Express the acceleration vector
a''(1) as the sum of a vector parallel to a'(1) and a vector orthogonal to a'(1)
The Attempt at a Solution
I took the first two derivatives and calculated a'(t)=<2t, -4t^2, -4/(2-t)^(1/2)> and a''(t)=<2, 8/t^3, -2/(2-x)^(3/2)>. I figured the tangent vector field a'(1)/|a'(1)| will give me the vector parallel to a'(1). But how do I get the orthogonal vector? I was thinking the cross product between a'(1)/|a'(1)| and a''(1) would give me the orthogonal vector to a'(1), but the vectors ended up being linearly independent so I couldn't represent a''(1) as a sum of the other two.
I then tried calculating the principal normal vector field but I would need to take the derivative of the tangent vector field a'(t)/|a'(t)| and that ended up being incredibly messy and I'm sure that isn't the right way to do this.