This page discusses a very specific planar curve, the cycloid. We can learn a lot from the cycloid in terms of vector analysis, ...

A cycloid is a curve traced by tracking the vertical (y) position of a point on the outside of a wheel as the wheel rolls along the x-axis without slipping. This is what it looks like. A circle of radius 1 is rolling from the origin (left) toward the right. Those curves aren't semicircles, though.

Now we'd like to find an equation for the cycloid. It begs for a parameterized form for the location of any point on the curve, $P(t)$, which will have the form

$$P(t) = (x(t), y(t)),$$

where $x$ and $y$ are the usual 2D Cartesian coordinates and $t$ is the parameter. Now what exactly would $t$ be? We could use **time**, but what if our wheel moved faster or slower? It would still trace out the same curve, so perhaps time isn't ideal. We could also use the rolling **distance**, say $x$, but it turns out that the **angle** of a fixed radius with the wheel, staring at $\theta = 0$ will be better. So we'll look to fill in the details of

$$P(\theta) = (x(\theta), y(\theta)),$$

where $\theta$ is the angle of rotation starting from the origin, where our radius vector points straight down to $(0, 0)$.

Here's a picture of one curve of the cycloid. Notice that the vector $\vec{OP}$ from the origin to a point on the cycloid is the sum of vectors $\vec{OA}$, $\vec{AB}$, and $\vec{BP}$.

Now let's figure out those three vectors. First $\vec{OA}$. Notice that the distance from $O$ to $A$ is the same as the length of the arc of our wheel from $A$ to $P$. Recall that as long as we're working in radians, the arc length is simple:

We start with a proportion: the ratio of the angle of the arc $(\theta)$ to the full arc of a circle $(2 \pi)$ rad is the same as the ratio of the arc $s$ to the circumfrence of the circle, $2 \pi r$:

$$ \require{cancel} \frac{\theta}{\cancel{2 \pi}} = \frac{s}{\cancel{2 \pi r}} \; \longrightarrow \; s = r \theta$$

So if we're working in radians (which we generally should in math), the arc length is just the radius of the circle multiplied by the angle. That means our vector $\vec{OA}$ is just $\vec{OA}(r \theta, 0)$, where $r$ is the radius of our rolling circle.

Vector $\vec {AB}$ a little easier. It's just $\vec {AB} = (0, r)$.

Vector $\vec {BP}$ can be rationalized from this graph. Here I've made the angle $\theta$ acute to make things easier, but this diagram should generalize to any value of $\theta$.

So we have $\vec {BP} = (-r \, sin(\theta), -r \, cos(\theta))$. Now the vector we want, $\vec{OP}$ is just the vector sum:

$$ \begin{align} \vec{OP} &= \vec{OA} + \vec{AB} + \vec{BP} \\[5pt] &= (r \theta, 0) + (0, r) + (-r \cdot sin(\theta), -r \cdot cos(\theta)) \\[5pt] &= (r \theta - r \cdot sin(\theta), r - r \cdot cos(\theta)) \end{align}$$

Thus:

$$ \begin{align} x(\theta) &= r \theta - r \cdot sin(\theta) \\[5pt] y(\theta) &= r - r \cdot cos(\theta) \end{align}$$

Now to make things easier we can divide through by $r$ and just make our unit of length the number of circle radii:

$$P(\theta) = ( \theta - sin(\theta), 1 - cos(\theta))$$

A true "cusp, " the slope is indeterminate at the midpoint

A point with defineable slopes on either side, but still indeterminate in the middle

Fanciful and unlikely — a loop-the-loop

A smooth curve with determinate slopes all the way across

We'd like to find the slope of the cycloid at its low points, near $\theta = 0, \, 2 \pi, \, 4 \pi, \dots$. We can do this by invoking the Taylor series approximation (MacLaurin series, really) for sine and cosine near $\theta = 0$:

$$ \begin{align} f(\theta) &\approx f(0) + f'(0) \theta + f''(0) \frac{\theta^2}{2} + f'''(0) \frac{\theta^3}{6} + \dots \\[5pt] cos(\theta) &\approx 1 - \frac{\theta^2}{2} + \dots \\[5pt] sin(\theta) &\approx \theta - \frac{\theta^3}{6} + \dots \end{align}$$

Here we've just kept the first two non-zero terms of each approximation. If we plug those into our parametric equation, we have

$$ \begin{align} P(\theta) &= ( \theta - sin(\theta), 1 - cos(\theta)) \\[5pt] &= \left( \theta - \left( \theta - \frac{\theta^3}{6} \right), 1 - \left( 1 - \frac{\theta^2}{2} \right) \right) \\[5pt] &= \left( \frac{\theta^3}{6}, \, \frac{\theta^2}{2} \right) \end{align}$$

Now the slope of the cycloid in this region is then approximately

$$ \begin{align} m &= \frac{y}{x} = \frac{\frac{\theta^2}{2}}{\frac{\theta^3}{6}} \\[5pt] &= \frac{\theta^2}{2} \cdot \frac{6}{\theta^3} \\[5pt] &= \frac{3}{\theta} \end{align}$$

Now we can clearly see that as $\theta \rightarrow 0$, the slope goes to infinity (vertical or infinite slope). That comports with the first scenario above, a true **cusp**, a singular point at which the vertical direction of our moving point changes by 180˚.

Now let's imagine that our wheel is rolling along at a constant speed, so we'll use time as our parameter. the x-y location of our special point on the wheel in time is

$$P(t) = ( t - sin(t), 1 - cos(t))$$

Then let's use this form to calculate the velocity vector, $\vec v$. Recall that velocity is the first derivative of position with respect to time, so our velocity vector will be

$$ \begin{align} \vec v(t) &= \frac{d}{dt} P(t) \\[5pt] &= \left( \frac{dx(t)}{dt}, \frac{dy(t)}{dt} \right) \\[5pt] &= \left( \frac{d}{dt}(t - sin(t)), \frac{d}{dt}(1 - cos(t)) \right) \\[5pt] &= (1 - cos(t), sin(t)) \end{align}$$

Let's use this new velocity vector to calculate actual velocities at a couple of points along the cycloid trajectory. First at the origin (or even multiples of $\pi$). Recall that the actual velocity is the *length* of the velocity vector evaluated at that location (or time in this case).

$$ \begin{align} \vec v(t=0) &= (1 - cos(0), sin(0)) \\[5pt] &= \left( 1-1, sin(0) \right) \\[5pt] &= (0, 0) \end{align}$$

This vector has a length of zero, so the velocity is zero at the turning points, as we expect. We can approach this problem another way, of course, by calculating the length of the vector as a function of $t$ and then plugging in $t = 0$. That looks like this:

$$ \begin{align} |\vec v| &= \sqrt{(1-cos(t))^2 + (sin(t))^2} \\[5pt] &= \sqrt{1 - 2 cos(t) + cos^2(t) + sin^2(t)} \tag{1} \\[5pt] &= \sqrt{2 - 2 cos(t)} \end{align}$$

Notice that in (1) we used the Pythagorean identity, $cos^2(x) + sin^2(x) = 1.$

And again we have $|\vec v (t = 0)| = \sqrt{2 - 2 cos(0)} = 0$.

Let's also calculate the velocity at the maximum of the cycloid curve, say at $t = \pi$. It's

$$ \begin{align} |\vec v| &= \sqrt{(1-cos(t))^2 + (sin(t))^2} \\[5pt] |\vec v (\pi)| &= \sqrt{(1 - cos(\pi))^2 + sin^2(\pi)} \\[5pt] &= \sqrt{(1 + 1)^2 + 0} = 2 \end{align}$$

Notice that the velocity of the point tracing out the cycloid is twice the velocity of the rolling wheel at the topmost points.

Now acceleration is the second derivative of position with respect to time, so it's

$$ \begin{align} \vec a(t) &= \frac{d^2}{dt^2} P(t) \\[5pt] &= \left( \frac{d^2x(t^2)}{dt}, \frac{d^2y(t)}{dt^2} \right) \\[5pt] &= \left( \frac{d}{dt}(1 - cos(t)), \frac{d}{dt}(sin(t)) \right) \\[5pt] &= \left( sin(t), cos(t) \right) \end{align}$$

The acceleration vector at the low points is $(0, 1)$, an upward pointing vector, as we saw above. On its trip down to the cusp our tracing point is slowing down (accelerating upward), and on its upward trip it's speeding up in the upward direction.

**xaktly.com** by Dr. Jeff Cruzan is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. © 2012, Jeff Cruzan. All text and images on this website not specifically attributed to another source were created by me and I reserve all rights as to their use. Any opinions expressed on this website are entirely mine, and do not necessarily reflect the views of any of my employers. Please feel free to send any questions or comments to jeff.cruzan@verizon.net.