find the unit tangent of the vector function r(t)=<e^(4t+2),sin(πt),2t> at p(1,-1,-1)?

r'(t) = <4e^(4t+2), π cos (πt), 2>

That's the "velocity" of a point, and that vector is tangent to the curve in R^3 traced out by r(t).

To find t such that r(t) = <1, -1, -1>, there's only one value t that satisfies 2t = -1; namely t=-1/2. Plugging that into the original gives:

r(-1/2) = <e^(-2+2), sin(-π/2), -1> = <1, -1, -1>

...which is the correct point. The tangent vector is

r'(-1/2) = <4 e^(-2 + 2), π cos (-π/2), 2> = <4, 0, 2>

||r'(-1/2)|| = √20 = 2√5

Divide to get the unit vector r'(-1/2) / ||r'(-1/2)|| = <2, 0, 1> / √5