Parametric Equations, Polar Coordinates, and Vector-Valued Functions
← Back to Unit 9Solution:
1) The curve r = 3sin(2θ) is a 4-petal rose
2) One petal occurs when 0 ≤ θ ≤ π/2
3) Area of one petal: A = (1/2)∫0π/2 [3sin(2θ)]² dθ
4) A = (9/2)∫0π/2 sin²(2θ) dθ
5) Using sin²(u) = (1-cos(2u))/2: A = (9/4)∫0π/2 (1-cos(4θ)) dθ
6) A = (9/4)[θ - sin(4θ)/4]0π/2 = (9/4)(π/2) = 9π/8
7) Total area = 4 × (9π/8) = 9π/2
Answer: 9π/2
Solution:
1) dx/dt = eᵗcos(t) - eᵗsin(t) = eᵗ(cos(t) - sin(t))
2) dy/dt = eᵗsin(t) + eᵗcos(t) = eᵗ(sin(t) + cos(t))
3) (dx/dt)² + (dy/dt)² = e²ᵗ[(cos(t) - sin(t))² + (sin(t) + cos(t))²]
4) = e²ᵗ[cos²(t) - 2cos(t)sin(t) + sin²(t) + sin²(t) + 2sin(t)cos(t) + cos²(t)]
5) = e²ᵗ[2cos²(t) + 2sin²(t)] = 2e²ᵗ
6) L = ∫0π √(2e²ᵗ) dt = √2 ∫0π eᵗ dt
7) L = √2[eᵗ]0π = √2(e^π - 1)
Answer: √2(e^π - 1)