Particle on a Cardioid with Constant Speed

⚡ Key Takeaways

A point mass travels along a cardioid (a heart-shaped curve) while its speed $v = |\dot{\mathbf r}|$ remains constant. The problem couples the fixed geometry of the cardioid to the kinematic constraint $|\dot{\mathbf r}| = \text{const}$ , yielding non-uniform angular motion, a tangential acceleration that vanishes, and a centripetal acceleration that varies strongly with position. It appears in advanced calculus, classical mechanics, and differential-geometry courses as a showcase of arc-length parametrization and curvature. Studied as a teaching staple since the 18th–19th century alongside the cardioid itself (Castillon, 1741).

💡 The Core Insight

Constant speed means arc length advances uniformly with time: $s = v\,t$ . Geometry (the cardioid) determines how arc length maps onto position, while kinematics fixes how fast arc length is consumed. The entire problem reduces to (1) computing arc length along the cardioid and (2) reading off acceleration from curvature, since $a_T = \dot v = 0$ and $a_N = \kappa v^2$ .

DefinitionThe Cardioid

The cardioid of parameter $a > 0$ is the locus

1

\quad r(\theta) = a\,(1 + \cos\theta), \qquad \theta \in [0, 2\pi),

Derivation hint: trace a fixed point on a circle of radius $a/2$ rolling on the outside of an equal circle (an epicycloid with one cusp); convert the rolling-point coordinates to polar form. with Cartesian parametrization

2

\quad x(\theta) = a(1+\cos\theta)\cos\theta, \qquad y(\theta) = a(1+\cos\theta)\sin\theta.

Derivation hint: substitute $x = r\cos\theta$ , $y = r\sin\theta$ into Eq. [1]. Here $\theta$ is the polar angle, $r$ the radius, and $a$ sets the overall scale (the curve reaches its maximum $r = 2a$ at $\theta = 0$ and degenerates to a cusp at the origin when $\theta = \pi$ ).

DefinitionConstant-Speed Motion

A particle position $\mathbf r(t) = (x(\theta(t)), y(\theta(t)))$ moves with constant speed $v$ if

3

\quad v \equiv |\dot{\mathbf r}(t)| = \left|\frac{d\mathbf r}{ds}\right|\left|\frac{ds}{dt}\right| = \frac{ds}{dt} = \text{const},

Derivation hint: $|d\mathbf r/ds| = 1$ because $s$ is arc length; the chain rule then gives speed as the rate of arc-length change. so that the arc length satisfies $s(t) = s_0 + v\,t$ . The symbol $\dot{(\ )}$ denotes $d/dt$ and primes will denote $d/d\theta$ .

TheoremKinematics on the Cardioid at Constant Speed

For a particle on the cardioid $r=a(1+\cos\theta)$ moving with constant speed $v$ :

(a) The speed element (arc-length rate in $\theta$ ) is

4

\quad \frac{ds}{d\theta} = \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} = 2a\,\Bigl|\cos\tfrac{\theta}{2}\Bigr|.

Derivation hint: with $dr/d\theta = -a\sin\theta$ , compute $r^2 + r'^2 = a^2(2 + 2\cos\theta) = 4a^2\cos^2(\theta/2)$ via the half-angle identity.

(b) The angular velocity required to keep speed constant is

5

\quad \dot\theta = \frac{v}{ds/d\theta} = \frac{v}{2a\,|\cos(\theta/2)|}.

Derivation hint: rearrange $v = (ds/d\theta)\,\dot\theta$ from the chain rule.

(c) The tangential acceleration vanishes and the (centripetal) acceleration has magnitude

6

\quad a_T = \dot v = 0, \qquad a_N = \kappa\,v^2 = \frac{3v^2}{2\,(ds/d\theta)} = \frac{3v^2}{4a\,|\cos(\theta/2)|},

Derivation hint: $a_T = dv/dt = 0$ by constant speed; $a_N = \kappa v^2$ where the cardioid's curvature is $\kappa = 3/(2\sqrt{2a r}) = 3/(4a|\cos(\theta/2)|)$ . where $\kappa$ is the signed curvature of the cardioid.

ProofAcceleration in the Frenet Frame

Step 1 — Decompose velocity. Write the velocity as $\mathbf v = v\,\hat{\mathbf T}$ , where $\hat{\mathbf T} = d\mathbf r/ds$ is the unit tangent and $v = ds/dt$ .

Step 2 — Differentiate. By the product rule,

7

\quad \mathbf a = \frac{d\mathbf v}{dt} = \dot v\,\hat{\mathbf T} + v\,\frac{d\hat{\mathbf T}}{dt}.

Derivation hint: differentiate $\mathbf v = v\hat{\mathbf T}$ treating both $v$ and $\hat{\mathbf T}$ as time-dependent.

Step 3 — Use the Frenet relation. Since $d\hat{\mathbf T}/ds = \kappa\,\hat{\mathbf N}$ , the chain rule gives $d\hat{\mathbf T}/dt = (d\hat{\mathbf T}/ds)(ds/dt) = \kappa v\,\hat{\mathbf N}$ , so

8

\quad \mathbf a = \dot v\,\hat{\mathbf T} + \kappa v^2\,\hat{\mathbf N}.

Derivation hint: substitute the first Frenet–Serret equation into Eq. [7].

Step 4 — Apply the constraint. Constant speed means $\dot v = 0$ , eliminating the tangential term:

9

\quad \mathbf a = \kappa v^2\,\hat{\mathbf N}.

Derivation hint: set $\dot v=0$ in Eq. [8]; only the normal (centripetal) component survives.

Step 5 — Insert cardioid curvature. The polar curvature formula

10

\quad \kappa = \frac{\bigl|\,r^2 + 2r'^2 - r\,r''\,\bigr|}{\bigl(r^2 + r'^2\bigr)^{3/2}}

Derivation hint: standard curvature expression for a polar curve $r(\theta)$ , obtained by parametrizing $(r\cos\theta, r\sin\theta)$ and applying $\kappa = |x'y''-y'x''|/(x'^2+y'^2)^{3/2}$ . evaluates, for $r=a(1+\cos\theta)$ with $r'=-a\sin\theta$ , $r''=-a\cos\theta$ , to

11

\quad \kappa = \frac{3a^2(1+\cos\theta)}{\bigl(2a^2(1+\cos\theta)\bigr)^{3/2}} = \frac{3}{2\sqrt{2}\,a\sqrt{1+\cos\theta}} = \frac{3}{4a\,|\cos(\theta/2)|}.

Derivation hint: numerator $r^2+2r'^2-rr'' = a^2(1+\cos\theta)^2 + 2a^2\sin^2\theta + a^2(1+\cos\theta)\cos\theta = 3a^2(1+\cos\theta)$ ; then use $1+\cos\theta = 2\cos^2(\theta/2)$ . Combining Eqs. [9] and [11] gives Eq. [6]. $\blacksquare$

∎

Figure

A particle on the cardioid

r=a(1+\cos\theta)

moving at constant speed; the velocity vector keeps constant length while curvature (hence centripetal acceleration) grows toward the cusp at the origin.

mathematician · Giovanni Francesco Melfi Castillon | 1708–1791

The cardioid was named (cardioides, "heart-shaped") and studied in detail by Johann Castillon in a 1741 paper to the Royal Society, though the curve had been examined earlier as a special epicycloid by Ole Rømer (1674) and by Louis Carré and Koersma. Castillon's contemporaries, including the Bernoullis and later differential geometers, developed the arc-length and curvature formulas that make constant-speed motion on the cardioid a clean exercise. The curve's role in optics (as a caustic) and in mechanics (epicyclic motion) cemented it as a standard object well before modern kinematics formalized the Frenet frame in the 1840s–50s.

🧠 Intuition

Imagine driving a car around a heart-shaped track while holding the speedometer fixed. On the broad, gently-curving outer lobe (near $\theta=0$ , $r=2a$ ) you barely turn the wheel — curvature is small, so you feel little sideways push. As you approach the cusp at the origin, the track pinches to a point and curves infinitely sharply: to keep the same speed you must whip the steering wheel ever faster ( $\dot\theta\to\infty$ ) and you feel an unbounded centripetal force. Constant speed does not mean constant effort — the geometry hides all the drama in the direction of motion. Mathematically, arc length is the "fair clock" here: equal time intervals carve out equal arc lengths, but those equal arcs subtend wildly different angles $\Delta\theta$ depending on where you are.

CS

Constant speed ≠ uniform motion

A common physicist's reflex is to equate "constant speed" with "no acceleration." On a curved path this is false: the velocity vector turns, so there is always centripetal acceleration $a_N = \kappa v^2$ unless the path is locally straight ( $\kappa=0$ ). On the cardioid $\kappa$ never vanishes, so the particle is always accelerating despite constant speed — the force is purely a steering force, doing no work ( $\mathbf F\cdot\mathbf v = 0$ ).

ExampleTime to traverse the cardioid

Given: $a = 0.5$ m, constant speed $v = 2$ m/s. Find the total period $T$ to go around the cardioid once.

Step 1 — Total arc length. Integrate Eq. [4] over $\theta\in[0,2\pi]$ :

L = \int_0^{2\pi} 2a\,\bigl|\cos\tfrac{\theta}{2}\bigr|\,d\theta = 2a\!\left[\int_0^{\pi}\!\cos\tfrac{\theta}{2}\,d\theta - \int_{\pi}^{2\pi}\!\cos\tfrac{\theta}{2}\,d\theta\right].

Step 2 — Evaluate. Each integral gives $\bigl[2\sin(\theta/2)\bigr]$ . The result is the classic total length $L = 8a$ . With $a=0.5$ m: $L = 8(0.5) = 4$ m.

Step 3 — Period. Since $s = vt$ , $T = L/v = 4/2 = 2$ s.

Answer:

\boxed{L = 8a = 4\ \text{m}, \qquad T = \frac{8a}{v} = 2\ \text{s}.}

ExampleCentripetal acceleration at a specific point

Given: $a = 1$ m, $v = 3$ m/s. Find $a_N$ at (i) $\theta = 0$ (the far point $r=2a$ ) and (ii) $\theta = 2\pi/3$ .

Step 1 — Use Eq. [6]: $a_N = \dfrac{3v^2}{4a\,|\cos(\theta/2)|}$ .

(i) $\theta=0$ : $\cos(0)=1$ , so

a_N = \frac{3(3)^2}{4(1)(1)} = \frac{27}{4} = 6.75\ \text{m/s}^2.

(ii) $\theta=2\pi/3$ : $\cos(\pi/3)=\tfrac12$ , so

a_N = \frac{3(9)}{4(1)(0.5)} = \frac{27}{2} = 13.5\ \text{m/s}^2.

Check the cusp: as $\theta\to\pi$ , $\cos(\theta/2)\to 0$ and $a_N\to\infty$ — the acceleration diverges, confirming the cusp singularity.

Answer:

\boxed{a_N(\theta{=}0) = 6.75\ \text{m/s}^2, \qquad a_N(\theta{=}2\pi/3) = 13.5\ \text{m/s}^2.}

⚠️ Common mistake

"Constant speed implies constant angular velocity $\dot\theta$ ." This is wrong for any non-circular path. Eq. [5] shows $\dot\theta = v/(2a|\cos(\theta/2)|)$ , which varies from a minimum $v/(2a)$ at $\theta=0$ to $+\infty$ at the cusp. Conflating uniform speed with uniform angular rate is the single most frequent error; only on a circle (where $ds/d\theta = r = \text{const}$ ) do the two coincide.

Counterexample

Suppose someone claims the acceleration is constant because "speed is constant and $a$ is constant." Take $a=1$ m, $v=2$ m/s. Then $a_N(\theta=0) = 3(4)/4 = 3$ m/s², while $a_N(\theta=\pi/2)$ uses $\cos(\pi/4)=\tfrac{\sqrt2}{2}$ giving $a_N = 12/(4\cdot 0.707) \approx 4.24$ m/s². The acceleration magnitude changes along the path — and it diverges at the cusp — so it is decidedly not constant, refuting the claim.

connection

The constant-speed cardioid sits at the crossroads of several topics. It is an instance of motion on a Plane Curve in Arc-Length Parametrization, where setting $|\dot{\mathbf r}|=v$ is exactly choosing arc length as a rescaled time. The acceleration decomposition uses the Frenet–Serret Frame (here in 2D, with torsion zero). The curve itself is a degenerate Epicycloid (a circle rolling on an equal circle) and a Caustic of a circle under point-source illumination. The divergence of $\dot\theta$ at the cusp links to the general theory of Singular Points of Plane Curves. Finally, computing $L=8a$ is a canonical exercise in Arc Length in Polar Coordinates.

🔬 Where this appears

⏳ History

- 1674 — Ole Rømer studies epicycloids (including the one-cusp case) while designing better gear teeth, implicitly encountering the cardioid.

- 1708 — Philippe de La Hire computes the length of the cardioid as

8a

, an early arc-length result.

- 1741 — Johann Castillon publishes on the curve in the Philosophical Transactions, coining the name cardioid from the Greek for "heart-shaped."

- 1750s — Leonhard Euler develops general curvature and arc-length machinery for plane curves, providing tools applicable to constant-speed motion.

- 1781 — Pierre-Simon Laplace and contemporaries study caustics by reflection, recognizing the cardioid as the caustic of a circle (a catacaustic).

- 1847 — Jean Frédéric Frenet derives the moving-frame equations; 1851, Joseph Serret independently completes them, giving the modern

a_T,a_N

decomposition.

- 1860s–70s — Standard mechanics texts adopt the tangential/normal acceleration split, making "particle on a curve at constant speed" a routine pedagogical problem.

- 1930s — The cardioid microphone pattern is engineered (ribbon/dynamic combinations), turning the abstract curve into a manufactured device.

- 20th century onward — Numerical path-planning and CNC machining revive the constant-speed traversal problem in engineering practice.

open-problem

The cusp at $\theta=\pi$ is where idealization breaks down. Strictly, constant-speed motion through the cusp is impossible: $\dot\theta\to\infty$ and $a_N\to\infty$ , requiring infinite force. A physical particle either (a) reverses direction at the cusp (the tangent direction is undefined/discontinuous there), (b) slows down — violating constant speed — or (c) leaves the curve. Open modeling questions include: how should one regularize the cusp (rounding it to radius $\varepsilon$ , then taking $\varepsilon\to0$ )? Does a least-action or minimal-jerk smoothing select a unique limiting trajectory? In control-theoretic path planning, finding the time-optimal speed profile that respects a bound $a_N \le a_{\max}$ near a cusp (necessarily slowing down) is a nontrivial constrained-optimization problem with no closed-form general solution. These issues generalize to motion through any singular point of a plane curve.

Particle on a Cardioid with Constant Speed

Formal Statement

Why It Holds / Derivation

Building Intuition

Worked Examples

Common Misconceptions

Connections

Real-World Applications

Historical Development

Subtleties & Limits

See Also