MATH U101 · Multivariable Calculus · Module 1

Polar Coordinates

Thomas' Calculus (15th ed.) §11.3 §11.4* §11.5 — a new way to name points, the curves it makes easy, and how to measure area and length in it.

Start here — a fact worth knowing

Polar coordinates are not in the CBSE 11th/12th syllabus. Almost everyone in the lecture hall — BITSAT or SAT channel — is meeting this topic for the first time. The only background this module leans on is trigonometry: unit-circle values, the graphs of sin and cos, and (for §11.5) one identity and basic integration. That's the whole entry fee.

Seeing this material for the very first time? This page is the compressed revision map. Read the first-time lesson first — same syllabus, taught slowly from zero — then come back here to revise and drill.

1 · The idea: name a point by distance and direction

Cartesian coordinates answer "how far right, how far up?" Polar coordinates answer a different, often more natural pair of questions: how far from the origin, and at what angle?

Fix the origin O (now called the pole) and the positive x-axis (now the initial ray). A point gets coordinates (r, θ):

r = directed distance from the pole; θ = angle from the initial ray (radians, counterclockwise positive).

Two features make polar coordinates behave differently from Cartesian — and both are exam favourites:

Negative r means "walk backwards"

(−2, π/6) means: face the direction π/6, then walk 2 units the opposite way — you land in the third quadrant. Negative r is legal and routinely appears when graphing.

Names are not unique

The same point has infinitely many polar addresses. Start from (2, π/6):

(2, π/6 + 2π) is the same point, because adding 2π is one full extra turn — you finish facing exactly the same direction before you walk.
(−2, π/6 + π) is the same point too: the extra π turns you to face the opposite direction, and the −2 walks you backwards. Two reversals cancel, and you land where you started.

In Cartesian coordinates a point has exactly one name; in polar it has infinitely many — add any multiple of 2π to θ, or flip the sign of r and add π. Every "find the intersections of two curves" problem turns on this.

2 · Converting between polar and Cartesian §11.3

Four equations do all the work. The first two convert polar → Cartesian; the second two go back.

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

Worked example 1Identify the curve r = 2 cos θ

The equation mixes r and θ. To use the conversion table we want r² and r cos θ, so multiply both sides by r: r² = 2r cos θ.
Substitute: x² + y² = 2x.
Complete the square: (x − 1)² + y² = 1 — a circle of radius 1 centred at (1, 0).
Unpack this step

Move the 2x across: x² − 2x + y² = 0. Take half the coefficient of x (half of −2 is −1), square it (getting 1), and add that to both sides: (x² − 2x + 1) + y² = 1. The bracket is exactly (x − 1)². If this move feels rusty, it's in the prerequisite kit (§6 below) — worth ten minutes of drill on its own.

The "multiply by r" move is the single most-used trick in this section. Memorise the outcome too: r = 2a cos θ is always a circle of radius a through the pole, sitting on the x-axis (with sin, on the y-axis).

Classic trap

tan θ = y/x does not pin down θ by itself — tan has period π, so it can't tell quadrant II from IV. When converting a point like (x, y) = (−1, 1), always sketch first: the point is in quadrant II, so θ = 3π/4, not −π/4.

Three more equations worth recognising on sight, because they look exotic in polar and are trivial in Cartesian: θ = α is a straight line through the pole at angle α; r = a sec θ means r cos θ = a, i.e. the vertical line x = a; and r = a csc θ means r sin θ = a, the horizontal line y = a. Tutorial problems love pairing one of these lines with a circle.

3 · Graphing r = a ± b sin θ and r = a ± b cos θ §11.4*

This family is called the limaçons (lee-ma-sonn — French for snail). Your syllabus names this exact form, so expect it on quizzes. The whole game is: watch r grow and shrink as θ sweeps around.

The method that always works: a table of values

Take r = 1 + cos θ. March θ through the standard angles and record r:

θ	0	π/3	π/2	2π/3	π	4π/3	3π/2	5π/3	2π
r	2	3/2	1	1/2	0	1/2	1	3/2	2

Reading the story: the point starts 2 units out along θ = 0, spirals inward, touches the pole at θ = π, then swells back out. The heart-shaped result is a cardioid.

Watch it being drawn

The table tells the story in nine frames — here it is continuously. The dashed ray points in the direction θ; the dot sits at distance r along it. Switch to the inner-loop curve and watch what happens when r goes negative: the ray points one way, the dot walks the other.

The shape depends only on the ratio a/b

All four shapes below are drawn exactly (these are computed curves, not sketches):

r = 1 + 2 sin θ
Inner loop (a/b < 1)

r = 1 + sin θ
Cardioid (a/b = 1)

r = 3 + 2 sin θ
Dimpled (1 < a/b < 2)

r = 5 + 2 sin θ
Convex (a/b ≥ 2)

Orientation is read straight off the equation: sin versions are symmetric about the y-axis, cos versions about the x-axis; a + bulges toward the positive axis, a − toward the negative.

Three shortcuts before you plot anything

Symmetry tests (the three from §11.4 — tutorial sheets ask these directly). x-axis: the equation survives θ → −θ (all cos versions). y-axis: survives θ → π − θ (all sin versions). Origin: survives r → −r (e.g. every r² = … curve) or, equivalently, θ → θ + π. Any two of the three force the third. One subtlety worth a mark: because a point has many polar names, a curve can be symmetric even when the test fails — the tests prove symmetry, they never disprove it. The tutorial companion works Sheet 2's two symmetry questions in full.
Where is r largest/smallest? Just ask where sin or cos hits ±1. Maximum r = a + b, and the curve reaches the pole only if r = 0 has a solution (needs a ≤ b).
Solve r = 0 first. For r = 1 + 2 sin θ: sin θ = −½ at θ = 7π/6, 11π/6. Between those angles r is negative — that's exactly the inner loop, drawn "backwards" through the pole.

Classic trap

When r goes negative, students either skip those angles (losing the inner loop) or plot the point on the wrong side. Negative r points go in the opposite direction to θ. Also: keep your calculator in radians — a degrees-mode table produces garbage that looks plausibly like a spiral.

One video, if you want one

Graphing Polar Equations — Cardioids, Limaçons & Rose Curves (The Organic Chemistry Tutor, YouTube) walks through the table-of-values method at whiteboard pace. Watch it once, before or after the §11.4 lecture — and stop when he reaches rose curves, which are not on your syllabus. One watch is enough; more YouTube is not more learning.

4 · Slope of a polar curve §11.4*

Why this matters

A polar curve is secretly a parametric curve: as θ runs, the point (x, y) = (r cos θ, r sin θ) moves, with r = f(θ). So the slope of the curve is an ordinary Cartesian slope, reached through θ: dy/dx = (dy/dθ) / (dx/dθ). Applying the product rule to r cos θ and r sin θ gives the formula:

dydx = r′ sin θ + r cos θr′ cos θ − r sin θ where r′ = drdθ

Worked example 2Slope of the cardioid r = 1 + cos θ at θ = π/2

Differentiate the radius: r′ = −sin θ.
Evaluate everything at θ = π/2: r = 1 + 0 = 1, r′ = −1, sin θ = 1, cos θ = 0.
Numerator: r′ sin θ + r cos θ = (−1)(1) + (1)(0) = −1.
Denominator: r′ cos θ − r sin θ = (−1)(0) − (1)(1) = −1.
dy/dx = (−1)/(−1) = 1 — at its topmost point (0, 1) the cardioid climbs at 45°. Sketch it and check that looks right; it does.

Classic trap

dr/dθ is not the slope. It measures how fast the radius grows as the angle sweeps — a different thing entirely from the steepness of the curve in the plane. The slope always needs the full fraction above. (This section is starred in the handout, but horizontal and vertical tangents are examined: the 2025 mid-sem spent 16 marks on one. The routine is in section 6.)

5 · Area and arc length in polar §11.5

Why this matters

In Cartesian, area under a curve is a sum of thin rectangles. In polar, the natural sliver is a thin pie slice of radius r and angle dθ, whose area is ½r² dθ (fraction dθ/2π of a full disc πr²). Summing slices gives the formula — understanding this derivation is worth more than memorising it.

A = 12 ∫_α^β r² dθ L = ∫_α^β √(r² + (dr/dθ)²) dθ

For the region between an outer curve R(θ) and an inner curve r(θ), subtract the pie-slice areas: A = ½ ∫ (R² − r²) dθ — square each radius first, then subtract (it is not ½∫(R − r)²). Multi-step problems in this section — shared areas, inner loops, curve-plus-line regions — are worked through shape by shape in the problem clinic.

Worked example 3Area enclosed by the cardioid r = 1 + cos θ

The curve is traced once as θ runs 0 → 2π, so A = ½ ∫₀^2π (1 + cos θ)² dθ.
Expand: (1 + cos θ)² = 1 + 2 cos θ + cos² θ.
The only new tool needed: the half-angle identity cos² θ = (1 + cos 2θ)/2. You cannot integrate cos² without it — it appears in nearly every problem in this section.
Integrate over a full period: ∫ 1 = 2π, ∫ cos θ = 0, ∫ cos² θ = π.
Unpack this step

∫₀^2π cos θ dθ = [sin θ]₀^2π = 0 — a full period of sin or cos always integrates to zero. For the square, use the identity from step 3: ∫₀^2π cos² θ dθ = ∫₀^2π (1 + cos 2θ)/2 dθ = [θ/2 + (sin 2θ)/4]₀^2π = π.
A = ½ (2π + 0 + π) = 3π/2.

The trap that costs the most marks

Wrong θ-limits silently double (or halve) your area. The circle r = 2 cos θ is traced completely as θ runs from −π/2 to π/2. Integrate 0 → 2π and you trace it twice and get double the area. Before integrating, always ask: over which θ-interval is this curve drawn exactly once? Sketch first, integrate second — never the reverse.

Classic trap — arc length's two killers

Nearly every wrong arc-length answer comes from one of two slips. (1) Differentiating before simplifying: if r contains a square root, look for a perfect square inside first — 1 ± cos 2θ is 2 cos²θ / 2 sin²θ, and 1 ± sin 2θ is (sin θ ± cos θ)²; differentiate the ugly original and the algebra never recovers. (2) Losing the outer √: after r² + (r′)² collapses to a constant like 2, the integrand is √2 — the root was in the formula all along. The full step-by-step recipe, with worked examples, is in the problem clinic.

Intersections: the pole is sneaky

To find where two polar curves meet, solving r₁(θ) = r₂(θ) is not enough, because the same point has many names. In particular both curves may pass through the pole at different values of θ — an intersection the algebra never shows. The reliable routine: solve the equation, and check the pole separately, and confirm against a sketch.

6 · Horizontal and vertical tangents, and lines in polar §11.4* §11.5

Why this matters

This is the combination the mid-sem likes: find where a polar curve has a horizontal (or vertical) tangent, draw that tangent line, then find an area bounded by the line and the curve. The 2025 paper spent 16 of its 50 marks on exactly this (Q3). It needs only the slope formula from section 4 and the area formula from section 5, plus one new idea: writing a straight line in polar form.

The tangent routine

Treat the curve as the parametric curve x = r cos θ, y = r sin θ with r = f(θ). The product rule gives the two pieces of the slope separately:

dydθ = r′ sin θ + r cos θ dxdθ = r′ cos θ − r sin θ

The slope is (dy/dθ) / (dx/dθ), so read off the special directions:

Horizontal tangent: dy/dθ = 0 and dx/dθ ≠ 0 (slope 0).
Vertical tangent: dx/dθ = 0 and dy/dθ ≠ 0 (slope undefined).
Both zero: the formula says 0/0 and decides nothing. Take the limit of the slope as θ approaches that angle. This happens at a cusp at the pole, where r = 0.

Straight lines in polar form

Substitute y = r sin θ into y = c and solve for r:

y = c ⟶ r = c csc θ x = c ⟶ r = c sec θ

(A line through the pole is simpler still: θ = constant.)

Once the line is written as r = R(θ), a region between the line and a curve is an ordinary "between two curves" area. Every ray from the pole leaves the inner curve and hits the line, so the pie-slice rule applies unchanged: A = ½∫(R_line² − r_curve²) dθ, as long as the line really is the outer boundary over the whole θ-interval (step 5 below checks this).

Worked example 4 · 2025 mid-sem Q3The cardioid r = 4(1 − sin θ) and its double tangent

"Draw a straight line L parallel to the X-axis which is tangent to the curve at two distinct points P and Q. (a) Find the polar coordinates of P and Q. [6M] (b) Sketch the region enclosed by the straight line L and outside the given polar curve. Find the area of this region. [10M]" (full question)

The two derivatives. r′ = −4 cos θ. Then dy/dθ = −4 cos θ sin θ + 4(1 − sin θ) cos θ = 4 cos θ(1 − 2 sin θ), and dx/dθ = −4 cos²θ − 4(1 − sin θ) sin θ = 4(2 sin θ + 1)(sin θ − 1).
Unpack the dx/dθ factorisation

Replace cos²θ by 1 − sin²θ: −4(1 − sin²θ) − 4 sin θ + 4 sin²θ = 4(2 sin²θ − sin θ − 1). With s = sin θ, the quadratic 2s² − s − 1 factors as (2s + 1)(s − 1); multiply back out to check.
Horizontal-tangent candidates: dy/dθ = 0 when cos θ = 0 (θ = π/2, 3π/2) or 1 − 2 sin θ = 0, i.e. sin θ = ½ (θ = π/6, 5π/6).
Check dx/dθ ≠ 0 at each candidate. At π/6 and 5π/6: 4(2·½ + 1)(½ − 1) = −4 ≠ 0 ✓, and r = 4(1 − ½) = 2, giving the points (√3, 1) and (−√3, 1). At 3π/2: 4(−2 + 1)(−1 − 1) = 8 ≠ 0 ✓, r = 8, the bottom point (0, −8). That's a genuine horizontal tangent, but it touches the curve at one point only. At π/2: sin θ − 1 = 0, so both derivatives vanish. This is the cusp at the pole (r = 0), and the slope near it grows without bound (about −667 at θ = π/2 + 0.001): the tangent there is vertical, not horizontal.
So L is y = 1, the only horizontal line tangent at two distinct points: P = (2, π/6) and Q = (2, 5π/6) in polar coordinates. In polar form L is r = csc θ.
The line is the outer boundary between P and Q. For θ ∈ [π/6, 5π/6], s = sin θ ≥ ½ > 0, so multiplying by s keeps the inequality's direction: 4(1 − s) ≤ 1/s ⟺ 4s − 4s² ≤ 1 ⟺ (2s − 1)² ≥ 0, which is always true. The curve dips inside the line and touches it only at P and Q (where s = ½).
Set up the area of the shaded notch below: A = ½∫_π/6^5π/6 (csc²θ − 16(1 − sin θ)²) dθ.
The line part: ½∫csc²θ dθ = ½[−cot θ]_π/6^5π/6 = ½(√3 + √3) = √3.
Unpack the two antiderivatives

d/dθ(−cot θ) = csc²θ, and cot(5π/6) = −√3, cot(π/6) = √3. For the square, sin²θ = (1 − cos 2θ)/2 integrates to θ/2 − (sin 2θ)/4. Both are in the prerequisite kit.
The curve part: expand 16(1 − sin θ)² = 16(1 − 2 sin θ + sin²θ). Over [π/6, 5π/6]: ∫1 = 2π/3, ∫sin θ = [−cos θ] = √3/2 + √3/2 = √3, ∫sin²θ = [θ/2 − (sin 2θ)/4] = π/3 + √3/4. So ½ · 16 (2π/3 − 2√3 + π/3 + √3/4) = 8(π − 7√3/4) = 8π − 14√3.
Subtract: A = √3 − (8π − 14√3) = 15√3 − 8π ≈ 0.848. It's positive and small, matching the thin notch in the sketch.

Computed from r = 4(1 − sin θ). The shaded notch is the region of part (b); the single-point tangent at the bottom shows why L must be y = 1.

Classic trap

1) Calling the cusp a horizontal tangent. At θ = π/2, dy/dθ = 0, but dx/dθ = 0 too, so the test says nothing, and the true tangent there is vertical. 2) Skipping the dx/dθ ≠ 0 check on every candidate. It's the step that separates a tangent from a cusp, and examiners look for it. 3) Integrating (csc θ − r)² instead of csc²θ − r²: square each radius, then subtract, exactly as in section 5.

7 · Your minimal prerequisite kit

Everything this module assumes, in one list. If any line feels shaky, patch just that line — nothing else is needed:

Unit-circle values of sin and cos at 0, π/6, π/4, π/3, π/2 and their reflections in all quadrants — instant recall, in radians. Drill it with the unit-circle flashcards.
The graphs of sin θ and cos θ (to predict where r grows and shrinks).
Half-angle identities: cos²θ = (1+cos 2θ)/2, sin²θ = (1−cos 2θ)/2 — drill these (and the rest of the kit) with the formula flashcards.
Completing the square (for identifying circles).
Basic integrals: ∫ cos kθ dθ, ∫ sin kθ dθ, and integrating constants (∫ c dθ = cθ — the step arc-length problems quietly rely on).
The product rule (the §11.4 slope formula is two product rules stacked).
One extra integral for line-and-circle areas: ∫ csc²θ dθ = −cot θ + C — the mirror of ∫ sec²θ dθ = tan θ; check it by differentiating back.

8 · What to practise in Thomas

Odd-numbered exercises have answers in the back — do them in this order, checking as you go:

Skill	Where	How many
Plotting points, negative r, converting points & equations	§11.3 exercises	8–10
Graphing limaçons/cardioids from a table + symmetry	§11.4 exercises	6–8 (sketch every one by hand)
Slope dy/dx at a given θ	§11.4 exercises	3–4
Horizontal/vertical tangents; a line and a curve bounding an area (section 6)	§11.4 exercises on horizontal/vertical tangents + 2025 mid-sem Q3	2–3
Areas inside one curve; area between two curves; arc length	§11.5 exercises	8–10, always sketching first

If the §11.5 problems keep collapsing partway — pieces fine, whole thing wrong — that's a known stage with a known fix: the checklists and worked twins in the problem clinic, plus the randomized polar pipeline drill for daily reps.

The handout's notes say exactly what the star on §11.4* excludes: "graphing other polar curves" — roses, lemniscates, spirals. Limaçons and cardioids, the symmetry tests, and the slope formula are all in.