1
Polar Coordinates
Polar coordinate system: a pole (fixed point) and a polar axis (directed ray with endpoint at pole).
Relation to rectangular coordinates
⇒ 𝑥 2 + 𝑦 2 = 𝑟 2 , tan 𝜃 =
● 𝑥 = 𝑟 cos 𝜃 , 𝑦 = 𝑟 sin 𝜃
𝑟 = √𝑥 2 + 𝑦 2 , 𝜃 = 𝑎𝑟𝑐 tan
𝑦
𝑥
𝑦
𝑥
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
2
The equation r = 0 is the pole. Thus, keep only the other equation: r + 5 cos θ = 0
Problem : Transform the equation r = 4sin θ to Cartesian coordinate form. What is the graph? Describe it fully!!!
√𝑥 2 + 𝑦 2 = 4
𝑥 2 + 𝑦2 = 4 𝑦
𝑦
√𝑥 2 +𝑦 2
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)Π, 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.
TI – 84 plus
CLEAR RAM follow these key strokes: "2nd", "+", "7","1","2"
Graphing polar equations
1.
2.
3.
4.
5.
graph the polar equation r = 1- sinθ
Hit the MODE key.
Arrow down to where it says Func and then use the right arrow to choose Pol.
Hit ENTER
The calculator is now in parametric equations mode.
Hit the Y= key.
In the r1 slot, type r = 1- sin(θ)
Hit X,T,θ,n key for typing θ
• Press [WINDOW] and enter the following settings:
θmin = 0
θmax = 2π
θstep = π /24
Xmin = -3
Xmax = 3
Xscl = 1
Ymin = -3
Ymax = 1
Yscl = 1
With these settings the calculator will evaluate the function from θ = 0 to θ = 2 π in increments of π /24.
• Press [GRAPH].
The following graph will be displayed.
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EXAMPLES:
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
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
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
𝑟=
3
+ cos 𝜃
2
limacon with a dimple
𝑟 = 1 + cos 𝜃
carotid
𝑟=
1
+ cos 𝜃
2
limacon with an inner loop
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
5 π/3
0
7 π/4
–0.41
2π
–1
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Flowers
Petal Curve: r = cos 2 θ
The area enclosed by a polar curve r = r():
Petal Curves: r = a cos n θ,
r = a sin n θ
r = cos 4 θ
r = sin 3θ
• If n is odd, there are n petals.
• If n is even, there are 2n petals.
First and second derivative
r = r():
𝑑𝑥
x = r cos 
1. derivative
2. derivative:
𝑑𝑦
𝑑𝑥
𝑑2𝑦
𝑑2𝑥
=
=
𝑑𝜃
𝑑𝑦 𝑑𝜃
𝑑𝜃 𝑑𝑥
𝑑
𝑑𝑥
=
=
𝑑𝑟
𝑑𝜃
𝑑𝑦 1
𝑑𝜃 𝑑𝑥
𝑑𝜃
𝑑𝑦
𝑑
𝑑𝑥
𝑑𝜃
( ) =
y = r sin 
𝑐𝑜𝑠 𝜃 − 𝑟 sin 𝜃
=
𝑑𝑦
𝑑𝜃
𝑑𝑥
𝑑𝜃
𝑑𝑦 𝑑𝜃
( )
𝑑𝑥 𝑑𝑥
=
=
𝑑𝑟
𝑑𝜃
𝑑𝑟
𝑑𝜃
𝑑𝑦
𝑑𝜃
=
𝑑𝑟
𝑑𝜃
𝑠𝑖𝑛 𝜃 + 𝑟𝑐𝑜𝑠 𝜃
𝑠𝑖𝑛 𝜃 + 𝑟𝑐𝑜𝑠 𝜃
𝑐𝑜𝑠 𝜃 − 𝑟 sin 𝜃
𝑑 𝑑𝑦
( )
𝑑𝜃 𝑑𝑥
𝑑𝑥
𝑑𝜃
=
1
𝑑
𝑑𝑥
𝑑𝜃
𝑑𝜃
𝑑𝑟
𝑑𝜃
( 𝑑𝑟
𝑑𝜃
𝑠𝑖𝑛 𝜃 + 𝑟𝑐𝑜𝑠 𝜃
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 .
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Tangent line: y = m (x – x0) + y0
at point 0 of the r = r():
r0 = r(0), x0 = r0 cos 0
Horizontal tangent line: Horizontal tangent will occur where the derivative is zero:
at point 0 of the r = r():
𝑑𝑦
𝑑𝑥
𝑑𝑦
=0 →
𝑑𝜃
= 0 → 0 →
r0 = r(0)
y0 = r0 sin 0
Concavity:
𝑑𝑦
𝑑𝑥
=∞ →
𝑑𝑥
𝑑𝜃
= 0 → 0 →
x0 = r0 cos 0
r0 = r(0)
𝑑𝑦
𝑑𝑥
at r0 = r(0)
𝑑𝑥
=0
𝑑𝑥
(𝑑𝜃 |𝜃=𝜃
eq: y = y0
Vertical tangent line: Vertical tangents will occur where the derivative is not defined:
at point 0 of the r = r():
𝑑𝑦
y0 = r0 sin 0 m =
𝑑𝑦
𝑑𝑥
≠ 0)
0
=∞
𝑑𝑦
eq: x = x0
(𝑑𝜃 |𝜃=𝜃
0
≠ 0)
If second derivate is negative curve will be concave down, and for positive second derivate curve will be concave up.
𝑑2𝑦
Polar curve: at point 0 of the r = r():
𝑑2𝑥
at (r0, θ0)
Arc length
The arc length of polar form is :
2
𝑏
𝑏
𝑏
𝑑𝑟
S = ∫𝑎 𝑑𝑠 = ∫𝑎 √(𝑑𝑥)2 + (𝑑𝑦)2 = ∫𝑎 √ 𝑟 2 + (𝑑𝜃) 𝑑𝜃
2
2
𝑏
𝑏
𝑏
𝑑𝑥
𝑑𝑦
S = ∫𝑎 𝑑𝑠 = ∫𝑎 √(𝑑𝑥)2 + (𝑑𝑦)2 = ∫𝑎 √ (𝑑𝜃) + (𝑑𝜃) 𝑑𝜃
x = r cos 
𝑑𝑥
𝑑𝜃
=
𝑑𝑟
𝑑𝜃
𝑐𝑜𝑠 𝜃 − 𝑟 sin 𝜃
dx =
𝑑𝑥
𝑑𝜃
y = r sin 
𝑑𝜃
dy =
𝑑𝑦
𝑑𝜃
=
𝑑𝑦
𝑑𝜃
𝑑𝑟
𝑑𝜃
𝑑𝜃
𝑠𝑖𝑛 𝜃 + 𝑟𝑐𝑜𝑠 𝜃
𝑑𝑥 2
𝑑𝑦 2
𝑑𝑟 2
𝑑𝑟
𝑑𝑟
( ) + ( ) = ( ) [𝑐𝑜𝑠 2 𝜃 + 𝑠𝑖𝑛2 𝜃] + 𝑟 2 [ 𝑠𝑖𝑛2 𝜃 + 𝑐𝑜𝑠 2 𝜃] − 2𝑟
cos 𝜃 sin 𝜃 + 2𝑟
cos 𝜃 sin 𝜃
𝑑𝜃
𝑑𝜃
𝑑𝜃
𝑑𝜃
𝑑𝜃
The area enclosed by a polar curve r = r():
𝜃2
1
1
2
A = ∫𝜃 𝑑𝐴 =
𝜃
2
∫𝜃 𝑟 2 𝑑𝜃
1
1 ≤  ≤ 2
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limaçon: r = 0.5 + cos θ
table:

0
/6
/3
2/3
5/6

7/6
4/3
5/3
11/6
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.
4. Concavity table.
1
4𝜋/3
1
4𝜋/3
1
4𝜋/3
A = 2 ∫2𝜋/3 𝑟 2 𝑑𝜃 =
1 4𝜋/3
(0.5
∫
2 2𝜋/3
+ 𝑐𝑜𝑠 𝜃)2 𝑑𝜃
= 2 ∫2𝜋/3 (0.25 + cos 𝜃 + 𝑐𝑜𝑠 2 𝜃) 𝑑𝜃
= 2 ∫2𝜋/3 (0.25 + cos 𝜃 + 0.5 𝑐𝑜𝑠 2𝜃 + 0.5) 𝑑𝜃
r
1.5
1.37
1
0
-0.367
-0.5
-0.367
0
1
1.37
1.5
= 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
𝑑𝑦
2. 𝑑𝑥
=
𝑑𝑦
𝑑𝜃
𝑑𝑥
𝑑𝜃
A = 0.25  - 0.375 3
𝑑𝑥
x = r cos = 0.5 cos + cos2
𝑑𝜃
𝑑𝑦
y = r sin = 0.5 sin + cos sin
𝑑𝑦
𝑑𝑥
horizontal tangents: slope = 0
𝑑𝜃
𝑑𝑦
𝑑𝜃
=0 →
= – 0.5 sin – 2 cos sin
= 0.5 cos – sin2 + cos2 = 0.5 cos + 2cos2 – 1
𝑑𝑥
𝑑𝜃
=0 &
2cos2 + 0.5 cos – 1 = 0
≠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
𝑑𝑦
𝑑𝑥
vertical tangents: slope = ∞
=∞ →
𝑑𝑥
𝑑𝜃
=0 &
sin + 4 cos sin = 0
sin  = 0
3.
r = 0.5 + cos θ
slope =
.
=0
𝑑𝑦
𝑑𝑥
=
=
tangent: x = 1.5
0.5 cos + 2cos2 – 1
– 0.5 sin – 2 cos sin
≠0
sin(1 + 4 cos) = 0
the same for
 = 2/3
will have two tangents at point r = 0
𝑑𝑦
𝑑𝜃
𝑑𝑥
𝑑𝜃
𝑑𝑦
𝑑𝜃
cos  = - ¼
and  = 4/3
calculate that for both  = 2/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
and  = 4/3
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Problem : Decide whether each of the following polar graphs is a limacon, a rose curve, a spiral, a circle,
or none of these: a) r = 2 + cos(θ) ; b) r = 2 ; c) r = sin(3θ) ; d) r = 1 - cos(θ) ; e) r = 2θ .
a) r = 2 + cos(θ) is a limacon.
b) r = 2 is a circle.
c) r = sin(3θ) is a rose curve.
d) r = 1 - cos(θ) is a limacon.
e) r = 2θ is a spiral.