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Consider the curve $C=\bigl\{(t,|t|): t \in \mathbb R\bigr\}$.Show that if $\alpha:\bigr]a,b\bigr[ \to \mathbb R$ is a differentiable curve whose trace lies in C and $t_0 \in \bigr]a,b\bigr[$ is such that $\alpha(t_0)=(0,0)$ then $\alpha'(t_0)=(0,0)$.

How can one assure that there will be points in the neighborhood of $t_0$ such that $\alpha_1$ is negative and $\alpha_1$ is non-negative? If one can assure that, we can verify that $\alpha'(t_0)=(0,0)$ by studying the derivative of $\alpha$ at $t_0$ by the definition of limit and consider both cases with $\alpha_1$ non-negative or $\alpha_1$ positive.

J P
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  • What is $a_1$? The interval $]a,b[$ could be far from $0$, for example, it could be $t_0 \in ]1,2[$. In that case you definitely won't find negative numbers in a neighborhood of $t_0$. Fortunately for you, that's not what the problem is about. – jjagmath Jul 21 '23 at 19:43
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    @AnneBauval My comment was about the question before it was edited. – jjagmath Jul 21 '23 at 21:37
  • Your approach won't work because it could be the case that the coordinates of $\alpha$ are never negative. Consider, for example, $\alpha:; ]-1,1[ \to C$ given by $\alpha(t) = (t^2,t^2)$. – jjagmath Jul 21 '23 at 21:41

1 Answers1

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Let $\alpha=(\alpha_1,|\alpha_1|)$ and $\alpha'(t_0)=(a,b).$

As $t\to t_0,$

$$\alpha_1(t)=a(t-t_0)+o(t-t_0)\quad\text{and}\quad|\alpha_1(t)|=b(t-t_0)+o(t-t_0)$$

hence $$b\left(t-t_0\right)=|a|\left|t-t_0\right|+o(t-t_0),$$ or equivalently $$b=\lim_{t\to t_0}\left(|a|\frac{\left|t-t_0\right|}{t-t_0}\right),$$ which implies $$a=b=0.$$

Anne Bauval
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