Unit 4, lesson 48

Arc Length and Area of a Sector

How a central angle in radians gives the length of an arc and the area of a sector, and how the turning rate of a wheel sets the speed of a point on its rim.

  • Theory part 95
  • Practice part 96
  • Angles and Arcs

A slice of pizza, the patch of lawn a sprinkler reaches and the path of the valve on a bicycle wheel are all pieces of a circle cut out by an angle at its center. With the angle measured in radians, their sizes and speeds come from formulas short enough to remember. The same formulas come back in calculus, when arcs and areas of curved shapes are built from many small pieces.

Key terms

Central angle, arc and sector

Definition A central angle of a circle is an angle whose vertex is the center of the circle. Its two sides are radii, and each meets the circle at one point. The part of the circle that lies inside the angle, running from one of these points to the other, is the arc cut off, or subtended, by the angle. The region bounded by the two radii and the arc is a sector. Throughout, a central angle θ\theta satisfies 0<θ≤2π0 < \theta \le 2\pi in radians, or 0∘<θ≤360∘0^\circ < \theta \le 360^\circ. For a full turn, θ=2π\theta = 2\pi, the two radii coincide, the arc is the whole circle and the sector is the whole disk.

In plain words The central angle is the opening at the tip of a pizza slice. The arc is the crust and the sector is the whole slice.

Example On a clock face the hands at 1212 and 44 form a central angle of one third of a turn, which is 120∘120^\circ or 2π3\frac{2\pi}{3} radians. The arc is the rim from 1212 to 44, and the sector is the wedge of the face between the hands.

Arc length

Definition On a circle of radius rr, a central angle of θ\theta radians with 0<θ≤2π0 < \theta \le 2\pi cuts off an arc of length

s=rθ.s = r\theta.

This is the definition of radian measure, θ=sr\theta = \frac{s}{r}, solved for ss, and ss comes out in the same unit as rr. For an angle of dd degrees the same arc length is

s=d360⋅2πr=πrd180.s = \frac{d}{360} \cdot 2\pi r = \frac{\pi r d}{180}.

In plain words Arc length is radius times angle, as long as the angle is in radians. Using degrees in s=rθs = r\theta is the most common mistake here.

Example On a circle of radius 88 centimeters a central angle of 3π4\frac{3\pi}{4} cuts off

s=8⋅3π4=6π≈18.85 centimeters.\begin{aligned} s &= 8 \cdot \frac{3\pi}{4} = 6\pi \\[0.8em] &\approx 18.85 \text{ centimeters.} \end{aligned}

An angle of 40∘40^\circ on a circle of radius 1515 meters first becomes 40π180=2π9\frac{40\pi}{180} = \frac{2\pi}{9} radians, so

s=15⋅2π9=10π3≈10.47 meters.s = 15 \cdot \frac{2\pi}{9} = \frac{10\pi}{3} \approx 10.47 \text{ meters.}

Plugging in 4040 instead would give 600600 meters, which is longer than the whole circle.

Sector area

Definition A sector of a circle of radius rr with central angle θ\theta radians, 0<θ≤2π0 < \theta \le 2\pi, has area

A=12r2θ.A = \frac{1}{2}r^2\theta.

The reason is that the sector is the fraction θ2π\frac{\theta}{2\pi} of the whole disk, so its area is θ2π⋅πr2\frac{\theta}{2\pi} \cdot \pi r^2. Using the arc length s=rθs = r\theta, the same area is A=12rsA = \frac{1}{2}rs. For an angle of dd degrees the area is d360πr2\frac{d}{360}\pi r^2.

In plain words A sector gets the same share of the disk’s area as its angle gets of a full turn.

Example A sprinkler reaches 1212 meters and sweeps back and forth through 135∘135^\circ, which is 3π4\frac{3\pi}{4} radians. The watered area is

A=12⋅122⋅3π4=54π≈169.65 square meters.\begin{aligned} A &= \frac{1}{2} \cdot 12^2 \cdot \frac{3\pi}{4} = 54\pi \\[0.8em] &\approx 169.65 \text{ square meters.} \end{aligned}

As a check, 135∘135^\circ is 38\frac{3}{8} of a turn, and 38\frac{3}{8} of 144π144\pi is also 54π54\pi.

Angular speed

Definition Suppose an object moves on a circle, or a body rotates about a center, and the angle swept out grows at a constant rate. If it turns through the angle θ\theta in time tt, its angular speed is

ω=θt.\omega = \frac{\theta}{t}.

It is measured in radians per unit of time, or in revolutions or degrees per unit of time. One revolution is 2π2\pi radians, so nn revolutions per minute equals 2πn2\pi n radians per minute.

In plain words Angular speed says how fast something turns, without caring how big it is.

Example A wheel turns at 300300 revolutions per minute. In radians,

ω=300⋅2π=600π≈1885 radians per minute,\begin{aligned} \omega &= 300 \cdot 2\pi = 600\pi \\[0.4em] &\approx 1885 \text{ radians per minute,} \end{aligned}

which is 10π≈31.4210\pi \approx 31.42 radians per second.

Linear speed

Definition Suppose an object moves along a circular path at a constant rate. If it covers an arc of length ss in time tt, its linear speed is

v=st,v = \frac{s}{t},

measured in units of length per unit of time. It is the ordinary speed of the object along its path.

In plain words Linear speed is how fast the object actually travels, the number a speedometer would show if it rode along.

Example A runner on a circular track covers 4545 meters of the curve in 99 seconds. The linear speed is 459=5\frac{45}{9} = 5 meters per second, whatever the radius of the track.

Linear speed from angular speed

Definition A point at distance rr from the center of rotation, turning with angular speed ω\omega in radians per unit of time, has linear speed

v=rω.v = r\omega.

This follows from s=rθs = r\theta, since

v=st=rθt=rω.v = \frac{s}{t} = \frac{r\theta}{t} = r\omega.

The formula needs ω\omega in radians per unit of time. Radians have no unit, so vv comes out in the unit of rr per unit of time.

In plain words Every point of a turning wheel has the same angular speed, but points farther from the center travel farther in each turn, so they move faster.

Example A bicycle wheel of radius 0.350.35 meters turns 44 times per second, so ω=8π\omega = 8\pi radians per second. A point on the rim moves at

v=0.35⋅8π=2.8π≈8.80 meters per second,\begin{aligned} v &= 0.35 \cdot 8\pi = 2.8\pi \\[0.4em] &\approx 8.80 \text{ meters per second,} \end{aligned}

about 31.731.7 kilometers per hour. A point on a spoke 0.10.1 meters from the hub has the same ω\omega but moves at only 0.8π≈2.510.8\pi \approx 2.51 meters per second.

Check yourself

  1. Find the length of the arc cut off by a central angle of 150∘150^\circ on a circle of radius 66 inches.
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The angle is 150⋅π180=5π6150 \cdot \frac{\pi}{180} = \frac{5\pi}{6} radians, so

s=6⋅5π6=5π≈15.71 inches.s = 6 \cdot \frac{5\pi}{6} = 5\pi \approx 15.71 \text{ inches.}
  1. A sector of a circle of radius 1010 centimeters has an arc of length 77 centimeters. Find its central angle and its area.
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The angle is θ=710=0.7\theta = \frac{7}{10} = 0.7 radians. The area is

A=12rs=12⋅10⋅7=35 square centimeters.\begin{aligned} A &= \frac{1}{2}rs = \frac{1}{2} \cdot 10 \cdot 7 \\[0.8em] &= 35 \text{ square centimeters.} \end{aligned}
  1. A Ferris wheel of radius 2525 meters makes one full turn every 44 minutes. Find its angular speed in radians per minute and the linear speed of a seat on the rim.
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One turn is 2π2\pi radians, so ω=2π4=π2\omega = \frac{2\pi}{4} = \frac{\pi}{2} radians per minute. A seat on the rim moves at

v=rω=25⋅π2=25π2≈39.27 meters per minute.\begin{aligned} v &= r\omega = 25 \cdot \frac{\pi}{2} = \frac{25\pi}{2} \\[0.8em] &\approx 39.27 \text{ meters per minute.} \end{aligned}

How this lesson connects