Where Will the Runners Meet?Date: 03/29/99 at 05:09:43 From: Jason Mize Subject: Runners and their Intersection Point Two runners, A and B, start 90 degrees away from each other on a circular track and run at the same speed. If Runner B decides to cut across the track, where will they meet? I have been given that the radius of the circular track is one mile. What do I need to do to figure out where on the track they will meet? I understand that C = 2(pi)r, in this case, 2(pi). Therefore, each quarter or quadrant's arc is equal to .5(pi). Can you help me solve this one? Date: 03/29/99 at 10:37:41 From: Doctor Rob Subject: Re: Runners and their Intersection Point I'll start by drawing a picture: A _..--+--.._ .-' | `-. ,' | `. ,' | `. / | \ / | \ . | . | |O | +--------------+-------------,+ B | / \ ,-' | . / \ ,-' . \ / \ ,-' / \ / ,-o / `. / ,-' Q ,' `. / ,-' .' `/._ _.-' P ''-----'' Measure all angles in radians. Then let <AOP = theta <BOP = 3*Pi/2 - theta <BOQ = <POQ = 3*Pi/4 - theta/2 BQ = QP = sin(3*Pi/4 - theta/2) arc(AP) = theta BP = 2*sin(3*Pi/4 - theta/2) According to the conditions of the problem arc(AP) = BP, so theta = 2*sin(3*Pi/4 - theta/2) theta/2 = sin(3*Pi/4 - theta/2) Now I let x = theta/2. So, x = sin(3*Pi/4 - x) This is not solvable explicitly for x. One can, of course, solve it numerically. One pretty quick way is to guess the value of x (I started with x[1] = 1.), and use the above formula to compute the next guess: x[n + 1] = sin(3*Pi/4 - x[n]), n > 0. This kind of iteration is easy to do on a pocket calculator. Continue until x[n + 1] - x[n] is smaller than the error you can tolerate. Then the value of x[n + 1] is equal to the solution x, to within the tolerance. This converges pretty rapidly to the solution x, and then theta = 2*x gives you the distance in miles each traveled. If you put the origin of your coordinate system at the center of the circle, with A = (0, 1) and B = (1, 0), then the place they meet is P = (cos(theta + Pi/2), sin(theta + Pi/2)) = (-sin(theta), cos(theta)) - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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