Ornithologists have determined that some species of birds tend to avoid flights over large bodies of water during daylight hours.

Question

It is believed that more energy is required to fly over water than land because air generally rises over land and falls over water during the day. A bird with these tendencies is released from an island that is 5 km from the nearest point B on a straight shoreline, flies to a point C on the shoreline, and then flies along the shoreline to its nesting area D. Assume that the bird instinctively chooses a path that will minimize its energy expenditure. Points B and D are 13 km apart.

(a) In general, if it takes 1.4 times as much energy to fly over the water as land, how far should point C be from B in order to minimize the total energy expended in returning to its nesting area?_____km

(b) Let W and L denote the energy (in joules) per kilometer flown over water and land respectively. Determine the ratio W/L corresponding to the minimum energy expenditure given that the bird flies to the shore at a point x kilometers from B. (Your answer may depend only on x.)

W/L =_____

(c) If the ornithologists observe that birds of a certain species reach the shore at a point 2 km from B, how many times more energy does it take this species to fly over water than land? _____

Answer 1

Let

h = 5 km be the distance from the island to the point B,

d = 13 km be the distance from B to D,

x (to be calculated) be the distance from B to C,

W (unknown) be the amount of energy to fly 1 km over water,

L (unknown) be the amount of energy to fly 1 km over land.

We are given W = 1.4L

The distance from the island to C is √(h² + x²) and

the distance from C to D is d - x.

The amount of energy spent during that trip is

f(x) = W√(h² + x²) + L(d - x)

First calculate where the derivative f'(x) = 0

f'(x) = Wx/√(h² + x²) - L

Solve

Wx/√(h² + x²) - L = 0

Rewrite as

Wx = L√(h² + x²)

Square both sides

W²x² = L²(h² + x²)

x² = L²h²/(W² - L²)

x = Lh/√(W² - L²) (1)

Substitute actual numbers

x = L×5/√((1.4L)² - L²) = 5/√((1.4)² - 1) = 5.1 km

The minimum occurs either at domain boundary or at a critical point.

(The domain is [0,d]).

As the function f(x) is defines and differentiable everywhere, the only critical

point is where the derivative is 0.

Therefore we have three candidates for minimum:

f(0) = W√(h² + 0²) + L(d - 0) = Wh + Ld = 1.4Lh + Ld = 22L

f(d) = W√(h² + d²) + L(d - d) = 1.4L√(5² + 13²) = 19.5L

f(5.1) = W√(h² + 5.1²) + L(d - 5.1) = 1.4L√(5² + 5.1²) + L(13 - 5.1) = 17.9L

Therefore the minimum occurs at the critical point when C is 5.1 km from B.

(b)

Let k = W/L

Rewrite (1)

x = Lh/√(k²L² - L²) = h/√(k² - 1)

Express k in term of x

k = √(h²/x² + 1) (2)

(c)

Into equation (2) substitute x = 2

k = √(5²/2² + 1) = 2.7