Unit Tangent Vector and Unit Normal Vector for a given value of t 
To Find - unit tangent vector $T$ and the unit normal vector $N$ for $t=0$; I 
  know the following 

$\mathbf r(t)=\mathbf i(t)+\frac{1}{2}\mathbf j(t^2)+\frac{1}{3}\mathbf k(t^3)$
$T=\frac{r'(t)}{|r'(t)|}$

and

$N=\frac{T'(t)}{|T'(t)|}$

now,

$\mathbf r'(t)=\mathbf i+\mathbf j(t)+\mathbf k(t^2)$

and
  $\Vert\mathbf r'(t)\Vert= \sqrt{1+t^2+t^4}$
therefore,

$T(t)=\frac{\mathbf i}{\sqrt{1+t^2+t^4}}+\frac{\mathbf j(t)}{\sqrt{1+t^2+t^4}}+\frac{\mathbf k(t^2)}{\sqrt{1+t^2+t^4}}$
$T '(t)=\mathbf i\frac{-t(2t^2+1)}{({t^4+t^2+1})^\frac{3}{2}}+\mathbf j\frac{1-t^4}{(t^4+t^2+1)^\frac{3}{2}}+\mathbf k\frac{t(t^2+2)}{(t^4+t^2+1)^\frac{3}{2}}$
$\Vert T '(t)\Vert=\sqrt{\frac{(-2t^3-t)^2+(1-t^4)^2+ (t^3+2t)^2}{(t^4+t^2+1)^3}}
$

This is far as I got!


\begin{align*}
  \mathbf{v} &= \dot{s} \mathbf{T} \\
  &=(1,t,t^{2}) \\
  \dot{s} &= \sqrt{1+t^{2}+t^{4}} \\
  \ddot{s} &= \frac{t(1+2t^{2})}{\sqrt{1+t^{2}+t^{4}}} \\
  \mathbf{v}(0) &= (1,0,0) \\
  \dot{s}(0) &= 1 \\
  \mathbf{T}(0) &= \mathbf{i} \\
  \mathbf{a} &= \ddot{s} \mathbf{T}+\kappa \dot{s}^{2}\mathbf{N} \\
  &=(0,1,2t) \\
  \mathbf{a}(0) &= (0,1,0) \\
  \ddot{s}(0) &= 0 \\
  \mathbf{N}(0) &= \mathbf{j}
\end{align*}
