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August 8th, 2017, 02:43 PM   #1
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Arrow 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.
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August 15th, 2017, 05:41 AM   #2
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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.
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August 15th, 2017, 01:50 PM   #3
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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.
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August 19th, 2017, 11:05 AM   #4
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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
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August 22nd, 2017, 04:26 AM   #5
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Quote:
Originally Posted by greg1313 View Post
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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