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 Post subject: Curves and a known result Posted: Wed Jul 13, 2016 6:35 pm
 Administrator  Joined: Sat Nov 07, 2015 6:12 pm
Posts: 841
Location: Larisa
Let $p, \; q$ be two points and let $\gamma$ be a curve passing through those points. ( $\gamma(a)=p, \; \gamma(b)=q$ ).

Show that if $u$ is an arbitrary unit vector then: $\gamma'(t)\cdot u \leq \left \| \gamma'(t) \right \|$.

Taking $\displaystyle u=\frac{q-p}{\left \| q-p \right \|}$ prove that the segment's lenght between the curve $\gamma$ and the two points $p, \; q$ has a length at least equal to the line distance $\left \| q-p \right \|$.

What is the conclusion you extract from the exercise?

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Imagination is much more important than knowledge. Top   Post subject: Re: Curves and a known result Posted: Wed Jul 13, 2016 6:39 pm
 Team Member Joined: Mon Nov 09, 2015 1:36 am
Posts: 460
Location: Ioannina, Greece
Hi Tolis.

There is at least one obscurity in this exercise. Writing that "the segment's length between the curve $\gamma$ and the two points $p$ and $q$" is meaningless! Perhaps you mean "the segment's length of the curve $\gamma$ between the points $p$ and $q$."

So, allow me rephrase the second part of the exercise:

2) Using that for an arbitrary unit vector $u$ we have that: $\gamma'(t)\cdot u \leqslant \left \| \gamma'(t) \right \|$ and taking $u=\frac{q-p}{ \| q-p \|}$ , prove that the segment of the curve $\gamma$ between the points $p$ and $q$ has length at least equal to the line distance $\| q-p \|$.

Proof: We give a proof of 2)* for an arbitary continuously differentiable parametric curve $\overrightarrow{\gamma}: [a,b]\subset\mathbb{R\longrightarrow \mathbb{R}}^m$:

We use that $\overrightarrow{q}-\overrightarrow{p}=\overrightarrow{\gamma}(b)-\overrightarrow{\gamma}(a)=\int_{a}^{b}{\overrightarrow{\gamma}\,'(t)\,dt}\quad (1)\,.$ Also, for an arbitrary unit vector $\overrightarrow{u}$ we have that: ${\overrightarrow{\gamma}}\,'(t)\cdot\overrightarrow{u} \leqslant \| \overrightarrow{\gamma}\,'(t) \|\,.$ So
$\int_{a}^{b}{\overrightarrow{\gamma}\,'(t)\cdot\overrightarrow{u}\,dt} \leqslant \int_{a}^{b} {\| \overrightarrow{\gamma}\,'(t) \|\,dt}=L(\overrightarrow{\gamma})\quad (2)\,,$ where $L(\overrightarrow{\gamma})$ is the segment's length of the curve $\overrightarrow{\gamma}$ between the points $\overrightarrow{p}=\overrightarrow{\gamma}(a)$ and $\overrightarrow{q}=\overrightarrow{\gamma}(b)$.
Taking $\overrightarrow{ u}=\frac{\overrightarrow{q}-\overrightarrow{p}}{ \|\overrightarrow{q}-\overrightarrow{p} \|}$ we have that \begin{align*}
\|{\overrightarrow{q}-\overrightarrow{p}}\|&= \frac{\|{\overrightarrow{q}-\overrightarrow{p}}\|^2}{\|{\overrightarrow{q}-\overrightarrow{p}}\|} \\
&=\bigl({\overrightarrow{q}-\overrightarrow{p}}\bigr)\cdot\frac{\overrightarrow{q}-\overrightarrow{p}}{\|{\overrightarrow{q}-\overrightarrow{p}}\|} \\
&= \bigl({\overrightarrow{\gamma}(b)-\overrightarrow{\gamma}(a)}\bigr)\cdot\overrightarrow{ u}\\
&\stackrel{(1)}{=}\int_{a}^{b}{\overrightarrow{\gamma}\,'(t)\,dt}\cdot\overrightarrow{u}\\
&=\int_{a}^{b}{\overrightarrow{\gamma}\,'(t)\cdot\overrightarrow{u}\,dt} \\
&\stackrel{(2)}{\leqslant} \int_{a}^{b} { \| \overrightarrow{\gamma}\,'(t) \|\,dt}\\
&\stackrel{(2)}{=}L(\overrightarrow{\gamma})\,.
\end{align*}

(*) The first part, i.e. for an arbitrary unit vector $\overrightarrow{u}$ we have that: $\overrightarrow{\gamma}\,'(t)\cdot\overrightarrow{u} \leqslant \| \overrightarrow{\gamma}\,'(t) \|$ comes out easily from the definition of the dot product.

P.S. The meaning of the exercise is to use the given suggestion to prove the desired inequality for the two lengths.
Can you give another proof of the inequality for the two lengths without this suggestion?

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Grigorios Kostakos

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