Ask Question
9 February, 10:34

Find the length of the curve given by ~r (t) = 1 2 cos (t 2) ~i + 1 2 sin (t 2) ~j + 2 5 t 5/2 ~k between t = 0 and t = 1. Simplify! The final result is quite simple

+1
Answers (1)
  1. 9 February, 10:57
    0
    The length of the curve is

    L ≈ 0.59501

    Step-by-step explanation:

    The length of a curve on an interval a ≤ t ≤ b is given as

    L = Integral from a to b of √[ (x') ² + (y') ² + (z') ²]

    Where x' = dx/dt

    y' = dy/dt

    z' = dz/dt

    Given the function r (t) = (1/2) cos (t²) i + (1/2) sin (t²) j + (2/5) t^ (5/2)

    We can write

    x = (1/2) cos (t²)

    y = (1/2) sin (t²)

    z = (2/5) t^ (5/2)

    x' = - tsin (t²)

    y' = tcos (t²)

    z' = t^ (3/2)

    (x') ² + (y') ² + (z') ² = [-tsin (t²) ]² + [tcos (t²) ]² + [t^ (3/2) ]²

    = t² (-sin² (t²) + cos² (t²) + 1)

    ...

    But cos² (t²) + sin² (t²) = 1

    => cos² (t²) = 1 - sin² (t²)

    ...

    So, we have

    (x') ² + (y') ² + (z') ² = t²[2cos² (t²) ]

    √[ (x') ² + (y') ² + (z') ²] = √[2t²cos² (t²) ]

    = (√2) tcos (t²)

    Now,

    L = integral of (√2) tcos (t²) from 0 to 1

    = (1/√2) sin (t²) from 0 to 1

    = (1/√2) [sin (1) - sin (0) ]

    = (1/√2) sin (1)

    ≈ 0.59501
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “Find the length of the curve given by ~r (t) = 1 2 cos (t 2) ~i + 1 2 sin (t 2) ~j + 2 5 t 5/2 ~k between t = 0 and t = 1. Simplify! The ...” in 📗 Mathematics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers