Expert: Paul Klarreich - 5/25/2007

Question
I haven't been able to find an example to help me with this problem.  A woman at a point A on the shore of a circular lake with radius 2mi wants to arrive at the point C diametrically opposite A on the other side of the lake in the shortest time possible.  She can walk at the rate of 4mi/hr and row a boat at 2mi/hr.  How should she proceed?  Thanks!!!

Answer
Questioner:   Jill
Category:  Calculus
 
Subject:  optimization
Question:  I haven't been able to find an example to help me with this problem.  

>> Neither have I.  I'm sure I answered it for someone else a while back but I can't find it.  Ah, well....

A woman at a point A on the shore of a circular lake with radius 2mi wants to arrive at the point C diametrically opposite A on the other side of the lake in the shortest time possible.  She can walk at the rate of 4mi/hr and row a boat at 2mi/hr.  

>> She's in good shape.  I'd like to meet her.

How should she proceed?  Thanks!!!
........................................
Hi, Jill,

I suggest that we set it up like this: (You will have to draw the diagram according to these instructions.  The site does not allow me to send you a picture.  Bummer.)

The center of the circle is at the origin (0,0), and its radius is 2.
A is the point  (-2,0).
C is the point   (2,0).
AC = 4 miles.
She rows in a straight line from A to some point B on the circle, then walks along  arc BC.

Call angle BAC = t.   (t stands for theta)
Note that angle ABC = 90 degrees.

She rows AB at 2 mph, so that takes time  t1 = AB/2.  We must compute the length AB.

She walks arc BC at 4 mph. That takes time  t2 = arc BC / 4.  We must compute the length of arc BC.

The sum  T = t1 + t2  is the time we want to minimize.  We will attempt to make this a function of angle t.

[Note, of course -- capital T is not the same as small t.]

...........................

In triangle  ABC, the hypotenuse is AC = 4.  AB is the leg adjacent to angle t, so

AB = 4 cos t.
t1 = AB/2 = 2 cos t

...........................
In the circle, arc length = radius * central angle.  The central angle for arc BC is angle  BOC and that is equal to 2t.  [Draw isosceles triangle ABO, then use that old theorem: An exterior angle of a triangle is equal to the sum of the remote interior angles.]

So  arc BC = 2 * 2t = 4t

t2 = 4t/4 = t.

..............................
Getting there:

T = t1 + t2 = 2 cos t + t

Interval of meaningfulness: [I just made that word up.]

t = 0 (all rowing) to t = 90 degrees, or pi/2 (all walking)

Ready to do our maximizing.  Differentiate:

dT/dt = - 2 sin t + 1

Set that equal to zero:

sin t = 1/2

t = 30 degrees, or pi/6, is a critical point:

Checking:  We have three critical points; that one and the endpoints.

T(30) = 2 cos 30 + pi/6
= 2 sqrt(3)/2 + pi/6
= sqrt(3) + pi/6 = 2.2556495831671761666045535720525

T(0) = 2 cos 0 + 0 = 2

T(90) = 2 cos 90 + pi/2 = 1.5707963267948966192313216916398

Aha!  The critical point at 30 degrees is not a minimum.  The minimum is at 90 degrees, meaning she should skip the rowing and just walk all the way.