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Let $p,q∈R^n$ and let $\gamma$ be a curve such that $\gamma(a) = p, \gamma(b) = q$, where $a$ < $b$.

(a) Show that, if $\mathbf u$ is a unit vector, then $$\dot\gamma \cdot \mathbf u\leq \|\dot\gamma\|$$

(b) Show that $$(q - p)\cdot\mathbf u ≤ \int_b^a\|\dot\gamma\|\,dt$$ (c) Show that the arc length of $\gamma$ from $\gamma(a)$ to $\gamma(b)$ is at least $\|q - p\|$, with equality when $\gamma$ is a straight line.

This is what I have worked out so far, for (b): $$(q - p)·\mathbf u = (\gamma(a) - \gamma(b))·\mathbf u$$ $$=\int_b^a\dot\gamma\cdot\mathbf u\,dt$$ and thus, using part (a): $$\int_b^a\dot\gamma\cdot\mathbf u\,dt\leq \int_b^a\|\dot\gamma\|\,dt$$

and for (c), as $\mathbf u$ is a unit vector we use the equation: $$\mathbf u = \frac{(q - p)}{\|q - p\|}$$ from here you can see that arc length of $\gamma$ from $\gamma(a)$ to $\gamma(b)$ is at least $\|q - p\|$, but I'm not sure what working to show or if I'm even doing this right.

Thomas Andrews
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Emily
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  • It seems like you've got your $a,b$ mixed up in the problem, given that $a<b$ at the top. – Thomas Andrews Aug 12 '14 at 14:16
  • In particular, $\int_b^a |\dot\gamma|,dt\leq 0$ if $b<a$, but we can easily pick $\mathbf u$ so that $(q-p)\cdot \mathbf u>0$ if $q\neq p$. – Thomas Andrews Aug 12 '14 at 14:20
  • Sorry to revive this question, but can OP (or someone else) explain how exactly (c) was answered? I'm confused how we use that value of u to determine the arc length is at least $||q-p||$ – Vishwa Iyer Sep 05 '16 at 23:52

1 Answers1

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The first parts look good (up to Thomas Andrews' comment). Are you asking if you're done? If so, you should just notice that the length of the line from $\gamma(a)$ to $\gamma(b)$ is exactly $\lVert \gamma(a) - \gamma(b) \rVert$. You've shown that the length of any curve has to be at least this number so in particular, any curve has to be at least as long as the line.

fuglede
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  • OP is missing the part that proves that only a line gives equality. – Thomas Andrews Aug 12 '14 at 14:31
  • @ThomasAndrews: There is no "only" in the wording of the problem though. If I would have to guess, the problem comes from Exercise 1.2.4 of Pressley's book "Elementary Differential Geometry". Judging by the author's own solution manual, leaving it out is intended. – fuglede Aug 12 '14 at 14:42
  • Ah, yes, I read it as "only" since that's the more interesting result :) – Thomas Andrews Aug 12 '14 at 14:56