Final answer:
To find the Frenet-Serret frame consisting of tangent (t), normal (n), and binormal (b) vectors at a point on the curve r(t) = (t², 2t³, t), we calculate the derivatives of r(t) and use them to compute the required vectors at the specified point.
Step-by-step explanation:
The question involves finding the tangent (t), normal (n), and binormal (b) vectors at a specific point on a curve represented by the position vector r(t) = (t², 2t³, t). These three vectors describe the Frenet-Serret frame of the curve at a particular point.
To find the t vector, we must first compute the derivative of r(t) to find the velocity vector v(t) and then normalize it:
- v(t) = d/dt r(t) = (2t, 6t², 1)
- At the point (4, 16/3, 2), we have t = 2, yielding v(2) = (4, 24, 1). To normalize, divide by its magnitude, which gives the tangent vector t.
To find the normal vector n, we differentiate v(t) to obtain the acceleration vector a(t), and then normalize the component of a(t) orthogonal to v(t).
- a(t) = d/dt v(t) = (2, 12t, 0)
- At t = 2, a(2) = (2, 24, 0). Then, we subtract the projection of a(t) on v(t) and normalize to get n.
The binormal vector b is found by taking the cross product v(t) × a(t) and normalizing.
Lastly, we compute these vectors at t = 2 using the formulas and methods described above to find the Frenet-Serret frame at the given point on the curve.