Parametric and Polar Curves

Section IParametric equations

A parametric curve gives both coordinates as functions of a third variable, the parameter, usually t for time:

x = x(t), y = y(t), a ≤ t ≤ b

As t runs from a to b, the point (x(t), y(t)) moves and traces the curve. So a parametric curve has a direction, and it can loop, stop or cross itself. For example x = cos t, y = sin t, 0 ≤ t ≤ 2π, goes once round the unit circle, anticlockwise from (1, 0).

To find an ordinary equation, eliminate the parameter. Here cos² t + sin² t = 1 gives x² + y² = 1. For x = t + 1, y = t², solve t = x − 1 and substitute: y = (x − 1)², a parabola. You lose the timing, but you see the shape.

A point on the rim of a rolling wheel of radius 1 traces the cycloid x = t − sin t, y = 1 − cos t, where t is the angle the wheel has turned. It is in the picture above, and it comes back in the history section.

Section IISlopes and the second derivative

In a short time Δt, the point moves about Δx ≈ (dx/dt)Δt across and Δy ≈ (dy/dt)Δt up. The slope is rise over run, and the Δt cancels:

dy over dx = dy/dt over dx/dt, provided dx/dt ≠ 0

The tangent is horizontal where dy/dt = 0 and dx/dt ≠ 0, and vertical where dx/dt = 0 and dy/dt ≠ 0. If both are 0, the point stops for an instant and you need to look more closely. The cycloid does this each time the rim touches the ground, making a sharp point called a cusp.

For the cycloid, dy/dx = sin t/(1 − cos t). At t = π, the top of the arch, this is 0/2 = 0: a horizontal tangent, as you would expect.

The second derivative, in brief. d²y/dx² is the derivative of dy/dx with respect to x. But dy/dx is a function of t, so use the same trick again:

d²y over dx² = d/dt (dy/dx) over dx/dt

A common slip is to write (d²y/dt²)/(d²x/dt²). That is not the second derivative. Its sign tells you concavity, just as for y = f(x).

Section IIIArc length and speed

In time Δt the point moves along a tiny, nearly straight piece of the curve with length √((Δx)² + (Δy)²) ≈ √((dx/dt)² + (dy/dt)²) Δt. The square root is the speed of the point. Adding up the pieces gives the arc length:

L = ∫_a^b √((dx/dt)² + (dy/dt)²) dt

Check it on the circle x = cos t, y = sin t: the speed is √(sin² t + cos² t) = 1, so the length for 0 ≤ t ≤ 2π is 2π, the circumference. The formula counts every bit of path the point travels, so if the curve is traced twice, the length counts twice.

In polar form (Section V), the same idea gives L = ∫ √(r² + (dr/dθ)²) dθ. For the cardioid r = 1 + cos θ this gives a total length of 8.

Section VPolar coordinates

Instead of “across and up”, describe a point by “how far, and which way”. Its polar coordinates (r, θ) are its distance r from the origin, called the pole, and the angle θ measured anticlockwise from the positive x-axis.

Right-angled triangles give the conversions:

x = r cos θ, y = r sin θ, r² = x² + y², tan θ = y/x

Going from Cartesian to polar, find r first, then choose θ in the right quadrant: tan θ = y/x alone cannot tell (1, 1) from (−1, −1). The same point has many names: (2, π/3), (2, 7π/3) and (−2, 4π/3) all describe it, because a negative r means “go the other way”.

A polar curve r = f(θ) gives the distance for each direction. r = 2 is a circle; r = θ is a spiral; r = 1 + cos θ is a heart-shaped cardioid; r = cos 3θ is a three-petal rose. To convert a polar equation, use r cos θ = x, r sin θ = y and r² = x² + y², often after multiplying through by r.

Section VISlopes of polar curves

A polar curve is a parametric curve in disguise, with θ as the parameter: x = f(θ) cos θ, y = f(θ) sin θ. Differentiate with the product rule, writing r′ = dr/dθ, and divide:

dy over dx = r′ sin θ + r cos θ over r′ cos θ − r sin θ

Watch out: dr/dθ is not the slope of the curve. It tells you how fast the distance changes as you turn, not which way the curve is pointing. Horizontal and vertical tangents come from the top and bottom of the fraction being 0, just as before. At the pole, where r = 0, the formula reduces to tan θ: the curve leaves the pole in the direction θ.

Section VIIPolar area

Areas in polar coordinates come from thin wedges, not thin rectangles. A wedge of a circle of radius r with angle Δθ has area (1/2)r² Δθ. A thin wedge of a polar curve is almost such a slice, so adding them up gives

A = (1/2)∫_α^β r² dθ

For the whole cardioid r = 1 + cos θ:

Choose the limits so the curve is traced exactly once. The rose r = cos 3θ is traced once as θ goes from 0 to π; integrating from 0 to 2π counts every petal twice.

Section IXCommon mistakes

• Dividing the wrong way. dy/dx = (dy/dt)/(dx/dt), not (dx/dt)/(dy/dt).
• Dividing second derivatives. d²y/dx² is not (d²y/dt²)/(d²x/dt²).
• Reading dr/dθ as the slope. Use the polar slope formula instead.
• Picking θ from tan θ alone. Check the quadrant of the point.
• Tracing a curve twice. In length and area integrals, choose limits that cover the curve exactly once.
• Forgetting the 1/2. Polar area is (1/2)∫ r² dθ, not ∫ r dθ.

Section XIIThe cycloid and the fastest slide

The cycloid was named by Galileo in 1599. He tried to find the area under an arch by weighing metal cut-outs, and got a ratio close to 3 times the area of the rolling circle, but guessed that it was not exactly 3. Gilles de Roberval proved in 1634 that it is exactly 3: for a wheel of radius a, the area under one arch is 3πa². In 1658, answering a challenge from Blaise Pascal, Christopher Wren found its length: an arch is 8a long, the result of Example 3.

The curve kept surprising people. Christiaan Huygens showed that a bead sliding on an upside-down cycloid takes the same time to reach the bottom wherever it starts, the tautochrone property, and published it in his Horologium oscillatorium of 1673, using it to design a pendulum clock.

In June 1696 Johann Bernoulli challenged the mathematicians of Europe in the journal Acta Eruditorum: along which curve does a bead slide from one point to a lower one in the shortest time? This is the brachistochrone problem. Galileo, in 1638, had wrongly thought the answer was an arc of a circle. Five solutions came in, from Newton, Jacob Bernoulli, Leibniz, de L’Hôpital and Johann Bernoulli himself, and the answer was again the cycloid. According to his nephew-in-law John Conduitt, Newton received the problem after a day at the Royal Mint and had solved it by four the next morning; his solution was published anonymously in the Philosophical Transactions in January 1697. The methods used grew into the calculus of variations.

Polar coordinates also grew out of special problems. Julian Coolidge traced their first use to Bonaventura Cavalieri, who used them to find the area inside an Archimedean spiral. Newton was the first to treat them as a way to fix any point in the plane, alongside Cartesian and other systems, to show how to find tangents. Jacob Bernoulli went further, writing the radius of curvature of a curve given in polar form.

NotesSources and credits

• J. J. O’Connor and E. F. Robertson, MacTutor History of Mathematics, University of St Andrews: Cycloid, on Galileo’s naming in 1599 and his weighing, Roberval’s area in 1634, Pascal’s 1658 challenge, Wren’s arc length 8a and Huygens’ tautochrone in Horologium oscillatorium (1673).
• MacTutor: The brachistochrone problem, on Johann Bernoulli’s challenge of June 1696, Galileo’s circular arc of 1638, the five solutions, Conduitt’s account of Newton and the anonymous January 1697 solution in the Philosophical Transactions.
• J. L. Coolidge, “The Origin of Polar Coordinates”, The American Mathematical Monthly 59(2), 1952, 78–85; extract on MacTutor, on Cavalieri, Newton and Jacob Bernoulli.

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