Expert: Paul Klarreich - 11/27/2005

Question
A woman at a Point A on the shore of a circular lake with radius 2 mi wants to arrive at point C diametricallly opposite A on the other side of the lake in the shortest possible time. She can walk at the rate of 4 mi/h and row a boat at 2 mi/h. how should she proceed?

Answer
Hi, Scott,

When you sent this question last November, I wrote to you:
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This is an oldie, but you probably didn't recognize it because in its original version, it was a man. (Women were not supposed to row boats alone.)

Anyway, set it up this way:

1. Make your circle with the center at the origin, O.  Put A at (-2,0) and C at (2,0), and x^2 + y^2 = 4 is the equation of the circle.

2. Pick B(x,y) on the circumference.  Hypothetically, let's put it in the first quadrant.  Label the angle BOC as theta.  (I will write t for theta.)

Now for the rowing part, directly from A to B. Use the distance formula from A(-2,0) to B(x,y):

AB = sqrt( (x + 2)^2 + y^2) = sqrt( x^2 + 4x + 4 + y^2)
  = sqrt(4x + 4 + 4)  -- remember, x^2 + y^2 = 4
  = sqrt(4x + 8) = 2sqrt(x + 2)

Rowing time = AB/2 mph. = sqrt(x + 2)

Now the walking part, along the arc from B to C.  Use the arc-length formula, s = tr.
(remember, t is theta)
What's theta in terms of x?  cos t = x/2, so  t = arccos(x/2)

So arc BC = 2t, and walking time = arc BC/4 = t/2 = (1/2)arccos(x/2)

Total time = rowing plus walking

T = sqrt(x + 2) + (1/2)arccos(x/2)

Ready to go, now.  Differentiate.  Use the formula:  D(arccos(x/a)) = -1/sqrt(a^2 - x^2)

dT       1              1
-- = ---------- - --------------
dx   2sqrt(x+2)   2sqrt(4 - x^2)

Keeping in mind that 4 - x^2 = (2 + x)(2 - x), we can combine over an LCD:

     sqrt(2 - x) - 1
  = ----------------
    2 sqrt(4 - x^2)

Just set the numerator equal to zero and solve:

sqrt(2 - x) = 1
2 - x = 1
x = 1
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That was the end of it, but it is incorrect, as another reader of this site has pointed out.  I should have continued by noting:

However, x = 1 might not actually be a minimum; it could be a maximum, so we should now check the values of the function at the endpoints.  Which, by the way, are at x=-2 and x=2.  There is no reason to assume that the point B lies in the first quadrant.

T(1) = sqrt(3) + 1/2 arccos(1/2)
 = sqrt(3) + (1/2) pi/3
 = sqrt(3) + pi/6
~~ 1.732 + 0.52 = 2.23 hours

Now to check the endpoints:  

Suppose x = 2, the right endpoint (she rows all the way; no walking)

T(2) = sqrt(4) + 1/2 arccos(1) = 2 + 0 = 2 hours

Suppose x = -2, the left endpoints(she walks all the way; no rowing)

T(-2) = sqrt(0) + 1/2 arccos(-1) = 0 + pi/2 = 1.62 hours

Yes the minimum value occurs at one of the endpoints.

Sorry for the incorrect answer the first time.

Paul