• Calculus > Cycloidal Curves

The Cycloidal Curves

Circles Rolling on Top of Curves

Suppose a circle with radius r rolls along a curve f(x), and the circle starts at a point where f crosses the y-axis. A point on the circle traces a curve given by parametric equations C(x, y). What are the equations that define this path for the curve C? We know that when f(x) = 0, the curve is the cycloid. But we want to extend this to all curves f(x).

A circle of any arbitrary radius cannot roll on all curves without slipping. For example, a circle of radius 1 cannot roll on the inside of a parabola \( y = x^{2} \) because it does not fit near the vertex. However, we can still trace this curve as if it did.

Step 1: Path of the Center of the Circle

We will find the path of the curve for a circle that “rolls on top” of the curve. First, we must find the curve P(x, y) that is traced by the center of the circle. This curve is parallel to f. (There are 2 circles shown at 2 different points. For the path of the center of the circle, consider any of these 2 circles.) A tangent line has been drawn at point (x_t, f(x_t)). The angle φ will enable us to define the curve P. The angle created by the tangent and the x-axis is equal to φ. Therefore, the measure of φ is given by:

(i) \(\varphi = \arctan(f'(x_t))\)

From (i), we can determine the position of point P. First, note that when the tangent has a slope of 0, then x_p = x_t. When slope of the tangent is negative (depicted in the picture as the green circle), then φ is negative; therefore, \(x_p = x_t - r\sin\varphi\). When the slope of the tangent is positive (depicted in the picture as the red circle), then φ is positive; therefore, \(x_p = x_t - r\sin\varphi\) still holds true. Thus, the point x_p is given by: \(x_p = x_t - r\sin\varphi =\) \(x_t - r\sin[\arctan(f'(x_t))] =\) \(x_t - \frac{r\cdot f'(x_t)}{\sqrt{1+[f'(x_t)]^2}}\).

(ii) \(x_p = x_t - \frac{r\cdot f'(x_t)}{\sqrt{1+[f'(x_t)]^2}}\)

Now, finding y_p is similar to finding x_p. Without difficulty, y_p can be determined to be:

(iii) \(y_p = f(x_t) + r\cos\left[\arctan(f'(x))\right] = f(x_t) + \frac{r}{\sqrt{1+[f'(x_t)]^2}}\)

The coordinates of points P = \( (x_{p}, y_{p}) \) traces the center of the circle rolling on the function \( f(x) \), and the parametric equation of this curve that runs parallel to \( f(x) \) is: \( x(t) = t - \frac{r\cdot f'(t)}{\sqrt{1+[f'(t)]^2}} \) and \( y(t) = f(t) + \frac{r}{\sqrt{1+[f'(t)]^2}} \).

Step 2: Measure of Angle \( \theta \)

Next, we must find the measure of angle \( \theta \), which is the measure of arc P₁C. This will allow us to find the coordinates of C, which is the point on the circle that creates the cycloidal curve on f. There are 2 cases to consider. First case if when the circle is tangent to f at a negative slope and the other case is when it is tangent to f as a positive slope. The difference is in the calculation of \( \arctan(f'(x)) \). When the slope is negative, \( \arctan(f'(x)) \lt 0 \) and when the slope is positive, \( \arctan(f'(x)) \gt 0 \). The third case is trivial, when the slope is 0 and \( \arctan(f'(x)) = 0 \). Let’s look at when the slope is positive first.

Let the angle \( \theta_{L} \) equal to the angle the circle has traveled from point F to point C. The arc FC is the distance the circle has travelled on f. This angle overlaps with \( \varphi \). The angle \( \varphi \) itselft is equal to \( \arctan(f'(x)) \) and is positive since the tangent line has a positive slope. From this setup, we have the following relationship:

(iv) \( 2\pi = \theta + \theta_{L} - \varphi + \frac{\pi}{2} \)

(v) \( \theta = \frac{3\pi}{2} + \varphi - \theta_{L}\)

For the second case where the tangent creates a negative slope, \( \arctan(f'(x)) \) will be negative. However, we are working with adding positive angle measures to equal the measure of the whole circle, \( 2\pi \). Since \( \varphi\) is intrinsically negative in the second case, we need to subtract the value. Hence, in both cases, equation (v) holds true and there is no need to worry about the sign of \( \arctan(f'(x)) \).

Measure of angle \( \theta_{L} \) can be found by noticing that the length of the arc FC, of the circle (going clockwise) is equal to the arc length of the curve f from x = 0 to x = x_t. In other words: \(r\theta_{L} = \int_{0}^{x_t}\sqrt{1+[f'(x)]^2}\,dx\). Therefore, we have the relationship:

(viii) \(\theta_{L} = \frac{1}{r}\int_{0}^{x_t}\sqrt{1+[f'(x)]^2}\,dx\)

This gives us the relationship for \( \theta \) that we need:

(ix) \(\theta = \frac{3\pi}{2} + \arctan(f'(x)) - \) \(\frac{1}{r}\int_{0}^{x_t}\sqrt{1+[f'(x)]^2}\,dx\)

Step 3: Putting It All Together

Knowing θ, we can find the coordinates C(x_c, y_c). We can easily relate the following: \(x_c = x_p + r\cos\theta\) and \(y_c = y_p + r\sin\theta\). Using the sum identity of sine and cosine, we can simplify cos(θ) and sin(θ) to remove the 3π/2 as: \(\cos\theta = \sin\left( \arctan(f'(x_t)) - \frac{1}{r}\int_{0}^{x_t}\sqrt{1+[f'(x)]^2} \, dx \right)\) and \(\sin\theta = -\cos\left( \arctan(f'(x_t)) - \frac{1}{r}\int_{0}^{x_t}\sqrt{1+[f'(x)]^2}\,dx \right)\).

(Note the change in variable x_t to t for simplicity.)

Finding the equations of the curve depends highly on being able to evaluate the integral for the arc length. Many curves give integrals that can be evaluated. Some curves, like the sine curve, do not result in a nice integral. I was able to graph the function, however, with the integral using Desmos. This curve is shown later in the page.

The Cycloid

We already know that when f(x) = 0, the curve should reduce to a cycloid. We will demonstrate that this does, in fact, occur:

\(x(t) = t - \frac{r\cdot 0}{1+[0]^2} + \) \( r\sin\left( \arctan(0) - \frac{1}{r}\int_{0}^{t}\sqrt{1+[0]^2}\,dx \right) = \) \(t-r\sin\frac{t}{r}\), and

\(y(t) = 0 + \frac{r}{1+[0]^2} - \) \(r\cos\left( \arctan(0) - \frac{1}{r}\int_{0}^{t}\sqrt{1+[0]^2}\,dx \right) = \) \(r - r\cos\frac{t}{r}\).

The equations above are equivalent to the cycloid equations \(x(\theta) = r\theta - r\sin\theta\) and \(y(\theta) = r - r\cos\theta\), and it is demonstrated below for a circle of radius 2.

Other Curves

Slanted Cycloid

When f(x) = x, a line with slope of 1, we have the equations:

\(x(t) = t - \frac{r}{\sqrt{2}} + r\sin\left(\frac{\pi}{4}-\frac{\sqrt{2}}{r}t\right)\) and

\(y(t) = t + \frac{r}{\sqrt{2}} - r\cos\left(\frac{\pi}{4}-\frac{\sqrt{2}}{r}t\right)\).

This is the cycloid generated on the line y = x. The graph for this curve is shown below (for r = 1).

Naturally, the cycloidal curve and the line above have common points. Since the line has a slope of 1, it is easy to find the common points using right-angle geometry. We can also set the x value of the line and the cycloidal curve to be equal and find the t where this occurs.

(i) \(t - \frac{r}{\sqrt{2}} + r\sin\left(\frac{\pi}{4}-\sqrt{2}t\right) = t \)

(ii) \(- \frac{1}{\sqrt{2}} + \sin\left(\frac{\pi}{4}-\sqrt{2}t\right) = 0 \)

(iii) \(\sin\left(\frac{\pi}{4}-\sqrt{2}t\right) = \frac{\sqrt{2}}{2} \)

The solutions to t in (iii) that apply to our case are \( \sqrt{2}\pi n \), where n is an integer. The graph below shows the common points and the line segment which equals the cicurmference of the circle.

The General Slanted Cycloid

For a cycloid on any given line defined by the equation \( y = ax \), the parametric equations are:

\( x(t) = \frac{r}{\sqrt{1+a^2}}(t-a) - r\sin(t - \arctan a ) \)

\( y(t) = \frac{r}{\sqrt{1+a^2}}(at+1) - r\cos(t - \arctan a ) \)

Rolling on a Parabola

Let’s roll on a horizontal parabola first. When \(f(x) = \sqrt{x}\), then:

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

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

The graph is shown below (for r = 1):

When \( f(x) = ax^{2} \), we have:

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

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

The graph is shown below (for r = 1 and a = 1/2).

Remember we mentioned above not all circles can roll on a curve without slipping. The cusp formed on the inside of the parabola represents the fact that the circle does not fit inside the parabola near the vertex.

I have created a new page for Parabolic Cycloids for a closer look, since these curves have a couple curious properties.

Trigonometric Functions

The curve below was generated for f(x) = sin(x).

\(x(t) = t-\frac{r\cos t}{\sqrt{1+\cos^{2}t}}+ \) \(r\sin\left( \arctan(\cos t) - \frac{1}{r}\int_{0}^{t}\sqrt{1+\cos^{2}x}\,dx \right)\)

\(y(t) = \sin t+\frac{r}{\sqrt{1+\cos^{2}t}}- \) \(r\cos\left( \arctan(\cos t) - \frac{1}{r}\int_{0}^{t}\sqrt{1+\cos^{2}x}\,dx \right)\)

Because the integral has no closed form, we have left the integral as is. The image is below:

Because of the curvature of the sine curve around 3π/2, there a small loop created.

If we let r ≈ 0.608, then the circumference of the circle is about C ≈ 3.8202. The circumference of this circle is approximately equal to the arc length of the sine curve from 0 to π. The following graph shows the curve that corresponds to this situation:

The cusps match quite nicely at the x-intercepts of the sine curve.

Logarithmic Curves

When \(f(x) = \ln(x)\), then the cycloidal curve is given by:

\(x(t) = t - \frac{r}{\sqrt{1+t^2}} + \) \( r\sin\left( \arctan\frac{1}{t} - \frac{1}{r}\sqrt{1+t^2} + \frac{1}{r}\sinh^{-1}\frac{1}{t} \right)\) and

\(y(t) = \ln t + \frac{rt}{\sqrt{1+t^2}} - \) \(r\cos\left( \arctan\frac{1}{t} - \frac{1}{r}\sqrt{1+t^2} + \frac{1}{r}\sinh^{-1}\frac{1}{t} \right)\)

The graph below shows the logarithm curve, the curve traced by the center of the circle and the point on the circle that traces the cycloidal curve.

The Geogebra activity below allows you to roll a circle on the logarithmic curve. You can also turn the radius negative and you will see the circle roll on the bottom of the curve.

When \(f(x) = \ln(\cos x)\), \( f'(x) = -\tan(x) \). The arclength integral is \( \int_{0}^{x} \sqrt{1+\tan(t)^{2}} \,dt = \ln\left| \sec x+\tan x \right| \). After changing some signs around for sine and cosine being odd and even respectively, the curve is given by equations:

\(x(t) = t + r\sin t - r\sin\left( t + \frac{1}{r}\ln\left| \sec x+\tan x \right| \right)\) and

\(y(t) = \ln(\cos t) + r\cos t - r\cos\left(t + \frac{1}{r}\ln\left| \sec x+\tan x \right| \right)\)

The graph is shown below:

Exponential Curves

Circle rolling on \(f(x) = e^x\) curve:

\(x(t) = t - \frac{r\cdot e^t}{\sqrt{1+e^{2t}}} + \) \( r\sin\left( \arctan(e^t) - \frac{1}{r}\sqrt{1+e^{2t}} + \frac{1}{2r}\ln\left| \frac{1+\sqrt{1+e^{2t}}}{1-\sqrt{1+e^{2t}}} \right| \right)\) and

\(y(t) = e^t + \frac{r}{\sqrt{1+e^{2t}}} - \) \(r\cos\left( \arctan(e^t) - \frac{1}{r}\sqrt{1+e^{2t}} + \frac{1}{2r}\ln\left| \frac{1+\sqrt{1+e^{2t}}}{1-\sqrt{1+e^{2t}}} \right| \right)\)

The arclength integral of \( e^{x} \) is a tough one because of the domain. The Wolfram solution is given as: \( \int \sqrt{1 + (e^{x})^{2}} \, dx = \sqrt{1+e^{2x}} - \tanh^{-1}(\sqrt{1+ e^{2x} }) +C \). However, the domain of the inverse hyperbolic tangent is (-1, 1), so it does not cover the domain of \( e^{x} \). Hence, we have to represent it as the logarithm with absolute value bars.

Verifying with Arc Length

We can verify the equations by actually calculating the arc length of the curve between the common points. Each time the point on the circle touches the curve, the circle has traveled its circumference. So the arc length should equal \( 2\pi r \) between these intervals. To simplify calculations, we will let the radius equal 1.

Let’s take the parabola \( y = x^2 \) as an example. The arc length is give by the integral \( \int_{a}^{b} \sqrt{1+4x^2} \, dx \) for the interval between a and b.

Finding the common points of the cycloidal curve and the parabola is a feat in itself. I used Wolfram Alpha to find the x-coordinate of 2 common points: 2.36537451633093... and 3.43304179425415... Note that 0 is also a common point, so we will find the arc length of these two intervals.

(i) \( \int_{0}^{2.365} \sqrt{1+4x^2} \, dx ≈ 6.28319... = 2\pi \)

We used radius of 1, so the arc length equals twice of \( \pi \), which is what we expected. You can see the result here: Wolfram.

Using the next interval (values truncated) to find the arc length, we get:

(i) \( \int_{2.365}^{3.433} \sqrt{1+4x^2} \, dx ≈ 6.28319... = 2\pi \)

The arc length equals twice of \( \pi \) again, which is also what we expected. You can see the result here: Wolfram. The graph below shows the intervals created by the cycloidal curve on the parabola.

For the natural logarithm curve, three consecutive common points are at an x-coordinates t ≈ 0.6627434193491..., 6.361620974339386..., and 12.606013463... The arc length over these intervals is:

(i) \( \int_{0.662}^{6.361} \sqrt{1+\left(\frac{1}{x}\right)^{2}} \, dx ≈ 6.28319... = 2\pi \)

(i) \( \int_{6.361}^{12.606} \sqrt{1+\left(\frac{1}{x}\right)^{2}} \, dx ≈ 6.28319... = 2\pi \)

The curve with the common points is graphed below.

Activities

Below is a Desmos activity that you can play with. You can change the function \( f(x) \) and see the result. Move the a variable slider to see the circle roll on the curve. You can also change the radius of the circle.

We cannot find the arc length integral for all functions. This presents a problem with certain CAS software. Particularly, Geogebra cannot evaluation arc length integrals for any arbitrary function, even when the arc length is defined as an integral of the form \( L(x) = \int_{0}^{x} \sqrt{1+f'(t)^{2}}\, dt \).

Desmos, however, can evaluate this integral without a problem. This allows us to roll the circle on any function, such as \( \sin(x) \) or \( \frac{1}{1+x^2} \) and see some neat results. The activity above allows you to change the function and the radius of the circle. Despite the benefits of Desmos, it lacks in being able to customize the embedded canvas and does not allow sliders to be placed on the canvas (they remain in the menu).

Simplifying the Cycloidal Equations

The equations of the curves we got for the simple cycloid using this method was slightly different than the well-known equation of the cycloid. But by changing the parameter a little, we can simpify the equation. As long as the range in the new parameter covers all the values required in the domain of \( x(t) \) and \( y(t) \), then this will work.

Let’s simplify the slanted cycloid. If we let \( \frac{\sqrt{2}}{r}t = u \), then \( t = \frac{\sqrt{2}}{2}ru \). And the equations below

\(x(t) = t - \frac{r}{\sqrt{2}} + r\sin\left(\frac{\pi}{4}-\frac{\sqrt{2}}{r}t\right)\) and

\(y(t) = t + \frac{r}{\sqrt{2}} - r\cos\left(\frac{\pi}{4} - \frac{\sqrt{2}}{r}t\right)\)

become

\(x(u) = \frac{\sqrt{2}}{2}ru - \frac{\sqrt{2}}{2}r + r\sin\left(\frac{\pi}{4}-u\right)\) and

\(y(u) = \frac{\sqrt{2}}{2}ru + \frac{\sqrt{2}}{2}r - r\cos\left(\frac{\pi}{4}-u\right)\)

The equation above may seem more complicated but at least we took the radius out of the sine and cosine functions, which makes them easier to handle.

The cycloid on the exponential curve becomes a bit better than the original also. We let \( u = e^{t} \). Then,

\(x(t) = t - \frac{r\cdot e^t}{\sqrt{1+e^{2t}}} + \) \( r\sin\left( \arctan(e^t) - \frac{1}{r}\sqrt{1+e^{2t}} + \frac{1}{2r}\ln\left| \frac{1+\sqrt{1+e^{2t}}}{1-\sqrt{1+e^{2t}}} \right| \right)\) and

\(y(t) = e^t + \frac{r}{\sqrt{1+e^{2t}}} - \) \(r\cos\left( \arctan(e^t) - \frac{1}{r}\sqrt{1+e^{2t}} + \frac{1}{2r}\ln\left| \frac{1+\sqrt{1+e^{2t}}}{1-\sqrt{1+e^{2t}}} \right| \right)\)

becomes

\(x(u) = \ln u - \frac{r\cdot u}{\sqrt{1+u^{2}}} + \) \( r\sin\left( \arctan(u) - \frac{1}{r}\sqrt{1+u^{2}} + \frac{1}{2r}\ln\left| \frac{1+\sqrt{1+u^{2}}}{1-\sqrt{1+u^{2}}} \right| \right)\) and

\(y(u) = u + \frac{r}{\sqrt{1+u^{2}}} - \) \(r\cos\left( \arctan(u) - \frac{1}{r}\sqrt{1+u^{2}} + \frac{1}{2r}\ln\left| \frac{1+\sqrt{1+u^{2}}}{1-\sqrt{1+u^{2}}} \right| \right)\)

This simplifies better because the exponential function is removed from the radicals.

For the parabolic cycloid, a substitution of \( u = 2at \) in

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

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

becomes

\(x(u) = \frac{u}{2a} - \frac{ru}{\sqrt{1+u^{2}}} + \) \( r\sin\left( \arctan(u) - \frac{1}{4ar}\sinh^{-1}(u)-\frac{u}{4ar}\sqrt{1+u^{2}} \right)\) 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 horizontal parabola can be simplified with the substitution \( u = 2\sqrt{t} \) from:

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

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

to

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

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

The absolute value bars were necessary because the domain of \( 2\sqrt{t} \) does not include negative numbers.

Anirdesh.com Last Updated: May 14 2025, 23:33:42.