User Name Remember Me? Password

 Real Analysis Real Analysis Math Forum

 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. Tags geometry, line, normal Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post najaa Calculus 4 October 21st, 2012 08:20 AM dataxa Geometry 4 April 27th, 2012 06:26 AM master555 Applied Math 1 January 13th, 2012 05:04 AM master555 Applied Math 0 December 2nd, 2011 05:56 PM sixflags11804 Geometry 4 September 24th, 2008 12:12 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2019 My Math Forum. All rights reserved.      