Minimum DistanceDate: 02/07/98 at 22:52:04 From: Jacob Putnam Subject: Minimum Distance What is the minimum possible distance between a point on the graph of x^2+y^2 = 1 and a point on the graph of xy = 1? Jacob Putnam Date: 02/25/98 at 14:07:38 From: Doctor Sonya Subject: Re: Minimum Distance Hi Jacob. This is a very good question - the easiest way that I can see to solve it involves some geometry and calculus. First, remember that the distance between any two points (x1, y1) and (x2, y2) in the plane is: sqrt[(x1 - x2)^2 + (y1 - y2)^2] Note that sqrt means 'square root of' and x^2 means x to the power 2. Now (and this is clearer if you draw a diagram) the distance from (u,v) to the given circle is the length of the shortest perpendicular line segment from (u,v) to the circle. But if this line segment is perpendicular to the circle, it will also pass through the origin. Make sure you see why this is so. What this means is that we can find the distance from the point (u, v) to the origin, and then subtract 1 from it to get the distance from (u, v) to the circle. From this you should be able to figure out the distance from an arbitrary point to the unit circle. However, you want more! You want the point to be on the graph of the function xy = 1. Now, if xy = 1, the y = 1/x. So plug the point (x, 1/x) into the equation you derived for the distance between any point and the unit circle. Set this value equal to f(x), and you'll have a function the describes the distance from the graph xy = 1 to the unit circle. Not bad, eh? Now that you have this function f(x), you need to find its minimum value; this will be the minimum distances you're looking for. Here's a hint: the function has two minimum values, one of which is x = 1. -Doctors Sonya and Martin, The Math Forum Check out our Web site http://mathforum.org/dr.math/ |
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