Ana Kuzle

A cycloid is the curve defined by the path of a point on the edge of circular wheel as the wheel rolls along a straight line.


O' is the origin (point of mass), A' point on the circle and phi the angle between O'A' and the y-axis.

I. Construction of a cycloid using GSP

1. Define coordinate system in the Graph menu.

2. Graph function y(x)=1. Graph of function f is a line y=1.

3. Create a circle of a unit radius with center O and point A that is movable.

4. Create point O' on that line that is movable

5. Translate vector O'A' by vector OO' using Mark vector and Translate vector from the Transform menu

6. Select points O' and A and animate them simultaneously where O' moves forward on the line and A clockwise on the circle.

7. Trace point A'.

8. The generated curve is a cycloid.


Looking at the GSP file we can see that generating of this curve is based on two movements: translation of point O' along line y and rotation of point A' clockwise on the circle.

II. Deducing the parametric equations of cycloid

Instead of looking at the equations online I tried to get to the equations on my own using GSP. Same can be done in the classroom as a meaning of using all features technology provides us.


1. First step: Coordinates of point A' are (0,2). Hence, x=0 and y=2.


2. Second step: Coordinates of point A' are (2.57, 1.02). Hence, x=2.57 and y=1.02.

3. Third step: Coordinates of point A' are (3.16, 0.00). Hence, x=3.16 and y=0.00. This made me think that x=pi. Let us go back to the step #2. The first coordinate was 2.57 that is approximately pi/2. Thus, I can conjecture further that x is related to pi when y is an integer. Let us explore more.

4. Forth step: Coordinates of point A' are (3.75, 1.02). Hence, x=3.75 and y=1.02.

5. Fifth step: Coordinates of point A' are (6.32, 2.00). Hence, x=6.32 and y=2.00. Probably precision error occurred somewhere on the way. But, we can see that x=2*pi.

6. Next few steps:



From this we can say even more. Since cycloid is generating after certain period and previous observation regarding pi we can conjecture that x is a linear combining of the angle phi of trigonometric functions (sine, cosine) and the angle phi.

III. Obtaining the formula through table analysis

Let us put the data obtained in the table in order to get the parametric equations of the cycloid.


Angle phi
x-coordinate of point A'
y-coordinate of point A'

With a knowledge on differential geometry I was able to get to the equations having in mind my previous conjecture that coordinates are linear combination of angle phi and sine or cosine of phi.

I got this:

x(t)=t+sin t

y(t)=1+cos t

and it fits into my table.

Remark: phi :=t

Note: The center of a circle was at the point (0,1). If it was at the origin of the coordinate system the parametric equations would look a bit differently:

x(t)=t-sin t

y(t)=1-cos t.

IV. Graphing cycloid with the graphic calculator and further explorations

a) How does multiplying both equations by a scalar influences graph of a cycloid?

From the picture above we can see that scalar influences on growth/shrinking of a cycloid. As scalar increases, cycloid ``grows''. Also, we can see that the scalar affects the maximum x-value reached and the maximum y-value reached, what is logical since coefficient t is in the y equation as well as the x equation.


b) How does multiplying coefficient t with a scalar influence the graph of a cycloid?


From the several pictures given above we can observe that changing parameter by multiplying it with a scalar n changes number of bumps. As n grows the number of bumps increases as well.

Hence, scalar n = number of bumps. However, it does not influence on the size of bumps.


V. Polar coordinates of a cycloid and its graph

In order to get polar coordinates of a cycloid we start with its parametric equations.

I have chosen, out of simplicity, to graph cycloid given its polar coordinates in GSP. Here is what I got.