I've been trying to independently learn differential geometry/vector calculus, and I've been experimenting with evolutes and curvature. I came up with the following problem that I need help solving.
Let $\alpha:I \mapsto \Bbb R^2$ be a parametrized curve defined as $\alpha(t)=(x(t),y(t))$, for some interval $I\subseteq\Bbb R$.
Question:
Is it possible to find the definitions of $x(t)$ and $y(t)$ such that the evolute of $\alpha(t)$ is given by the trace of $E(t)=(t,1/t)$?
Attempt:
Since the unit tangent is given by $$T(t)=\frac{\alpha'(t)}{|\alpha'(t)|}$$
And the unit normal is given by $$N(t)=\frac{T'(t)}{|T'(t)|}$$
We know that the evolute would be $$E(t)=\rho(t)N(t)+\alpha(t)$$ where the curvature it given by $$\rho(t)=\frac{|\alpha'(t)|}{|T'(t)|}$$ Thus, $$E(t)=\frac{|\alpha'(t)|}{|T'(t)|}N(t)+\alpha(t)=\frac{|\alpha'(t)|T'(t)}{T'(t)\cdot T'(t)}+\alpha(t)=(t,1/t)$$ And since $\alpha(t)=(x(t),y(t))$, $$(t,1/t)=(\frac{|\alpha'(t)|}{(\frac{d}{dt}\frac{x'(t)}{|\alpha'(t)|})^2+ (\frac{d}{dt}\frac{y'(t)}{|\alpha'(t)|})^2}(\frac{d}{dt}\frac{x'(t)}{|\alpha'(t)|})+x(t), \frac{|\alpha'(t)|}{(\frac{d}{dt}\frac{x'(t)}{|\alpha'(t)|})^2+ (\frac{d}{dt}\frac{y'(t)}{|\alpha'(t)|})^2}(\frac{d}{dt}\frac{y'(t)}{|\alpha'(t)|})+y(t))$$
Which I don't know how to deal with.
Best Answer
I guess that in your question you assume that the parametrisation of the evolute $E(t)$ is $(t,\frac{1}{t})$. If we only know that the evolute is given by the trace of $(t,\frac{1}{t})$ then the question is more difficult.
Let $\alpha\colon I\subset \mathbb{R}^+ \to \mathbb{R}^2$ be a unit speed parametrised curve and assume $E(t)$ is the parametrisation of its evolute. Then $$ \left(t,\frac{1}{t}\right) = \alpha(t) + \frac{1}{\kappa}N(t). $$ Differentiate and use the Frenet formula $N'(t)=-\kappa(t) T(t)$: $$\begin{align*} \left(1,-\frac{1}{t^2}\right)&=T(t) + \left(\frac{1}{\kappa(t)}\right)'N(t) - \frac{\kappa(t)}{\kappa(t)}T(t)\\ &= \left(\frac{1}{\kappa(t)}\right)'N(t). \end{align*}$$ We may assume that $\kappa(t)>0$ everywhere; if not, then there are points where $\kappa$ vanishes, and then the evolute is not well-defined. Therefore, taking the norm of previous equation gives $$ \left(\frac{1}{\kappa}\right)'=\sqrt{1+\frac{1}{t^4}}\\ \frac{1}{\kappa} = \int \sqrt{1+\frac{1}{t^4}}\,dt.$$ In this case it difficult to express the integral in terms of elementary functions. Moreover, I have written an indefinite integral, but we should specify the value of $\kappa(0)$ in the question, in order to set the integration constant.
Next, since $N(t)$ has unit length, we also know $$ N(t) = \frac{(1,-\frac{1}{t^2})}{\sqrt{1+\frac{1}{t^4}}}$$
Now substitute the expressions for $\frac{1}{\kappa}$ and $N(t)$ into the very first equation and obtain $$ \alpha(t) = \left(t,\frac{1}{t}\right) -\left( \int \sqrt{1+\frac{1}{t^4}}\,dt\right)\frac{(1,-\frac{1}{t^2})}{\sqrt{1+\frac{1}{t^4}}}.$$ I would like to add that a curve $\alpha$ who's evolute is $E(t)$ is called an involute of $E(t)$. More information on involutes and there geometric interpretation can be found here.