Section 1.1 Angle Measure and Arc Length Section 1.1 Angle Measure and Arc Length.
Section 6.1 Angle Measure
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Section 6.1 Angle Measure An angle AOB consists of two rays R 1 and R 2 with a common vertex O (see the Figures below). We often interpret an angle as a rotation of the ray R 1 onto R 2 . In this case, R 1 is called the initial side, and R 2 is called the terminal side of the angle. If the rotation is counterclockwise, the angle is considered positive, and if the rotation is clockwise, the angle is considered negative. Angle Measure The measure of an angle is the amount of rotation about the vertex required to move R 1 onto R 2 . Intuitively, this is how much the angle “opens.” One unit of measurement for angles is the degree. An angle of measure 1 degree is formed by rotating the initial side 1 360 of a complete revolution. In calculus and other branches of mathematics, a more natural method of measuring angles is used — radian measure. The amount an angle opens is measured along the arc of a circle of radius 1 with its center at the vertex of the angle. The circumference of the circle of radius 1 is 2π and so a complete revolution has measure 2π rad, a straight angle has measure π rad, and a right angle has measure π/2 rad. An angle that is subtended by an arc of length 2 along the unit circle has radian measure 2 (see the Figures below). Since a complete revolution measured in degrees is 360 ◦ and measured in radians is 2π rad, we get the following simple relationship between these two methods of angle measurement. 1
Transcript of Section 6.1 Angle Measure
Section 6.1 Angle Measure
An angle AOB consists of two rays R1 and R2 with a common vertex O (see the Figuresbelow). We often interpret an angle as a rotation of the ray R1 onto R2. In this case, R1 iscalled the initial side, and R2 is called the terminal side of the angle. If the rotation iscounterclockwise, the angle is considered positive, and if the rotation is clockwise, the angleis considered negative.
Angle Measure
The measure of an angle is the amount of rotation about the vertex required to move R1
onto R2. Intuitively, this is how much the angle “opens.” One unit of measurement for anglesis the degree. An angle of measure 1 degree is formed by rotating the initial side 1
360of a
complete revolution. In calculus and other branches of mathematics, a more natural methodof measuring angles is used — radian measure. The amount an angle opens is measured alongthe arc of a circle of radius 1 with its center at the vertex of the angle.
The circumference of the circle of radius 1 is 2π and so a complete revolution has measure 2πrad, a straight angle has measure π rad, and a right angle has measure π/2 rad. An angle thatis subtended by an arc of length 2 along the unit circle has radian measure 2 (see the Figuresbelow).
Since a complete revolution measured in degrees is 360◦ and measured in radians is 2π rad, weget the following simple relationship between these two methods of angle measurement.
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To get some idea of the size of a radian, notice that
1 rad ≈ 57.296◦ and 1◦ ≈ 0.01745 rad
EXAMPLE:
(a) Express 60◦ in radians. (b) Expressπ
6rad in degrees.
Solution:
(a) 60◦ = 60( π
180
)rad =
π
3rad (b)
π
6rad =
(π
6
) (180
π
)= 30◦
EXAMPLE:
(a) Express 40◦ in radians. (b) Express3π
4rad in degrees.
Solution:
(a) 40◦ = 40( π
180
)rad =
2π
9rad (b)
3π
4rad =
(3π
4
)(180
π
)= 135◦
Angles in Standard Position
An angle is in standard position if it is drawn in the xy-plane with its vertex at the origin andits initial side on the positive x-axis. The Figures below give examples of angles in standardposition.
(a) (b) (c) (d)
Two angles in standard position are coterminal if their sides coincide. The angles in Figures(a) and (c) above are coterminal.
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EXAMPLE:
(a) Find angles that are coterminal with the angle θ = 30◦ in standard position.
(b) Find angles that are coterminal with the angle θ =π
3in standard position.
Solution:
(a) To find positive angles that are coterminal with θ, we add any multiple of 360◦ to 30◦. Thus
30◦ + 360◦ = 390◦, 30◦ + 720◦ = 750◦, etc.
are coterminal with θ = 30◦. To find negative angles that are coterminal with θ, we subtractany multiple of 360◦ from 30◦. Thus
30◦ − 360◦ = −330◦, 30◦ − 720◦ = −690◦, etc.
are coterminal with θ.
(b) To find positive angles that are coterminal with θ, we add any multiple of 2π to π/3. Thus
π
3+ 2π =
7π
3,
π
3+ 4π =
13π
3, etc.
are coterminal with θ = π/3. To find negative angles that are coterminal with θ, we subtractany multiple of 2π from π/3. Thus
π
3− 2π = −5π
3,
π
3− 4π = −11π
3, etc.
are coterminal with θ.
EXAMPLE:
(a) Find angles that are coterminal with the angle θ = 62◦ in standard position.
(b) Find angles that are coterminal with the angle θ =5π
6in standard position.
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EXAMPLE:
(a) Find angles that are coterminal with the angle θ = 62◦ in standard position.
(b) Find angles that are coterminal with the angle θ =5π
6in standard position.
Solution:
(a) To find positive angles that are coterminal with θ, we add any multiple of 360◦ to 62◦. Thus
62◦ + 360◦ = 422◦, 62◦ + 720◦ = 782◦, etc.
are coterminal with θ = 62◦. To find negative angles that are coterminal with θ, we subtractany multiple of 360◦ from 62◦. Thus
62◦ − 360◦ = −298◦, 62◦ − 720◦ = −658◦, etc.
are coterminal with θ.
(b) To find positive angles that are coterminal with θ, we add any multiple of 2π to 5π/6. Thus
5π
6+ 2π =
17π
6,
5π
6+ 4π =
29π
6, etc.
are coterminal with θ = 5π/6. To find negative angles that are coterminal with θ, we subtractany multiple of 2π from 5π/6. Thus
5π
6− 2π = −7π
6,
5π
6− 4π = −19π
6, etc.
are coterminal with θ.
EXAMPLE: Find an angle with measure between 0◦ and 360◦ that is coterminal with the angleof measure 1290◦ in standard position.
Solution: We can subtract 360◦ as many times as we wish from 1290◦, and the resulting anglewill be coterminal with 1290◦. Thus, 1290◦ − 360◦ = 930◦ is coterminal with 1290◦, and so isthe angle 1290◦ − 2(360)◦ = 570◦.
To find the angle we want between 0◦ and 360◦, we subtract 360◦ from 1290◦ as many timesas necessary. An efficient way to do this is to determine how many times 360◦ goes into 1290◦,that is, divide 1290 by 360, and the remainder will be the angle we are looking for. We seethat 360 goes into 1290 three times with a remainder of 210. Thus, 210◦ is the desired angle(see the Figures below).
EXAMPLE:
(a) Find an angle between 0 and 2π that is coterminal with 100.
(b) Find an angle between 0◦ and 360◦ that is coterminal with −3624◦.
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EXAMPLE:
(a) Find an angle between 0 and 2π that is coterminal with 100.
(b) Find an angle between 0◦ and 360◦ that is coterminal with −3624◦.
Solution:
(a) We have100 − 30π ≈ 5.7522
(b) We have−3624◦ + 11 · 360◦ = −3624◦ + 3960◦ = 336◦
EXAMPLE: Find an angle between 0 and 2π that is coterminal with88π
3.
Solution: We have88π
3− 28π =
4π
3
EXAMPLE: Find an angle with measure between 0◦ and 360◦ that is coterminal with the angleof measure 1635◦ in standard position.
Solution: We have1635◦ − 4 · 360◦ = 1635◦ + 1440◦ = 195◦
Length of a Circular Arc
An angle whose radian measure is θ is subtended by an arc that is thefraction θ/(2π) of the circumference of a circle. Thus, in a circle of radiusr, the length s of an arc that subtends the angle θ (see the Figure on theright) is
s =θ
2π× circumference of circle =
θ
2π(2πr) = θr
Solving for θ, we get the important formula
θ =s
r
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The formula above allows us to define radian measure using a circle of any radius r: The radianmeasure of an angle θ is s/r, where s is the length of the circular arc that subtends θ in a circleof radius r (see the Figures below).
EXAMPLE:
(a) Find the length of an arc of a circle with radius 10 m that subtends a central angle of 30◦.
(b) A central angle θ in a circle of radius 4 m is subtended by an arc of length 6 m. Find themeasure of θ in radians.
Solution:
(a) We know that 30◦ =π
6. So the length of the arc is
s = rθ = (10)π
6=
5π
3m
(b) By the formula θ = s/r, we have
θ =s
r=
6
4=
3
2rad
EXAMPLE:
(a) Find the length of an arc of a circle with radius 21 m that subtends a central angle of 15◦.
(b) A central angle θ in a circle of radius 9 m is subtended by an arc of length 12 m. Find themeasure of θ in radians.
Solution:
(a) We know that 15◦ =π
12. So the length of the arc is
s = rθ = (21)π
12=
7π
4m
(b) By the formula θ = s/r, we have
θ =s
r=
12
9=
4
3rad
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Area of a Circular Sector
The area of a circle of radius r is A = πr2. A sector of this circle with centralangle θ has an area that is the fraction θ/(2π) of the area of the entire circle(see the Figure on the right). So the area of this sector is
A =θ
2π× area of circle =
θ
2π(πr2) =
1
2r2θ
EXAMPLE: Find the area of a sector of a circle with central angle 60◦ if the radius of the circleis 3 m.
Solution: To use the formula for the area of a circular sector, we must find the central angle ofthe sector in radians:
60◦ = 60( π
180
)rad =
π
3rad
Thus, the area of the sector is
A =1
2r2θ =
1
2(3)2
(π
3
)=
3π
2m2
EXAMPLE: Find the area of a sector of a circle with central angle 4◦ if the radius of the circleis 45 m.
Solution: To use the formula for the area of a circular sector, we must find the central angle ofthe sector in radians:
4◦ = 4( π
180
)rad =
π
45rad
Thus, the area of the sector is
A =1
2r2θ =
1
2(45)2
( π
45
)=
45π
2m2
EXAMPLE: A sprinkler on a golf course fairway is set to spray water overa distance of 70 feet and rotates through an angle of 120◦ (see the Figureon the right). Find the area of the fairway watered by the sprinkler.
Solution: First convert 120◦ to radian measure as follows.
θ = 120◦ = 120( π
180
)rad =
2π
3rad
Therefore
A =1
2r2θ =
1
2(70)2
(2π
3
)=
4900π
3≈ 5131 ft2
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Circular Motion
Suppose a point moves along a circle as shown in the Figure on the right.There are two ways to describe the motion of the point — linear speed andangular speed. Linear speed is the rate at which the distance traveledis changing, so linear speed is the distance traveled divided by the timeelapsed. Angular speed is the rate at which the central angle θ is changing,so angular speed is the number of radians this angle changes divided by thetime elapsed.
EXAMPLE: A boy rotates a stone in a 3-ft-long sling at the rate of 15revolutions every 10 seconds. Find the angular and linear velocities ofthe stone.
Solution: In 10 s, the angle θ changes by 15 · 2π = 30π radians. So the angular speed of thestone is
ω =θ
t=
30π rad
10 s= 3π rad/s
The distance traveled by the stone in 10 s is s = 15 · 2πr = 15 · 2π · 3 = 90π ft. So the linearspeed of the stone is
v =s
t=
90π ft
10 s= 9π ft/s
EXAMPLE: A disk with a 12-inch diameter spins at the rate of 45 revolutions per minute.Find the angular and linear velocities of a point at the edge of the disk in radians per secondand inches per second, respectively.
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EXAMPLE: A disk with a 12-inch diameter spins at the rate of 45 revolutions per minute.Find the angular and linear velocities of a point at the edge of the disk in radians per secondand inches per second, respectively.
Solution: In 60 s, the angle θ changes by 45 · 2π = 90π radians. So the angular velocity is
ω =θ
t=
90π rad
60 s=
3
2π rad/s
The distance traveled by the point in 60 s is s = 45 · πd = 45 · π · 12 = 540π in. So the linearspeed of the point is
v =s
t=
540π in
60 s= 9π in/s
EXAMPLE: The second hand of a clock is 10.2 centimeters long, asshown in the Figure on the right. Find the linear speed of the tip of thissecond hand as it passes around the clock face.
Solution: In one revolution, the arc length traveled is
s = 2πr = 2π(10.2) = 20.4π cm
The time required for the second hand to travel this distance is 60 seconds. So, the linear speedof the tip of the second hand is
v =s
t=
20.4π cm
60 s≈ 1.068 cm/s
EXAMPLE: A Ferris wheel with a 50-foot radius (see the Figure on theright) makes 1.5 revolutions per minute.
(a) Find the angular speed of the Ferris wheel in radians per minute.
(b) Find the linear speed of the Ferris wheel.
Solution:
(a) Because each revolution generates 2π radians, it follows that the wheel turns (1.5)(2π) = 3πradians per minute. In other words, the angular speed is
ω =θ
t=
3π rad
1 min= 3π rad/min
(b) The linear speed is
v =s
t=
rθ
t=
50(3π) ft
1 min≈ 471.2 ft/min
REMARK: Notice that angular speed does not depend on the radius of the circle, but onlyon the angle θ. However, if we know the angular speed ω and the radius r, we can find linearspeed as follows:
v =s
t=
rθ
t= r
(θ
t
)= rω
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EXAMPLE: A woman is riding a bicycle whose wheels are 26 inches in diameter. If the wheelsrotate at 125 revolutions per minute (rpm), find the speed at which she is traveling, in mi/h.
Solution: The angular speed of the wheels is 2π · 125 = 250π rad/min. Since the wheels haveradius 13 in. (half the diameter), the linear speed is
v = rω = 13 · 250π ≈ 10, 210.2 in./min
Since there are 12 inches per foot, 5280 feet per mile, and 60 minutes per hour, her speed inmiles per hour is
10, 210.2 in./min × 60 min/h
12 in./ft × 5280 ft/mi=
612, 612 in./h
63, 360 in./mi≈ 9.7 mi/h
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