AP Calculus BC Practice Quiz: The Arc Length of a Smooth, Planar Curve and Distance Traveled (BC ONLY)

Correct Answer: A

The formula for the arc length of a function y=f(x) from x=a to x=b is L = ∫ₐᵇ √(1 + (dy/dx)²) dx. For y = ln(cos(x)), the derivative is dy/dx = (1/cos(x)) * (-sin(x)) = -tan(x). Squaring the derivative gives (dy/dx)² = (-tan(x))² = tan²(x). Substituting this into the arc length formula yields L = ∫₀^(π/4) √(1 + tan²(x)) dx.

Correct Answer: C

First, find the derivative: dy/dx = (2/3) * (3/2)x^(1/2) = x^(1/2). Then, set up the arc length integral: L = ∫₀³ √(1 + (x^(1/2))²) dx = ∫₀³ √(1 + x) dx. To evaluate, let u = 1+x, so du = dx. The new limits of integration are u(0)=1 and u(3)=4. The integral becomes ∫₁⁴ u^(1/2) du = [(2/3)u^(3/2)] from 1 to 4. Evaluating this gives (2/3)(4^(3/2) - 1^(3/2)) = (2/3)(8 - 1) = 14/3.

Correct Answer: D

The total distance traveled is the integral of the speed. Speed is the magnitude of the velocity vector, ||v(t)|| = √((dx/dt)² + (dy/dt)²). Here, speed = √((3t²)² + (3t)²) = √(9t⁴ + 9t²) = √(9t²(t²+1)) = 3t√(t²+1) for t≥0. The distance is L = ∫₀² 3t√(t²+1) dt. Using u-substitution with u=t²+1, du=2t dt, the integral becomes (3/2)∫₁⁵ u^(1/2) du = (3/2)[(2/3)u^(3/2)] from 1 to 5 = [u^(3/2)] from 1 to 5 = 5^(3/2) - 1^(3/2) = 5√5 - 1.

Correct Answer: B

The arc length for a parametric curve is L = ∫ₐᵇ √((dx/dt)² + (dy/dt)²) dt. Using the product rule, dx/dt = eᵗcos(t) - eᵗsin(t) and dy/dt = eᵗsin(t) + eᵗcos(t). Squaring and adding these gives (dx/dt)² + (dy/dt)² = (eᵗ(cos(t)-sin(t)))² + (eᵗ(sin(t)+cos(t)))² = e²ᵗ((cos²(t)-2sin(t)cos(t)+sin²(t)) + (sin²(t)+2sin(t)cos(t)+cos²(t))) = e²ᵗ(1 + 1) = 2e²ᵗ. The integrand is √(2e²ᵗ) = √2 eᵗ. The integral is ∫₀^π √2 eᵗ dt.

Correct Answer: A

The arc length of a polar curve r(θ) is given by L = ∫ₐᵝ √(r² + (dr/dθ)²) dθ. For r = 3sin(θ), we have dr/dθ = 3cos(θ). The expression inside the square root is r² + (dr/dθ)² = (3sin(θ))² + (3cos(θ))² = 9sin²(θ) + 9cos²(θ) = 9(sin²(θ) + cos²(θ)) = 9. Therefore, the integrand is √9 = 3. The integral for the arc length is ∫₀^π 3 dθ.

Correct Answer: C

The arc length is L = ∫ₐᵇ √(1 + (dy/dx)²) dx. First, find the derivative using the chain rule: dy/dx = (1/3) * (3/2)(x² + 2)^(1/2) * (2x) = x√(x² + 2). Next, square the derivative: (dy/dx)² = (x√(x² + 2))² = x²(x² + 2) = x⁴ + 2x². Now, add 1: 1 + (dy/dx)² = 1 + x⁴ + 2x² = (x² + 1)². The integrand simplifies: √((x² + 1)²) = x² + 1. The integral is L = ∫₀³ (x² + 1) dx = [x³/3 + x] from 0 to 3 = (3³/3 + 3) - 0 = (9 + 3) = 12.

Correct Answer: B

The formula for arc length with respect to y is L = ∫ₐᵇ √(1 + (dx/dy)²) dy. For x = y⁴/4 + 1/(8y²), the derivative is dx/dy = y³ - 1/(4y³). Squaring this gives (dx/dy)² = y⁶ - 2(y³)(1/(4y³)) + 1/(16y⁶) = y⁶ - 1/2 + 1/(16y⁶). Then, 1 + (dx/dy)² = 1 + y⁶ - 1/2 + 1/(16y⁶) = y⁶ + 1/2 + 1/(16y⁶). This expression is a perfect square: (y³ + 1/(4y³))². The integral becomes ∫₁² √((y³ + 1/(4y³))²) dy = ∫₁² (y³ + 1/(4y³)) dy, since the term is positive on the interval [1, 2].