Final answer:
To find the unit tangent vector, calculate the derivative of the given position function to get the velocity vector, then normalize it by dividing by its magnitude.
Step-by-step explanation:
The question asks us to find the unit tangent vector of a given curve represented by the position function r(t) = (5+10t²)i + (8+11t²)j + (5+2t²)k. To find the unit tangent vector T(t), we first need to find the derivative of r(t), which will give us the velocity vector v(t). Then, we normalize v(t) to get the unit tangent vector by dividing it by its magnitude.
The velocity vector v(t) is the first derivative of the position vector r(t), which can be calculated as v(t) = (20t)i + (22t)j + (4t)k. The magnitude of v(t) is given by the square root of the sum of the squares of its components: |v(t)| = √(20t)² + (22t)² + (4t)²).
To get the unit tangent vector T(t), divide v(t) by its magnitude: T(t) = v(t) / |v(t)|. This question is generally approached by students in a calculus or vector analysis course in college.