📐 Concept diagram

05-10 - Parametric Equations and Polar Coordinates

Phase: 5 | Subject: 05-10 Prerequisites: 02-07-unit-circle-and-radians.md (radians), 02-03-polygons-and-circles.md (arc length, sector area) Next subject: 06-01-functions-of-several-variables.md

Learning Objectives

By the end of this subject, you will be able to:

1. Understand parametric equations and convert to Cartesian form
2. Differentiate parametric equations
3. Calculate arc length of parametric curves
4. Understand polar coordinates and convert between polar and Cartesian
5. Calculate area in polar coordinates
6. Calculate arc length in polar coordinates

Core Content

Parametric Equations

Instead of y = f(x), we describe curves using a parameter t:

x = f(t), y = g(t)

Example: Circle of radius r: x = r·cos(t), y = r·sin(t)

Converting to Cartesian

Eliminate the parameter.

Example: x = t², y = t³ From x = t²: t = √x Substitute: y = (√x)³ = x^(3/2)

Derivatives

$dy/dx = (dy/dt) / (dx/dt)
$

Example: x = t², y = t³ dx/dt = 2t, dy/dt = 3t² dy/dx = 3t²/2t = 3t/2

At t = 1: dy/dx = 3/2.

Arc Length

$L = ∫[a to b] √((dx/dt)² + (dy/dt)²) dt
$

Polar Coordinates

A point is described by (r, θ) where: - r = distance from origin - θ = angle from positive x-axis (in radians)

Converting

Polar to Cartesian: x = r·cos(θ), y = r·sin(θ) r² = x² + y², tan(θ) = y/x

Cartesian to Polar: r = √(x² + y²), θ = tan⁻¹(y/x)

Polar Curves

Example: r = 2 (circle of radius 2 centred at origin) Example: r = θ (Archimedean spiral) Example: r = 1 + cos(θ) (cardioid) Example: r = cos(2θ) (four-petal rose)

Area in Polar Coordinates

$A = (1/2) ∫[α to β] r² dθ
$

Example: Area of circle r = 2 from θ = 0 to 2π: A = (1/2) ∫[0 to 2π] 4 dθ = 2[θ][0 to 2π] = 4π ✓

Arc Length in Polar Coordinates

$L = ∫[α to β] √(r² + (dr/dθ)²) dθ
$

Key Terms

• 05 10 Parametric Equations And Polar Coordinates
• Arc Length
• Arc Length in Polar Coordinates
• Area in Polar Coordinates
• Converting
• Converting to Cartesian
• Correct: A)
• Correct: B)
• Derivatives
• Example 1: Parametric derivative
• Example 2: Polar area
• Example 3: Arc length of a parametric curve

Worked Examples

Example 1: Parametric derivative

x = cos(t), y = sin(t) (unit circle) dx/dt = -sin(t), dy/dt = cos(t) dy/dx = cos(t)/(-sin(t)) = -cot(t)

At t = π/4: dy/dx = -1 (tangent slope is -1, 45° downward).

Example 2: Polar area

Area inside r = 2sin(θ) from θ = 0 to π. This is a circle of radius 1 centred at (0, 1).

A = (1/2) ∫[0 to π] 4sin²(θ)dθ = 2 ∫[0 to π] (1 - cos(2θ))/2 dθ = ∫[0 to π] (1 - cos(2θ))dθ = [θ - (1/2)sin(2θ)][0 to π] = π - 0 = π

Area = π (circle of radius 1). ✓

Example 3: Arc length of a parametric curve

Find the arc length of x = t³, y = t² from t = 0 to t = 1.

dx/dt = 3t², dy/dt = 2t

L = ∫[0 to 1] √((3t²)² + (2t)²) dt = ∫[0 to 1] √(9t⁴ + 4t²) dt = ∫[0 to 1] t√(9t² + 4) dt

u = 9t² + 4, du = 18t dt, t dt = du/18 Limits: t = 0 → u = 4; t = 1 → u = 13

= (1/18) ∫[4 to 13] √u du = (1/18) · (2/3)[u^(3/2)][4 to 13] = (1/27)(13^(3/2) − 4^(3/2)) = (1/27)(13√13 − 8)

Practice Problems

Problem 1: Convert x = 3cos(t), y = 3sin(t) to Cartesian

Problem 2: dy/dx for x = t², y = t³ at t = 2

Problem 3: Area inside r = 1 from 0 to 2π

Problem 4: Convert (3, 4) to polar

Problem 5: Arc length of x = t, y = t² from t = 0 to 1

Summary

Key takeaways:

• Parametric: x = f(t), y = g(t)
• dy/dx = (dy/dt)/(dx/dt)
• Polar: (r, θ), x = rcos(θ), y = rsin(θ)
• Polar area: A = (1/2)∫r²dθ
• Many curves simpler in polar (circles, roses, cardioids)

Pitfalls

• Mishandling dy/dx for parametric curves. dy/dx = (dy/dt)/(dx/dt). Students often invert the ratio, writing (dx/dt)/(dy/dt), or forget to check that dx/dt ≠ 0 at the point of interest.
• Forgetting the 1/2 in the polar area formula. The polar area formula is A = (1/2)∫r²dθ. Dropping the factor of 1/2 and writing ∫r²dθ is one of the most common errors in polar calculus. The 1/2 plays the same role as the 1/2 in the formula for a circular sector.
• Using the wrong angular limits for polar area. For r = 1 + cos(θ) (cardioid), the full curve is traced from θ = 0 to 2π. But for r = cos(2θ) (four-petal rose), one petal corresponds to θ from −π/4 to π/4. Misidentifying the angular range gives dramatically wrong areas.
• Confusing polar and Cartesian arc length formulas. The polar arc length formula is L = ∫√(r² + (dr/dθ)²)dθ with r² inside the radical, not 1. It is NOT analogous to the Cartesian √(1 + (dy/dx)²) with an extra r factor — the r² term comes from the squared magnitude of the position vector derivative.
• Converting incorrectly between coordinate systems. r = √(x² + y²) and θ = arctan(y/x), but θ requires quadrant adjustment. A point (−1, 1) in Cartesian has θ = 3π/4, not arctan(−1) = −π/4. Forgetting quadrant correction is common and leads to wrong angle values.

Quiz

Q1: For parametric x = t², y = t³, dy/dx equals:

A) 3t/2 B) 2t/3 C) t/2 D) t²/3

Q2: Converting r = 4 to Cartesian gives:

A) x² + y² = 4 B) x² + y² = 16 C) x + y = 4 D) r = 4 is already Cartesian

Q3: The area inside r = 2 from θ = 0 to 2π is:

A) 2π B) 4π C) π D) 2

Q4: In polar coordinates, x = r·cos(θ) and y = r·sin(θ). Then r² equals:

A) x + y B) x² + y² C) x² - y² D) (x + y)²

Q5: For parametric x = cos(t), y = sin(t), the curve is:

A) A line B) A circle C) A parabola D) An ellipse

Next Steps

Next up: 06-01-functions-of-several-variables.md