Your vector-valued function, r(t), has been presented to you as the ordered triple, (x(t),y(t),z(t)).
r(t)=(x(t),y(t),z(t))
In your case,
x(t) = -2t
y(t) = e^(-t)
z(t) = √(t+5) - t = (t+5)^(1/2) - t (always express square roots as exponentiation by 1/2 when taking derivatives)
To get the tangent vector, r'(t), just take the derivatives of each of the component functions:
r'(t) = (x'(t),y'(t),z'(t))
You can then substitute in 2 for t to find r'(2).
To get the unit tangent vector, T(t), you'll want to divide r'(t) by its magnitude, ||r'(t)||.
||r'(t)|| = √(x'(t)^2 + y'(t)^2 + z'(t)^2)
T(t) = r'(t)/||r'(t)||
T(t) is a vector pointing in the direction of r'(t) and having a magnitude of 1.
You can substitute in t=2 to find the T(2), the tangent vector at t=2.
