1

POLAR COORDINATES

Polar coordinate system: a pole (fixed point) and a polar axis (directed ray with endpoint at pole).

rectangular coordinates ⇒ polar coordinates polar coordinates ⇒ rectangular coordinates

𝑟 = √𝑥2 + 𝑦2, 𝜃 = 𝑎𝑟𝑐 𝑡𝑎𝑛𝑦

𝑥 𝑥 = 𝑟 𝑐𝑜𝑠 𝜃 𝑦 = 𝑟 𝑠𝑖𝑛 𝜃

The angle, θ, is measured from the polar axis to a line that passes through the point and the pole. If the angle is measured in a counterclockwise direction, the angle is positive. If the angle is measured in a clockwise direction, the angle is negative. The directed distance, r, is measured from the pole to point P. If point P is on the terminal side of angle θ, then the value of r is positive. If point P is on the opposite side of the pole, then the value of r is negative.

Problem : P (x, y) = (1, 3). Express it in polar coordinates (r, θ) two different ways such that 0≤ θ < 2

(r, θ) = (2, /3), (- 2, 4/3) .

Problem : P(x, y) = (-4, 0). Express it in polar coordinates (r, θ) two different ways such that 0≤ θ < 2.

(r, θ) = (4, ),(- 4, 0) .

Problem : P (x, y) = (-7, -7), express it in polar coordinates (r, θ) two different ways such that 0≤ θ < 2.

(r, θ) = (98, 5/4),(- 98, /4) .

Problem : Given a point in polar coordinates (r, θ) = (3, /4), express it in rectangular coordinates (x, y) . (x, y) = (3√2/2, 3√2/2) .

Problem : How many different ways can a point be expressed in polar coordinates such that r > 0 ?

An infinite number. (r, θ) = (r, θ +2n) , where n is an integer.

Problem : Transform the equation x2 + y2 + 5x = 0 to polar coordinate form.

x2 + y2 + 5x = 0 r2 + 5(r cos θ) = 0

r ( r + 5 cos θ) = 0 The equation r = 0 is the pole. Thus, keep only the other equation: r + 5 cos θ = 0

The location of a point can be named using many different pairs of polar coordinates. ← three different sets of polar coordinates for the point P (5, 60°).

The distance r and the angle 𝜃 are both directed--meaning that they represent the

distance and angle in a given direction. It is possible, therefore to have negative values for

both r and 𝜃. However, we typically avoid points with negative r , since they could just as

easily be specified by adding to 𝜃 .

2

Problem : Transform the equation r = 4sin θ to Cartesian coordinate form. What is the graph? Describe it fully!!!

√𝑥2 + 𝑦2 = 4𝑦

√𝑥2+𝑦2

𝑥2 + 𝑦2 = 4 𝑦 𝑥2 + (𝑦 − 2)2 = 22 circle: r = 2 C(0, 2)

Problem : What is the maximum value of | r| for the following polar equations:

a) r = cos(2 θ) ;

b) r = 3 + sin(θ) ;

c) r = 2 cos(θ) - 1 .

a) The maximum value of | r| in r = cos(2 θ) occurs when θ = n/2 where n is an integer and | r| = 1 .

b) The maximum value of | r| in r = 3 + sin(θ) occurs when θ = /2+2n where n is an integer and | r| = 4 .

c) The maximum value of | r| in r = 2 cos(θ) - 1 occurs when θ = (2n + 1) where n is an integer and | r| = 3 .

Problem : Find the intercepts and zeros of the following polar equations: a) r = cos(θ) + 1 ; b) r = 4 sin(θ) .

a) Polar axis intercepts: (r, θ) = (2, 2n),(0, (2n + 1)) , where n is an integer.

Line θ = /2 intercepts: (r, θ) = (1, /2 + n) , where n is an integer. r = cos(θ) + 1 = 0 for θ = (2n + 1) , where n is an integer.

b) Polar axis intercepts: (r, θ) = (0, n) where n is an integer. Line θ = /2 intercepts: (r, θ) = (4, /2 +2n) where n is an integer.

r = 4 sin(θ) = 0 for θ = n, where n is an integer.

Problem : Sketch

Spiral of Archimedes: r = θ, θ ≥ 0

The curve is a nonending spiral.

Here it is shown in detail from θ = 0 to θ = 2π

Problem : Sketch Lima¸cons (Snail): 𝑟 = 1 − cos 𝜃

θ 0 π/4 π/3 π/2 2 π/3 3 π/4 π 5 π/4 4 π/3 3 π/2 5 π/3 7 π/4 2 π

r – 1 –0.41 0 1 2 2.41 3 2.41 2 1 0 –0.41 –1

3

Problem : Sketch Lima¸cons (Snail): 𝑟 = 𝑎 + 𝑏 cos 𝜃

The general shape of the curve depends on the relative magnitudes of |a| and |b|.

𝑟 = 3 + cos 𝜃 𝑟 =3

2+ cos 𝜃 𝑟 = 1 + cos 𝜃 𝑟 =

1

2+ cos 𝜃

convex limacon limacon with a dimple carotid limacon with an inner loop

Problem : Sketch Cardioids (Heart-Shaped): r = 1 ± cosθ , r = 1 ± sinθ

𝑟 = 1 + cos 𝜃 𝑟 = 1 + sin 𝜃 𝑟 = 1 − cos 𝜃 𝑟 = 1 − sin 𝜃

Flowers

Problem : Sketch Petal Curve: r = cos 2 θ

Problem : Sketch Petal Curves: r = a cos n θ, r = a sin n θ

r = sin 3θ r = cos 4 θ

• If n is odd, there are n petals. • If n is even, there are 2n petals.

4

First and second derivative r = r():

𝑻𝒂𝒏𝒈𝒆𝒏𝒕 𝒍𝒊𝒏𝒆 𝑎𝑡 𝑝𝑜𝑖𝑛𝑡 𝜽 = 𝜽𝟎 𝑜𝑓 𝑎 𝑐𝑢𝑟𝑣𝑒 𝒓 = 𝒓(𝜽): 𝑦 = 𝑚 (𝑥 – 𝑥0) + 𝑦0

𝑟0 = 𝑟(𝜃0) ⟹𝑥0 = 𝑟0𝑐𝑜𝑠 𝜃0

𝑦0 = 𝑟0𝑠𝑖𝑛 𝜃0 𝑎𝑛𝑑 𝑚 =

𝑑𝑦

𝑑𝑥 𝑎𝑡 𝑟0 = 𝑟(𝜃0)

𝑵𝒐𝒓𝒎𝒂𝒍 𝒍𝒊𝒏𝒆: 𝑦 = − 1

𝑚 (𝑥 – 𝑥0) + 𝑦0

𝑯𝒐𝒓𝒊𝒛𝒐𝒏𝒕𝒂𝒍 𝒕𝒂𝒏𝒈𝒆𝒏𝒕 𝒍𝒊𝒏𝒆: 𝑤𝑖𝑙𝑙 𝑜𝑐𝑐𝑢𝑟 𝑤ℎ𝑒𝑟𝑒 𝑡ℎ𝑒 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 𝑖𝑠 𝑧𝑒𝑟𝑜 ⟹ 𝑑𝑦

𝑑𝑥= 0

𝑑𝑦

𝑑𝑥= 0 ⟹

𝑑𝑦

𝑑𝜃= 0 ⟹ 𝜃0 (𝑐ℎ𝑒𝑐𝑘

𝑑𝑥

𝑑𝑡|

𝜃0

≠ 0) ⟹ 𝑟0 = 𝑟(𝜃0) ⟹ 𝑦0

= 𝑟0 𝑠𝑖𝑛𝜃0 𝑒𝑞: 𝑦 = 𝑦0

𝑽𝒆𝒓𝒕𝒊𝒄𝒂𝒍 𝒕𝒂𝒏𝒈𝒆𝒏𝒕 𝒍𝒊𝒏𝒆: 𝑤𝑖𝑙𝑙 𝑜𝑐𝑐𝑢𝑟 𝑤ℎ𝑒𝑟𝑒 𝑡ℎ𝑒 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 𝑖𝑠 𝑛𝑜𝑡 𝑑𝑒𝑓𝑖𝑛𝑒𝑑: 𝑑𝑦

𝑑𝑥= ∞

𝑑𝑦

𝑑𝑥= ∞ ⟹

𝑑𝑥

𝑑𝜃= 0 ⟹ 𝜃0 (𝑐ℎ𝑒𝑐𝑘

𝑑𝑦

𝑑𝑡|

𝜃0

≠ 0) ⟹ 𝑟0 = 𝑟(𝜃0) ⟹ 𝑥0 = 𝑟0 𝑐𝑜𝑠𝜃0 𝑒𝑞: 𝑥 = 𝑥0

𝑪𝒐𝒏𝒄𝒂𝒗𝒊𝒕𝒚 𝒂𝒕 𝒑𝒐𝒊𝒏𝒕 (𝒙𝟏, 𝒚𝟏) 𝑜𝑟 𝒕𝟏 𝑜𝑟 𝜽𝟏 ∶ 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑒 𝑑2𝑦

𝑑𝑥2 𝑎𝑡 𝑡ℎ𝑎𝑡 𝑝𝑜𝑖𝑛𝑡.

𝐼𝑓 𝑑2𝑦

𝑑𝑥2< 0 → 𝑐𝑢𝑟𝑣𝑒 𝑖𝑠 𝑐𝑜𝑛𝑐𝑎𝑣𝑒 𝑑𝑜𝑤𝑛. 𝐼𝑓

𝑑2𝑦

𝑑𝑥2> 0 → 𝑐𝑢𝑟𝑣𝑒 𝑖𝑠 𝑐𝑜𝑛𝑐𝑎𝑣𝑒 𝑢𝑝

𝑥 = 𝑟𝑐𝑜𝑠𝜃 ⟹ 𝑑𝑥

𝑑𝜃=

𝑑𝑟

𝑑𝜃 𝑐𝑜𝑠 𝜃 − 𝑟 sin 𝜃

𝑦 = 𝑟𝑠𝑖𝑛𝜃 ⟹ 𝑑𝑦

𝑑𝜃=

𝑑𝑟

𝑑𝜃 𝑠𝑖𝑛 𝜃 + 𝑟𝑐𝑜𝑠 𝜃

𝒅𝒚

𝒅𝒙=

𝑑𝑦𝑑𝜃𝑑𝑥𝑑𝜃

=

𝑑𝑟𝑑𝜃

𝑠𝑖𝑛 𝜃 + 𝑟𝑐𝑜𝑠 𝜃

𝑑𝑟𝑑𝜃

𝑐𝑜𝑠 𝜃 − 𝑟 sin 𝜃

𝑑2𝑦

𝑑𝑥2=

𝑑

𝑑𝑥[𝑑𝑦

𝑑𝑥] =

𝑑𝑑𝑡

[𝑑𝑦𝑑𝑥

]

𝑑𝑥𝑑𝑡

𝒅𝟐𝒚

𝒅𝟐𝒙=

𝑑

𝑑𝑥 (

𝑑𝑦

𝑑𝑥) =

1

𝑑𝑥𝑑𝜃

𝑑

𝑑𝜃 (

𝑑𝑟𝑑𝜃

𝑠𝑖𝑛 𝜃 + 𝑟𝑐𝑜𝑠 𝜃

𝑑𝑟𝑑𝜃

𝑐𝑜𝑠 𝜃 − 𝑟 sin 𝜃 )

and now good luck

Note that rather than trying to remember this formula it would probably be easier to

remember how we derived it .

5

Area enclosed by a polar curve r = r():

For a very small 𝜃 (𝑑𝜃), the curve could be approximated by a straight line and the area could be found using the triangle formula:

𝑑𝐴 =1

2(𝑟 𝑑𝜃)𝑟 =

1

2𝑟2𝑑𝜃

1 ≤ ≤ 2

𝐴 = ∫ 𝑑𝐴 =1

2 ∫ 𝑟2 𝑑𝜃

𝜃2

𝜃1

𝜃2

𝜃1

Example: Find the area enclosed by:

example: Find the area of the inner loop of r = 2 + 4 cos θ

𝑟 = 2(1 + 𝑐𝑜𝑠𝜃)

6

7

8

Length of a Polar Curve:

𝑆 = ∫ 𝑑𝑠 = ∫ √(𝑑𝑥)2 + (𝑑𝑦)2 =𝑏

𝑎

𝑏

𝑎

∫ √ (𝑑𝑥

𝑑𝜃)

2

+ (𝑑𝑦

𝑑𝜃)

2

𝑑𝜃 𝜃2

𝜃1

𝜃1 𝑎𝑛𝑑 𝜃2 𝑎𝑟𝑒 𝑎𝑛𝑔𝑙𝑒𝑠 𝑐𝑜𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑡𝑜 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛𝑠 𝑎 𝑎𝑛𝑑 𝑏

𝑥 = 𝑟𝑐𝑜𝑠𝜃 ⟹ 𝑑𝑥

𝑑𝜃=

𝑑𝑟

𝑑𝜃 𝑐𝑜𝑠 𝜃 − 𝑟 sin 𝜃 𝑎𝑛𝑑 𝑦 = 𝑟𝑠𝑖𝑛𝜃 ⟹

𝑑𝑦

𝑑𝜃=

𝑑𝑟

𝑑𝜃 𝑠𝑖𝑛 𝜃 + 𝑟𝑐𝑜𝑠 𝜃

(𝑑𝑥

𝑑𝜃)

2

+ (𝑑𝑦

𝑑𝜃)

2

= (𝑑𝑟

𝑑𝜃)

2

𝑐𝑜𝑠2𝜃 − 2𝑟𝑑𝑟

𝑑𝜃𝑐𝑜𝑠𝜃 𝑠𝑖𝑛𝜃 + 𝑟2𝑠𝑖𝑛2𝜃 + (

𝑑𝑟

𝑑𝜃)

2

𝑠𝑖𝑛2𝜃 + 2𝑟𝑑𝑟

𝑑𝜃𝑐𝑜𝑠𝜃 𝑠𝑖𝑛𝜃 + 𝑟2𝑐𝑜𝑠2𝜃 = (

𝑑𝑟

𝑑𝜃)

2

+ 𝑟2

𝑆 = ∫ 𝑑𝑠 = ∫ √(𝑑𝑥)2 + (𝑑𝑦)2 = ∫ √ 𝑟2 + (𝑑𝑟

𝑑𝜃)

2

𝑑𝜃 𝜃2

𝜃1

𝑏

𝑎

𝑏

𝑎

9

10

EXAMPLE: limaçon: r = 0.5 + cos θ

1. Find the area of the inner circle. 2. Find all vertical and horizontal tangents. 3. Find the points with two tangent lines. Find tangents.

= 0.375 (4/3 − 2/3) + 0.5(𝑠𝑖𝑛 4/3 – 𝑠𝑖𝑛 2/3) + 0.125 (𝑠𝑖𝑛 8/3 – 𝑠𝑖𝑛 4/3)

= 0.25 − 0.5 3 + 0.125 3 𝑨 = 𝟎. 𝟐𝟓 − 𝟎. 𝟑𝟕𝟓 √𝟑

2. 𝑑𝑦

𝑑𝑥=

𝑑𝑦𝑑𝜃𝑑𝑥𝑑𝜃

𝑥 = 𝑟 𝑐𝑜𝑠 = 0.5 𝑐𝑜𝑠 + 𝑐𝑜𝑠2 𝑑𝑥

𝑑𝜃 = – 0.5 𝑠𝑖𝑛 – 2 𝑐𝑜𝑠 𝑠𝑖𝑛

𝑦 = 𝑟 𝑠𝑖𝑛 = 0.5 𝑠𝑖𝑛 + 𝑐𝑜𝑠 𝑠𝑖𝑛 𝑑𝑦

𝑑𝜃 = 0.5 𝑐𝑜𝑠 – 𝑠𝑖𝑛2 + 𝑐𝑜𝑠2𝜃 = 0.5 𝑐𝑜𝑠 + 2𝑐𝑜𝑠2𝜃 – 1

ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑠: 𝑑𝑦𝑑𝑥

= 0 → 𝑑𝑦𝑑𝜃

= 0 & 𝑑𝑥𝑑𝜃

≠ 0

2cos2 + 0.5 cos – 1 = 0 cos = (– 0.5 ± √8.25)/4

cos = 0.593 1 = 0.936 rad 2 = 5.347 rad

cos = – 0.843 3 = 2.572 rad 4 = 3.710 rad

each equation is: y = r sin

𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑠: 𝑑𝑦

𝑑𝑥= ∞ →

𝑑𝑥

𝑑𝜃= 0 &

𝑑𝑦

𝑑𝜃 ≠ 0

𝑠𝑖𝑛 + 4 𝑐𝑜𝑠 𝑠𝑖𝑛 = 0 𝑠𝑖𝑛(1 + 4 𝑐𝑜𝑠) = 0

sin = 0 = 0 tangent: x = 1.5 the same for cos = - ¼

3. r = 0.5 + cos θ will have two tangents at point r = 0 = 2/3 and = 4/3

slope = . 𝑑𝑦

𝑑𝑥=

𝑑𝑦

𝑑𝜃𝑑𝑥

𝑑𝜃

= 0.5 cos + 2cos2 – 1

– 0.5 sin – 2 cos sin calculate that for both = 2/3 and = 4/3

first tangent line at r = 0 = 2/3 (y1 = 0, x1 = 0) y = y’ (x – x1) + y1 second tangent line at r = 0 = 4/3 (y1 = 0, x1 = 0) y = y’ (x – x1) + y1

table:

r

0 1.5

/6 1.37

/3 1

2/3 0

5/6 -0.367

-0.5

7/6 -0.367

4/3 0

5/3 1

11/6 1.37

2 1.5

𝐴 =1

2∫ 𝑟2𝑑𝜃

4𝜋/3

2𝜋/3

= 1

2∫ (0.5 + 𝑐𝑜𝑠 𝜃)2𝑑𝜃

4𝜋/3

2𝜋/3

=1

2∫ (0.25 + cos 𝜃 + 𝑐𝑜𝑠2𝜃) 𝑑𝜃

4𝜋/3

2𝜋/3

=1

2∫ (0.25 + cos 𝜃 + 0.5 𝑐𝑜𝑠 2𝜃 + 0.5) 𝑑𝜃

4𝜋/3

2𝜋/3