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 October 11th, 2016, 04:14 AM #1 Newbie   Joined: Nov 2014 From: math Posts: 27 Thanks: 1 geometry - normal line Problem: Let $\displaystyle \alpha:I\to \Bbb R^2$ be a regular curve and $a$ is not a point on the curve. If there exists $\displaystyle t_0\in I$ such that $\displaystyle |\alpha(t)-a|\geq |\alpha(t_0)-a|$ for each $\displaystyle t\in I$, prove that the straight line joining the point $\displaystyle a$ with $\displaystyle \alpha(t_0)$ is the normal line of $\displaystyle a$ at $\displaystyle t_0$. The same is true if we reverse the inequality. I tried to parametrise the straight line joining the point $\displaystyle a$ with $\displaystyle \alpha(t_0)$ and the normal line of $\displaystyle a$ at $\displaystyle t_0$ and then argue that they are the same. I got $\displaystyle \beta(t)=a+t(\alpha(t_0)-a)$ and $\displaystyle \gamma(t)=a+t\alpha'(t_0)$. How can I prove that they are the same? Also, if we reverse the inequality ($\displaystyle |\alpha(t)-a|\leq |\alpha(t_0)-a|$ for each $\displaystyle t\in I$?), I don't see how the proposition is still true. I don't even know how the position of $\displaystyle \alpha(t_0)$ relates to $\displaystyle a$. Please shed some light on this, thanks!! [Sorry if the question is too elementary to be put under the thread of real analysis, I should've put it under calculus.] Last edited by ach4124; October 11th, 2016 at 04:27 AM.
 October 11th, 2016, 12:50 PM #2 Global Moderator   Joined: May 2007 Posts: 6,805 Thanks: 716 What is $\displaystyle \alpha '(t_0)$?
 October 11th, 2016, 07:16 PM #3 Newbie   Joined: Nov 2014 From: math Posts: 27 Thanks: 1 I'm not sure how to find the parametrisation of the normal line of $\displaystyle \alpha$. After sketching a graph, I can see that the normal line should be $\displaystyle \gamma(t)=a+t(\alpha(t_0)-a)$.
 October 12th, 2016, 08:45 AM #4 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 Once you see the notation, it's pretty straight-forward. $\displaystyle \alpha$(t)=r(t), position vector Eq of curve: r=x(t)i+y(t)j Distance from a to curve: |r-a| Distance from a to curve is a minimum when (d/dt)|r-a|=0 |r-a|=sqrt[(r-a).(r-a)] (dr/dt).(r-a)=0 and solve for t0. dr(t0)/dt is tangent vector to curve so r(t0)-a is normal vector to curve. Thanks from ach4124 Last edited by zylo; October 12th, 2016 at 08:48 AM.

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