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Chapter 10: Problem 12

Given the cylinder defined by \(x^{2}+y^{2}=4\), find the path of shortest length connecting the given points. a. \((2,0,0)\) and \((0,2,5)\) b. \((2,0,0)\) and \((2,0,5)\)

Short Answer

Expert verified
The shortest path for path a is \(\sqrt{(π)^2 + 25}\) and for path b is 5.

Step by step solution

01

Path a

First, solve for the path a from \((2,0,0)\) to \((0,2,5)\). We need to solve for the shortest distance between these two points in terms of the angular component \(θ\) and the height \(z\). To get from point a to b we need to go an angle of \(π/2\) around the cylinder and up 5 units. Thus, the shortest path for a is \(d = \sqrt{((π/2)*2)^2 + 5^2} = \sqrt{(π)^2 + 25}\)
02

Path b

Next, solve for the path b from \((2,0,0)\) to \((2,0,5)\). Here, there is no change in the angular component \(θ\). The shortest path is thus only due to a change in the height \(z\), giving a shortest path \(d = 5\).
03

Final minimization conclusion

Thus the shortest path for each point in the cylinder is a distance of \(\sqrt{(π)^2 + 25}\) for path a and \(5\) for path b.

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