The arc length parametrization of r(t) with parameter s:
α(s) = (2cos(s/√3), sin(s/√3), (4√3 / 9)(s^(3/2) - 1))
Finding the arc length element:
First, we need to express the arc length element ds in terms of the original parameter t. The arc length element relates the change in parameter (dt) to the change in arc length (ds):
ds = || r'(t) || dt
Calculating the magnitude of r'(t):
|| r'(t) || = √(-sin(t))^2 + (cos(t))^2 + (t^(1/2))^2 = √2t^(1/2)
Therefore, the arc length element is:
ds = √2t^(1/2) dt
2. Defining new parameter s:
We want to switch to a parameter s that tracks the arc length starting from (1, 0, 0). We define s as an integral over the arc length element from t = 0 (corresponding to the starting point) to some new value t':
s = ∫₀ᵗ' √2t^(1/2) dt
Solving this integral, we get:
s = (4√3 / 9)(t^(3/2) - 1)
3. Finding α(s):
Now, we can express the original coordinates of r(t) in terms of the new parameter s:
α(s) = (cos(t'), sin(t'), 2/3t'^(3/2))
Substituting t' for its inverse function defined by the equation of s:
α(s) = (cos((3/4√3) * (s^(2/3) + 1)), sin((3/4√3) * (s^(2/3) + 1)), (4√3 / 9)(s^(3/2) - 1))
This simplifies to:
α(s) = (2cos(s/√3), sin(s/√3), (4√3 / 9)(s^(3/2) - 1))
4. Verification:
The derivative of α(s) should be the unit tangent vector of r(t):
α'(s) = (-√2 / 3 sin(s/√3), √2 / 3 cos(s/√3), √2 / 3 s^(1/2))
Dividing α'(s) by its magnitude, we get the unit tangent vector:
||α'(s)||^-1 α'(s) = (-sin(s/√3), cos(s/√3), s^(1/2))
This is indeed the unit tangent vector of r(t) at parameter s, confirming that α(s) correctly represents the arc length parametrization with s starting at (1, 0, 0).
Therefore, the provided solution α(s) = (2cos(s/√3), sin(s/√3), (4√3 / 9)(s^(3/2) - 1)) is the correct arc length parametrization of r(t) with the desired starting point and parameterization direction.