1
POLAR COORDINATES
Polar coordinate system: a pole (fixed point) and a polar axis (directed ray with endpoint at pole).
rectangular coordinates ⇒ polar coordinates
𝑟 = √𝑥 2 + 𝑦 2 , 𝜃 = 𝑎𝑟𝑐 𝑡𝑎𝑛
𝑦
𝑥
polar coordinates ⇒ rectangular coordinates
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.
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 .
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
r ( r + 5 cos θ) = 0
r2 + 5(r cos θ) = 0
The equation r = 0 is the pole. Thus, keep only the other equation: r + 5 cos θ = 0
2
Problem : Transform the equation r = 4sin θ to Cartesian coordinate form. What is the graph? Describe it fully!!!
√𝑥 2 + 𝑦 2 = 4
𝑦
√𝑥 2 +𝑦 2
𝑥 2 + 𝑦2 = 4 𝑦
𝑥 2 + (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
r
0
–1
π/4
–0.41
π/3
0
π/2
1
2 π/3
2
3 π/4
2.41
π
3
5 π/4
2.41
4 π/3
2
3 π/2
1
5 π/3
0
7 π/4
–0.41
2π
–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
convex limacon
3
+ cos 𝜃
2
limacon with a dimple
𝑟=
𝑟 = 1 + cos 𝜃
carotid
1
+ cos 𝜃
2
limacon with an inner loop
𝑟=
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
• If n is odd, there are n petals.
• If n is even, there are 2n petals.
4
r = sin 3θ
r = cos 4 θ
First and second derivative r = r():
𝑥 = 𝑟𝑐𝑜𝑠𝜃 ⟹
𝑑𝑥
𝑑𝑟
=
𝑐𝑜𝑠 𝜃 − 𝑟 sin 𝜃
𝑑𝜃
𝑑𝜃
𝑦 = 𝑟𝑠𝑖𝑛𝜃 ⟹
𝑑𝑦
𝑑𝑟
=
𝑠𝑖𝑛 𝜃 + 𝑟𝑐𝑜𝑠 𝜃
𝑑𝜃
𝑑𝜃
𝑑 𝑑𝑦
[ ]
𝑑𝑡 𝑑𝑥
=
[ ] =
2
𝑑𝑥
𝑑𝑥
𝑑𝑥 𝑑𝑥
𝑑𝑡
𝑑2 𝑦
𝑑𝑟
1 𝑑 𝑑𝜃 𝑠𝑖𝑛 𝜃 + 𝑟𝑐𝑜𝑠 𝜃
=
( )=
(
)
𝑑𝑥 𝑑𝜃 𝑑𝑟
𝑑𝑥 𝑑𝑥
𝒅𝟐 𝒙
𝑐𝑜𝑠 𝜃 − 𝑟 sin 𝜃
𝑑𝜃
𝑑𝜃
𝒅𝟐 𝒚
𝑑𝑦
𝑑𝑟
𝑠𝑖𝑛 𝜃 + 𝑟𝑐𝑜𝑠 𝜃
𝑑𝜃
𝑑𝜃
=
=
𝑑𝑥
𝑑𝑟
𝒅𝒙
𝑐𝑜𝑠 𝜃 − 𝑟 sin 𝜃
𝑑𝜃
𝑑𝜃
𝒅𝒚
𝑥0 = 𝑟0 𝑐𝑜𝑠 𝜃0
𝑎𝑛𝑑
𝑦0 = 𝑟0 𝑠𝑖𝑛 𝜃0
𝑵𝒐𝒓𝒎𝒂𝒍 𝒍𝒊𝒏𝒆: 𝑦 = −
1
𝑚
𝑑
𝑑𝑦
and now good luck 
Note that rather than trying to remember
this formula it would probably be easier to
remember how we derived it .
𝑻𝒂𝒏𝒈𝒆𝒏𝒕 𝒍𝒊𝒏𝒆 𝑎𝑡 𝑝𝑜𝑖𝑛𝑡 𝜽 = 𝜽𝟎 𝑜𝑓 𝑎 𝑐𝑢𝑟𝑣𝑒 𝒓 = 𝒓(𝜽):
𝑟0 = 𝑟(𝜃0 ) ⟹
𝑑 𝑑𝑦
𝑚=
𝑦 = 𝑚 (𝑥 – 𝑥0 ) + 𝑦0
𝑑𝑦
𝑎𝑡 𝑟0 = 𝑟(𝜃0 )
𝑑𝑥
(𝑥 – 𝑥0 ) + 𝑦0
𝑯𝒐𝒓𝒊𝒛𝒐𝒏𝒕𝒂𝒍 𝒕𝒂𝒏𝒈𝒆𝒏𝒕 𝒍𝒊𝒏𝒆: 𝑤𝑖𝑙𝑙 𝑜𝑐𝑐𝑢𝑟 𝑤ℎ𝑒𝑟𝑒 𝑡ℎ𝑒 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 𝑖𝑠 𝑧𝑒𝑟𝑜 ⟹
𝑑𝑦
𝑑𝑥
=0 ⟹
𝑑𝑦
𝑑𝜃
= 0 ⟹ 𝜃0 (𝑐ℎ𝑒𝑐𝑘
𝑑𝑥
𝑑𝑡
|
𝜃0
≠ 0) ⟹ 𝑟0 = 𝑟(𝜃0 ) ⟹ 𝑦0 = 𝑟0 𝑠𝑖𝑛𝜃0
𝑽𝒆𝒓𝒕𝒊𝒄𝒂𝒍 𝒕𝒂𝒏𝒈𝒆𝒏𝒕 𝒍𝒊𝒏𝒆: 𝑤𝑖𝑙𝑙 𝑜𝑐𝑐𝑢𝑟 𝑤ℎ𝑒𝑟𝑒 𝑡ℎ𝑒 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 𝑖𝑠 𝑛𝑜𝑡 𝑑𝑒𝑓𝑖𝑛𝑒𝑑:
𝑑𝑦
𝑑𝑥
=∞ ⟹
𝑑𝑥
𝑑𝜃
= 0 ⟹ 𝜃0 (𝑐ℎ𝑒𝑐𝑘
𝑑𝑦
𝑑𝑡
|
𝑒𝑞: 𝑦 = 𝑦0
𝑑𝑦
=∞
𝑑𝑥
≠ 0) ⟹ 𝑟0 = 𝑟(𝜃0 ) ⟹ 𝑥0 = 𝑟0 𝑐𝑜𝑠𝜃0
𝑒𝑞: 𝑥 = 𝑥0
𝜃0
𝑪𝒐𝒏𝒄𝒂𝒗𝒊𝒕𝒚 𝒂𝒕 𝒑𝒐𝒊𝒏𝒕 (𝒙𝟏 , 𝒚𝟏 ) 𝑜𝑟 𝒕𝟏 𝑜𝑟 𝜽𝟏 ∶ 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑒
𝐼𝑓
𝑑𝑦
=0
𝑑𝑥
𝑑2𝑦
𝑎𝑡 𝑡ℎ𝑎𝑡 𝑝𝑜𝑖𝑛𝑡.
𝑑𝑥 2
𝑑2𝑦
𝑑2 𝑦
<
0
→
𝑐𝑢𝑟𝑣𝑒
𝑖𝑠
𝑐𝑜𝑛𝑐𝑎𝑣𝑒
𝑑𝑜𝑤𝑛.
𝐼𝑓
> 0 → 𝑐𝑢𝑟𝑣𝑒 𝑖𝑠 𝑐𝑜𝑛𝑐𝑎𝑣𝑒 𝑢𝑝
𝑑𝑥 2
𝑑𝑥 2
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
1
(𝑟 𝑑𝜃)𝑟 = 𝑟 2 𝑑𝜃
2
2
1 ≤  ≤ 2
𝜃2
𝐴 = ∫ 𝑑𝐴 =
𝜃1
1 𝜃2 2
∫ 𝑟 𝑑𝜃
2 𝜃1
Example: Find the area enclosed by:
𝑟 = 2(1 + 𝑐𝑜𝑠𝜃)
example: Find the area of the inner loop of r = 2 + 4 cos θ
6
7
8
Length of a Polar Curve:
𝑏
𝑏
𝜃2
𝑑𝑥 2
𝑑𝑦 2
𝑆 = ∫ 𝑑𝑠 = ∫ √(𝑑𝑥)2 + (𝑑𝑦)2 = ∫ √ ( ) + ( ) 𝑑𝜃
𝑑𝜃
𝑑𝜃
𝑎
𝑎
𝜃1
𝜃1 𝑎𝑛𝑑 𝜃2 𝑎𝑟𝑒 𝑎𝑛𝑔𝑙𝑒𝑠 𝑐𝑜𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑡𝑜 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛𝑠 𝑎 𝑎𝑛𝑑 𝑏
𝑥 = 𝑟𝑐𝑜𝑠𝜃 ⟹
𝑑𝑥
𝑑𝑟
=
𝑐𝑜𝑠 𝜃 − 𝑟 sin 𝜃
𝑑𝜃
𝑑𝜃
𝑎𝑛𝑑
𝑦 = 𝑟𝑠𝑖𝑛𝜃 ⟹
𝑑𝑦
𝑑𝑟
=
𝑠𝑖𝑛 𝜃 + 𝑟𝑐𝑜𝑠 𝜃
𝑑𝜃
𝑑𝜃
𝑑𝑥 2
𝑑𝑦 2
𝑑𝑟 2
𝑑𝑟
𝑑𝑟 2
𝑑𝑟
𝑑𝑟 2
( ) + ( ) = ( ) 𝑐𝑜𝑠 2 𝜃 − 2𝑟
𝑐𝑜𝑠𝜃 𝑠𝑖𝑛𝜃 + 𝑟 2 𝑠𝑖𝑛2 𝜃 + ( ) 𝑠𝑖𝑛2 𝜃 + 2𝑟
𝑐𝑜𝑠𝜃 𝑠𝑖𝑛𝜃 + 𝑟 2 𝑐𝑜𝑠 2 𝜃 = ( ) + 𝑟 2
𝑑𝜃
𝑑𝜃
𝑑𝜃
𝑑𝜃
𝑑𝜃
𝑑𝜃
𝑑𝜃
𝑏
𝑏
𝜃2
𝑑𝑟 2
𝑆 = ∫ 𝑑𝑠 = ∫ √(𝑑𝑥)2 + (𝑑𝑦)2 = ∫ √ 𝑟 2 + ( ) 𝑑𝜃
𝑑𝜃
𝑎
𝑎
𝜃1
9
10
EXAMPLE: limaçon: r = 0.5 + cos θ
table:

0
/6
/3
2/3
5/6
r
1.5
1.37
1
0
0.367
-0.5

7/6 0.367
4/3 0
5/3 1
11/6 1.37
1.5
2
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.
1 4𝜋/3 2
1 4𝜋/3
𝐴= ∫
𝑟 𝑑𝜃 = ∫
(0.5 + 𝑐𝑜𝑠 𝜃)2 𝑑𝜃
2 2𝜋/3
2 2𝜋/3
=
1 4𝜋/3
∫
(0.25 + cos 𝜃 + 𝑐𝑜𝑠 2 𝜃) 𝑑𝜃
2 2𝜋/3
=
1 4𝜋/3
∫
(0.25 + cos 𝜃 + 0.5 𝑐𝑜𝑠 2𝜃 + 0.5) 𝑑𝜃
2 2𝜋/3
= 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
𝑑𝑦
𝑑𝑦
𝑑𝜃
2.
=
𝑑𝑥
𝑑𝑥
𝑑𝜃
𝑑𝑥
𝑥 = 𝑟 𝑐𝑜𝑠 = 0.5 𝑐𝑜𝑠 + 𝑐𝑜𝑠 2 
𝑑𝜃
𝑑𝑦
= 0.5 𝑐𝑜𝑠 – 𝑠𝑖𝑛2  + 𝑐𝑜𝑠 2 𝜃 = 0.5 𝑐𝑜𝑠 + 2𝑐𝑜𝑠 2 𝜃 – 1
𝑑𝜃
𝑦 = 𝑟 𝑠𝑖𝑛 = 0.5 𝑠𝑖𝑛 + 𝑐𝑜𝑠 𝑠𝑖𝑛
𝑑𝑦
𝑑𝑥
ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑠:
=0 →
𝑑𝑦
𝑑𝜃
= – 0.5 𝑠𝑖𝑛 – 2 𝑐𝑜𝑠 𝑠𝑖𝑛
=0 &
2cos2 + 0.5 cos – 1 = 0
𝑑𝑥
𝑑𝜃
≠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
𝑑𝑦
=∞ →
=0 &
≠0
𝑑𝑥
𝑑𝜃
𝑑𝜃
𝑠𝑖𝑛 + 4 𝑐𝑜𝑠 𝑠𝑖𝑛 = 0
sin  = 0
3.
r = 0.5 + cos θ
slope =
.
=0
𝑠𝑖𝑛(1 + 4 𝑐𝑜𝑠) = 0
tangent: x = 1.5
will have two tangents at point r = 0
𝑑𝑦
𝑑𝑥
=
𝑑𝑦
𝑑𝜃
𝑑𝑥
𝑑𝜃
=
0.5 cos + 2cos2 – 1
– 0.5 sin – 2 cos sin
the same for
 = 2/3
cos  = - ¼
and  = 4/3
calculate that for both  = 2/3
and  = 4/3
11
first tangent line at
second tangent line at
r=0
r=0
 = 2/3 (y1 = 0, x1 = 0)
 = 4/3 (y1 = 0, x1 = 0)
y = y’ (x – x1) + y1
y = y’ (x – x1) + y1