Source: Homework Set 3, §13.1–13.4
Tags: cycloid, cardioid, parametric circles, rolling wheel, parametrisation, arc length, space curves, MATH 215, Calculus III
Difficulty: Intermediate to Challenging | Prerequisites: Parametric equations, trig functions, arc length integral (§13.3), vector addition.
Parametrising curves is a core skill in multivariable calculus, and not every curve comes pre-packaged as r(t) = ⟨cos(t), sin(t)⟩. This section covers curves that arise from geometric constructions: circles in 3D space with specific intersection constraints, cycloids (traced by a point on a rolling wheel), and cardioids (traced by a point on a circle rolling around another circle). The key technique is decomposing motion into simpler vector components and adding them. If you can parametrise a circle and add vectors, you can build these.
Circles in 3D space need a centre, a radius, and a plane to live in. Cycloids and cardioids come from decomposing the motion of a rolling wheel into the translation of the centre plus the rotation of the marked point. Arc length for these curves uses trig identities heavily.
Cycloid
The curve traced by a point on the rim of a circle (wheel) as the circle rolls along a straight line without slipping. Think of it as the path a reflector on a bicycle tyre traces as the bike moves forward.
Cardioid
The curve traced by a point on the rim of a circle as that circle rolls around the outside of a fixed circle of the same radius. The resulting shape resembles a heart. In simple terms, it is the "cycloid" equivalent for rolling around a circle instead of along a line.
Parametric circle in space
A circle in R³ described by parametric equations. Unlike a circle in the plane, you need to specify the plane the circle lies in (equivalently, two perpendicular direction vectors in that plane), a centre, and a radius.
No-slip condition (rolling without slipping)
The constraint that the arc length traced on the rolling circle equals the distance travelled along the surface it rolls on. This is what links the rotation angle to the translation distance. In a wheel of radius R rolling a distance d, the wheel has rotated through angle θ = d/R.
Vector decomposition of rolling motion
The technique of writing the position of a point on a rolling circle as the sum of two vectors: one from the origin to the centre of the circle, and one from the centre to the point on the rim.
A circle in the plane is r(t) = (a + R cos(t), b + R sin(t)). A circle in space needs a centre point, a radius, and two orthogonal unit vectors that span the plane it lies in.
General form: r(t) = C + R cos(t) · u + R sin(t) · v, where C is the centre, R is the radius, and u, v are perpendicular unit vectors in the plane of the circle.
The two given intersection points are (0, 2, −1) and (0, 0, 3).
A clever approach: place both circles in coordinate planes passing through the x = 0 plane.
Both circles must pass through both points and have the same radius.
One clean solution: put both circles in planes that contain the line segment between the two points. For instance, C₁ in the plane x = 0 and C₂ in another plane.
The midpoint of the two intersection points is (0, 1, 1). The distance between them is √((2−0)² + (−1−3)²) = √(4 + 16) = √20 = 2√5.
Any circle through both points has a chord of length 2√5. The centre lies on the perpendicular bisector of this chord, within its plane.
The simplest approach: use two different coordinate planes that both contain the two points. Since both points have x = 0, you can place one circle entirely in the yz-plane and choose another circle in a plane that also passes through both points but tilts into the x-direction.
The cycloid is built in two steps, using a wheel of radius R with a marked point starting at the bottom (the origin).
Step 1: Centre of the wheel.
As the wheel rolls a distance d along the x-axis, its centre moves from (0, R) to (d, R).
If the wheel has rolled through angle θ (measured from the starting position), the no-slip condition gives d = Rθ.
Centre position: (Rθ, R).
Step 2: From centre to the marked point.
The marked point starts at the bottom of the wheel (directly below the centre).
After rolling through angle θ, the marked point has rotated clockwise by angle θ relative to the centre.
The displacement from centre to marked point: (−R sin(θ), −R cos(θ)). The minus signs come from the fact that the point starts at the bottom and the wheel rolls clockwise relative to the ground contact.
Combined: the cycloid parametrisation.
r(θ) = (Rθ − R sin(θ), R − R cos(θ)) = R(θ − sin(θ), 1 − cos(θ)).
One full revolution: θ from 0 to 2π.
r'(θ) = R(1 − cos(θ), sin(θ)).
|r'(θ)|² = R²[(1 − cos(θ))² + sin²(θ)] = R²[1 − 2cos(θ) + cos²(θ) + sin²(θ)] = R²[2 − 2cos(θ)].
|r'(θ)| = R√(2 − 2cos(θ)) = R√(2) · √(1 − cos(θ)).
Using the half-angle identity: 1 − cos(θ) = 2sin²(θ/2), so |r'(θ)| = 2R|sin(θ/2)|.
On [0, 2π], sin(θ/2) ≥ 0, so the absolute value drops.
L = ∫ from 0 to 2π of 2R sin(θ/2) dθ = 2R[−2cos(θ/2)] from 0 to 2π = 2R(2 + 2) = 8R.
The arc length of one arch of the cycloid is 8R, exactly 8 times the radius of the wheel.
A cardioid is traced by a point on a circle of radius R rolling around the outside of a fixed circle of radius R (same radius). The homework requires you to use the rolling decomposition method, not polar coordinates.
Step 1: Centre of the rolling circle.
The fixed circle is centred at the origin with radius R.
The rolling circle's centre stays at distance 2R from the origin (R for the fixed radius plus R for the rolling radius).
As the rolling circle has rolled through angle θ (measured from the starting position on the fixed circle), its centre is at (2R cos(θ), 2R sin(θ)).
Step 2: From centre to the marked point.
The marked point starts at the point of tangency (distance R from the origin, on the positive x-axis if the rolling starts there).
The no-slip condition: the arc on the rolling circle equals the arc on the fixed circle, so the rolling circle rotates through the same angle θ (since both radii are R).
The displacement from the rolling circle's centre to the marked point: (R cos(−θ), R sin(−θ)) rotated appropriately. Accounting for the orientation, the marked point is at (R cos(2θ), R sin(2θ)) relative to the rolling centre's angular position, giving displacement (−R cos(θ + θ), −R sin(θ + θ)) from the centre. The exact signs depend on your starting configuration.
Combined: the cardioid parametrisation.
r(θ) = (2R cos(θ) − R cos(2θ), 2R sin(θ) − R sin(2θ)).
This can also be written using the identity: x = R(2cos(θ) − cos(2θ)), y = R(2sin(θ) − sin(2θ)).
r'(θ) = R(−2sin(θ) + 2sin(2θ), 2cos(θ) − 2cos(2θ)).
|r'(θ)|² = R²[4sin²(θ) − 8sin(θ)sin(2θ) + 4sin²(2θ) + 4cos²(θ) − 8cos(θ)cos(2θ) + 4cos²(2θ)].
Simplify using Pythagorean identities and the product-to-sum formula cos(A − B) = cos(A)cos(B) + sin(A)sin(B):
= R²[4 + 4 − 8cos(θ)] = 8R²(1 − cos(θ)).
|r'(θ)| = 2R√2 · √(1 − cos(θ)) = 4R|sin(θ/2)| (same half-angle identity as the cycloid).
For one full revolution, θ from 0 to 2π:
L = ∫ from 0 to 2π of 4R sin(θ/2) dθ = 4R[−2cos(θ/2)] from 0 to 2π = 4R(2 + 2) = 16R.
Cycloid:
\mathbf{r}(\theta) = \langle R(\theta - \sin\theta),\; R(1 - \cos\theta) \rangleCycloid arc length (one arch): L = 8R.
Cardioid:
\mathbf{r}(\theta) = \langle R(2\cos\theta - \cos 2\theta),\; R(2\sin\theta - \sin 2\theta) \rangleCardioid arc length (full curve): L = 16R.
Half-angle identity used in both: 1 − cos(θ) = 2sin²(θ/2).
Cycloids appear in the brachistochrone problem: the fastest slide between two points under gravity follows a cycloid, not a straight line. This was one of the great problems of early calculus. Gears with cycloidal tooth profiles mesh smoothly, which is why the curve matters in mechanical engineering. Cardioids appear in antenna radiation patterns and in the shape of microphone pickup zones (hence "cardioid microphone").
Students often get the direction of rotation wrong for the marked point. On a wheel rolling to the right, the marked point rotates clockwise relative to the ground. Drawing a picture before writing equations prevents sign errors.
Forgetting the no-slip condition and treating the rolling angle and the distance travelled as independent. They are linked: d = Rθ.
Attempting to use polar coordinates for the cardioid when the problem explicitly requires the rolling-circle decomposition. The polar form r = 2R(1 + cos(θ)) is elegant but does not answer the question being asked.
Dropping absolute values in the arc length computation. On [0, 2π], sin(θ/2) ≥ 0, so the absolute value drops cleanly. On other intervals, it might not.
⚠️ The rolling-circle decomposition is a transferable technique. Any problem involving a point on a rolling or rotating object can be handled the same way: find the centre, then find the displacement from centre to the point.
⚠️ The half-angle identity 1 − cos(θ) = 2sin²(θ/2) appears in nearly every cycloid and cardioid arc length problem. Memorise it.
⚠️ The clean answers (8R for the cycloid, 16R for the cardioid) are worth remembering as sanity checks.
The cycloid parametrisation is r(θ) = ___.
True or false: The no-slip condition relates the arc length on the rolling circle to the distance rolled. ___
The arc length of one arch of a cycloid with wheel radius R is ___.
The half-angle identity used in these arc length problems is: 1 − cos(θ) = ___.
True or false: You can use polar coordinates for the cardioid on Question 3c of this homework. ___
Answers: 1. R(θ − sinθ, 1 − cosθ). 2. True. 3. 8R. 4. 2sin²(θ/2). 5. False (the problem specifically forbids it).
Q: Describe in words the two-step decomposition for parametrising a cycloid.
A: First, find the position of the wheel's centre as a function of how far the wheel has rolled (translation). Second, find the displacement from the centre to the marked point on the rim (rotation). Add the two vectors.
Q: Why does the marked point on the cycloid start at the bottom of the wheel, and what does this mean for the parametrisation?
A: Because the wheel starts with the marked point at the origin (the contact point). This means that at θ = 0, the displacement from the centre to the marked point is (0, −R), pointing straight down.
Q: Compute |r'(θ)| for the cycloid r(θ) = R(θ − sinθ, 1 − cosθ).
A: r'(θ) = R(1 − cosθ, sinθ). |r'(θ)| = R√(2 − 2cosθ) = 2R|sin(θ/2)|.
Q: For the cardioid, why is the rolling circle's rotation angle equal to θ when both circles have radius R?
A: The no-slip condition says the arc on the fixed circle equals the arc on the rolling circle. Arc on fixed = Rθ. Arc on rolling = Rφ. So φ = θ.
The rolling-circle technique generalises to epicycloids (rolling outside a larger circle) and hypocycloids (rolling inside a circle), which appear in gear design and planetary motion models. Arc length computations here reinforce the integral techniques from §13.3. The cardioid also appears in polar coordinates (r = 2R(1 + cosθ)) in Calculus II, and understanding both representations deepens your grasp of coordinate systems.
Cycloid, cardioid, epicycloid, hypocycloid, rolling circle, no-slip condition, parametric circle in space, vector decomposition, arc length, half-angle identity, brachistochrone, tautochrone, MATH 215, Calculus III, §13.1, §13.3, parametrisation