• Calculus > Parabolic Cycloid

The Parabolic Cycloid

The Cycloidal Curves page was getting too long so I broke the discussion about Parabolic Cycloid on a separate page. This page is dedicated to the circle rolling on a parabola.

For the parabolic cycloid, a substitution of \( u = 2at \) in the above equations can simplify the equations:

\(x(u) = \frac{u}{2a} - \frac{ru}{\sqrt{1+u^{2}}} + \) \( r\sin\left( \arctan(u) - \frac{1}{4ar}\sinh^{-1}(u)-\frac{u}{4ar}\sqrt{1+u^{2}} \right)\) and

\(y(u) = \frac{u^{2}}{4a} + \frac{r}{\sqrt{1+u^{2}}} - \) \( r\cos\left( \arctan(u) - \frac{1}{4ar}\sinh^{-1}(u)-\frac{u}{4ar}\sqrt{1+u^{2}} \right)\)

The Slipping of the Circle

Let’s investigate a circle rolling down a parabola. When the circle rolls on the outside of the parabola, the parabola is curving away from the circle so we have a good roll without the path intersecting. When the parabola rolls on the inside, this presents an obstacle if the circle is too large.

If an actual circle rolled down the inside of an actual parabola, it may get stuck and never make it to the vertex if the circle is too big. The image here shows the circle overlapping with the parabola. Since the rolling is simply theoretical with equations, we can make it roll down regardless of whether it is possible or not. What happens to the path when a circle too large to fit theoretically rolls down a parabola? The path created by the center of the circle backtracks and crosses the path again as seen in the image above.

Moreover, when the circle is too large, the circle also crosses the parabola and enters the third and fourth quadrants. In the next image, the red curve is the parabola \( y = \frac{1}{4}x \), which looks like a flat line because it has been zoomed-in significantly. The blue path is the cycloid, and we can see it cross the x-axis from second quadrant into the third quadrant, then into the fourth quadrant. Then, it returns to touch the origin, then goes into the third quadrant before proceeding into the first quadrant via fourth quadrant.

This happens if the circle is too large to roll freely down the parabola.

Largest Circle That Does Not Backtrack

The question now becomes, what is the largest size of the circle that can roll down without this backtracking and crossing the x-axis? Looking at the graph, we can deduce that if the center of the circle passes the y-axis only once, then the curve will be smooth at this point and hence, will not backtrack. Moreover, there is only 1 slope at this point, which is 0. This is the mininum in the path.

This zoomed-in image shows the path created by the circle’s center at its minimum where the radius is set to \(r = \frac{1}{2a} \) for the parabola \( y = ax^2 \). We will show when the radius is less than \( \frac{1}{2a} \), the circle will roll smoothly down the parabola without any backtracking. The maximum radius is twice the focal distance!

We will find the derivative of the path created by the circle and evaluate for r. Let’s remember the parametric equation for the circle’s center path is: \( x(t) = t - \frac{r\cdot f'(t)}{\sqrt{1+[f'(t)]^2}} \) and \( y(t) = f(t) + \frac{r}{\sqrt{1+[f'(t)]^2}} \). For our parabola, \( f(t) = at^2 \) and \( f'(t) = 2at \).

The center of the circle rolling down a parabola is given by the parametric equations: \( x(t) = t - \frac{2art}{\sqrt{1+4a^{2}t^{2}}} \) and \( y(t) = at^{2} + \frac{r}{\sqrt{1+4a^{2}t^{2}}} \).

The derivative of the x position is: \( \frac{d}{dt}(x(t)) = \frac{(4 a^2 t^2 + 1)^{3/2} - 2 a r}{(4 a^2 t^2 + 1)^{3/2}} \) and the derivative of the y position is: \( \frac{d}{dt}(y(t)) = \frac{2at(4 a^2 t^2 + 1)^{3/2}-4a^{2}rt}{(4 a^2 t^2 + 1)^{3/2}} \).

Therefore, the derivative of this path is: \( \frac{dy}{dx} = \frac{2at(4 a^2 t^2 + 1)^{3/2}-4a^{2}rt}{(4 a^2 t^2 + 1)^{3/2} - 2 a r} \). The only t where this equation equals 0 is t = 0.

Now, let’s solve for the radius where the slope is 0:

(i) \( \frac{dy}{dx} = 0 = \frac{2at(4 a^2 t^2 + 1)^{3/2}-4a^{2}rt}{(4 a^2 t^2 + 1)^{3/2} - 2 a r} \)

(ii) \( 2at(4 a^2 t^2 + 1)^{3/2}-4a^{2}rt = 0 \)

(iii) \( (4 a^2 t^2 + 1)^{3/2} = 2ar \)

(iv) \( r = \frac{(4 a^2 t^2 + 1)^{3/2}}{2a} \)

Now, let t = 0 since the slope is 0 when t = 0.

(v) \( r = \frac{(1)^{3/2}}{2a} = \frac{1}{2a} \)

Therefore, when the radius of the circle is \( \frac{1}{2a} \), the circle’s center will not backtrack.

You can interact with the circle rolling on the parabola in the Geogebra activity below. Increase or decrease the coefficient, a of the parabola, the radius, r, of the circle, or the point, p, on the parabola. When the radius is less than \( \frac{1}{2a} \), the circle will only roll forward and never backward for increasing p.

No Kinks: The Unsolved Problem

The Problem

In the above section, we learned that if a circle with radius \( r \leq \frac{1}{2a} \) rolls down a parabola, the path created by the circle is smooth. If \( r \gt \frac{1}{2a} \), then the path reverses before proceeding forward.

Now, we focus on the parabolic cycloid itself, which has its own criss-crossing. The parabolic cycloid crosses its path if the circle is large. In this image, the parabola is green. The orange path is the cycloid. When the circle rolls down, it crosses the y-axis at X₁. Then, it reaches the relative maximum x-value at M₁. This is the largest x value, so the derivative will be 0 there. Then, it touches the origin at X₂. Then, it reaches its relative minimum x-value at M₂ before crossing itself again at X₃. Both X₁ and X₂ are the same point but have a different t value.

So, here is our problem: What would the radius of the circle have to be to eliminate the crossing at points labeled X₁ and X₂?

Background for the Proof

To solve this, let’s understand how the x-values alone vary as the parameter t varies. We do not need to involve \( y(t) \) in this. Refer to the image above.

The curve in red is the \( x(t) \) curve graphed alone to see how the x-values of the cycloid fluctuate. The maximum of \( x(t) \) at N₁ corresponds to M₁. The minimum of \( x(t) \) at N₂ corresponds to M₂.

The blue curve \( x'(t) \) is the derivative of \( x(t) \). We can see the max and min are the zeros of \( x'(t) \).

If we want to get rid of the kink, the max M₁ and min M₂ have to disappear, meaning the max and min on \( x(t) \) have to be eliminated and the \( x'(t) \) cannot have any zeros. We have to find a and r which will eliminate these.

I have not found the proof but it seems the radius has to be \( r \le \frac{1}{4a} \) or \( ar \le \frac{1}{4} \) to make the kink disappear.

This image shows a circle with radius \( r = \frac{1}{4a} \). The circle’s center path in yellow is smooth and there is no kinking at the origin. The center of the circle will coincide with the focus when it rolls down, and the point on the circle will coincide with the origin. The Geogebra activity below allows you to play with the circle rolling. You can change a, r to see the effects. You can also zoom in to see the crisscrossing, since it is difficult to see that when zoomed out.

The radius is set to a fourth of the parabola coefficient. You can change it and zoom in to view the kinks. Remember that \( r \leq \frac{1}{2a} \) to see a smooth path traced by the circle’s center, and \( r \leq \frac{1}{4a} \) to not see any kinks on the cycloid path near the origin.

Anirdesh.com Last Updated: July 10 2025, 13:27:34.