Length is independent of parametrization

Question

Length is independent of parametrization: To illustrate that the length of a smooth space curve does not depend on the parametrization you use to compute it, calculate the length of one turn of the helix in example 1 with the following parametrization.

a> r(t) = (cos 4t)i + (sin 4t)j + 4tk, 0<=t<=(pi/2)

b> r(t) = [cos (t/2)]i + [sin (t/2)]j + (t/2)k, 0<=t<= (4 pi)

c> r(t) = (cos t)i - (sin t)j -tk, -2 pi <= t <= 0

 

 

Thank you! I really appreciate your help!

Answer 1

a. (dL)² = [d(cos 4t)]² + [d(sin 4t)]² +[d(4t)]² = [(-4 sin 4t)² + (4 cos 4t)² + 4²] (dt)² = 32(dt)²

     L = ∫₀^π/2 4√2 dt = 2π√2

b. (dL)² = [d(cos t/2)]² + [d(sin t/2)]² +[d(t/2)]² = [(-(1/2) sin t/2)² + ((1/2) cos t/2)² + (1/2)²] (dt)² = (1/2)(dt)²

     L = ∫₀^4π (√2 / 2) dt = 2π√2

c. (dL)² = [d(cos t)]² + [d(sin t)]² +(dt)² = [(-sin t)² + (cos t)² + 1²] (dt)² = 2(dt)²

     L = ∫_-2π⁰ √2 dt = 2π√2