Math Problem Statement

answer in one line:

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.

Solution

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 T^(t)\hat{T}(t) is (12sin2(t)cos(t)j+12cos2(t)sin(t)k)12\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=0t = 0 to t=π6t = \frac{\pi}{6} is 22.

Would you like further details on the steps?

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 v(t)\mathbf{v}(t) is crucial.

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Math Problem Analysis

Mathematical Concepts

Unit Tangent Vector
Arc Length
Parametric Curves
Vector Calculus

Formulas

Unit Tangent Vector: T(t) = v(t) / |v(t)|
Arc Length: S = ∫ |v(t)| dt
Parametric Curve Velocity: v(t) = dr(t)/dt

Theorems

Vector Differentiation
Arc Length Formula for Parametric Curves

Suitable Grade Level

University Level (Calculus II or Multivariable Calculus)