• Calculus > Cycloidal Curves

The Cycloid and Its Properties and Related Curves

If a point lies at a factor of f, where 0 ≤ f ≤ 1, along the radius of the circle, then the equation of the curve is:

$\left\{\begin{matrix} x(t)= rt - fr\sin(t)\\ y(t)= r-fr\cos(t)\\ \end{matrix}\right.$.

The Geogebra interaction above shows a circle with radius 1 rolling along the x-axis. Other points shown are the halfway point the radium, quarter, and an eigth of the radiu. You can drag the angle slider left and right to make the circle roll and see the points trace the respective curves. You may zoom in or out and move the canvas around to see the full curve.

Area

The area, A, under one arch of the cycloid can be found using the formula: $A = \int_{a}^{b}y\text{ }dx$.

We substitute $y=r(1-\cos(t))$ and $dx=r(1-\cos(t)) dt $to obtain:

(i) $A = \int_{0}^{2\pi} r(1-\cos(t))\cdot r(1-\cos(t))dt =$ $r^2\int_{0}^{2\pi} (1-\cos(t))^2 dt$.

Expanding and substituting the double-angle identity of cosine:

(ii) $A = r^2\int_{0}^{2\pi} 1-2\cos(t)+\frac{\cos(2t)+1}{2}dt =$ $r^2\left [\frac{3t}{2} -2\sin(t)+\frac{\sin(2t)}{4}\right ]_{0}^{2\pi}= $ $r^2\left [ \frac{3(2\pi)}{2} \right ] = 3\pi r^2$.

Arc Length

To find the arc length, L, of one arch of the cycloid, we use the arc length formula for parametric equations: $L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2} dt$.

We make the substitutions: $\frac{dx}{dt} = r(1-\cos(t))$ and $\frac{dy}{dt} = r\sin(t)$:

(i) $L = \int_{0}^{2\pi} \sqrt{r^2\sin^2t+r^2(1-\cos(t))^2} dt =$ $r\int_{0}^{2\pi} \sqrt{2-2\cos(t)} dt =$ $\sqrt{2}r\int_{0}^{2\pi} \sqrt{1-\cos(t)} dt$

(ii) $\sqrt{2}r\int_{0}^{2\pi} \sqrt{1-\cos(t)} dt =$ $\sqrt{2}r\int_{0}^{2\pi} \sqrt{2}\sin(t/2) dt =$$-4r\left[\cos(t/2)\right]_{0}^{2\pi} = -4r(-1-1) = 8r$

It’s interesting that the area formula has a π factor like the area of a circle but the arc length formula does not have a π factor like the circumference of a circle.

Volume

The volume, V, of a solid generated by revolving once arch of the cycloid about the x-axis can be found by using the formula: $V = \pi\int_{a}^{b}y^2\text{ }dx$.

We arrive at the following integral and evaluation: $\pi r^3\int_{0}^{2\pi}(1-\cos(t))^3dt = 5\pi^2r^3$.

If the cycloid is revolved about the y-axis, the volume generated by one arch is given by the integral with an evaluation of: $2\pi r\int_{0}^{2\pi}(t-\sin(t))(1-\cos(t))^2dt=6(\pi r)^3$.

Cycloid Cup

What is the volume of a cycloid revolved about its axis of symmetry and shaped like a cup? I have come up with the following volume using the disc method. The axis of symmetry is the line x = π.

Take a horizontal cross section from 0 to π for the area of the circle that would generate by revolution. Subtract the x value from π to get the radius of this cicle. The radius, r, would be $R\pi - R(t-\sin(t))$, where R is the radius of the circle. The change in y would be $\frac{dy}{dt} = R\sin(t) dt$. Our interval of integration of 0 to π. Putting it together we have:

(i) $V = \int_{0}^{\pi} \pi[R\pi - R(t-\sin(t))]^2(R\sin(t))dt =$ $\pi R^3\int_{0}^{\pi} [\pi - (t-\sin(t))]^2(\sin(t))dt$

Using Wolfram Alpha, this evaluates to $V = \pi R^3\left (\frac{3\pi^2}{2}-\frac{8}{3}\right)$ ≈ 38.13R³. I have not been able to verify this formula as I have not seen the cycloid rotation into a cup. However, this formula seems to confirm when we find the volume using the equations of the cycloid that is upside down and has the y-axis as the line of symmetry:

$\left\{\begin{matrix} x(t)= Rt + R\sin(t)\\ y(t)= R-R\cos(t)\\ \end{matrix}\right.$

The volume is given by: $V = \pi\int_{0}^{\pi}(Rt+R\sin(t))^2\cdot R\sin(t)dt$. This evaluates to the same volume as above. This volume would seem reasonable if it is less than the volume of the cylinder created by a radius of πR, which is the “length” the circle travels to the cusp of the cycloid. The volume of this cylinder would be $(\pi R)^2\pi\cdot 2R = 2\pi^3 R^3 \approx 62R^3$. The volume of the cycloid cup is more than half of the volume of the cylinder and this seems reasonable.

The Right Triangle

There’s a right angle hiding in the cycloid curve. The angle occurs at the intersection of the cycloid and the circle. The diametere of the circle is the hypotenuse.

In the image above, APC is a right triangle. To prove this, we can prove that the orthogonal of the tangent to the cycloid passes through the circle where it is tangent to the x-axis. First, let’s find the slope of the tangent to the cycloid at t.

(i) $\frac{dy}{dx} = \frac{\frac{dy}{dt}(r-r\cos t)}{\frac{dx}{dt}(rt-r\sin t)} = $ $\frac{r\sin t}{r - r\cos t} =$ $\frac{\sin t}{1 - \cos t}$

The slope of the orthogonal would be the negative inverse: $-\frac{dx}{dy} = -\frac{1-\cos t}{\sin t}$. At t, the point on the cycloid is $(rt - r\sin t, r - r\cos t)$. Therefore, the equation of the orthogonal is:

(ii) $y - (r - r\cos t) = \frac{\cos t - 1}{\sin t}[x - (rt - r\sin t)]$

The point at which the circle is tangent to the x-axis is rt because t is really the angle the circle travels along the x-axis. Therefore, we can find the y value when x equal rt by substituting x = rt in the equation of the orthogonal. (There’s no need to simplify the equation of the line in (ii).)

(iii) $y - (r - r\cos t) = \frac{\cos t - 1}{\sin t}[rt - (rt - r\sin t)]$

(iv) $y - (r - r\cos t) = \frac{\cos t - 1}{\sin t}(r\sin t)$

(v) $y - (r - r\cos t) = r\cos t - r$

(vi) $y = 0$

Therefore, for any t, the orthogonal will always intersect the circle at its tangent point to the x-axis. And since the orthogonal, by definition, is a right angle, segment PA has to intersect the circle at point A, making AC the diameter of the circle. If we let y = 0 and solve for x, we would get x = rt.

The Cycloid and the Sine Curve

The sine curve has a special relationship with the cycloid. If we graph the translated sine curve $y = r\sin(x - \frac{\pi}{2}) + r$. This is the exact sine curve only scaled by r. The shape is the same as if we had graphed y = sin x. The cycloid and the sine curve are shown below.

If we take a point P on the cycloid and draw a parallel at P, then it intersects the sine curve at D and E and the cycloid again at G. The segments PD and EG are congruent because of symmetry and equal to rsin t.

We can show that DO is perpendicular to the x-axis. Therefore, the sum of segments PD and EG is equal to length the chord of the circle PK for all t. Or PK is twice the length of PD. This means the area enclosed by the cycloid and the sine curve is equal to the area of the circle. And the area below the sine curve is equal to twice the area of the circle since the area under the cycloid is equal to 3 times the circle.

We can show by integration the area below the sine curve is 2πr:

(i) $\int_{0}^{2\pi} r\sin(x-\frac{\pi}{2}) + r\text{ }dx =$ $\left [-r\cos(x-\frac{\pi}{2}) + rx \right ]_{0}^{2\pi}$ = $\left [-r\cos(2\pi-\frac{\pi}{2}) + 2\pi r \right ] - \left [-r\cos(0-\frac{\pi}{2}) + 0 \right ] = $ $2\pi r$

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