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Let $c: [0,1] \rightarrow \mathbb{R}^{n}$ be a $C^1$ curve. Suppose $c$ passes through two points $c(x_1), c(x_2) \in \mathbb{R}^{n}$ with $0 < x_1 < x_1 < 1$. Does there exist a point $x_0 \in (x_1, x_2)$ such that $Dc(x_0)$ lies on the line $c(x_2)-c(x_1)$? I'm not sure if my question makes sense as written, let me know if it doesn't! I will provide some intuition/example for what I want below.

This requirement seems much less strict than let's say the false, directly generalized MVT in this question: counterexample for direct generalization of the one-dimensional Mean Value Theorem. In the case of the circle, as in the example in the answer to the above question, one could just slide the secant line determined by $c(x_1)$ and $c(x_2)$ until it became a tangent line, which would correspond to the point $x_0$ we are looking for. In $\mathbb{R}^{3}$ this becomes a bit more confusing. For instance, the curve $c$ can move away from being "above" the segment $c(x_1)c(x_2)$. However, in that case what I'd like to say is something along the lines of: suppose the origin, $c(x_1)$, and $c(x_2)$ are not collinear. Then the origin and $c(x_1)$, the origin and $c(x_2)$ determine a plane. Does there exist a time point $x_0$ such that $Dc(t_0)$ lies in that plane? In general, the idea is similar to the mean value theorem, but most generalizations deal with the magnitude of the derivative, whereas here we just want the tangent to be parallel to the secant.

This is probably very simple but given the discussion above, I have trouble formulating exactly what I want.

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  • $\begingroup$ "two points "$c(x_1),c(x_1)$" -- typo? Second subscript $2$ instead? $\endgroup$ – coffeemath Mar 8 at 12:25
  • $\begingroup$ @coffeemath thanks I fixed it $\endgroup$ – forgotcalc Mar 8 at 20:22
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For $\mathbb R^3$ this is true. Pick wlog basis in which endpoints of your $C^1$ curve lie in $z=0$ plane. In this case the 'tangent parallel to the plane' condition is equivalent to just having last component zero which is true by e.g. Rolle's theorem.

For $\mathbb R^n$ for $n>3$ it is not true which is easy to see from the previous example. Take curve with endpoints $(1,0,0,0)$, $(0,1,0,0)$ and in the first two coordinates you take this curve to follow rectlinear motion between the endpoints. In the last two coordinates pick two functions with different locations of stationary points and zeros at endpoints giving you the counterexample.

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Consider the helix $$\gamma:\quad t\mapsto c(t):=(\cos t,\sin t, t)\qquad(0\leq t\leq 2\pi)\ .$$ Then $c(2\pi)-c(0)=(0,0,2\pi)$, but there is no point on the curve $\gamma$ with $c'(t)=(-\sin t\cos t, 1)$ parallel to the $z$-axis.

A remark: One sentence in your question says: "Does $c'(t)$ lie in that plane for some $t\,$?" Note that $c'(t)$ lying in some plane is asking much less than asking for $c'(t)$ being parallel to some vector.

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  • $\begingroup$ Thanks for the counterexample! Yes, what if I want containment in a plane? In the case of this example, I'm not sure which plane I would like exactly (since the two points $c(2\pi)$ and $c(0)$ are collinear with $0$. But suppose they weren't, could we get containment in the plane? $\endgroup$ – forgotcalc Mar 8 at 20:19

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