of 22 /22
Section 9.3 Polar Coordinates 2010 Kiryl Tsishchanka Polar Coordinates DEFINITION: The polar coordinate system is a two-dimensional coordinate system in which each point P on a plane is determined by a distance r from a ﬁxed point O that is called the pole (or origin) and an angle θ from a ﬁxed direction. The point P is represented by the ordered pair (r, θ) and r, θ are called polar coordinates. REMARK: We extend the meaning of polar coordinates (r, θ) to the case in which r is negative by agreeing that the points (-r, θ) and (r, θ) lie in the same line through O and at the same distance |r| from O, but on opposite sides of O. If r> 0, the point (r, θ) lies in the same quadrant as θ; if r< 0, it lies in the quadrant on the opposite side of the pole. EXAMPLE: Plot the points whose polar coordinates are given: (a) (1, 5π/4) (b) (2, 3π) (c) (2, -2π/3) (d) (-3, 3π/4) Solution: REMARK: In the Cartesian coordinate system every point has only one representation, but in the polar coordinate system each point has many representations. For instance, the point (1, 5π/4) in the Example above could be written as (1, -3π/4) or (1, 13π/4) or (-1,π/4): EXAMPLE: Find all the polar coordinates of the point P (2,π/6). 1

Embed Size (px)

### Transcript of Polar Coordinates

Section 9.3 Polar Coordinates 2010 Kiryl Tsishchanka

Polar Coordinates

DEFINITION: The polar coordinate system is a two-dimensional coordinate system in which eachpoint P on a plane is determined by a distance r from a fixed point O that is called the pole (or origin)and an angle θ from a fixed direction. The point P is represented by the ordered pair (r, θ) and r, θ arecalled polar coordinates.

REMARK: We extend the meaning of polar coordinates (r, θ) to the case in which r is negative by agreeingthat the points (−r, θ) and (r, θ) lie in the same line through O and at the same distance |r| from O, buton opposite sides of O. If r > 0, the point (r, θ) lies in the same quadrant as θ; if r < 0, it lies in thequadrant on the opposite side of the pole.

EXAMPLE: Plot the points whose polar coordinates are given:

(a) (1, 5π/4) (b) (2, 3π) (c) (2,−2π/3) (d) (−3, 3π/4)

Solution:

REMARK: In the Cartesian coordinate system every point has only one representation, but in the polarcoordinate system each point has many representations. For instance, the point (1, 5π/4) in the Exampleabove could be written as (1,−3π/4) or (1, 13π/4) or (−1, π/4):

EXAMPLE: Find all the polar coordinates of the point P (2, π/6).

1

Section 9.3 Polar Coordinates 2010 Kiryl Tsishchanka

EXAMPLE: Find all the polar coordinates of the point P (2, π/6).

Solution: We sketch the initial ray of the coordinate system, draw the ray from the origin that makes anangle of π/6 radians with the initial ray, and mark the point (2, π/6). We then find the angles for the othercoordinate pairs of P in which r = 2 and r = −2.

For r = 2, the complete list of angles is

π

6,

π

6± 2π,

π

6± 4π,

π

6± 6π, . . .

For r = −2, the angles are

−5π

6, −5π

6± 2π, −5π

6± 4π, −5π

6± 6π, . . .

The corresponding coordinate pairs of P are(2,

π

6+ 2nπ

), n = 0, ±1, ±2, . . .

and (−2, −5π

6+ 2nπ

), n = 0, ±1, ±2, . . .

When n = 0, the formulas give (2, π/6) and (−2,−5π/6). When n = 1, they give (2, 13π/6) and (−2, 7π/6),and so on.

The connection between polar and Cartesian coordinates can be seen from the figure below and describedby the following formulas:

r2 = x2 + y2 tan θ =y

x

x = r cos θ y = r sin θ

EXAMPLE:

(a) Convert the point (2, π/3) from polar to Cartesian coordinates.

(b) Represent the point with Cartesian coordinates (1,−1) in terms of polar coordinates.

2

Section 9.3 Polar Coordinates 2010 Kiryl Tsishchanka

EXAMPLE:

(a) Convert the point (2, π/3) from polar to Cartesian coordinates.

(b) Represent the point with Cartesian coordinates (1,−1) in terms of polar coordinates.

Solution:

(a) We have:

x = r cos θ = 2 cosπ

3= 2 · 1

2= 1 y = r sin θ = 2 sin

π

3= 2 ·

√3

2=

√3

Therefore, the point is (1,√

3) in Cartesian coordinates.

(b) If we choose r to be positive, then

r =√

x2 + y2 =√

12 + (−1)2 =√

2 tan θ =y

x= −1

Since the point (1,−1) lies in the fourth quadrant, we can choose θ = −π/4 or θ = 7π/4. Thus one possibleanswer is (

√2,−π/4); another is (

√2, 7π/4).

EXAMPLE: Express the equation x = 1 in polar coordinates.

Solution: We use the formula x = r cos θ.

x = 1

r cos θ = 1

r = sec θ

EXAMPLE: Express the equation x2 = 4y in polar coordinates.

Solution: We use the formulas x = r cos θ and y = r sin θ.

x2 = 4y

(r cos θ)2 = 4r sin θ

r2 cos2 θ = 4r sin θ

r = 4sin θ

cos2 θ= 4 sec θ tan θ

3

Polar Curves

The graph of a polar equation r = f(θ), or more generally F (r, θ) = 0, consists of all points P thathave at least one polar representation (r, θ) whose coordinates satisfy the equation.

EXAMPLE: Sketch the polar curve θ = 1.

Solution: This curve consists of all points (r, θ) such that the polar angle θ is 1 radian. It is the straightline that passes through O and makes an angle of 1 radian with the polar axis. Notice that the points(r, 1) on the line with r > 0 are in the first quadrant, whereas those with r < 0 are in the third quadrant.

EXAMPLE: Sketch the following curves:

(a) r = 2, 0 ≤ θ ≤ 2π.

(b) r = 2 cos θ, 0 ≤ θ ≤ π.

4

Section 9.3 Polar Coordinates 2010 Kiryl Tsishchanka

EXAMPLE: Sketch the curve r = 2, 0 ≤ θ ≤ 2π.

Solution: We have

-2 -1 1 2

-2

-1

1

2

r=2, theta=Pi�6

-2 -1 1 2

-2

-1

1

2

r=2, theta=2 Pi�6

-2 -1 1 2

-2

-1

1

2

r=2, theta=3 Pi�6

-2 -1 1 2

-2

-1

1

2

r=2, theta=4 Pi�6

-2 -1 1 2

-2

-1

1

2

r=2, theta=5 Pi�6

-2 -1 1 2

-2

-1

1

2

r=2, theta=6 Pi�6

-2 -1 1 2

-2

-1

1

2

r=2, theta=7 Pi�6

-2 -1 1 2

-2

-1

1

2

r=2, theta=8 Pi�6

-2 -1 1 2

-2

-1

1

2

r=2, theta=9 Pi�6

-2 -1 1 2

-2

-1

1

2

r=2, theta=10 Pi�6

-2 -1 1 2

-2

-1

1

2

r=2, theta=11 Pi�6

-2 -1 1 2

-2

-1

1

2

r=2, theta=12 Pi�6

5

Section 9.3 Polar Coordinates 2010 Kiryl Tsishchanka

EXAMPLE: Sketch the curve r = 2 cos θ, 0 ≤ θ ≤ π.

Solution: We have

-0.5 0.5 1.0 1.5 2.0

-1.0

-0.5

0.5

1.0

r=2cosHthetaL, theta=Pi�12

-0.5 0.5 1.0 1.5 2.0

-1.0

-0.5

0.5

1.0

r=2cosHthetaL, theta=2 Pi�12

-0.5 0.5 1.0 1.5 2.0

-1.0

-0.5

0.5

1.0

r=2cosHthetaL, theta=3 Pi�12

-0.5 0.5 1.0 1.5 2.0

-1.0

-0.5

0.5

1.0

r=2cosHthetaL, theta=4 Pi�12

-0.5 0.5 1.0 1.5 2.0

-1.0

-0.5

0.5

1.0

r=2cosHthetaL, theta=5 Pi�12

-0.5 0.5 1.0 1.5 2.0

-1.0

-0.5

0.5

1.0

r=2cosHthetaL, theta=6 Pi�12

-0.5 0.5 1.0 1.5 2.0

-1.0

-0.5

0.5

1.0

r=2cosHthetaL, theta=7 Pi�12

-0.5 0.5 1.0 1.5 2.0

-1.0

-0.5

0.5

1.0

r=2cosHthetaL, theta=8 Pi�12

-0.5 0.5 1.0 1.5 2.0

-1.0

-0.5

0.5

1.0

r=2cosHthetaL, theta=9 Pi�12

-0.5 0.5 1.0 1.5 2.0

-1.0

-0.5

0.5

1.0

r=2cosHthetaL, theta=10 Pi�12

-0.5 0.5 1.0 1.5 2.0

-1.0

-0.5

0.5

1.0

r=2cosHthetaL, theta=11 Pi�12

-0.5 0.5 1.0 1.5 2.0

-1.0

-0.5

0.5

1.0

r=2cosHthetaL, theta=12 Pi�12

6

EXAMPLE: Express the polar equation r = 2 cos θ in rectangular coordinates.

Solution: We use the formulas r2 = x2 + y2 and x = r cos θ. We have

r = 2 cos θ

r2 = 2r cos θ

x2 + y2 = 2x

x2 − 2x + y2 = 0

x2 − 2x + 1 + y2 = 1

(x − 1)2 + y2 = 1

EXAMPLE: Express the polar equation in rectangular coordinates. If possible, determine the graph ofthe equation from its rectangular form.

(a) r = 5 sec θ (b) r = 2 sin θ (c) r = 2 + 2 cos θ

7

EXAMPLE: Express the polar equation in rectangular coordinates. If possible, determine the graph ofthe equation from its rectangular form.

(a) r = 5 sec θ (b) r = 2 sin θ (c) r = 2 + 2 cos θ

Solution: We use the formulas r2 = x2 + y2, x = r cos θ and y = r sin θ.

(a) We have

r = 5 sec θ

r cos θ = 5

x = 5

(b) We have

r = 2 sin θ

r2 = 2r sin θ

x2 + y2 = 2y

x2 + y2 − 2y = 0

x2 + y2 − 2y + 1 = 1

x2 + (y − 1)2 = 1

(c) We have

r = 2 + 2 cos θ

r2 = 2r + 2r cos θ

x2 + y2 = 2r + 2x

x2 + y2 − 2x = 2r

(x2 + y2 − 2x)2 = 4r2

(x2 + y2 − 2x)2 = 4(x2 + y2)

8

Section 9.3 Polar Coordinates 2010 Kiryl Tsishchanka

EXAMPLE: Sketch the curve r = 1 + sin θ, 0 ≤ θ ≤ 2π (cardioid).

Solution: We have

-1.5 -1.0 -0.5 0.5 1.0 1.5

-0.5

0.5

1.0

1.5

2.0

r=1+sinHthetaL, theta=Pi�6

-1.5 -1.0 -0.5 0.5 1.0 1.5

-0.5

0.5

1.0

1.5

2.0

r=1+sinHthetaL, theta=2 Pi�6

-1.5 -1.0 -0.5 0.5 1.0 1.5

-0.5

0.5

1.0

1.5

2.0

r=1+sinHthetaL, theta=3 Pi�6

-1.5 -1.0 -0.5 0.5 1.0 1.5

-0.5

0.5

1.0

1.5

2.0

r=1+sinHthetaL, theta=4 Pi�6

-1.5 -1.0 -0.5 0.5 1.0 1.5

-0.5

0.5

1.0

1.5

2.0

r=1+sinHthetaL, theta=5 Pi�6

-1.5 -1.0 -0.5 0.5 1.0 1.5

-0.5

0.5

1.0

1.5

2.0

r=1+sinHthetaL, theta=6 Pi�6

-1.5 -1.0 -0.5 0.5 1.0 1.5

-0.5

0.5

1.0

1.5

2.0

r=1+sinHthetaL, theta=7 Pi�6

-1.5 -1.0 -0.5 0.5 1.0 1.5

-0.5

0.5

1.0

1.5

2.0

r=1+sinHthetaL, theta=8 Pi�6

-1.5 -1.0 -0.5 0.5 1.0 1.5

-0.5

0.5

1.0

1.5

2.0

r=1+sinHthetaL, theta=9 Pi�6

-1.5 -1.0 -0.5 0.5 1.0 1.5

-0.5

0.5

1.0

1.5

2.0

r=1+sinHthetaL, theta=10 Pi�6

-1.5 -1.0 -0.5 0.5 1.0 1.5

-0.5

0.5

1.0

1.5

2.0

r=1+sinHthetaL, theta=11 Pi�6

-1.5 -1.0 -0.5 0.5 1.0 1.5

-0.5

0.5

1.0

1.5

2.0

r=1+sinHthetaL, theta=12 Pi�6

9

Section 9.3 Polar Coordinates 2010 Kiryl Tsishchanka

EXAMPLE: Sketch the curve r = 1 − cos θ, 0 ≤ θ ≤ 2π (cardioid).

Solution: We have

-2.0 -1.5 -1.0 -0.5 0.5

-1.5

-1.0

-0.5

0.5

1.0

1.5r=1-cosHthetaL, theta=Pi�6

-2.0 -1.5 -1.0 -0.5 0.5

-1.5

-1.0

-0.5

0.5

1.0

1.5r=1-cosHthetaL, theta=2 Pi�6

-2.0 -1.5 -1.0 -0.5 0.5

-1.5

-1.0

-0.5

0.5

1.0

1.5r=1-cosHthetaL, theta=3 Pi�6

-2.0 -1.5 -1.0 -0.5 0.5

-1.5

-1.0

-0.5

0.5

1.0

1.5r=1-cosHthetaL, theta=4 Pi�6

-2.0 -1.5 -1.0 -0.5 0.5

-1.5

-1.0

-0.5

0.5

1.0

1.5r=1-cosHthetaL, theta=5 Pi�6

-2.0 -1.5 -1.0 -0.5 0.5

-1.5

-1.0

-0.5

0.5

1.0

1.5r=1-cosHthetaL, theta=6 Pi�6

-2.0 -1.5 -1.0 -0.5 0.5

-1.5

-1.0

-0.5

0.5

1.0

1.5r=1-cosHthetaL, theta=7 Pi�6

-2.0 -1.5 -1.0 -0.5 0.5

-1.5

-1.0

-0.5

0.5

1.0

1.5r=1-cosHthetaL, theta=8 Pi�6

-2.0 -1.5 -1.0 -0.5 0.5

-1.5

-1.0

-0.5

0.5

1.0

1.5r=1-cosHthetaL, theta=9 Pi�6

-2.0 -1.5 -1.0 -0.5 0.5

-1.5

-1.0

-0.5

0.5

1.0

1.5r=1-cosHthetaL, theta=10 Pi�6

-2.0 -1.5 -1.0 -0.5 0.5

-1.5

-1.0

-0.5

0.5

1.0

1.5r=1-cosHthetaL, theta=11 Pi�6

-2.0 -1.5 -1.0 -0.5 0.5

-1.5

-1.0

-0.5

0.5

1.0

1.5r=1-cosHthetaL, theta=12 Pi�6

10

Section 9.3 Polar Coordinates 2010 Kiryl Tsishchanka

EXAMPLE: Sketch the curve r = 2 + 4 cos θ, 0 ≤ θ ≤ 2π.

Solution: We have

1 2 3 4 5 6

-3

-2

-1

1

2

3

r=2+4cosHthetaL, theta=Pi�6

1 2 3 4 5 6

-3

-2

-1

1

2

3

r=2+4cosHthetaL, theta=2 Pi�6

1 2 3 4 5 6

-3

-2

-1

1

2

3

r=2+4cosHthetaL, theta=3 Pi�6

1 2 3 4 5 6

-3

-2

-1

1

2

3

r=2+4cosHthetaL, theta=4 Pi�6

1 2 3 4 5 6

-3

-2

-1

1

2

3

r=2+4cosHthetaL, theta=5 Pi�6

1 2 3 4 5 6

-3

-2

-1

1

2

3

r=2+4cosHthetaL, theta=6 Pi�6

1 2 3 4 5 6

-3

-2

-1

1

2

3

r=2+4cosHthetaL, theta=7 Pi�6

1 2 3 4 5 6

-3

-2

-1

1

2

3

r=2+4cosHthetaL, theta=8 Pi�6

1 2 3 4 5 6

-3

-2

-1

1

2

3

r=2+4cosHthetaL, theta=9 Pi�6

1 2 3 4 5 6

-3

-2

-1

1

2

3

r=2+4cosHthetaL, theta=10 Pi�6

1 2 3 4 5 6

-3

-2

-1

1

2

3

r=2+4cosHthetaL, theta=11 Pi�6

1 2 3 4 5 6

-3

-2

-1

1

2

3

r=2+4cosHthetaL, theta=12 Pi�6

11

Section 9.3 Polar Coordinates 2010 Kiryl Tsishchanka

EXAMPLE: Sketch the curve r = cos(2θ), 0 ≤ θ ≤ 2π (four-leaved rose).

Solution: We have

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=cosH2thetaL, theta=Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=cosH2thetaL, theta=2 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=cosH2thetaL, theta=3 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=cosH2thetaL, theta=4 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=cosH2thetaL, theta=5 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=cosH2thetaL, theta=6 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=cosH2thetaL, theta=7 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=cosH2thetaL, theta=8 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=cosH2thetaL, theta=9 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=cosH2thetaL, theta=10 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=cosH2thetaL, theta=11 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=cosH2thetaL, theta=12 Pi�6

12

Section 9.3 Polar Coordinates 2010 Kiryl Tsishchanka

EXAMPLE: Sketch the curve r = sin(2θ), 0 ≤ θ ≤ 2π (four-leaved rose).

Solution: We have

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH2thetaL, theta=Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH2thetaL, theta=2 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH2thetaL, theta=3 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH2thetaL, theta=4 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH2thetaL, theta=5 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH2thetaL, theta=6 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH2thetaL, theta=7 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH2thetaL, theta=8 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH2thetaL, theta=9 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH2thetaL, theta=10 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH2thetaL, theta=11 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH2thetaL, theta=12 Pi�6

13

Section 9.3 Polar Coordinates 2010 Kiryl Tsishchanka

EXAMPLE: Sketch the curve r = sin(3θ), 0 ≤ θ ≤ π (three-leaved rose).

Solution: We have

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH3thetaL, theta=Pi�12

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH3thetaL, theta=2 Pi�12

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH3thetaL, theta=3 Pi�12

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH3thetaL, theta=4 Pi�12

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH3thetaL, theta=5 Pi�12

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH3thetaL, theta=6 Pi�12

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH3thetaL, theta=7 Pi�12

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH3thetaL, theta=8 Pi�12

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH3thetaL, theta=9 Pi�12

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH3thetaL, theta=10 Pi�12

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH3thetaL, theta=11 Pi�12

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH3thetaL, theta=12 Pi�12

14

Section 9.3 Polar Coordinates 2010 Kiryl Tsishchanka

EXAMPLE: Sketch the curve r = sin(4θ), 0 ≤ θ ≤ 2π (eight-leaved rose).

Solution: We have

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH4thetaL, theta=Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH4thetaL, theta=2 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH4thetaL, theta=3 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH4thetaL, theta=4 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH4thetaL, theta=5 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH4thetaL, theta=6 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH4thetaL, theta=7 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH4thetaL, theta=8 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH4thetaL, theta=9 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH4thetaL, theta=10 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH4thetaL, theta=11 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH4thetaL, theta=12 Pi�6

15

Section 9.3 Polar Coordinates 2010 Kiryl Tsishchanka

EXAMPLE: Sketch the curve r = sin(5θ), 0 ≤ θ ≤ 2π (five-leaved rose).

Solution: We have

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH5thetaL, theta=Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH5thetaL, theta=2 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH5thetaL, theta=3 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH5thetaL, theta=4 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH5thetaL, theta=5 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH5thetaL, theta=6 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH5thetaL, theta=7 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH5thetaL, theta=8 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH5thetaL, theta=9 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH5thetaL, theta=10 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH5thetaL, theta=11 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH5thetaL, theta=12 Pi�6

16

Section 9.3 Polar Coordinates 2010 Kiryl Tsishchanka

EXAMPLE: Sketch the curve r = sin(6θ), 0 ≤ θ ≤ 2π (twelve-leaved rose).

Solution: We have

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH6thetaL, theta=Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH6thetaL, theta=2 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH6thetaL, theta=3 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH6thetaL, theta=4 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH6thetaL, theta=5 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH6thetaL, theta=6 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH6thetaL, theta=7 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH6thetaL, theta=8 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH6thetaL, theta=9 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH6thetaL, theta=10 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH6thetaL, theta=11 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH6thetaL, theta=12 Pi�6

17

Section 9.3 Polar Coordinates 2010 Kiryl Tsishchanka

EXAMPLE: Sketch the curve r = sin(7θ), 0 ≤ θ ≤ 2π (seven-leaved rose).

Solution: We have

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH7thetaL, theta=Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH7thetaL, theta=2 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH7thetaL, theta=3 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH7thetaL, theta=4 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH7thetaL, theta=5 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH7thetaL, theta=6 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH7thetaL, theta=7 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH7thetaL, theta=8 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH7thetaL, theta=9 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH7thetaL, theta=10 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH7thetaL, theta=11 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=sinH7thetaL, theta=12 Pi�6

18

Section 9.3 Polar Coordinates 2010 Kiryl Tsishchanka

EXAMPLE: Sketch the curve r = 1 +1

10sin(10θ), 0 ≤ θ ≤ 2π.

Solution: We have

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=1+sinH10thetaL�10, theta=Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=1+sinH10thetaL�10, theta=2 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=1+sinH10thetaL�10, theta=3 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=1+sinH10thetaL�10, theta=4 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=1+sinH10thetaL�10, theta=5 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=1+sinH10thetaL�10, theta=6 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=1+sinH10thetaL�10, theta=7 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=1+sinH10thetaL�10, theta=8 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=1+sinH10thetaL�10, theta=9 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=1+sinH10thetaL�10, theta=10 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=1+sinH10thetaL�10, theta=11 Pi�6

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

r=1+sinH10thetaL�10, theta=12 Pi�6

19

Section 9.3 Polar Coordinates 2010 Kiryl Tsishchanka

EXAMPLE: Match the polar equations with the graphs labeled I-VI:

(a) r = sin(θ/2) (b) r = sin(θ/4)

(c) r = sin θ + sin3(5θ/2) (d) r = θ sin θ

(e) r = 1 + 4 cos(5θ) (f) r = 1/√

θ

20

Section 9.3 Polar Coordinates 2010 Kiryl Tsishchanka

Tangents to Polar Curves

To find a tangent line to a polar curve r = f(θ) we regard θ as a parameter and write its parametricequations as

x = r cos θ = f(θ) cos θ y = r sin θ = f(θ) sin θ

Then, using the method for finding slopes of parametric curves and the Product Rule, we have

dy

dx=

dy

dθdx

=

dr

dθsin θ + r cos θ

dr

dθcos θ − r sin θ

(1)

EXAMPLE:

(a) For the cardioid r = 1 + sin θ, find the slope of the tangent line when θ = π/3.

(b) Find the points on the cardioid where the tangent line is horizontal or vertical.

21

Section 9.3 Polar Coordinates 2010 Kiryl Tsishchanka

EXAMPLE:

(a) For the cardioid r = 1 + sin θ, find the slope of the tangent line when θ = π/3.

(b) Find the points on the cardioid where the tangent line is horizontal or vertical.

Solution: Using Equation 1 with r = 1 + sin θ, we have

dy

dx=

drdθ

sin θ + r cos θdrdθ

cos θ − r sin θ=

cos θ sin θ + (1 + sin θ) cos θ

cos θ cos θ − (1 + sin θ) sin θ

=cos θ(1 + 2 sin θ)

1 − 2 sin2 θ − sin θ=

cos θ(1 + 2 sin θ)

(1 + sin θ)(1 − 2 sin θ)

(a) The slope of the tangent at the point where θ = π/3 is

dy

dx

∣∣∣∣θ=π/3

=cos(π/3)(1 + 2 sin(π/3))

(1 + sin(π/3))(1 − 2 sin(π/3))=

12(1 +

√3)

(1 +√

3/2)(1 −√

3)

=1 +

√3

(2 +√

3)(1 −√

3)=

1 +√

3

−1 −√

3= −1

(b) Observe thatdy

dθ= cos θ(1 + 2 sin θ) = 0 when θ =

π

2,

2,

6,

11π

6

dx

dθ= (1 + sin θ)(1 − 2 sin θ) = 0 when θ =

2,

π

6,

6

Therefore there are horizontal tangents at the points (2, π/2), (12, 7π/6), (1

2, 11π/6) and vertical tangents

at (32, π/6) and (3

2, 5π/6). When θ = 3π/2, both dy/dθ and dx/dθ are 0, so we must be careful. Using

l’Hospital’s Rule, we have

limθ→(3π/2)−

dy

dx=

(lim

θ→(3π/2)−

1 + 2 sin θ

1 − 2 sin θ

)(lim

θ→(3π/2)−

cos θ

1 + sin θ

)

= −1

3lim

θ→(3π/2)−

cos θ

1 + sin θ= −1

3lim

θ→(3π/2)−

− sin θ

cos θ= ∞

By symmetry,

limθ→(3π/2)+

dy

dx= −∞

Thus there is a vertical tangent line at the pole.

REMARK: Instead of having to remember Equation 1, we could employ the method used to derive it. Forinstance, in the above Example we could have written

x = r cos θ = (1 + sin θ) cos θ = cos θ +1

2sin 2θ

y = r sin θ = (1 + sin θ) sin θ = sin θ + sin2 θ

Then we would havedy

dx=

dy/dθ

dx/dθ=

cos θ + 2 sin θ cos θ

− sin θ + cos 2θ=

cos θ + sin 2θ

− sin θ + cos 2θ

which is equivalent to our previous expression.

22