Find the vectors T, N, and B at hte given point. r(t) = 6 cos t, 6 sin t, 6 ln cos t, (6,0,0)?

B = (r' × r'')/|r' × r'' First find what shall t equal as may not exist at all -1 = sin(t) - cos(t) , 1 = sin(t) + cos(t) , 3 = 3 which we know the third is true at least 1 = cos(t) - sin(t) , 1 = sin(t) + cos(t) as you can guess it must be t = 2pi*n as the algebra way would be to long r'(t) = ((sin(t) + cos(t)) , (cos(t) - sin(t)) , 0); r''(t) = ((cos(t) - sin(t)) , (-sin(t) - cos(t)) , 0) (r' × r'') = (0,0,-2); |r' × r''| = 2 so the final answer is B = (0, 0 ,-1).

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