Date: 10/29/96 at 0:35:16 From: Andy Clark Subject: Minimum Distance My question deals with general applications of differentiation: At noon, ship A is 100 miles due east of ship B. Ship A is sailing west at 12 miles per hour and ship B is sailing south at 10 miles per hour. At what time are the ships nearest to each other? How close are they? I tried drawing a diagram, but my diagram is lacking information (or so it seems). Please help! Andy Clark
Date: 10/29/96 at 6:6:58 From: Doctor Kate Subject: Re: Minimum Distance Andy, I'll get you started. First, remember that what you're looking for is the time when distance is at a minimum. You should immediately look for a diagram that uses some 'x' in hours to express distances. This way you will be able to make an equation using time to express distance. So I let 'x' be the number of hours since noon. Here's my diagram: <-------------------100------------------> *<-----100-12x------>A-------------------* ^ / | / | / | / 10x / | / | / | / ' / B I'll explain the markings. The *'s represent the original positions of the boats. I've marked on the top that they were originally 100 miles apart. But 'x' hours have passed, and A has moved west and B has moved south. Now, B is 10x miles south of its original position (because it is moving at 10 miles per hour). Also, A is 12x miles west of its original position, or 100-12x miles east of B's original position (marked on the diagram) because it is moving at 12 miles per hour. Don't worry that 100 - 12x might get negative! It will just mean that A is west of B's original position (remember -2 miles east = 2 miles west). Using the Pythagorean theorem: distance^2 = (10x)^2 + (100 - 12x)^2 Wait! Don't square root - there's no reason to! You want a minimum distance, and since the distance will never be negative between A and B, distance will be a minimum when distance^2 is a minimum, and (10x)^2 + (100 - 12x)^2 is a lot easier to differentiate than the square root of all that stuff. You should be able to do the minimization stuff from here without my help, but feel free to ask if anything causes trouble for you. Hope this has helped, -Doctor Kate, The Math Forum Check out our web site! http://mathforum.org/dr.math
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