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### Find the Path and the Time Taken

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Date: 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
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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

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/
```
Associated Topics:
High School Calculus

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