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 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: 2,966 Thanks: 807 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: 7,761 Thanks: 1009 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: 1,290 Thanks: 93 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
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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.

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