝒓=𝟒



Example: 𝒓=𝟐

𝟏−𝒄𝒐𝒔𝜽

Plotting Polar Curves

(𝒓 ,𝜽)

𝜽≠𝟎 𝜽=𝝅𝟔

𝜽=𝝅𝟐

𝒓=𝒖𝒏𝒅 . 𝒓=𝟏𝟒 .𝟗𝟑

𝒓=𝟐 𝜽=𝝅𝒓=𝟏

𝜽=𝝅𝟑𝒓=𝟒


Special Polar Curves𝒓=𝒂+𝒃𝒄𝒐𝒔 𝜽

Limacon w/ a dimple

𝒂>𝒃𝒓=𝒂+𝒃𝒄𝒐𝒔 𝜽𝒂=𝒃

Cardioid Limacon

w/ a loop

𝒓=𝒂+𝒃𝒄𝒐𝒔 𝜽𝒂<𝒃

𝒓=𝒃𝒄𝒐𝒔 𝜽

Circle

11.4 – Graphing in Polar Coordinates

Special Polar Curves𝒓=𝒂𝜽

Spiral

out

𝒓=𝒂𝒄𝒐𝒔 (𝒏𝜽)

Lemniscate

Roses

Parabola

𝒓=𝒃

𝒂𝜽

Spiral in

𝒓𝟐=𝒂𝒄𝒐𝒔 (𝟐𝜽)

11.4 – Graphing in Polar Coordinates

𝒓=𝟐

𝒂−𝒂𝒄𝒐𝒔 𝜽

Area of a Polar Curve

𝑐𝑖𝑟𝑐𝑙𝑒 𝑎𝑟𝑒𝑎=𝜋 𝑟2

11.5 – Area and Lengths in Polar Coordinates

𝑎𝑟𝑒𝑎𝑜𝑓 𝑤𝑒𝑑𝑔𝑒=𝜋𝑟2( 𝜃2𝜋 )

𝑎𝑟𝑒𝑎𝑜𝑓 𝑤𝑒𝑑𝑔𝑒=12 𝑟

2𝜃

r

𝑎𝑟𝑒𝑎𝑜𝑓 𝑎𝑝𝑜𝑙𝑎𝑟 𝑐𝑢𝑟𝑣𝑒=∫𝛼

𝛽 12𝑟 2𝑑𝜃 𝑎𝑟𝑒𝑎𝑜𝑓 𝑎𝑝𝑜𝑙𝑎𝑟 𝑐𝑢𝑟𝑣𝑒=∫

𝛼

𝛽 12𝑓 (𝜃)2𝑑𝜃

Area of a Polar Curve11.5 – Area and Lengths in Polar Coordinates

𝐴=18.850

𝑟=2+2𝑐𝑜𝑠𝜃 𝑎𝑟𝑒𝑎𝑜𝑓 𝑎𝑝𝑜𝑙𝑎𝑟 𝑐𝑢𝑟𝑣𝑒=∫𝛼

𝛽 12𝑟 2𝑑𝜃

𝐴=∫0

2𝜋 12(2+2𝑐𝑜𝑠𝜃)2𝑑𝜃

𝐴=∫0

2𝜋 12(2(1+𝑐𝑜𝑠 𝜃))2𝑑𝜃

𝐴=∫0

2𝜋

2(1+𝑐𝑜𝑠𝜃)2𝑑𝜃

𝐴=2∫0

𝜋 12(2+2𝑐𝑜𝑠𝜃)2𝑑𝜃

𝐴=18.850

0≤ 𝜃≤2𝜋Cardioid


𝐴=19.635

𝑟=5cos (3 𝜃)

𝐴=∫𝛼


𝐴=∫0

𝜋 12(5cos (3𝜃))2𝑑𝜃

𝐴=∫0

𝜋 12(5𝑐𝑜𝑠(3𝜃))2𝑑𝜃

Area of a one petalArea of entire region

𝐴=19.6353

𝐴=6.545

0≤ 𝜃≤𝜋Rose w/3 petals


𝐴=6.545

𝑟=5cos (3 𝜃)

𝐴=∫𝛼


𝐴=2∫0

𝜋612(5cos (3 𝜃))2𝑑𝜃

𝐴=∫𝜋6

𝜋212(5𝑐𝑜𝑠(3𝜃))2𝑑𝜃

Area of a one petal𝑟=5cos (3 𝜃)

0=5cos (3𝜃)

3 𝜃=𝜋2 ,3𝜋2 , 5𝜋2 , 7𝜋2 ,

𝜃=𝜋6 ,

𝜋2 ,5𝜋6 , 7𝜋6 ,

𝐴=∫𝜋2

5 𝜋612(5𝑐𝑜𝑠(3 𝜃))2𝑑𝜃

0≤ 𝜃≤𝜋Rose w/3 petals


𝐴=0.381

𝐴=∫𝛼


𝐴=2 ∫0.848

𝜋212(3−4 𝑠𝑖𝑛𝜃)2𝑑𝜃

𝐴= ∫0.848

2.294 12(3−4 𝑠𝑖𝑛𝜃)2𝑑𝜃

Area of the inner loop

𝑟=3−4sin 𝜃0=3−4 𝑠𝑖𝑛𝜃

𝜃=𝜋 −0.848𝜃=0.848

0≤ 𝜃≤2𝜋

𝑟=3−4sin θ

Limacon w/ a loop

𝜃=2.294



𝐴=7.653

𝐴=∫𝛼

𝛽 12 (𝑟22−𝑟12)𝑑𝜃

𝐴= ∫0.524

2.618 12

((4 𝑠𝑖𝑛𝜃 )2−22 )𝑑𝜃

Area of the region outside r = 2 and inside r = 4sin

4 𝑠𝑖𝑛𝜃=2𝑠𝑖𝑛𝜃=

12

𝜃=0.524

𝜃=𝜋6 ,5𝜋6

𝑟=2Need Pts. Of Intersection

𝜃=2.618

𝑟=4sin θ


𝐴=2.657

𝐴=∫𝛼

𝛽 12 (𝑟22−𝑟12)𝑑𝜃

𝐴= ∫0

0.785 12

( (2+2𝑐𝑜𝑠𝜃 )2− (2+2𝑠𝑖𝑛𝜃 )2 )𝑑𝜃

Calculate the area of the region outside , inside , and in the first quadrant.

2+2 𝑠𝑖𝑛𝜃=2+2𝑐𝑜𝑠𝜃2𝑠𝑖𝑛𝜃=2𝑐𝑜𝑠𝜃

𝜃=𝜋4 =0.785

𝑡𝑎𝑛𝜃=1

Points Of Intersection

11.5 – Area and Lengths in Polar CoordinatesArc Length of a Polar Curve


𝐿=∫𝛼

𝛽

√𝑟 2+( 𝑑𝑟𝑑𝜃 )2

𝑑𝜃 𝑟=5sin θ0≤ 𝜃≤𝜋

𝑑𝑟𝑑𝜃=5 cosθ

𝐿=∫0

𝜋

√(5 𝑠𝑖𝑛𝜃 )2+(5𝑐𝑜𝑠𝜃 )2𝑑𝜃

𝐿=∫0

𝜋

√25 𝑠𝑖𝑛2𝜃+25𝑐𝑜𝑠2𝜃𝑑𝜃

𝐿=∫0

𝜋

5𝑑𝜃

𝐿=5 𝜃|𝜋0

𝐿=5𝜋=15.708

Find the arc length of the polar curve between the given angle interval.

𝐿=∫0

𝜋

√25 (𝑠𝑖𝑛¿¿ 2𝜃+𝑐𝑜𝑠2𝜃)𝑑𝜃 ¿


𝐿=∫𝛼

𝛽

√𝑟 2+( 𝑑𝑟𝑑𝜃 )2

𝑑𝜃 𝑟=𝑒𝜃

0≤ 𝜃≤𝜋

𝑑𝑟𝑑𝜃=𝑒𝜃

𝐿=∫0

𝜋

√(𝑒𝜃 )2+ (𝑒𝜃 )2𝑑𝜃

𝐿=∫0

𝜋

√𝑒2𝜃+𝑒2𝜃𝑑𝜃

𝐿=√2𝑒𝜋−√2𝑒0

𝐿=√2𝑒𝜃|𝜋0

𝐿=31.312

𝐿=∫0

𝜋

√2𝑒2𝜃 𝑑𝜃

𝐿=√2∫0

𝜋

𝑒𝜃 𝑑𝜃

Find the arc length of the polar curve between the given angle interval.


𝐿=∫𝛼

𝛽

√𝑟 2+( 𝑑𝑟𝑑𝜃 )2

𝑑𝜃 𝑟=2𝑐𝑜𝑠θ+1

Find the arc length of half of the inside loop

𝑑𝑟𝑑𝜃=−2sin θ𝐿=∫

2 𝜋 3

𝜋

√ (2𝑐𝑜𝑠𝜃+1 )2+ (−2𝑠𝑖𝑛𝜃 )2𝑑𝜃

𝐿=∫2 𝜋 3

𝜋

√4 𝑐𝑜𝑠2𝜃+4𝑐𝑜𝑠 𝜃+1+4 𝑠𝑖𝑛2𝜃 𝑑𝜃

𝐿=1.34 1

𝜃=0𝑟=2𝑐𝑜𝑠0+1=3𝜃=𝜋𝑟=2𝑐𝑜𝑠𝜋+1=−10=2𝑐𝑜𝑠θ+1

𝜃=2𝜋 3 , 4𝜋 3

𝐿=∫2 𝜋 3

𝜋

√4 𝑐𝑜𝑠𝜃+4𝑐𝑜𝑠2𝜃+4 𝑠𝑖𝑛2𝜃+1𝑑𝜃

𝐿=∫2 𝜋 3

𝜋

√4 𝑐𝑜𝑠𝜃+4+1𝑑𝜃

𝐿=∫2 𝜋 3

𝜋

√4 𝑐𝑜𝑠𝜃+5𝑑𝜃


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11.4 – Graphing in Polar Coordinates