Final answer:
To find T, N, and B at t = 1 for the position function r(t) = (t, 3/4 t^(3/2), t^2), one must calculate the velocity and acceleration vectors, normalize these to get T and N, and take the cross product of T and N for B.
Step-by-step explanation:
The student is asking to find the tangent vector T, the normal vector N, and the binormal vector B of the curve defined by the position function r(t) = (t, 3/4 t3/2, t2) at the point corresponding to t = 1. These vectors represent the Frenet-Serret frame which describes how the curve is situated in space.
To find the T vector, we first calculate the derivative of r(t) to get the velocity vector v(t), and then we normalize it:
- v(t) = r'(t) = (1, 3/2*t1/2, 2t)
- At t=1, v(1) = (1, 3/2, 2)
- T = v(1)/||v(1)|| = (1, 3/2, 2)/||(1, 3/2, 2)||
Next, we take the derivative of v(t) to get a(t), the acceleration vector, and use it to find the N vector:
- a(t) = v'(t) = (0, 3/4*t-1/2, 2)
- At t=1, a(1) = (0, 3/4, 2)
- N = a(1)/||a(1)|| = (0, 3/4, 2)/||(0, 3/4, 2)||
Last, the B vector is found by taking the cross product of T and N:
It's important to note that after getting the numerical values for v(1) and a(1), we would need to calculate their magnitudes to complete the normalization for T and N, and perform the cross product for B.