Find the Path and the Time TakenDate: 10/17/2001 at 07:29:43 From: Mike Poulton Subject: Vectors and calculus A man wants to cross a river 500m wide. His rowing speed (relative to the water) is 3000 m/hr. The river flows at a speed of 1200 m/hr. If the man's walking speed on shore is 5000m/hr, find (a) the path (combined rowing and waliking) he should take to get to the point directly opposite his starting point in the shortest possible time, and (b) how long does it take? Answer: 29 degrees upstream and 11 minutes. I have done this problem and arrived at the answers from the text as given above. But, using simple vectors, an angle of 23.6 degrees upstream would produce a resultant velocity at right angles across the river, arriving at the destination without having to walk! I cannot see my mistake. Cheers. Mike Date: 10/17/2001 at 15:23:57 From: Doctor Peterson Subject: Re: Vectors and calculus Hi, Mike. I don't know just what method you used to get the answers, but I get your angle. However, you seem to have used the wrong method to get the right answer, and the right method to get the wrong answer! I took my variable to be theta, the angle of the heading of the boat relative to a course across the river (with a positive angle going upstream), and found the time for the crossing (in hours), including rowing and walking from the landing point to the target point, to be T = sec(theta)/6 + |sec(theta)/25 - tan(theta)/10| = (25 + |6 - 15 sin(theta)|) sec(theta)/150 The absolute value is needed because the walk may be in either direction from the landing point, but the time is positive either way. I'm wondering if you and the book may have neglected this, because this oversight explains everything you told me. Using the derivative, I find that T is minimized at theta=arcsin(15/31) for the case where the argument of the absolute value is positive, and at theta=arcsin(-15/19) when it is negative. But each of these minima is outside the region where it is valid; T decreases monotonically to the point where the absolute value is zero, then abruptly changes direction and increases from there. In other words, the actual minimum is where the function is not differentiable, when 6 - 15 sin(theta) = 0, so that theta=arcsin(2/5). This gives your 23.6 degrees, with a time of 10.9 minutes. At the false minimum of arcsin(15/31) = 28.9 degrees, the time would be 12 minutes using the absolute value, but 10.9 minutes using the function for which it is a minimum. The problem with this "minimum" is that it requires a negative time for walking. So it looks as if they (and you when you followed their method) used that wrong minimum. This is equivalent to the common error of neglecting to check the ends of the domain for minima (or even to recognize the restriction of the domain). But it is not obvious without doing the calculus that the shortest route is in fact the one that involves no walking. Once you know that, simple vector work gives the correct answer; but there's more to it than that. That's why I say that your method is not strictly correct, though it gives the correct result. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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