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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
- Parametric slope: dy/dx = (dy/dt)/(dx/dt).
- Second derivative: d²y/dx² = [d/dt(dy/dx)] / (dx/dt).
- Speed is the magnitude of the velocity vector.
- Polar area uses wedges: A = ½∫ r² dθ.
- Polar slope converts with x = r cos θ, y = r sin θ.
Formulas and rules
Common mistakes
- Dividing d/dt(dy/dx) by dt instead of dx/dt for the second derivative.
- Forgetting the ½ in polar area.
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.
x = t², y = t³. Find dy/dx at t = 2.
- 6
- 1/3
- 3
- 12
Show the answer and steps
Answer: C) 3
- dx/dt = 2t, dy/dt = 3t².
- dy/dx = 3t²/(2t) = 3t/2.
- At t = 2: 3.
A particle moves with x = 7t, y = 24t. Find its speed.
- 17
- 31
- 25
- 625
Show the answer and steps
Answer: C) 25
- Speed = √((dx/dt)² + (dy/dt)²) = √(7² + 24²).
- = √625 = 25.
Area inside r = 2.
- π
- 4π
- 8π
- 2π
Show the answer and steps
Answer: B) 4π
- A = ½∫₀^2π r² dθ = ½∫ 4 dθ.
- = ½·4·2π = 4π.