This section covers:
Around the time you’re learning about radian measures, you may have to work with Linear and Angular Speeds, and also Area of Sectors and Lengths of Arcs.
Note: For the purposed of this section, we will talk about linear speed and angular speed (scalar quantities), as opposed to linear velocity and angular velocity (vector quantities).
We discussed radians and how they related to degrees of central angles here in the Angles and the Unit Circle section. Before we talk about linear and angular speeds, let’s go over how radians relate back to the circumference of a circle and also revolutions of a circle.
Note that \(\theta \) is the central angle of rotation in radians, \(r\) is the radius of the circle, \(s\) is the arc length, or intercepted arc (part of the circumference) of the circle, and a revolution (or rotation) is when an object has gone all the way around a circle (or the circle returns back to where it started). These units can all relate back to the circumference of a circle: \(2\pi r\), where \(r\) is the radius. (Remember in a Unit Circle, the circumference is just \(2\pi \), since \(r=1\)).
Note that the words radian and radius are related, since there are \(2\pi r\) (radians) in a revolution, and \(2\pi r\) (radius) is the measurement of the circumference.
One of the most important concepts is that the length of an intercepted arc is the radius times the radian measure of that arc’s central angle. To see this, set up a proportion comparing this arc to the whole circumference:
\(\displaystyle \require{cancel} \frac{{\text{Length of arc}}}{{\text{Circumference}}}=\frac{{\text{Arc }\!\!’\!\!\text{ s Angle}}}{{\text{Total Angle Measurement in Circle}}}:\,\,\,\,\frac{s}{{\cancel{{2\pi }}r}}=\frac{\theta }{{\cancel{{2\pi }}}}\)
From here you can see that \(s=r\theta \).
I’m giving you some basic formulas, but honestly, when I do most of these problems, I just use Unit Multipliers (Dimensional Analysis) to get to the units needed for the answer of the problem!!
Linear and Angular Speed Formulas:
We’ll use these formulas in some of the linear/angular speed problems below.
Remember that you typically use radius with linear speed, and radians with angular speed in \(2\pi r\).
Linear Speed
Linear speed is the speed at which a point on the outside of the object travels in its circular path around the center of that object. The units can be any usual speed units, such as miles per hour, meters per second, and so on.
We remember that \(\text{Distance}=\text{Rate}\times \text{Time}\), or \(\displaystyle \text{Rate (Speed)}=\frac{{\text{Distance}}}{{\text{Time}}}\). We’ll first talk about how fast an object along the circumference of a circle is changing.
Think of a car that drives around in a circle on a track with arc length (the actual length of the curvy part – part of the circumference) \(s\). The formula for the speed around a circle, or the linear speed is \(\displaystyle v=\frac{s}{t}\), where \(s\) is the arc length and \(t\) is the time.
Note that linear speed needs to have a circumference (or radius) in the problem!
Here’s a type of problem that you might have. Note that we have to use Unit Multipliers (Dimensional Analysis) when the units don’t match.
Linear Speed Problem  Solution 
A car travels at a constant speed around a track that is circular and has a circumference of 4 miles.
If the car completes 8 laps in 10 minutes, what is the linear speed of the car in miles per hour? 
Let’s first calculate the distance that the car travels. If the circle has a circumference of 4 miles and the car goes 8 laps, the car goes a total of 32 miles:
\(\require{cancel} \displaystyle 8\cancel{{\text{laps}}}\cdot \frac{{4\,\,\text{miles}}}{{\text{1 }\cancel{{\text{lap}}}}}=32\,\,\text{miles}\)
Now let’s get the linear speed using the formula \(\displaystyle v=\frac{s}{t}\). We know \(s=32\) miles, and \(t=10\) minutes. But we have to add some unit multipliers since we have our time in minutes, and we want to end up with miles per hour:
\(\displaystyle v=\frac{s}{t}=\frac{{32\,\,\text{miles}}}{{10\,\,\cancel{{\text{minutes}}}}}\cdot \frac{{60\,\,\cancel{{\text{minutes}}}}}{{1\,\,\text{hour}}}=192\,\,\text{miles/hour}\)
The linear speed of the car is 192 miles per hour. See how we could have figured this out without the formula, but simply using unit multipliers? 
A wheel with a diameter of 10 inches rotates at a constant rate of 2 revolutions per second.
Find the linear speed of the wheel in miles per hour (1 mile is approximately 5,280 feet). 
We want to end with miles per hour, so let’s put that at the end of our unit multiplier equation; we know we need miles on the top and hours on the bottom somewhere):
\(\displaystyle …\,\,\frac{?}{\text{?}}\cdot \frac{?}{?}\cdot \frac{?}{{\text{?}\,\,\text{hours}}}\cdot \frac{{?\,\,\,\text{miles}}}{?}\,\,…=?\,\,\text{miles per hour}\)
Let’s create ratios of everything else we know, making sure we can cross out anything we don’t need. We need to know that one revolution of a circle is the equivalent of \(\pi d\) inches. You may want to start out with a shorter equation, and add multipliers slowly:
\(\displaystyle …\,\,\frac{{2\,\,\cancel{{\text{revolutions}}}}}{{\text{1}\,\,\text{second}\,}}\cdot \frac{{\pi d}}{{1\,\,\cancel{{\text{revolution}}}}}\cdot \frac{?}{{\text{?}\,\,\text{hours}}}\cdot \frac{{?\,\,\,\text{miles}}}{?}\,\,…=?\,\,\text{miles per hour}\) \(\displaystyle \frac{{2\,\,\cancel{{\text{revolutions}}}}}{{\text{1}\,\,\cancel{{\text{second}}}\,}}\cdot \frac{{\pi \left( {10} \right)\,\,\cancel{{\text{inches}}}}}{{1\,\,\cancel{{\text{revolution}}}}}\cdot \frac{{60\,\,\cancel{{\text{min}}}}}{{\text{1}\,\,\,\text{hour}}}\cdot \frac{{60\,\,\cancel{{\text{seconds}}}}}{{1\,\,\cancel{{\min }}}}\cdot \frac{{1\,\,\,\text{mile}}}{{5280\,\,\cancel{{\text{ft}}}}}\cdot \frac{{1\,\,\cancel{{\text{ft}}}}}{{12\,\,\cancel{{\text{inches}}}}}\)
\(\displaystyle \,=\frac{{2\cdot \pi \left( {10} \right)\cdot 60\cdot 60}}{{5280\cdot 12}}\,\,\text{miles per hour}\approx 3.57\,\,\text{miles per hour}\)
The linear speed of the wheel is 3.57 miles per hour. 
Angular Speed
Angular speed is the rate at which the object turns, described in units like revolutions per minute, degrees per second, radians per hour, and so on.
Angular speed has to do with how fast the central angle of a circle is changing, as opposed to the circumference of the circle.
Again, think of a car that drives around in a circle on a track with central angle \(\theta \). The formula for the speed around a circle in terms of this angle, or the angular speed is \(\displaystyle \omega =\frac{\theta }{t}\), where \(\theta \) is in radians, and \(t\) is the time.
Note that angular speed does NOT need a circumference (or radius) in the problem!
Here’s a type of problem that you might have. Note that we have to use Unit Multipliers (Dimensional Analysis) again when the units don’t match.
Angular Speed Problem  Solution 
Heather flashes a flashlight and turns in a circle in a constant speed.
If Heather’s flashlight completes one rotation (revolution) every 15 seconds, what is the angular speed of the light coming from the flashlight in radians per minute? 
Since the problem includes information about the time it takes for one rotation (revolution), we need to use the fact that there are \(2\pi \) radians in one revolution.
Now let’s use our angular speed equation \(\displaystyle \omega =\frac{{\theta \text{ (radians)}}}{t}\) with some unit multipliers. Start out with what we know and end up with what we need to know (we need to have radians in answer): \(\require{cancel} \displaystyle \omega =\frac{\theta }{t}=\frac{{\cancel{{1\text{ rotation}}}}}{{15\,\,\cancel{{\text{seconds}}}}}\cdot \frac{{2\pi \text{ radians}}}{{\cancel{{1\text{ rotation}}}}}\cdot \frac{{60\,\,\cancel{{\text{seconds}}}}}{{1\,\,\text{minute}}}=\,8\pi \,\,\text{radians/minute}\)
The angular speed of the light is \(8\pi \) radians per minute. Again, see how we could have figured this out without the formula, but simply using unit multipliers?

Here’s an example that shows the difference between finding angular speeds and linear speeds. Notice with angular speeds, we ignore the radius, since we are just dealing with rotation of an angle. Also notice that:
\(\text{Linear Speed}=\text{Radius}\times \text{Angular Speed}\), or \(\displaystyle \text{Radius}=\frac{{\text{Linear Speed}}}{{\text{Angular Speed}}}\)
Linear and Angular Speed Problems  Solution 
Janie sits on a carousel and she is 6 feet from the center.
Find her angular and linear speeds if the carousel is moving at 5 revolutions per minute.
Note that: \(\displaystyle \text{radius}=\frac{{\text{linear speed}}}{{\text{angular speed}}}\) 
The equation for the angular speed is \(\displaystyle \omega =\frac{{\theta \text{ (in radians)}}}{t}\), but we are already given a speed in terms of revolutions per minute. Convert this speed to an angular speed by using unit multipliers (knowing that there are \(2\pi \) radians in one revolution):
\(\require{cancel} \displaystyle \frac{{5\text{ }\cancel{{\text{revolutions}}}}}{{\text{minute}}}\cdot \frac{{2\pi \text{ radians}}}{{1\text{ }\cancel{{\text{revolution}}}}}=10\pi \,\,\text{radians per minute}\). Notice that we don’t use the radius, since we’re just interested in an angular speed.
To get the linear speed, let’s use unit multipliers again, but now we have to look at the radius (since we need the distance around the circumference of the circle with this radius). You can also remember that \(s\) (arc length, or part of the circumference) is \(2\pi r\) for one revolution: \(\displaystyle \begin{align}\frac{{5\text{ }\cancel{{\text{revolutions}}}}}{{\text{minute}}}\cdot \frac{{2\pi r\text{ feet}}}{{1\text{ }\cancel{{\text{revolution}}}}}&=\frac{{5\text{ }\cancel{{\text{revolutions}}}}}{{\text{minute}}}\cdot \frac{{2\pi \left( 6 \right)\text{ feet}}}{{1\text{ }\cancel{{\text{revolution}}}}}\\&=60\pi \,\text{feet per minute}\end{align}\)
(We can also use the formula \(\text{Linear Speed}=\text{Radius}\times \text{Angular Speed}\) to get \(60\pi \).)

Here are more problems with linear and angular speeds, and rotations/revolutions:
Linear and Angular Speed Problems  Solution 
Brie wants to jump on a moving carousel with a diameter of 50 feet and traveling at 3 revolutions per minute.
How fast must Brie run to match the carousel’s speed to jump on (in feet per second)? 
Since we want to know how fast Brie must run, we are looking for a linear speed. But we are given a speed of 3 revolutions per minute, so let’s just relate the speeds to each other.
We can start with what we know, and using unit multipliers, start filling in units, putting units in numerator or denominator, depending on whether or not we want them to stay or cross them out. Remember that for every rotation or revolution, the linear distance or arc length \(s=2\pi r\), where \(\displaystyle r=\frac{{\text{diameter}}}{2}\): \(\require{cancel} \displaystyle \begin{align}\frac{{3\text{ revolutions}}}{{1\text{ minute}}}\cdot \frac{?}{?}\cdot \frac{{2\pi \left( {25} \right)\,\,\text{feet}}}{{1\text{ revolution}}}&=\frac{{\text{feet}}}{{\text{second}}}\\\frac{{3\,\,\cancel{{\text{revolutions}}}}}{{1\,\,\cancel{{\text{minute}}}}}\cdot \frac{{1\,\,\cancel{{\text{minute}}}}}{{60\,\,\text{seconds}}}\cdot \frac{{50\pi \,\text{ feet}}}{{1\,\,\,\cancel{{\text{revolution}}}}}&=\frac{{3\cdot 50\pi \text{ }}}{{\text{60 }}}\frac{{\text{feet}}}{{\text{second}}}\\&=5\pi \,\,\text{feet per second}\\&\approx \text{14 ft/sec}\end{align}\)
Brie needs to run about 14 feet per second to jump on the carousel. That’s fast! 
What is the rotation in revolutions per minute given an angular speed of 50 radians per hour?  We have an angular speed of 50 radians per hour, but we want a rotation speed in revolutions per minute.
We can start with what we know, and using unit multipliers, start filling in units, putting units in numerator or denominator, depending on whether or not we want them to stay or cross them out: \(\begin{align}\frac{{50\text{ radians}}}{{1\text{ hour}}}\cdot \frac{?}{?}\cdot \frac{?}{?}&=\frac{{\text{revolutions}}}{{\text{minute}}}\\\frac{{50\text{ }\cancel{{\text{radians}}}}}{{\cancel{{1\text{ hour}}}}}\cdot \frac{{\text{1 revolution}}}{{2\pi \,\,\,\cancel{{\text{radians}}}}}\cdot \frac{{\cancel{{1\,\,\text{hour}}}}}{{\text{60 minutes}}}&=\frac{{50}}{{2\pi \cdot 60}}\frac{{\text{revolutions}}}{{\text{minute}}}\\&=\frac{5}{{12\pi }}\,\,\,\text{revolutions per minute}\end{align}\)
The rotation is \(\displaystyle \frac{5}{{12\pi }}\), or .133 revolutions per minute. 
What is the radius of a circle with linear speed of 100 inches per hour and .1 revolutions per minute?  Let’s use the formula that relates the radius with the linear and angular speeds: \(\displaystyle \text{radius}=\frac{{\text{linear speed}}}{{\text{angular speed}}}=\frac{v}{\omega }\). We also have to use unit multipliers to change the units since have both hours and minutes. We also know that one revolution is \(2\pi \) (and radians can be unitless). We have:
\(\displaystyle \begin{align}\text{radius}&=\frac{v}{\omega }=\frac{{\frac{{100\,\,\text{inches}}}{{1\,\,\text{hour}}}}}{{\frac{{.1\,\,\text{revolution}}}{{1\,\,\text{minute}}}}}\\&=\frac{{100\,\,\text{inches}}}{{1\,\,\cancel{{\text{hour}}}}}\cdot \frac{{1\,\,\cancel{{\text{minute}}}}}{{.1\,\,\cancel{{\text{revolution}}}}}\cdot \frac{{1\,\,\cancel{{\text{hour}}}}}{{60\,\,\cancel{{\text{minutes}}}}}\cdot \frac{{1\,\,\cancel{{\text{revolution}}}}}{{2\pi }}\\&=\frac{{100}}{{.1\cdot 60\cdot 2\pi }}\end{align}\)
The radius is \(\displaystyle \frac{{25}}{{3\pi }}\), or about 2.65 inches.
Note: We could have also done this without the formula, using unit multipliers. Start out with what we know, and end up with what we need to know (we need to have inches in answer, so we need inches on top): \(\displaystyle \frac{{100\,\,\text{inches}}}{{1\,\,\cancel{{\text{hour}}}}}\cdot \frac{{\cancel{{1\text{ revolution}}}}}{{2\pi \text{ radians}}}\cdot \frac{{1\text{ }\cancel{{\text{minute}}}}}{{.1\text{ }\cancel{{\text{revolution}}}}}\cdot \frac{{1\,\,\cancel{{\text{hour}}}}}{{60\,\,\cancel{{\text{minutes}}}}}=\,\frac{{25}}{{3\pi }}\,\,\text{inches}\) 
Area of Sectors
Let’s get the area of a sector of a circle based on the radius and a central angle in radians. We know from geometry that a sector of a circle is like a pizza slice; it’s a region bounded by a central angle and its intercepted arc.
You may have seen in geometry how to get the area of a sector based on radius and degrees of the sector: \(\displaystyle \frac{{\text{area of the sector}}}{{\text{area of the whole circle}}}=\frac{{\text{degrees of the sector}}}{{360{}^\circ }}\).
Let’s use the fact that the area of a circle is \(\pi {{r}^{2}}\), the arc length of a sector is \(r\theta \), and the arc length of a whole circle is \(2\pi r\). Now we can solve for the area of a sector given a radius and central angle: \(\displaystyle \frac{{\text{area of the sector }(A)}}{{\pi\,{{r}^{2}}}}=\frac{{r\theta }}{{2\pi }};\,\,\,\,\,\,\,\,\,A=\frac{1}{2}{{r}^{2}}\theta \).
Length of Arcs
You may have seen in geometry how to get the length of an arc based on radius and degrees of the sector:
\(\displaystyle \frac{{\text{length of intercepted arc}}}{{\text{ arc length (circum}\text{.) of circle (2}\pi \text{r)}}}=\frac{{\text{degrees of the central angle of the arc}}}{{360{}^\circ }}=\frac{{\text{radians of the central angle of the arc}}}{{2\pi r}}\)
From the first proportion above and the last one, we can derive what we already know: \(s=r\theta \), where \(s=\) the length of the intercepted arc, \(r\) is the radius of the circle, and \(\theta \) is the intercepted arc. It’s as simple as that!
Let’s get the area of a sector and length of intercepted arc, given the radius and central angle:
Radius/Central Angle  Area of Sector  Length of Intercepted Arc 
Since the equation for the area of a sector is \(\displaystyle A=\frac{1}{2}{{r}^{2}}\theta \), we have:
\(\displaystyle A=\frac{1}{2}{{r}^{2}}\theta =\frac{1}{2}{{\left( {3.5} \right)}^{2}}\left( {\frac{{2\pi }}{5}} \right)=7.7\,\,\text{cm}\) 
Since the equation for the length of an intercepted arc is \(\displaystyle s=r\,\theta \), we have:
\(\displaystyle s=r\theta =\left( {3.5} \right)\left( {\frac{{2\pi }}{5}} \right)=4.4\,\,\text{cm}\) 

The equation for the area of a sector is \(\displaystyle A=\frac{1}{2}{{r}^{2}}\theta \), but we have to convert our degrees into radians (there are \(\pi \) radians in 180°).
We have: \(\displaystyle 260\cancel{{\text{degrees}}}\cdot \frac{{\pi \,\,\text{radians}}}{{180\,\,\cancel{{\text{degrees}}}}}=\frac{{13\pi }}{9}\,\text{radians}\)
Now we can use the equation: \(\displaystyle A=\frac{1}{2}{{r}^{2}}\theta =\frac{1}{2}{{\left( {15} \right)}^{2}}\left( {\frac{{13\pi }}{9}} \right)=510.51\,\,\text{ft}\) 
The equation for the area of a sector is \(\displaystyle s=\,r\theta \), but we have to convert our degrees into radians (there are \(\pi \) radians in 180°).
We have: \(\displaystyle 260\,\,\cancel{{\text{degrees}}}\cdot \frac{{\pi \,\,\text{radians}}}{{180\,\,\cancel{{\text{degrees}}}}}=\frac{{13\pi }}{9}\,\text{radians}\).
Now we can use the equation: \(\displaystyle s=r\theta =\left( {15} \right)\left( {\frac{{13\pi }}{9}} \right)=68.1\,\,\text{ft}\)

Understand these problems, and practice, practice, practice!
On to Graphs of Trig Functions – you’re ready!