Homing pigeons avoid flying over large bodies of water, preferring to fly around them instead. Assume that a pigeon released from a boat 1 mile from the shore of a lake flies first to point P on the shore and then along the straight edge of the lake to reach its home at L. If L is 2 miles from point A, the point on the shore closest to the boat, and if a pigeon needs 10/9 as much energy per mile to fly over water as over land, find the location of point P, which minimizes energy used?
if the distance AP = x, the total distance flown is
d = √(1+x^2) + (2-x) where x<2.
If flying over land requires 1 unit of energy, then the energy cost is
c = 10/9 √(1+x^2) + 1(2-x)
Now just find the minimum of c.
5.6
8.9
To find the location of point P that minimizes energy used, we need to determine the shortest distance for the pigeon to travel over water while avoiding flying over large bodies of water.
Let's start by visualizing the scenario. Draw a diagram with the boat located 1 mile from the shore, the lake, the point P on the shore, and the home location L, which is 2 miles from point A. It will look something like this:
Boat ----1 mile----- A ------P------------------------------L
Now, let's work on optimizing the path for the pigeon. We can consider two cases:
Case 1: P is located on the same line as A and L.
In this case, the pigeon will fly straight from point A to point L without flying over water. Let's consider this line as the "direct path" from A to L. The pigeon will need to fly 2 miles along this direct path.
Case 2: P is located on a line perpendicular to the line through A and L.
In this case, the pigeon will fly from the boat to point P, then continue flying along the shore until it reaches point L. This path allows the pigeon to avoid flying over water.
Let's calculate the distance of this path. The distance from the boat to point P will be the shortest distance it can cover over water.
By Pythagoras' theorem, we know that the shortest distance from point A to the direct path (line AL) is a line perpendicular to AL that meets AL at point X. Since AX and PX are perpendicular, we can apply Pythagoras' theorem to find PX.
Let's denote PX as x. We know that AX = 2 miles, so AXP forms a right-angled triangle with sides AX and PX. Applying Pythagoras' theorem:
AX^2 + PX^2 = AP^2
2^2 + x^2 = AP^2
4 + x^2 = AP^2
The distance from P to L by traveling along the shore is 2 - x (since P is x miles from A, and L is 2 miles from A).
Now, let's find the distance from P to L over water:
The distance the pigeon needs to fly over water is the shortest distance from P to the direct path (AL). This can be calculated as the difference between the straight-line distance of AP and the distance over land (2 - x).
The energy required to fly over water is given to be 10/9 times the energy required to fly over land (which we will assume is 1 unit of energy per mile). So, the new energy required to fly over water is (10/9) units of energy per mile.
Therefore, the energy required to fly over water is (10/9) * (2 - x).
The total energy used is the energy required over land (2 units, as L is 2 miles from A) plus the energy required over water.
So, the total energy is 2 + (10/9) * (2 - x).
To minimize the energy used, we need to differentiate this expression with respect to x and set it equal to zero. The critical point will give us the location of P that minimizes energy used.
Taking the derivative:
d(energy)/dx = - (10/9)
Setting this equal to zero:
- (10/9) = 0
This equation has no solution, which means that there is no critical point (maximum or minimum).
Since there is no critical point, we need to examine the boundary points. In this case, the boundary points are x = 0 (the boat location) and x = 2 (same line as A and L).
Let's calculate the values of energy used at these boundary points:
When x = 0:
Energy = 2 + (10/9) * (2 - 0) = 2 + (10/9) * 2 = 2 + 20/9 = 38/9 units of energy
When x = 2:
Energy = 2 + (10/9) * (2 - 2) = 2 + (10/9) * 0 = 2 units of energy
Comparing the energy used at the boundary points, we can see that the energy used at x = 2 is lower than at x = 0. Therefore, the optimal point P that minimizes energy used is when P is located on the same line as A and L.