Shortest Distance between a Point and a CircleDate: 1/22/96 at 10:14:18 From: Anonymous Subject: Calculus Problem: Find the shortest distance between the point (4,5) and the circumference of x^2 + y^2 = 9. The problem involves finding the derivative and then letting that equal zero to find the minimum. Where I'm stuck is I don't know how to set up the relationship between the circle and the point so that I get an equation to derive. Please help! Date: 5/30/96 at 14:54:23 From: Doctor Charles Subject: Re: Calculus Well you CAN use calculus. Here's the technique: Use polar coordinates to cut down x and y to one variable. Any point on the circle has coordinates (3 cos theta, 3 sin theta) for some theta. Then let L be the square of the distance of (4,5) from this point: L = (3*(cos theta) - 4)^2 + (3*(sin theta) - 5) ^2 Using (cos^2 theta + sin^2 theta = 1) you should get: L = 50 - 24 cos theta - 30 sin theta Differentiate with respect to theta and get that tan theta = 4/5. This will actually give you two values of theta. One is the point on the circle furthest from (4,5) and the other gives you the one nearest (4,5). Substitute these into the formula of L to get the square of the distances. Choose the smallest! Without using calculus: Note that (4,5) is sqrt(4^2 + 5^2) units away from the origin. The circle is the set of points 3 units from the origin. If you draw a line from the origin to (4,5) then it will cut the circle at the point closest to (4,5). The length of the line from the circle to (4,5) must be: sqrt(4^2 + 5^2) - 3 = sqrt(41) - 3 The corresponding point on the circle has coordinates (a,b) given by: (a,b) = (3/sqrt(41)*4,3/sqrt(41)*5) = (12/sqrt(41) , 15/sqrt(41)) This is much simpler than the calculus method, although both will give the same answer. -Doctor Charles, The Math Forum |
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