Polar Coordinates, taught from zero

1 · Why invent another way to name points?

Suppose you're standing in a flat field and a friend asks where the treasure is. You could say: "walk 3 km east, then 4 km north." That's Cartesian thinking — two perpendicular distances. But nobody talks like that. You'd point and say: "it's 5 km away, in that direction." One distance, one direction. That is a polar coordinate.

Both descriptions pin down the same spot; they just answer different questions. And some shapes strongly prefer one language. A circle centred on you is a nightmare in Cartesian (x² + y² = 25) but trivial in polar: "everything 5 km away" — r = 5. Spirals, flower-petal curves, orbits, radar sweeps: all natural in polar, all ugly in Cartesian. That's the entire reason this system exists.

The setup takes one sentence: fix a centre point O called the pole (it will sit at the Cartesian origin) and a reference direction, the initial ray (the positive x-axis). Then any point gets a pair (r, θ): θ is the angle from the initial ray, measured counterclockwise in radians, and r is how far out along that direction you walk. (The notes page has a labelled figure if you want the picture.)

Plot a few by hand — it's the whole skill

To plot (2, π/6): face 30° above the initial ray, walk 2 units. Done — a point in the first quadrant. To plot (3, 3π/4): face 135° (up and to the left), walk 3 — second quadrant. The recipe never changes: turn, then walk.

2 · The two strange rules (this is where polar differs)

Cartesian coordinates are boringly well-behaved: every point has exactly one name. Polar gives up that property, on purpose, and both quirks show up in exams.

Quirk 1: negative r means "walk backwards"

What could (−2, π/6) mean? Take the recipe literally: face the direction π/6… then walk minus 2 — that is, 2 units backwards, without turning around. You end up on the opposite side of the pole, in the third quadrant. Strange but consistent, and it's not a curiosity: when we graph curves in section 4, the formula itself will hand us negative r values, and they draw real parts of the curve.

Quirk 2: every point has infinitely many names

Start with (2, π/6) and watch how many other names reach the same spot:

• (2, π/6 + 2π) — turn one extra full circle before walking. You face the same way, so: same point.
• (−2, π/6 + π) — turn to face the opposite direction, then walk backwards. Two reversals cancel: same point again.

Add any multiple of 2π, or flip the sign of r while adding π, and the point doesn't move. In Cartesian, "same point" means "same coordinates"; in polar it doesn't, and one exam consequence is worth planting now: when you hunt for intersections of two polar curves by setting their equations equal, the algebra can miss intersections where the curves pass through the same point wearing different names. The notes page returns to this; for now, just internalise that names are not unique.

3 · Translating between polar and Cartesian §11.3

Drop a point (r, θ) onto the Cartesian plane and draw the vertical line from it down to the x-axis. You've made a right triangle: hypotenuse r, angle θ at the origin. School trigonometry reads the sides straight off: adjacent = r cos θ, opposite = r sin θ. But the adjacent side is the x-coordinate and the opposite side is y:

That's polar → Cartesian, and it always works. Going the other way, Pythagoras on the same triangle gives r² = x² + y², and dividing the two equations gives tan θ = y/x. Four equations, one right triangle — that's the entire conversion machinery of §11.3.

Worked slowly, both directions:

Polar → Cartesian for (4, π/3): x = 4 cos(π/3) = 4 · ½ = 2; y = 4 sin(π/3) = 4 · (√3/2) = 2√3. Point: (2, 2√3). Mechanical — this direction never has surprises.

Cartesian → polar for (−1, 1): r = √(1 + 1) = √2, fine. But now tan θ = 1/(−1) = −1, and here is the direction that does have a surprise: tan is −1 at 3π/4 and at 7π/4 (tan repeats every π, so it can't tell those apart). Which is right? Look at the point: (−1, 1) is up-and-left — quadrant II — so θ = 3π/4. Always sketch the point first and let the quadrant choose. Answering −π/4 or 7π/4 here is the single most reliable mark-loser in §11.3.

Converting equations (not just points)

The same four formulas convert whole equations, and one small trick unlocks most of them. Identify the curve r = 2 cos θ:

The right side has a lone cos θ, but our dictionary only knows r cos θ (which is x) and r² (which is x²+y²). So multiply both sides by r to manufacture dictionary words: r² = 2r cos θ, which translates instantly to x² + y² = 2x. Complete the square (the notes page unpacks this step) and you get (x−1)² + y² = 1: a circle, radius 1, centred at (1, 0). A humble one-term polar equation was secretly a circle sitting on the x-axis.

4 · Graphing polar curves: watch r breathe §11.4*

Now the fun part, and the part your syllabus names explicitly: curves of the family r = a ± b sin θ and r = a ± b cos θ, called limaçons. Here is the one mental model that makes all of them easy:

Imagine a searchlight beam sweeping counterclockwise from the pole. The equation tells the beam how long to be at each angle. As θ sweeps from 0 to 2π, r grows and shrinks — breathes — and the tip of the beam draws the curve.

Take r = 1 + cos θ and narrate one sweep, using nothing but the shape of the cos graph:

• θ = 0: cos is 1, so r = 2 — the beam is longest, pointing right.
• θ → π/2: cos falls to 0, so r shrinks to 1 — the beam, now pointing up, is half as long.
• θ → π: cos reaches −1, so r = 0 — the beam shrinks to nothing: the curve touches the pole, pointing left.
• θ → 2π: cos climbs back, the beam regrows to 2 — the curve swells back out, mirror-image below.

The result is a heart shape: the cardioid. On the notes page there is an animation of exactly this sweep — go press play, watch one full revolution of the cardioid, and then switch it to the inner-loop curve and keep watching. What you'll see there is the payoff of Quirk 1: for r = 1 + 2 sin θ, there's a stretch of angles (between 7π/6 and 11π/6, where sin θ < −½) where the formula makes r negative — and the beam draws backwards through the pole, producing a small loop inside the big one. Students who skip negative r lose the whole inner loop; the animation makes it impossible to un-see.

For exams, the reliable method is exactly this narration made into a table: march θ through the standard angles, compute r at each, plot, connect smoothly. The notes page has the worked table. Two shortcuts make the table cheaper:

• Symmetry: cos curves are symmetric about the x-axis (because cos(−θ) = cos θ — the equation literally can't tell +θ from −θ); sin curves about the y-axis. So plot half, mirror the rest.
• Extremes first: the beam is longest where sin/cos hits +1 or −1 relative to the sign in the equation (max r = a + b), and touches the pole only if r = 0 is solvable — which needs a ≤ b.

And the whole family's shape is decided by one number, the ratio a/b: less than 1 → inner loop; equal to 1 → cardioid; between 1 and 2 → dimpled; at least 2 → convex. The notes page draws all four, computed exactly.

5 · Area in polar: pie slices, not rectangles §11.5

In Cartesian calculus you found area under a curve by stacking thin rectangles: width dx, height f(x). That shape is wrong for polar curves — a searchlight sweep doesn't cut rectangles, it cuts thin pie slices: one slice per sliver of angle dθ, with radius r.

How big is one slice? It's a fraction dθ/2π of a full disc of area πr²:

Add up the slices — that's what an integral is — and you get the section's headline formula:

Understanding that derivation is worth more than memorising the result: it tells you the ½ is not decoration, the integrand is r² not r, and the limits α, β are angles — the sweep of the beam, not x-values.

Let's use it on something we can verify: the circle r = 2 cos θ from section 3, which we know has radius 1 and hence area π. First question, and it's the section's famous trap: over which angles is this circle drawn exactly once? Watch the beam: at θ = −π/2 the beam has length 0; as θ sweeps up through 0 it grows to 2 and shrinks back to 0 at θ = +π/2. The full circle is already drawn — in half a revolution. Integrate 0 to 2π and you'd trace it twice and get double the area. So:

To integrate cos² you need the half-angle identity — there is no way around it, and it appears in nearly every problem of this section: cos² θ = (1 + cos 2θ)/2. Then:

So A = 2 · π/2 = π. The formula agrees with the geometry — which is exactly the kind of sanity check to run in an exam whenever one is available. For a full exam-style example on a curve you can't check by geometry (the cardioid, area 3π/2), see worked example 2 on the notes page.

6 · Arc length: how far did the beam's tip travel? §11.5

Same setup, different question: not how much area the sweep covers, but how long the drawn curve itself is. The formula (derived in Thomas from the conversion equations) is:

The intuition for the two terms: in a sliver dθ the tip moves sideways (along the sweep) by about r dθ and radially (in or out) by dr; Pythagoras on those two little legs gives the sliver's length, and the integral sums the slivers.

Sanity check first: for the circle r = 3 (constant, so dr/dθ = 0), one sweep gives L = ∫₀^2π √9 dθ = 6π — the circumference 2π·3. The formula tells the truth.

Now a real one — the cardioid r = 1 + cos θ, traced once over 0 to 2π:

1. dr/dθ = −sin θ, so the expression under the root is (1+cos θ)² + sin² θ = 1 + 2cos θ + cos²θ + sin²θ = 2 + 2 cos θ (the identity sin²+cos² = 1 collapsing things, as it constantly does).
2. To take the square root, reach for the half-angle identity in reverse: 2 + 2cos θ = 4 cos²(θ/2), so the root is 2|cos(θ/2)|.
3. Handle the absolute value with symmetry: the cardioid is symmetric about the x-axis, so compute twice the top half, where θ ∈ [0, π] makes cos(θ/2) ≥ 0: L = 2∫₀^π 2 cos(θ/2) dθ = 4 [2 sin(θ/2)]₀^π = 8.

A heart-shaped curve with a perfectly clean integer length — polar coordinates being elegant where Cartesian would be agony.

7 · You're ready — what to do next

That's the entire module, taught once: naming points, the two quirks, converting, graphing by watching r breathe, pie-slice area, and arc length. To convert understanding into marks:

• Hand-sketch six curves from tables of values (the catch-up plan schedules this in Week 2) — the sketching is the learning.
• Work the exercises from the table on the notes page, odd numbers, checking answers in the back.
• Revise from the notes page, not this one — it's the compressed map, with every classic trap flagged.