Finding the Unit Vector

Find the unit vector T(t) tangent to the graph of the vector function:
r(t) =

I know that the unit vector is the derivative of the vector function divided by the magnitude of that derivative. However, how would I further simplify the problem. I got my derivative to be <2t, 2, 1/t>, and my magnitude is 2t + 2 + 1/t.

Any help appreciated.

The magnitude is the square root of the sum of the squares of the components, not the sum of the components.

|dr/dt| = √(4t² + 4 + (1/t²)) = √((2t + (1/t))²)

Assuming t > 0, |dr/dt| = 2t + (1/t) so divide each component of dr/dt by (2t + (1/t)) to get the unit vector.

It looks a little less complicated if you then multiply the numerator and denominator of each component by t.

The magnitude is sqrt((2t)^2 + 2^2 + (1/t)^2)) = sqrt (4t^2 + 4 + 1/t^2)) which you can factor to get
sqrt ((2t+1/t)^2)) = (2t + 1/t)

So the unit tangent is 2t^2/(2t^2+1) i + 2t/(2t^2 + 1) j + 1/(2t^2 + 1) k

Hope that helps!