# Chapter 27: Parametric Equations and Polar Coordinates

> Curves driven by a parameter or an angle: parametric derivatives, speed, the polar toolkit — conversions, cardioids, limaçons, roses — polar tangents, and area via A = ½∫r² dθ.

**Canonical:** https://calculus.kitzo.work/chapter/parametric-polar  
**Level:** Advanced  
**Estimated time:** 50 minutes  
**Exercises:** 8 (auto-marked; answers stay server-side)

## Topics
- Parametric derivatives
- Speed
- Polar conversions
- Sketching polar curves
- Tangents to polar curves
- Polar area

**Why this matters**

Nature rarely moves in straight lines — planets on elliptical loops, beads on wires, radar dishes sweeping — so calculus needs curves that Cartesian y = f(x) cannot express. Parametric and polar equations are how calculus handles motion with a time or angle knob, and they power half of physics.

## 1 · Parametric equations — a parameter drives the point

A **parametric curve** is given by x = f(t), y = g(t). As t runs, the point traces a path; t is usually time but can be any variable. Eliminating t recovers an ordinary equation when possible, but the parametric form keeps *direction and pacing*, which the Cartesian equation alone throws away. One curve can have many parametrizations — speed differs, shape does not.

## 2 · Derivatives of parametric curves

By the chain rule, **dy/dx = (dy/dt)/(dx/dt)**, provided dx/dt ≠ 0. For second derivatives, differentiate dy/dx with respect to t and divide by dx/dt again: d²y/dx² = [d/dt(dy/dx)] ÷ (dx/dt). Horizontal tangents occur where dy/dt = 0 (and dx/dt ≠ 0); vertical tangents where dx/dt = 0 (and dy/dt ≠ 0).

**Worked example**

**Example 1 — x = t³ − 3t, y = t². Find dy/dx at t = 2, and the tangent line there.**

Step 1: dx/dt = 3t² − 3, dy/dt = 2t.

Step 2: dy/dx = 2t/(3t² − 3). At t = 2: dy/dx = 4/(12 − 3) = 4/9.

Step 3: the point is x = 8 − 6 = 2, y = 4. Tangent: y − 4 = (4/9)(x − 2).

## 3 · Speed

Velocity is the vector (dx/dt, dy/dt); **speed** is its magnitude: **√((dx/dt)² + (dy/dt)²)**. This is exactly the integrand for parametric arc length, L = ∫ √((x′)² + (y′)²) dt — so speed also measures distance traveled along the curve.

**Example 2 — x = 3cos t, y = 4sin t. Speed at t = π/2.**

Step 1: dx/dt = −3sin t, dy/dt = 4cos t.

Step 2: speed = √(9sin²t + 16cos²t).

Step 3: at t = π/2: √(9·1 + 16·0) = 3. The ellipse is traversed faster and slower along its length.

## 4 · Polar coordinates — and conversions

**Polar coordinates** label a point (r, θ): distance r from the origin and angle θ from the positive x-axis. Conversions both ways: `x = r cos θ, y = r sin θ`, `r² = x² + y²`, and tan θ = y/x (fix the quadrant). Polar tames spirals, circles through the origin and centered limaçons that look horrendous in Cartesian form.

## 5 · Sketching polar curves — cardioids, limaçons, roses

  - **Circles**: r = a (centered at pole), r = 2a cos θ (diameter on x-axis), r = 2a sin θ (y-axis).

  - **Cardioids**: r = a(1 ± cos θ) or a(1 ± sin θ) — heart-shaped, one cusp at the pole.

  - **Limaçons**: r = a + b cos θ. a ≥ b: dimpled or convex; a < b: *inner loop*.

  - **Roses**: r = a cos(nθ) — n petals for odd n, 2n petals for even n.

Sketch by plotting r at θ = 0, π/2, π, 3π/2 and using symmetry (θ → −θ: x-axis symmetry; θ → π − θ: y-axis; θ → θ + π: pole). Negative r flips the point through the origin — never discard those values.

## 6 · Tangents to polar curves

Treat r = r(θ) as parametric with parameter θ: x = r cos θ, y = r sin θ. Differentiate: dx/dθ = r′cos θ − r sin θ, dy/dθ = r′sin θ + r cos θ. Then **dy/dx = (r′ sin θ + r cos θ)/(r′ cos θ − r sin θ)**. Horizontal tangent: numerator zero. Vertical tangent: denominator zero. At the pole (r = 0) check whether the curve passes through with a defined tangent direction.

**Watch out**

r negative at some θ does NOT mean the point fails to exist — it plots in the opposite direction. And dy/dx for polar curves is not dr/dθ: differentiating a polar equation means treating θ as the parameter everywhere.

## 7 · Area enclosed by a polar curve

A thin wedge of radius r(θ) and angle dθ has area ½r²dθ, so summing gives the master formula **A = ½ ∫ₐᵇ r² dθ**. Between two curves: A = ½ ∫(r_out² − r_in²) dθ. Find limits by solving r₁ = r₂ (where the curves meet) or by sweeping exactly one petal/loop.

**Example 3 — Area inside r = sin θ.**

Step 1: r = sin θ traces the full circle once as θ runs 0 → π.

Step 2: A = ½ ∫₀^π sin²θ dθ = ½ · π/2 = π/4.

Step 3: check — r = sin θ is a circle of diameter 1 sitting on the x-axis; area π/4. ✓

**Chapter formula sheet**

dy/dx = (dy/dt)/(dx/dt) · speed = √((dx/dt)² + (dy/dt)²)

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

A = ½ ∫ₐᵇ r² dθ (polar area) · area between: ½ ∫(r_out² − r_in²) dθ

Rose r = a cos nθ: n petals (n odd), 2n petals (n even) · cardioid r = a(1 ± cos θ)

Polar tangent: dy/dx = (r′sin θ + r cos θ)/(r′cos θ − r sin θ)

## Common mistakes

  - Writing dy/dx = dy/dt instead of dividing by dx/dt.

  - Thinking dx/dt = 0 gives a horizontal tangent — it is vertical.

  - Converting with tan θ = y/x and ignoring the quadrant of θ.

  - Integrating r² dθ over 0 to 2π for a rose and double-counting petals.

  - Forgetting the factor ½ in A = ½ ∫ r² dθ.

  - Discarding negative r values when sketching — they belong on the graph.

## Practice exercises

1. For x = t², y = t³ , dy/dx equals:

- A. 3t/2
- B. 3t²
- C. 2t/3
- D. t

2. A particle moves as x = 3cos t, y = 4sin t . Find its speed at t = π/2 .

3. The curve r = 1 + cos θ is a … (one word).

4. Convert the Cartesian point (3, 4) to polar form. What is r?

5. The curve r = cos 3θ is a:

- A. cardioid
- B. three-petaled rose
- C. limaçon with inner loop
- D. circle

6. Find the area enclosed by the polar curve r = sin θ (a circle).

7. A curve of the form r = a + b cos θ with a (so it has an inner loop) is called a … (one word).

8. The area swept by ONE petal of r = cos 2θ is:

- A. π/8
- B. π/4
- C. π/2
- D. π

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