Space Curves --> Unit Tangent Vector and Curvature

[DeadxBunny]

Here is the original question:

Consider the space curve r(t) = (e^t)*cos(t)i + (e^t)*sin(t)j + k. Find the unit tangent vector T(0) and the curvature of r(t) at the point (0,e^(pi/2),1).

I believe I have found the unit tangent vector, T(0), correctly: (1/sqrt(2))i + (1/sqrt(2))j
Is this correct? Also, how do I find the curvature at that particular point?

Thanks!

[HallsofIvy]

Yes, your unit tangent vector, at T= 0, is correct.

As for the curvature, there are a variety of formulas that could be applied. What formulas do you know?

[DeadxBunny]

I know:

K=|T'(t)|/|r'(t)|

and

K=|r'(t) x r''(t)|/|r'(t)|^3