My Math Forum Optimization of lengths in space
 User Name Remember Me? Password

 Real Analysis Real Analysis Math Forum

 August 8th, 2017, 02:43 PM #1 Newbie   Joined: Aug 2017 From: Costa Rica Posts: 2 Thanks: 0 Optimization of lengths in space i've been presented a problem of calculus in R3, i need help ASAP ! here's the problem 1.- Let S be a cylindrical surface in space. Consider two different points, A and B in space, none of them in S. Find a point P over S so that the addition of the lengths of the segments AP and PB is minimal. 2.- Enunciate possible extentions to part 1, propose solutions.
 August 15th, 2017, 05:41 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,731 Thanks: 705 It looks like the setup, at least, is pretty straight forward. We can select a coordinate system in which the z-axis is the central axis of the cylinder and the intersection of the cylinder with the xy-plane is the circle $\displaystyle x^2+ y^2= R^2$. We can use the angle, $\displaystyle \theta$, as parameter for the circle to write $\displaystyle x= R cos(\theta)$ and $\displaystyle y= R sin(\theta)$. We can take $\displaystyle z= t$ as the other parametric equation. Now, let one of the two points not on that cylinder be $\displaystyle (x_0, y_0, z_0)$ and the other point $\displaystyle (x_1, y_1, z_1)$. The distance from the first point to a point on the cylinder is $\displaystyle \sqrt{(R cos(\theta)- x_0)^2+ (R sin(\theta)- y_0)^2+ (t- z_0)^2}$ and the distance from that point on the cylinder to the second point is $\displaystyle \sqrt{(R cos(\theta)- x_1)^2+ (R sin(\theta)- y-1)^2+ (t- z_1)^2}$. The total distance is the sum of those, $\displaystyle \sqrt{(R cos(\theta)- x_0)^2+ (R sin(\theta)- y_0)^2+ (t- z_0)^2}+ \sqrt{(R cos(\theta)- x_1)^2+ (R sin(\theta)- y_1)^2+ (t- z_1)^2}$. Find values of t and $\displaystyle \theta$ that minimize that.
 August 15th, 2017, 01:50 PM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,593 Thanks: 937 Math Focus: Elementary mathematics and beyond The shortest distance between two points is a straight line (triangle inequality). Choose P so that A, P and B are collinear and P is between A and B.
 August 19th, 2017, 11:05 AM #4 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,158 Thanks: 90 This is a problem in Lagrange multipiers. Minimize $\displaystyle f(x,y,z)=P_{0}P+PP_{1}$ subject to $\displaystyle g(x,y,z)=x^{2}+y^{2}-r^{2}=0$ let $\displaystyle h=f+\lambda g$ $\displaystyle \frac{\partial h}{\partial x}=0, \frac{\partial h }{\partial y}=0,\frac{\partial h}{\partial z}=0, g=0$ 4 eqs in 4 unknowns
August 22nd, 2017, 04:26 AM   #5
Math Team

Joined: Jan 2015
From: Alabama

Posts: 2,731
Thanks: 705

Quote:
 Originally Posted by greg1313 The shortest distance between two points is a straight line (triangle inequality). Choose P so that A, P and B are collinear and P is between A and B.
You might not be able to do that. It might be that no point on the cylinder lies on line segment AB.

 Tags lengths, optimization, space

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Inkaho Trigonometry 2 May 13th, 2015 08:03 PM Zman15 Geometry 2 February 22nd, 2015 03:21 PM zgonda Algebra 9 August 27th, 2010 02:18 PM c8h8r8i8s8 Algebra 1 March 1st, 2009 10:51 AM SecretSamurai Algebra 6 February 16th, 2009 04:21 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2017 My Math Forum. All rights reserved.