How do you find the unit tangent vector at a given point for the parametric function r(t)?

To find the unit tangent vector at a given point for the parametric function r(t) = (t^2, 2t, 1 + 3t^3 - 12t^2), we follow these steps:

1. First, we need to compute the derivative of the vector function r(t) with respect to t, denoted as r'(t).
2. Next, we evaluate r'(t) at the specific value of t that we are interested in.
3. Then, we find the magnitude of r'(t) to normalize it.
4. Finally, we divide r'(t) by its magnitude to obtain the unit tangent vector t(t).

Let’s illustrate this with an example. Assume we want to find the unit tangent vector at t = 1.

r(t) = (t^2, 2t, 1 + 3t^3 - 12t^2)

The first step is to take the derivative:

r'(t) = (2t, 2, 9t^2 - 24t)

Now, substituting t = 1:

r'(1) = (2*1, 2, 9*1^2 - 24*1) = (2, 2, -15)

Next, we calculate the magnitude of r'(1):

|r'(1)| = √(2^2 + 2^2 + (-15)^2) = √(4 + 4 + 225) = √233

Finally, we normalize r'(1) to get the unit tangent vector:

t(1) = r'(1) / |r'(1)| = (2 / √233, 2 / √233, -15 / √233)

Therefore, the unit tangent vector at t = 1 is (2/√233, 2/√233, -15/√233).