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

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