Finding the evolute of a parabola I previously tried to find the evolute of a parabola by using parameterisation by arc length. It didn't work. While I was hoping for an answer I kept working on it and came up with the following method (unfortunately, something is not quite right as it leads to a different result than the answer to my previous question). 

I would appreciate it if someone could tell me where I went wrong:

Given $\gamma (t) = (t,t^2)$ we find $\gamma'(t) = (1,2t)$. Normalising we find the unit tangent vector to be
$$ v(t) = {1 \over \sqrt{1 + 4 t^2}} \left (\begin{matrix} 1 \\ 2t \end{matrix}\right)$$
Since this vector is unit lenght, its derivative will be orthogonal to it so that $v'(t)$ should be the normal vector to the plane curve:
$$ n(t) := v'(t) = {1\over (1+4 t^2)^{3/2}}  \left (\begin{matrix} -4t \\ 2(1 + 4t^2)^{2} +8t^2 \end{matrix}\right)$$
The curvature is $\kappa (t) = \|n(t)\|$ hence the radius of curvature is $r = {1\over \kappa (t)}$. 
Hence the equation for the evolute should be
$$ e(t) = \gamma (t) + r n(t) =  \left (\begin{matrix} t \\ t^2 \end{matrix}\right) + {1\over \sqrt{16t^2 + (2(1+4t^2)^2 +8t^2)^2}} \left (\begin{matrix} -4t \\ 2(1+4t^2) + 8t^2 \end{matrix}\right)$$
(But the equation of the evolute (according to the answer by Chappers) should be 
$$x = -4t^3, \qquad y = \frac{1}{2}+3t^2$$)

Right, your method is not inherently flawed, but you have calculational mistakes: they first appear in the normal vector: differentiating the tangent vector should give
$$ -\frac{4t}{(1+4t^2)^{3/2}} \begin{pmatrix} 1 \\ 2t \end{pmatrix} + \frac{1}{(1+4t^2)^{1/2}}\begin{pmatrix} 0 \\ 2 \end{pmatrix} = \frac{1}{(1+4t^2)^{3/2}} \begin{pmatrix} -4t \\ -8t^2+2(1+4t^2) \end{pmatrix} \\
=\frac{2}{(1+4t^2)^{3/2}} \begin{pmatrix} -2t \\ 1 \end{pmatrix}
, $$
Now you do as you suggested, (and here is your second mistake, in that you should have $\hat{n}$ instead of $n$, for obvious reasons):
$$ e(t) = \gamma(t) + r\hat{n}(t) = \gamma(t) + \frac{1}{\lVert n(t) \rVert} \hat{n}(t), $$
and here, obviously
$$ \hat{n}(t) = \frac{1}{(1+4t^2)^{1/2}}\begin{pmatrix} -2t \\ 1 \end{pmatrix}, $$
so
$$ e(t) = \begin{pmatrix} t \\ t^2 \end{pmatrix} + \frac{(1+4t^2)^{3/2}}{2} \frac{1}{(1+4t^2)^{1/2}}\begin{pmatrix} -2t \\ 1 \end{pmatrix} \\
= \begin{pmatrix} t \\ t^2 \end{pmatrix} + \frac{(1+4t^2)}{2} \begin{pmatrix} -2t \\ 1 \end{pmatrix} \\
= \begin{pmatrix} -4t^3 \\ 1/2 + 3t^3 \end{pmatrix} $$
