This exercise is from Do Carmo book (ex $2$, p:$5$)
Let $\alpha(t)$ be a parametrized curve which does not pass through the origin. If $\alpha(t_0)$is a point of the trace of $\alpha$ closest to the origin and $\alpha^{'}(t)\neq 0$ , show that the position vector $\alpha(t_0)$ is orthogonal to $\alpha^{'}(t_0)$.
I have tried to solve it but I cannot understand , and also why we have to consider "$\alpha(t_0)$ is a point of the trace of $\alpha$ $\underline{ closest}$ to the origin" ? Thank you in advance.
If we have the point $t_0$ at which the distance to the origin is minimal, then the derivative of $\bigl\|\alpha(t)\bigr\|^2$ at $t_0$ is $0$. But$$\left(\bigl\|\alpha(t)\bigr\|^2\right)'=2\alpha(t).\alpha'(t).$$So, $\alpha(t_0).\alpha'(t_0)=0.$