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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.

Bernard
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Bernstein
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    Being closest means (roughly) it is trying to "turn around" at that point. To see why this matters, consider the case where $\alpha$ is a straight line avoiding the origin. Notice that the perpendicular from the origin to the line occurs where the line is closest to the origin. This problem is the curvy generalization of this fact. – Randall May 15 '18 at 16:40
  • thank you @Randall ,could you explain to me your answer better, because my english is a beginner . for me closest to the origin thats mean not far in a topological why ..., can you give me an examples where it works and where it does not work – Bernstein May 15 '18 at 16:56
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    Consider a closed loop. Let $ \alpha(t) \cdot \alpha^{'}(t) $ be dot product of position vector which is making an angle to tangent vector. When it vanishes it is either nearest or farthest from the origin when dot product vanishes. – Narasimham May 16 '18 at 03:15

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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, $\alpha(t_0).\alpha'(t_0)=0.$

José Carlos Santos
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