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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 $$x = -4t^3, \qquad y = \frac{1}{2}+3t^2$$)

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  • read the tag-wikis before you add inappropriate tags to you question. Your question does not seem to be about differential geometry. Thank you for your comprehension. – Asinomás May 07 '15 at 02:23
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    @Gamamal It is in fact about differential geometry. More precisely the differential geometry of curves (look up evolute). It's possible there is a more specific tag for this. – aes May 07 '15 at 04:27
  • the differential geometry of curves is not a part of differential geometry – Asinomás May 07 '15 at 04:32
  • As regards the problem itself: You're using a different parametrization than the answer by Chappers. So you shouldn't expect to get the same formula for the evolute, but rather a correspondingly different parametrization of it. I haven't checked whether that is in fact the case here (which it should be if you've made no errors). – aes May 07 '15 at 04:38
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    @Gamamal Yes it is. Undergraduate differential geometry courses often use do Carmo, starting with the differential geometry of curves and surfaces. This is part of the origins of the subject, and certainly still part of it. – aes May 07 '15 at 04:42

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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 $$ e(t) = \begin{pmatrix} t \\ t^2 \end{pmatrix} + \frac{(1+4t^2)^{3/2}}{2} \frac{1}{(1+4t^2)^{1/2}}\begin{pmatrix} -2t \\ 1 \end{pmatrix} \\ = \begin{pmatrix} t \\ t^2 \end{pmatrix} + \frac{(1+4t^2)}{2} \begin{pmatrix} -2t \\ 1 \end{pmatrix} \\ = \begin{pmatrix} -4t^3 \\ 1/2 + 3t^3 \end{pmatrix} $$

Chappers
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