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Calculus II (Integration & Series) Unit 7: Parametric, polar & vector-valued functions

Parametric, polar and vector-valued functions describe curves and motion that aren't simple y = f(x) graphs.

Covers: Derivatives & arc length of parametric curves, polar area, vector motion

Study guide

Parametric equations x(t), y(t) describe a curve traced over time. The slope of the curve is dy/dx = (dy/dt)/(dx/dt), and horizontal tangents occur where dy/dt = 0 (with dx/dt ≠ 0). The second derivative is d²y/dx² = [d/dt(dy/dx)]/(dx/dt). Arc length is ∫ √((dx/dt)² + (dy/dt)²) dt.

Vector-valued functions r(t) = ⟨x(t), y(t)⟩ treat parametric motion as a position vector. Velocity is v(t) = r′(t) = ⟨x′(t), y′(t)⟩, acceleration is a(t) = r″(t), and speed is the magnitude |v(t)| = √(x′(t)² + y′(t)²). The total distance traveled is the integral of speed. To find position from velocity, integrate each component and use the initial position.

Polar curves r = f(θ) can be treated as parametric curves with x = r cos θ and y = r sin θ, which gives their slopes. The area enclosed by a polar curve from θ = α to θ = β is A = ½∫ r² dθ, because thin sectors have area ½r²Δθ. The area between two polar curves is ½∫ (R² − r²) dθ, with limits found where the curves intersect. Watch for loops traced twice and choose limits that trace the region exactly once.

When θ changes with time, dr/dt = (dr/dθ)·(dθ/dt). The same chain rule applies to x = r cos θ and y = r sin θ. If r > 0 and dr/dθ > 0, points on the curve move away from the origin as θ increases. If r > 0 and dr/dθ < 0, they move toward the origin. The conclusions reverse when r < 0.

For a polar curve, dy/dx = (r′ sin θ + r cos θ)/(r′ cos θ − r sin θ), where r′ = dr/dθ. The arc length of r = f(θ) from θ = α to θ = β is ∫ √(r² + (dr/dθ)²) dθ.

A parametric curve has a vertical tangent where dx/dt = 0 and dy/dt ≠ 0. A particle moves left when dx/dt < 0 and down when dy/dt < 0.

Key ideas

Formulas and rules

dy/dx = (dy/dt)/(dx/dt)
Parametric arc length: ∫ √((dx/dt)² + (dy/dt)²) dt
Speed = √(x′(t)² + y′(t)²)
Polar area: A = ½∫ r² dθ

Common mistakes

Words to know

Slope of a parametric curve
dy/dx = (dy/dt)/(dx/dt)
Speed of a particle with position ⟨x(t), y(t)⟩
√(x′(t)² + y′(t)²)
Area inside a polar curve
½∫ r² dθ
Second derivative for parametric curves
[d/dt(dy/dx)] / (dx/dt)
Arc length of a parametric curve
∫ √((dx/dt)² + (dy/dt)²) dt
Parametric curve
x = x(t), y = y(t), traced as t changes.
Horizontal tangent (parametric)
Where dy/dt = 0 and dx/dt ≠ 0.
Velocity vector
v(t) = ⟨x′(t), y′(t)⟩
Acceleration vector
a(t) = ⟨x″(t), y″(t)⟩
Total distance (vector motion)
∫ speed dt = ∫ √(x′² + y′²) dt
Area between polar curves
½∫ (R² − r²) dθ

Sample practice questions

29 more questions like these, from level 1 to 5, are in the practice set.

Question 1 (level 1)

x = t², y = t³. Find dy/dx at t = 2.

  1. 6
  2. 1/3
  3. 3
  4. 12
Show the answer and steps

Answer: C) 3

  1. dx/dt = 2t, dy/dt = 3t².
  2. dy/dx = 3t²/(2t) = 3t/2.
  3. At t = 2: 3.
Question 2 (level 1)

A particle moves with x = 7t, y = 24t. Find its speed.

  1. 17
  2. 31
  3. 25
  4. 625
Show the answer and steps

Answer: C) 25

  1. Speed = √((dx/dt)² + (dy/dt)²) = √(7² + 24²).
  2. = √625 = 25.
Question 3 (level 3)

Area inside r = 2.

  1. π
  2. 4π
  3. 8π
  4. 2π
Show the answer and steps

Answer: B) 4π

  1. A = ½∫₀^2π r² dθ = ½∫ 4 dθ.
  2. = ½·4·2π = 4π.