Curves traced by a moving point: rolling wheels, roses and spirals in polar coordinates, and the area they sweep
Not every curve is the graph of a function. A point on a bicycle wheel loops and cusps, a spiral winds round and round, a rose doubles back through its centre. Describe such curves by where a moving point is at each moment, (x(t), y(t)), or by how far it is from the origin in each direction, r = f(θ), and calculus works on them just as well: velocities, lengths, slopes and areas.
Roll a wheel along a line, around a circle or inside one, and move the pen in or out. The pink arrow is the pen’s velocity: watch it shrink to nothing at every cusp.
The wheel has radius r = 1 and turns through angle t, so its centre is at (t, 1) and the pen at (t − d sin t, 1 − d cos t). Lengths are adaptive Simpson integrals of the speed √(x′² + y′²). Areas are signed: a loop traced clockwise counts negatively, and a region circled twice counts twice.
A wheel of radius r that has turned through angle t without slipping has rolled a distance rt, so its centre is at (rt, r). The pen, at distance d from the centre, has turned with it:
x(t) = rt − d sin t, y(t) = r − d cos t, velocity (r − d cos t, d sin t).
With the pen on the rim (d = r) the velocity is r(1 − cos t, sin t), which is zero when t = 0, 2π, 4π, …: the pen stops dead each time it touches the ground, which is why the cycloid has cusps rather than smooth bottoms. Its speed is r√(2 − 2cos t) = 2r|sin(t/2)|, so one arch has length
∫₀^2π 2r sin(t/2) dt = 8r, and area under it ∫ y dx = ∫₀^2π r²(1 − cos t)² dt = 3πr².
Gilles de Roberval found the area, three times the wheel’s, around 1634, and Christopher Wren the length, four times the wheel’s diameter, in 1658. Christiaan Huygens showed that a bead sliding on an upside-down cycloid reaches the bottom in the same time wherever it starts (the tautochrone), and in his Horologium Oscillatorium of 1673 he described a pendulum swinging between cycloidal cheeks so that its bob follows a cycloid and keeps time whatever the size of its swing.
In June 1696 Johann Bernoulli challenged the mathematicians of Europe in the Acta Eruditorum: along what curve does a bead slide from one point to a lower one in the least time? Solutions came from Johann himself, his brother Jakob, Leibniz, l’Hôpital and, anonymously, Newton, and the answer was again the cycloid. Jakob’s method grew into the calculus of variations.
Rolling around a circle of radius R gives epicycloids (outside) and hypocycloids (inside). With R/r a whole number k the curve closes with k cusps: the cardioid (k = 1), nephroid (2), deltoid (3, inside) and astroid (4, inside). A wheel rolling inside a circle twice its size draws a straight diameter, a device known as the Tusi couple after the 13th-century astronomer Nasir al-Din al-Tusi.
A ray sweeps round and the point at distance r = f(θ) traces the curve. Watch the ordinary graph of r against θ on the right: each hump becomes a petal, and each dip below zero a petal on the far side.
Type θ or theta; the letter k is a parameter with its own slider. Drag on the polar picture to turn the ray, or along the graph to scrub θ. With k odd, the rose cos(kθ) has k petals and θ from 0 to π already draws them all; running on to 2π traces each petal a second time with the sign of r flipped. With k even it has 2k petals. A half-integer k needs θ to run to 4π to close.
A polar point (r, θ) sits at x = r cos θ, y = r sin θ: turn to angle θ, then walk r. When r is negative you walk backwards, landing on the opposite ray, at angle θ + π. That convention is what lets a single formula like cos 3θ draw all three petals of a rose: on the stretches where cos 3θ < 0, the petal appears on the far side.
The graph of r against θ is the key to sketching. Where r = 0 the curve passes through the origin, heading along the ray at that angle; where |r| is largest it is furthest out; and a symmetry of the graph is a symmetry of the curve (f(−θ) = f(θ) means a mirror image across the x-axis).
Fill a polar region with thin wedges and add up their areas. Drag the two rays, refine the wedges, and drag the pink point to see the tangent line. Try the limaçon’s inner loop.
Each wedge takes the radius at its middle angle; when that radius is negative the wedge is drawn on the opposite ray and outlined in rose, but ½r² is positive either way. The slope comes from the exact derivative r′(θ) of your formula. At the origin (r = 0) the tangent is the ray itself, at angle θ.
In Cartesian coordinates we slice area into thin rectangles. In polar coordinates the natural slice is a thin wedge: a sector of a circle of radius r and angle Δθ, whose area is the fraction Δθ/2π of πr², that is ½r²Δθ. Adding the wedges and letting Δθ → 0 gives
A = ½∫ r² dθ from θ₁ to θ₂.
One petal of cos 3θ runs from θ = −π/6 to π/6, where r returns to 0, and has area ½∫ cos² 3θ dθ = π/12 (for cos kθ, π/(4k)). The cardioid 1 + cos θ gives ½∫₀^2π (1 + 2cos θ + cos² θ) dθ = 3π/2. The limits are the whole difficulty: find where r = 0 to know where a loop begins and ends, and check whether the sweep covers any region twice. For the limaçon 1/2 + cos θ, r is negative for θ between 2π/3 and 4π/3, and that sweep draws the inner loop, of area π/4 − 3√3/8. Integrating from 0 to 2π counts the inner loop twice.
Slopes come from the chain rule. With x = r cos θ and y = r sin θ, where r depends on θ,
dy/dx = (dy/dθ)/(dx/dθ) = (r′ sin θ + r cos θ)/(r′ cos θ − r sin θ).
The tangent is horizontal where the numerator is 0 and vertical where the denominator is 0. At the origin, where r = 0 and r′ ≠ 0, the formula collapses to tan θ: the curve leaves the origin along the ray itself.
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