Cycloid Arc Length And Motion

The Motion of a Cycloid

A cycloid is the curve traced by a point on the edge of a circle as the circle rolls along a straight line. Its repeated arches have a characteristic shape: each arch begins and ends at a cusp, with a smooth rise to a highest point between them. The parametrization describes both the shape and how the point moves along it.

We study the standard parametrization with a positive constant $a$. One complete arch is traced as $t$ increases from $0$ to $2\pi$. The parameter $t$ is often interpreted as the angle through which the rolling circle has turned.

Definition: For $a>0$, the cycloid is parametrized by $$ x(t)=a(t-\sin t),\qquad y(t)=a(1-\cos t). $$ The interval $0\le t\le 2\pi$ traces one arch, from $(0,0)$ to $(2\pi a,0)$.

Here $a$ is the radius of the rolling circle and has units of length. Consequently, the coordinates, speed, and arc length also have units of length (with speed understood as length per unit of $t$, or per unit of time if $t$ is time).

Velocity and Speed

Differentiate each coordinate with respect to $t$. The velocity vector records the direction and rate of motion; its magnitude is the speed.

\mathbf{v}(t) = \left(\frac{dx}{dt},\frac{dy}{dt}\right) = \bigl(a(1-\cos t),a\sin t\bigr).

Using the standard parametric speed formula, the speed is the square root of the sum of the squares of the velocity components. The resulting expression simplifies substantially.

\begin{aligned} v(t) &=\sqrt{a^2(1-\cos t)^2+a^2\sin^2 t}\\ &=a\sqrt{(1-\cos t)^2+\sin^2 t}\\ &=a\sqrt{1-2\cos t+\cos^2 t+\sin^2 t}\\ &=a\sqrt{2-2\cos t}\\ &=a\sqrt{4\sin^2(t/2)}\\ &=2a|\sin(t/2)|. \end{aligned}

The absolute value is part of the general simplification: $\sqrt{u^2}=|u|$, not $u$ for every real $u$. On the one-arch interval $0\le t\le2\pi$, we have $0\le t/2\le\pi$, so $\sin(t/2)\ge0$. Thus the speed on this arch is $2a\sin(t/2)$.

Key result: The speed along the cycloid is $$ v(t)=2a|\sin(t/2)|. $$ For one arch, where $0\le t\le2\pi$, this becomes $v(t)=2a\sin(t/2)$.

Deriving the Arc-Length Function

For a continuously differentiable parametrized curve, the distance accumulated from parameter value $0$ to $t$ is the integral of its speed. Write this accumulated distance as $s(t)$. On the cycloid’s first arch, the speed has no sign change to account for beyond the endpoint zeros, so we can integrate $2a\sin(t/2)$ directly.

Start with accumulated distance.
Use the speed integral $s(t)=\int_0^t v(u)\,du$ for $0\le t\le2\pi$.

Substitute the cycloid speed.
On this interval, $v(u)=2a\sin(u/2)$, so $s(t)=\int_0^t2a\sin(u/2)\,du$.

Integrate and evaluate the endpoints.
An antiderivative is $-4a\cos(u/2)$, giving $s(t)=4a(1-\cos(t/2))$.

The antiderivative follows because differentiating $-4a\cos(u/2)$ gives $2a\sin(u/2)$. Evaluating it at the limits gives the stated function:

\begin{aligned} s(t) &=\int_0^t2a\sin(u/2)\,du\\ &=\left[-4a\cos(u/2)\right]_0^t\\ &=-4a\cos(t/2)+4a\\ &=4a\bigl(1-\cos(t/2)\bigr). \end{aligned}

This derivation also explains why $s(t)$ measures distance traveled from the start, rather than merely giving a coordinate: its derivative is the nonnegative speed on this arch, and $s(0)=0$.

Theorem: The length of a continuously differentiable parametrized curve on an interval $[t_0,t_1]$ is the integral of its speed over that interval. In particular, the length of one cycloid arch is $$ L=s(2\pi)=4a\bigl(1-\cos\pi\bigr)=8a. $$

Worked Example: Finding the Speed and Length for a Given Radius

Suppose a cycloid is generated by a circle of radius $a=3$ units. Find its speed on one arch and the length of that arch.

First substitute $a=3$ into the speed formula. On $0\le t\le2\pi$, the sine is nonnegative, so

v(t)=2(3)\sin(t/2)=6\sin(t/2).

The arc length is the speed integral over the full interval:

\begin{aligned} L &=\int_0^{2\pi}6\sin(t/2)\,dt\\ &=\left[-12\cos(t/2)\right]_0^{2\pi}\\ &=-12\cos\pi+12\cos0\\ &=-12(-1)+12(1)\\ &=24\text{ units}. \end{aligned}

Check using $L=8a$: $8(3)=24$ units, agreeing with the integral. Differentiating the antiderivative $-12\cos(t/2)$ gives $6\sin(t/2)$, the speed used in the calculation.

Position and Motion During One Arch

The parametrization identifies several useful positions. At $t=0$, the point is at the starting cusp. At $t=\pi$, it reaches the top of the arch. At $t=2\pi$, it arrives at the next cusp. The arc-length function shows that the top is halfway along the arch.

\begin{aligned} t=0 &: \quad (x,y)=(0,0), \quad s=0,\\ t=\pi &: \quad (x,y)=(\pi a,2a), \quad s=4a,\\ t=2\pi &: \quad (x,y)=(2\pi a,0), \quad s=8a. \end{aligned}

For $0<t<2\pi$, the horizontal velocity component $a(1-\cos t)$ is positive. Therefore the point moves from left to right throughout the interior of the arch. Its vertical velocity is $a\sin t$: it is positive for $0<t<\pi$, zero at $t=\pi$, and negative for $\pi<t<2\pi$. The point rises to the top and then descends.

Differentiating velocity gives acceleration:

\mathbf{a}(t)=\mathbf{v}'(t)=\bigl(a\sin t,a\cos t\bigr).

At the endpoints, the speed is zero. These are stops in the parameterized motion and correspond to the cusps of the arch. At $t=\pi$, the velocity is $(2a,0)$: the point is moving horizontally to the right at the top, while its acceleration is $(0,-a)$.

Worked Example: Tracking the Motion Between Two Parameter Values

Let $a=2$, and follow the point from $t=\pi/3$ to $t=4\pi/3$. Find the positions, velocities, and distance traveled along the curve.

At $t=\pi/3$, use $\sin(\pi/3)=\sqrt{3}/2$ and $\cos(\pi/3)=1/2$:

\begin{aligned} x(\pi/3)&=2\left(\frac{\pi}{3}-\frac{\sqrt3}{2}\right) =\frac{2\pi}{3}-\sqrt3,\\ y(\pi/3)&=2\left(1-\frac12\right)=1. \end{aligned}

The velocity there is

\mathbf{v}(\pi/3) =\bigl(2(1-\tfrac12),2\tfrac{\sqrt3}{2}\bigr) =(1,\sqrt3).

Its magnitude is $\sqrt{1^2+(\sqrt3)^2}=2$, which agrees with $v(t)=4\sin(t/2)$ at $t=\pi/3$, since $4\sin(\pi/6)=2$.

At $t=4\pi/3$, use $\sin(4\pi/3)=-\sqrt{3}/2$ and $\cos(4\pi/3)=-1/2$:

\begin{aligned} x(4\pi/3)&=2\left(\frac{4\pi}{3}+\frac{\sqrt3}{2}\right) =\frac{8\pi}{3}+\sqrt3,\\ y(4\pi/3)&=2\left(1+\frac12\right)=3,\\ \mathbf{v}(4\pi/3)&=\bigl(2(1+\tfrac12),-2\tfrac{\sqrt3}{2}\bigr) =(3,-\sqrt3). \end{aligned}

The speed is $\sqrt{3^2+(-\sqrt3)^2}=2\sqrt3$. This also matches $4\sin(2\pi/3)=2\sqrt3$.

To find the distance traveled, subtract the arc-length values. Since the parameter increases along the arch, this difference is the distance along the curve:

\begin{aligned} s(4\pi/3)&=8\left(1-\cos(2\pi/3)\right) =8\left(1+\frac12\right)=12,\\ s(\pi/3)&=8\left(1-\cos(\pi/6)\right) =8\left(1-\frac{\sqrt3}{2}\right)=8-4\sqrt3,\\ \text{distance traveled} &=s(4\pi/3)-s(\pi/3)\\ &=12-(8-4\sqrt3)=4+4\sqrt3\text{ units}. \end{aligned}

As a check, the difference is positive, as required for forward motion along the arch. The displacement is a separate quantity: subtracting the two positions gives $(2\pi+2\sqrt3,2)$, not the distance traveled along the curve.

Common Mistake: Losing the Absolute Value

A frequent error is to simplify $2a|\sin(t/2)|$ to $2a\sin(t/2)$ without restricting the parameter interval. For example, at $t=3\pi$, the expression without the absolute value would give $2a\sin(3\pi/2)=-2a$, which cannot be a speed. The correct value is $2a|\sin(3\pi/2)|=2a$.

For one arch, the simplification without absolute value is valid precisely because $0\le t/2\le\pi$, where sine is nonnegative. Another interval error is to use $0\le t\le\pi$ as the full arch. That interval reaches the top but not the next cusp; it traces only half the arch, whose length is $s(\pi)=4a$. One full arch requires $0\le t\le2\pi$ and has length $8a$.

Takeaway: Keep the absolute value in the general speed formula. Remove it only after establishing that the sine has a known sign on the interval being considered. For one complete cycloid arch, that interval is $0\le t\le2\pi$.

Check Your Understanding

Use the speed and arc-length formulas to answer these questions.

For $a=5$, what is the length of one complete arch?
At which parameter value does the point reach the top of one arch, and what are its coordinates?
Why is the speed $2a\sin(t/2)$ on $0\le t\le2\pi$, but $2a|\sin(t/2)|$ is needed as the general expression?
For $a=1$, how far along the arch has the point traveled by $t=\pi$?
What are the velocity and acceleration vectors at $t=3\pi/2$ for $a=1$?

What You'll Learn

The Motion of a Cycloid

Velocity and Speed

Deriving the Arc-Length Function

Worked Example: Finding the Speed and Length for a Given Radius

Worked Example: Finding the Speed and Length for a Given Radius

Position and Motion During One Arch

Worked Example: Tracking the Motion Between Two Parameter Values

Worked Example: Tracking the Motion Between Two Parameter Values

Common Mistake: Losing the Absolute Value

Check Your Understanding

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