Answer:
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
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