I need to find the unit normal vector given:

r(t)= ti + (6/t) j where t= 3

I know I need to find the unit tangent vector first, which is r'(t) divided by the magnitude of r'(t).
r(t) can be rewritten as ti+ 6t^-1
r'(t) = i- 6t^-2 = i- [6/ (t^2)] j
For the magnitude of r'(t) I got the square root of (t^4 +36) all divided by t^2.
For T(t), I got (it^2 -6j)/ (sq. rt. t^4 +36)

Is this right? If so, I don't know how to go about finding T'(t) or the magnitude of T'(t). Please help! Also, where does the t= 3 come in?

To find the unit normal vector, you are correct in first finding the unit tangent vector. The unit tangent vector, T(t), is obtained by dividing r'(t) by the magnitude of r'(t). Let's review your calculation so far:

r(t) = ti + (6/t)j

To find r'(t), you differentiated the components of r(t) with respect to t:
r'(t) = i - [6/(t^2)]j

For the magnitude of r'(t), you correctly obtained:
| r'(t) | = √(t^4 + 36) / t^2

Now, to find T(t), we divide r'(t) by its magnitude:
T(t) = (i - [6/(t^2)]j) / (√(t^4 + 36) / t^2)

To find T'(t), we differentiate each component of T(t) with respect to t. The differentiation is straightforward since there are no t terms in the equation:

T'(t) = (0 - (-12/t^3)j) / (√(t^4 + 36) / t^2)

Simplifying T'(t), we get:
T'(t) = (-12/t^3)j * (t^2 / √(t^4 + 36))

Now, to find the magnitude of T'(t), we take the absolute value of T'(t). Since we only have the j-component, we ignore the i-component:
| T'(t) | = |-12/t^3| * |t^2 / √(t^4 + 36)|

Simplifying, we get:
| T'(t) | = 12t^2 / (t^4 + 36)^(1/2)

Finally, you mentioned t = 3. To find the unit normal vector at t = 3, substitute t = 3 into the expression for T(t):

T(3) = (i - [6/(3^2)]j) / (√(3^4 + 36) / 3^2)

Calculating this expression will give you the unit tangent vector at t = 3. Remember that the unit normal vector is perpendicular to the unit tangent vector, so you need to find the vector that is perpendicular to T(3) as well.