A cycloid is the locus of a point on a circle that
rolls along a line. Write parametric equations
for the cycloid and graph it. Consider also
a GSP construction of the cycloid.
Cycloid is a plane curve that can be defined
parametrically. Consider a circle rolling
along a line. In this case, we are faced
with two motions:
1) Translational motion: This is due to the motion of the center of mass of the circle, namely its center.
2) Rotational motion: This is due to the rotation of the circle about an axis perpendicular to the center of mass.
PART I - CYCLOID CONSTRUCTION:
STEP ONE: Let's define a cartesian coordinate system. In this way, we can also get a sense of the parametric equations of the cycloid.
STEP TWO: Create a circle of unit radius somewhere on the right upper corner of the window. (It is easier to work with unit circles)
STEP THREE: Label the center of this circle as O. Create a movable point on the circle and label it A.
STEP FOUR: Graph the function f(x)=1: This is the equation of the horizontal line y=1.
STEP FIVE: Now create a movable point O' on this line.
STEP SIX: Now highlight the center of the circle O and the movable point O'. Then choose mark vector option. This is where we want to translate the circle on the right upper corner.
STEP SEVEN: Now highlight the center O, the movable point A, and the circle. Then choose the translate option. In this way we get a copy of the circle on the line y=1.
STEP EIGHT: Select the point O' and the point A. Edit one animation button for these points. Be careful: We want to animate the center of mass point O' forward and the point A clockwise.
STEP NINE: Make sure that you enable trace
point option for the point A': This is the
point that will generate the cycloid.
CLICK THE FIGURE BELOW TO DOWNLOAD THIS ANIMATION FILE:
PART II- DERIVE PARAMETRIC EQUATIONS OF CYCLOID:
Now let's use the figure we created. First,
let's ask GSP to calculate the coordinates
of the point A' that generates the cycloid.
I want my students to make sense out of what
they are doing. Therefore, instead of simply
writing the parametric equations, let's go
step by step and derive those equations by
SNAPSHOT ONE: A' has coordinates (0,2). What does this mean? This means that when x=0, y=2. See figure below:
SNAPSHOT TWO: Animate the points once again. Now A' has coordinates (2.55,1). What does this mean? This means that when x=2.55, y=1. See figure below:
SNAPSHOT THREE: Animate the points once again. Now A' has coordinates (3.14,0). What does this mean? This means that when x=3.14, y=0. Does the x-coordinate look familiar to you? When x = pi , y = 1. It looks like when y is an integer, x is always a decimal number which is related to pi. Think about it! In this way, students realize that the x-coordinate could be a linear combination of the angle theta and a sine or cosine of the same angle theta.
SNAPSHOT FOUR: Animate the points once again. Now A' has coordinates (3.71,1). What does this mean? This means that when x=3.71, y=1.
SNAPSHOT FIVE: Animate the points once again. Now A' has coordinates (6.28,0). What does this mean? This means that when x=6.28, y=2. Remark that 6.28 =2*pi
In this way we completed one rotation. If
we define an angle theta between the line
segment [OA] and the vertical axis to be
positive when the rotation is clockwise,
then we can summarize the five snapshots
by this little table:
|x-coordinate of A'||0||2.55||3.14||3.71||6.28|
|y-coordinate of A'||2||1||0||1||2|
PART III - ANALYSIS OF THE DATA TABLE
In this part, we will analyze five snapshots and we will try to guess a reasonable parametric equation.
Simplify the notation first: Use t for the angle theta, x(t) for the x-coordinate of the trace point A', and y(t) for the y-coordinate.
GUESS AND CHECK TABLE FOR SNAPSHOT ONE (t=0):
|candidates for x(t)||t||sin[t]||1-cos[t]||t+sin[t]||t-sin[t]|
|candidates for y(t)||t+2||2+sin[t]||2cos[t]||1+cos[t]||1+sin[t]+cos[t]|
GUESS AND CHECK TABLE FOR SNAPSHOT TWO (t=pi/2):
Students try to write 2.55 as a function pi/2 now. They caluclate pi/2=1.57. We then realize that 2.55 is almost 1 more than pi/2. Therefore, they eliminate some of the x(t) candidates from the above guess and check table. The function x(t)=t is eliminated because x(pi/2)=pi/2=1.57. We want 2.57. The function x(t)=sin[t] is eliminated because sin[pi/2]=1. We want 2.57. How about x(t)=1-cos[t] ? x(pi/2)=1-cos[pi/2]=1: It's eliminated. The candidate x(t)=t+sin[t] is a good choice because x(pi/2)=(pi/2)+sin[pi/2]=(pi/2)+1=2.57. Then the last candidate is automatically eliminated. THEREFORE WITH THE HELP OF THE TEACHER, STUDENT ELIMINATES ALL x(t) EXCEPT x(t)=t+sin[t]
Students try to write 1 as a function pi/2 now. By a similar fashion, they eliminate some of the y(t) candidates from the above guess and check table. The function y(t)=t+2 is eliminated because y(pi/2)=2+pi/2=3.57. We want 1. The function y(t)=2+sin[t] is eliminated because 2+sin[pi/2]=3. We want 1. How about y(t)=2cos[t] ? y(pi/2)=2cos[pi/2]=0: It's eliminated. The candidate y(t)=1+cos[t] is a good choice because y(pi/2)=1+cos[pi/2]=1+0=1. Then the last candidate is automatically eliminated. THEREFORE WITH THE HELP OF THE TEACHER, STUDENT ELIMINATES ALL y(t) EXCEPT y(t)=1+cos[t]
Then the teacher asks them to repeat the table they did at the end of PART - II with x(t)=t+sin[t] and y(t)=1+cos[t]
This means we are done! The cycloid that we constructed in PART-I is parametrically defined by x(t) = t + sin[t] , y(t) = 1 + cos [t]
PART IV - GRAPH OF CYCLOID IN CARTESIAN COORDINATES: x(t) = t + sin[t] , y(t) = 1 + cos [t]
PART V - GRAPH OF CYCLOID IN POLAR COORDINATES:
First let's derive an equation of cycloid in polar coordinates:
x^2 = t^2 + ( sin [t] ) ^2 + 2 t sin[t]
y^2 = 1 + ( cos[t] ) ^2 + 2 cos[t]
Add these two equations side by side:
x^2 + y^2 = t^2 + ( sin [t] ) ^2 + 2 t sin[t] + 1 + ( cos[t] ) ^2 + 2 cos[t]
r^ 2 = t^2 + 2 + 2 t sin[t] + 2 cos[t]