
Curvature
Find the curvature of r(t) = 3/2 (e^2t + e^2t)i + 3/2 (e^2t  e^2t)j + 6tk at t = ln 2
I think I screwed up somewhere in my calculations.
I got r'(t) = (3e^2t  3e^2t)i + (3e^2t + 3e^2t)j + 6k
r"(t) = (6e^2t + 6e^2t)i + (6e^2t  6e^2t)j
r'(t) x r"(t) = 72e^2t
r'(t) = (18e^4t + 18 e^4t +36) ^ 3/2
Those answers just don't look right. Can anyone tell me what I'm doing wrong.

Once you have r ' and r '' go ahead and plug in t =ln(2)
it will make the calulation a lot simpler
r ' = (123/4) i + (12+3/4) j + 6 k = 45/4 i + 51/4 j + 6k
etc