Find the unit tangent and normal vectors for r(t)

[marek]

Hello,

Find T(t) and N(t) for r(t) = <ln t, t> at t=e.

I began the problem by finding T(t), which is simply r'(t), divided by the norm of r'(t)

My unit tangent vector at T(t) is:

<1, t> divided by sqrt(1 + t^2)

so at T(e):

<1, e> divided by sqrt(1 + e^2)


To find the unit normal vector, I have to work with the formula T'(t), divided by the norm of T'(t)

My results for T'(t) are:

< -t/[(1+t^2)^(3/2)], [1/[1 + t^2)^(1/2)] + [(t^2)/[1 + t^2)^(3/2)] >

My results for the norm of T'(t) are:

sqrt [ [(t^2) + (t^4)]/[(1+t^2)^3] + [1/(1+t^2)]


Obviously this looks extremely messy if I were to take the quotient. How would I go about simplifying this "mess"? Is there another way to represent N(t)?

 

[Ishuda]

Let's back up to T and T'
T(t) = \(\displaystyle \frac{1}{(1+t^2)^{1/2}} <1,\, t>\)
T'(t)=<\(\displaystyle \frac{-t}{(1+t^2)^{3/2}},\, \frac{1}{(1+t^2)^{1/2}}\, +\, \frac{-t^2}{(1+t^2)^{3/2}}\)>
=<\(\displaystyle \frac{-t}{(1+t^2)^{3/2}},\, \frac{1+t^2}{(1+t^2)^{3/2}}\, +\, \frac{-t^2}{(1+t^2)^{3/2}}\)>
=\(\displaystyle \frac{1}{(1+t^2)^{3/2}} <-t,\, 1+t^2\, -\, t^2>\)
=\(\displaystyle \frac{1}{(1+t^2)^{3/2}} <-t,\, 1>\)

Now if we were to let
x(t)= (1+t²)^-1/2
the formulations become even more compact
T(t) = x(t) <1, t>
T'(t) = x³(t) <t, -1>