Ornithologists have determined that some species of birds tend to avoid flights over large bodies of water during daylight hours. It is believed that more energy is required to fly over water than over 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. In general, if it takes 1.4 times as much energy to fly over water as it does over land, to what point C should the bird fly in order to minimize the total energy expended in returning to its nesting area?

A point 5.103km from B

Step-by-step explanation:

Let x be the distance from B to C. Thus, the distance from the island to C is given as;

√(x²+5²) = √(x²+25)

Mow, the distance from C to D will be equal to 13-x

Hence the total distance travelled will now be;

√(x²+25) + 13-x

Let's say k is the energy per km it takes to fly over land. Thus, total energy is given as;

E(x) = 1.4k√(x²+25) + k(13-x)

Mow, let's find the derivative of E(x);

E'(x) = 1.4k(1/2)(2x)[(x²+25)^(-1/2)] + k(-1)

Thus;

E'(x) = 1.4kx/(√(x²+25)) - k

Simplifying this to get;

E'(x) = [1.4kx - k(√(x²+25))]/[√(x²+25)]

Now, to find the distance, it will be at E'(x) = 0

Thus;

[1.4kx - k(√(x²+25))]/[√(x²+25)] = 0

Multiply both sides by [√(x²+25)] to obtain;

[1.4kx - k(√(x²+25))] = 0

(divide each term by k) to obtain ;

1.4x - (√(x²+25)) = 0

1.4x = (√(x²+25))

Square both sides;

1.4²x² = x²+25

1.96x² = x² + 25

1.96x² - x² = 25

0.96 x² = 25

x² = 25/0.96

x² = 26.042

x = √26.042

x = 5.103km