The following activity is a one day activity dealing with trigonometric functions. Before beginning this activity, students should have been introduced to sine and cosine.

Suppose a Ferris wheel with a radius of 20 feet makes a complete revolution in 10 seconds. In this activity, we want students to develop a mathematical model that describes the relationship between the height h of a rider above the bottom of a Ferris wheel (4 feet above the ground) and time t.

First, students should develop a table of t- and h- values. Assuming that the rider is at the bottom of the Ferris wheel when t = 0, students can develop values of h for t = 0 through t = 10. Values of h can be estimated from a scale drawing of the Ferris wheel (as shown below).

Time t (sec)    Height h       Height     
                  (ft)         above      
                               ground     
    0.00           0        4             
    1.25           10       14            
    2.50           20       24            
    3.75           30       34            
    5.00           40       44            
    6.25           30       34            
    7.50           20       24            
    8.75           10       14            
   10.00           0        4             

Students should plot their data and note the periodicity of the function. (See the below graph.) Each time the person goes around the Ferris wheel, the graph will "repeat itself". Hopefully, students will conjecture that the graph has a sinusoidal shape.

The below graph shows two revolutions around the Ferris wheel.

Students can now use right-triangle trigonometry and simple proportions (see below picture) to derive the parametric representation of a point (x(t),y(t)) on the rotating Ferris wheel as a function of time, thereby establishing that the height is a sinusoidal function of t.

Let's look at the right triangle with vertices W, Z, and P=(x(t),y(t)). Call angle (ZWP) . Students should have a good understanding of . If is between 0 and 2[[pi]], then is the measure of the angle swept out by point P. If is greater then 2[[pi]], we can think of the

angle being swept out by a point P that has moved counter-clockwise more than once around the circle. And if is negative, we think of the angle as being swept out by a point P moving in the opposite direction (with the Ferris wheel going backwards).

Sin = x(t)/20. This implies that x(t) = 20sin. Similarly, cos = (measurement of segment WZ)/20. WZ = 20 - y(t), so cos = (20-y(t))/20 implies that y(t) = -20cos + 20. Since one complete revolution has an angle of 2[[pi]], and a revolution takes 10 seconds, we can use the following ratio to find .

2[[pi]]/10 = /t implies that = ([[pi]]/5)t.

Plugging this in for , we get x(t) = 20sin[([[pi]]/5)t] and y(t) = -20cos[([[pi]]/5)t] + 20.

Since h(t) is the height above the bottom of the Ferris wheel, h(t) = y(t).

Students should be able to graph the equation h(t) = -20cos[([[pi]]/5)t] + 20 and then analyze their graphs.

This equation is graphed below. (Algebra Xpresser was used here.)

After graphing the equation, students should be able to find h-values for given t-values (and vice-versa). They should also be able to find the amount of time t required for a given number of revolutions.

Students should also explore the changes in the graph for a Ferris wheel that has a different radius of rate of revolution. For example, if the radius was 25 ft. and it took the Ferris wheel 12 seconds to complete a revolution, then the equations would be

h(t) = -25cos([[pi]]/6)t + 25. Students should compare this graph with there first graph and make conjectures about the relationships between Ferris wheels and their corresponding equations.

There are many other activities that could be looked at which deal with trigonometric functions. However, I think the Ferris wheel problem is a good demonstration of how sine and cosine can be used.

References

National Council of Teachers of Mathematics. (1989). Curriculum and Evaluation Standards for School Mathematics. Reston, VA: author.