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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
dθ
=
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,
3π
2,
7π
6,
11π
6
dx
dθ= (1 + sin θ)(1 − 2 sin θ) = 0 when θ =
3π
2,
π
6,
5π
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
