~The Bug-Frog Problem~
The Problem: A bug is
traveling north along the curve x = 0.25y2. This is a basic
parabola opening right with a vertex at (0,0). Since the y is squared,
there will be symmetry with respect to the x-axis. A hungry
frog is situated at the point (3,0) on the x-axis. This frog is very
lazy and will not move from this point. The frog needs to be patient
and wait for the bug to come within reach of his tongue, which is 2.9
inches. At this moment, the bug is located at the point (9,-6) on the
curve. Using the distance formula between two points, it can be
calculated to be 8.49 inches away from the frog. Therefore, well
out of reach of that hungry frog. However, as it travels upward
along this curve, will there be a point on the curve that the frog's
tongue captures the bug? Let's find out.
The problem reduces to the problem of finding the point(s) on the curve closest to (3,0), the position of the frog.
So, we need to set-up a general distance function that will give the
distance from any point (x,y) on the curve to the point (3,0), the
position of the frog. Let's call this distance s.
Again, use the formula for the distance between the two points (x,y) and (3,0).
We get, s = √[(x-3)2+ (y-0)2] which simply becomes s = √[(x-3)2+ y2].
We want to find the x and y value(s) the make s a minimum.
Since s is a positive distance, if we minimize s2, then that will
automatically make s a minimum. That way, we can make our work simpler
by eliminating the square root.
So, Let Q = s2. So, Q = (x-3)2 + y2
We need the right side in terms of one variable. No problem,
since x = 0.25y2 gives y2 = 4x. So substitute 4x for y2 in the
equation for Q.
We get, Q = (x-3)2 + 4x. Now, we are ready to take the
derivative of Q and find the critical value(s) where it is 0
dQ/dx = 2(x-3) + 4. Setting dQ/dx = 0, we get, 2x-6 +4 = 0
or x =1. Since the 2nd derivative is always positive (2), we know that
x = 1 produces a minimum Q, consequently, a minimum s, hence, the x
value on the curve closest to the point (3,0).
when x=1, y2 = 4, which gives y = 2 and -2. So, the points (1,
-2) and (1, 2) are the points on the curve (the bug's path) closest to
(3,0), the frog.
Since the bug is at (9,-6) presently and is safe, it needs to escape
being eaten by making it past (1,-2). If it survives this point, it
will not be eaten since no other point on the curve is closer.
So, let's use the distance formula between two points once again to find that distance from (1,-2) to (3,0)
we get, √[(1-3)2 + (-2-0)2] = √[4 + 4} = √(8) = 2.83
inches. Bad news for the bug, since the frog's tongue can reach
2.9 inches. So, the frog gets a meal.
Hope you enjoyed my problem.