Hi RKL, I just posted a solution to this problem at my site,
http://mathhelpnow.blogspot.com
, that avoids the use of quadratic
inequalities through the use of tangency and continuity property of
curves. Here is the stripped down, text-only version of what I posted:
The way I understand it, the situation being described can be visually
represented by the following diagram:
Where x is the distance he runs on land, and y is the distance he
swims through water.
So by Pythagoras' theorem, (150-x)^2 + 60^2 = y^2, or f(a,b) := (150-
a)^2 + 60^2 - b^2, and f(x,y) = 0.
On the other hand, we are told the lifeguard runs 8 m/s on land. This
implies that the time he spends on land is (Distance/Rate), which is x/
8.
Similarly, the time he spends in water is y/2.
Thus, his total time spent getting to the swimmer is T = x/8 + y/2.
Let g(a,b,T)= a/8 + b/2 - T.
Now, going from here, there are several ways to approach this. We
could solve for one variable then find the minimal value of a
quadratic equation, or we could use a more conceptual approach that
utilizes properties of curves, such as tangents. We will go with the
latter because it yields greater insight.
As stated before, the allowable values of (x,y) lie on the curve
f(x,y) = 0. In conics, we learn that this is a hyperbola (with one
"bowl" pointing up and one "bowl" pointing down). We are only
concerned with positive values of x and y, so we are only concerned
with the bowl that is pointing upwards. It looks something like this:
Now, we want the minimum value of T such that g(a,b,T) = 0 intersects
f(a,b) = 0. If we find this T, we can let (x,y) be one of the
intersections.
If the line is not tangent to the curve, then the line splits the
plane into two halves, and one of those halves will correspond to
lower values of T, and bits of the curve will be present on both sides
of the line. That is, we could lower T so slightly, that after
****fting the line correspondingly, it will still intersect the
hyperbola.
In other words, if T is a minimum then the line must be tangent to the
hyperbola.
You can compute the tangent vector to f(x,y)=0 by computing its
gradient, then finding the vector perpendicular to its gradient. If
you have not studied calculus before and are unfamiliar with the
gradient, that is okay, but this will give you an incentive to learn
it eventually!
So, taking the necessary partial derivatives, we see that the gradient
of f(x,y) is (2x-300, -2y) = (x-150. -y). The perpendicular to (a,b)
is (-b,a), thus the perpendicular to the gradient (a.k.a. the tangent)
is (y, x-150).
Let g_T(x,y) = g(x,y,T). The gradient of g_T(x,y) is (1/8,1/2), so the
tangent is (-1/2,1/8). We can multiply this by 8 to get another
tangent vector, (-4,1).
For g_T(x,y) to be tangent to f(x,y) means that the tangent vectors we
computed are multiples of each other, e.g. (-4*m, m) = (y, (x-150))
for some number m.
This means that y = -4m and x = m + 150. y > 0 implies that m <>.
Then we must have 0 = f(x,y) = f(-4m, 150 + m) = (150 - (150+m))^2 -
(-4m)^2 + 60^2
which is equivalent to
0 = m^2 - 16m^2 + 60^2
equivalent to
m^2 = 240. Remember, m <0.
Hence, m = -sqrt[240] = -4*sqrt[15]
Thus, the minimum time T must be:
T = x/8 + y/2 = (150 - m)/8 - 4m/2 = 15*(10-m)/8 = 15*(10 -
4*sqrt[15])/8
= 15*(5 - 2*sqrt[15])/4
On Apr 6, 3:17 pm, RKL <mary...@[EMAIL PROTECTED]
> wrote:
> Here's an interesting one:
>
> A lifeguard stationed on a beach with a straight shoreline sees a
> swimmer in trouble 150 meters down the beach and 60 meters out in the
> water.
>
> The lifeguard can run 8 m/s on the beach and swim 2 m/s in the water.
>
> What path down the shoreline and out into the water should the
> lifeguard take to reach the swimmer in the SHORTEST AMOUNT OF TIME?
>
> --
> He who waits to do a great deal of good at once, will never do anything.
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