Tangent of evolute and singed curvature This is an exercise from differential geometry textbook by Do Carmo. Let $\alpha:I\to \mathbb{R}^2$ be a regular parametrized plain curve (arbitrary parameter), define $n=n(t)$ and $k=k(t)$, where $k$ is the signed curvature. Assume that $k(t)\neq 0,t\in I$. In this situation, the curve $$\beta(t)=\alpha(t)+\frac{1}{k(t)}n(t),\quad t\in I,$$ is called the evolute of $\alpha$. Show that the tangent at $t$ of the evolute of $\alpha$ is the normal to $\alpha$ at $t$. My professor uses another book as our textbook. In that book, we only define positive curvature for curve. Hence I am a little confused about the sign. I can show the tangent of evolute is $\frac{(\frac{1}{k})^{'}}{| (\frac{1}{k})^{'} |}n(t)$, but I don't know how to cancel the absolute value.

You can restrict yourself to an arc length parametrization of the curve $\alpha$ (ask for details if it is not clear why this is!).

Then compute $$\beta'(t)=\alpha'(t)-\frac{k'(t)}{k(t)^2}n(t)+\frac{1}{k(t)}n'(t).$$ To answer your question you need to show $\langle \beta'(t),\alpha'(t)\rangle=0$. This follows directly if you use $\langle \alpha''(t),n(t)\rangle =- \langle \alpha'(t),n'(t)\rangle$ which follows from differentiating $\langle \alpha'(t),n(t)\rangle =-0$ with respect to $t$. Then use $\alpha''(t)=k(t)n(t)$.

EDIT: What do you mean by "the normal"? EDIT: Added missing sign.

EDIT: I obtain in my way $\beta'(t)=-\frac{k'}{k^2}n(t).$