Determine the quickest route for the lifeguard to rescue the swimmer. 
Aswimmer is drowning in the ocean. Her current position is 20m off the
coast and 50m east along the coast from the lifeguard. The lifegurd
will move toward her. Use the fact that lifeguard travels faster on
land than in water. There is an optimal point at whivh the lifeguard
should inter the water and then swim to the swimmer. Suppose the life
guard runs at 2m/s and swims at 1m/s.

1. Set up a formula for the total rescue time T, as a function of the
funning distance x. Plot this function t, for x in the interval
(0,50). Estimate the value of x in the graph that minimises the rescue
time. Use the derivative to determine the optimal running distance x.
2. Repeat the process for the didtance to the swimmer being 30m
offshore and 20m east of the lifeguard. What do you notice about the
best place to start swimming in these problems?
3. Now solve the problem in general. Let the speed on water be Vw, the
speed on land be V1, the swimmer's distance from the shore be a and
the along shore distance to the lifeguard be b.
4. Use tan(otheater)as a function of x. Use this to find the angle
which will minimise the time needed to reach the swimmer. What
constants does this angle depend on?

Dear lowt,


1.

Since the lifeguard moves at 2m/s on land, his running time is x/2. To
compute his swimming time, we consider the swimming distance, which is
the hypotenuse of a right triangle with legs of length 20 and 50-x. The
sum of these is the total lifesaving time.

  T(x)  =  x/2 + sqrt(x^2 - 100x + 2900)

The interval (0, 50) of T(x) is shown in the following plot. It appears
that T is minimized near the value x = 38.

http://plg.uwaterloo.ca/~mlaszlo/answers/lifeguard.1.png

To confirm this estimate, let us find the first derivative of T(x).

  T'(x)  =  1/2 + (x-50)/sqrt(x^2 - 100x + 2900)

Now let us set T'(x) to 0 and solve for x.

  1 + (2x-100)/sqrt(x^2 - 100x + 2900)  =  0

  100 - 2x  =  sqrt(x^2 - 100x + 2900)

  10000 - 4x^2  =  x^2 - 100x + 2900

  3x^2 - 300x + 7100  =  0

  x  =  (300 +/- sqrt(90000 - 4*3*7100))/6

The solutions of this quadratic equation are 61.55 and 38.45, of which
only the latter is feasible. Hence, the optimal running distance is
38.45 meters.


2.

The lifeguard's running time is again x/2. His swimming distance is now
the hypotenuse of a right triangle with legs of length 30 and 20-x.

  T(x)  =  x/2 + sqrt(x^2 - 30x + 1300)

The interval (0, 20) of T(x) is shown in the following plot. It appears
that T is minimized near the value x = 3.

http://plg.uwaterloo.ca/~mlaszlo/answers/lifeguard.2.png

The first derivative is the following.

  T'(x)  =  1/2 + (x-20)/sqrt(x^2 - 40x + 1300)

We set it to 0 and solve for x.

  1 + (2x-40)/sqrt(x^2 - 40x + 1300)  =  0

  40 - 2x  =  sqrt(x^2 - 40x + 1300)

  4x^2 - 160x + 1600  =  x^2 - 40x + 1300

  3x^2 - 120x + 300  =  0

  x  =  (120 +/- sqrt(14400 - 4*3*300))/6

The solutions are 37.32 and 2.68, of which only the latter is
feasible. The optimal running distance in this case is therefore 2.68
meters.

It appears that the best place to start swimming is one that strikes
the same angle with respect to the shore. Consider that

  20/(50-38.45)  =  1.73

is the slope of the swimming segment in the first case and

  30/(20-2.68)  =  1.73

in the second as well.


3.

In the general case, the lifeguard's running time is b/Vl. His swimming
distance is the hypotenuse of a right triangle with legs of length a
and b-x.

  T(x)  =  b/Vl + sqrt(x^2 - 2bx + b^2 + a^2)/Vw

The first derivative is as follows.

  T'(x)  =  1/Vl + (x-b)/(Vw*sqrt(x^2 - 2bx + b^2 + a^2))

Let us solve for T'(x) = 0.

  Vl*b - Vl*x  =  Vw * sqrt(x^2 - 2bx + b^2 + a^2)

  Vl^2*x^2 - 2Vl^2*bx + Vl^2*b^2  =  Vw^2*x^2 - 2Vw^2*bx + Vw^2*a^2 + Vw^2*b^2

  (Vl^2 - Vw^2)x^2 + 2(Vw^2 - Vl^2)bx - Vw^2*a^2 + (Vl^2 - Vw^2)b^2  =  0

To simplify this quadratic equation, let us introduce a constant

  k  =  Vl^2 - Vw^2.

We observe that the equation is of the form

  ax^2 + bx + c  =  0

where

  a  =  k

  b  =  -2kb

  c  =  kb^2 - Vw^2*a^2.

We can substitute these values into the expression

  (-b +/- sqrt(b^2 - 4ac))/2a

to obtain the solution

  (2kb +/- sqrt(4k^2*b^2 - 4k(kb^2 - Vw^2*a^2))/2k

  =  (2kb +/- sqrt(4k^2*b^2 - 4k^2*b^2 + 4kVw^2*a^2))/2k

  =  (2kb +/- 2Vw*a*sqrt(k))/2k

  =  b +/- Vw*a/sqrt(k).

Since the optimal value of x must not exceed b, the rescue time is
optimized by the following.

  x  =  b - Vw*a/sqrt(Vl^2 - Vw^2)


4.

The tangent of the swimming angle with respect to the shore is simply
the rise of the right triangle divided by its run.
  
  tan(theta)  =  a/(b-x)

We now substitute the optimal value of x and solve for theta.
  
  tan(theta)  =  a/(b - (b - Vw*a/sqrt(Vl^2 - Vw^2)))

  tan(theta)  =  a/(Vw*a/sqrt(Vl^2 - Vw^2))

  tan(theta)  =  sqrt(Vl^2 - Vw^2)/Vw

  theta  =  atan(sqrt(Vl^2 - Vw^2)/Vw)

So the angle depends only on Vl and Vw, which are the lifeguard's speed
on land and water respectively.
  
This verifies our earlier observation that the angle of departure from
the shore was the same in both cases. The exact angle was

  atan(sqrt(2^2 - 1^2)/1)  =  atan(sqrt(3)) 

  =  1.047 radians

  =  1.047/pi * 180  =  60 degrees.

  
If you find fault with any part of my answer, do let me know through
a Clarification Request so that I may fully meet your needs before you  
assign a rating.

Regards,
  
leapinglizard

Just a note to point out that this is a classic problem that is
usually encountered when learning about optics, specifically, Snell's
Law of refraction.

The lifeguard is an analog for a light ray, which, according to
Fermat's Least Time Principle, will take a path between two points
that minimizes the travel time.  If the light crosses a boundary
between two different media that have different refractive indices (n1
and n2)then:

    n1*sin(w1) = n2*sin(w2)

where w1 is the angle between the light ray measured and the normal to
the interface in medium 1, and w2 is the angle between the light ray
in the second medium and the normal to the interface in medium 2.

The index of refraction is a measure of the speed of light relative to
its speed in a vacuum:  n = c/v, where v is the speed of light in a
medium, and c is the speed of light in a vacuum.  Plugging this into
Snell's Law to express it in terms of velocities instead of refractive
indices yields:

    v2*sin(w1) = v1* sin(w2)

There are lots of web pages that discuss this analogy, i.e.,
<http://smccd.net/accounts/goth/courses_s_2003/phys_221/phys_221_lectures/phys_221_snell.pdf>.
 Do a Google search for Snell's Law lifeguard and you'll get quite a
few hits.