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To find the curve's unit tangent vector and the length of the indicated portion, here's the result in one line:

The unit tangent vector \( \hat{T}(t) \) is \( \frac{(-12 \sin^2(t) \cos(t) \, \mathbf{j} + 12 \cos^2(t) \sin(t) \, \mathbf{k})}{12} \), and the length of the curve from \( t = 0 \) to \( t = \frac{\pi}{6} \) is \( 2 \).

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Here are some related questions:
1. How is the unit tangent vector derived?
2. What is the process of calculating the arc length of a parametric curve?
3. Can the tangent vector help determine the curve's orientation?
4. How do you compute the derivative of parametric functions like these?
5. What role does the magnitude of the derivative play in the unit tangent vector?

**Tip:** For arc length calculations, the magnitude of the velocity vector \( \mathbf{v}(t) \) is crucial.

Find the curve's unit tangent vector. Also, find the length of the indicated portion of the curve. r(t) = (4cos^3(t))j + (4sin^3(t))k, 0 ≤ t ≤ π/6.

Learn how to find the unit tangent vector and calculate the arc length for the parametric curve r(t) = (4cos^3(t))j + (4sin^3(t))k from t = 0 to t = π/6. This solution involves vector calculus concepts and arc length computation.

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