Find the unit tangent vector T(t ) at the point with the given value of the parameter t.
r(t) = (t² - 2t,1 + 3t, 1/3t³ + 1/2t²), t = 2

Solution:

Given, the parametric curve is r(t) = t²- 2t, 1 + 4t, (1/3)t³ + (1/2)t²

We have to find the unit tangent vector at t = 2.

Now, r’(t0 = [x’(t), y’(t), z’(t)]

x’(t) = 2t - 2

y’(t) = 3

z’(t) = (1/3)3t² + (1/2)2t = t² + 1

So, r’(t) = [2t - 2, 3, t² + 1]

At t = 2,

r’(2) = [2(2) - 2, 3, (2)² + 1]

r’(2) = [4 - 2, 3, 4 + 1]

r’(2) = [2, 3, 5]

Also the modulus of the r’(2) vector is

|r’(2)| = \(\sqrt{(2)^{2}+(3)^{2}+(5)^{2}}=\sqrt{4+9+25}=\sqrt{38}\approx 6\)

Now, T(t) = r’(2)/|r’(4)|

= 2/6, 3/6, 5/6

= 1/3, 1/2, 5/6

Therefore, T(t) = 1/3, 1/2, 5/6.