for the past few days I've been studying Geometry and more specific curves in the euclidean plane. I got stuck with a problem which goes as follows:
Let $\alpha:\mathbb{R}_0^+ \rightarrow \mathbb{E}^2$ be a regular curve with strictly positive curvature $\kappa$. Let $\kappa (1) = 1$ and the evolut is given by $\gamma (t) = (cos(t),sin(t))$. Then proof that $\kappa(t) = \frac{1}{t}$.
I don't really have a clue how to approach this problem I thought of deriving the evolut to get the an equation for the derivative of the curvature but this didn't work out.
If anyone has a tip to further approach this problem it would be greatly appreciated :)).
Because you are given the equation of the evolute and asked about a property of the original curve with a certain regularity assumption, the problem suggests using the "dual" notion to evolute: an involute. Perhaps you have already heard about this concept in your study, due to its relation to that of a evolute, but otherwise the main content of the definition is summarised by saying:
A curve is the evolute of its involutes.
The reason why I wrote "dual" is precisely because the same way a curve only has one evolute, it does in general have a non-trivial family of involutes. We see precisely that in the formula for the involute $\eta(t)=(a(t),b(t))$ of a regular curve $\gamma:[a,b]\to \mathbb{R}^2, t\mapsto(x(t),y(t))$ with non-vanishing curvature:
$$a(t)=x(t)-\frac{x'(t)}{\sqrt{x'(t)^2+y'(t)^2}}\int_{t_0}^{t}\sqrt{x'(s)^2+y'(s)^2}ds$$
$$b(t)=y(t)-\frac{y'(t)}{\sqrt{x'(t)^2+y'(t)^2}}\int_{t_0}^{t}\sqrt{x'(s)^2+y'(s)^2}ds$$
with $t_0\in[a,b]$. We obtain a one-parameter family of involutes parametrised by $t_0$.
For your evolute $\gamma$, we can compute the curvature to check that it is non-vanishing (although you probably already know that for such a circle $\kappa=1$) via the formula:
$$\kappa(t)=\frac{x'(t)y''(t)-y'(t)x''(t)}{(x'(t)^2+y'(t)^2)^{3/2}}$$
Therefore, we can apply the formula for the involutes of your evolute and obtain:
$$\eta(t)=(cos(t)+(t-t_0)sin(t),sin(t) - (t-t_0)cos(t))$$
Computing the curvature of the involute of parameter $t_0$ we get:
$$\kappa_{\eta}(t)=\frac{1}{t-t_0}$$
Now it is simple: the condition $\kappa_{\eta}(1)=1$ means $t_0=0$ and this signals out one involute in the family that precisely satisfies what you asked for.
If your problem is in finding out the formula for the family of involutes, I would suggest directly searching "involute" online and you will get many good proofs and resources.