Question

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 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 6 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. points b and d are 11 km apart.
assume that the bird instinctively chooses a path that will minimize its energy expenditure. (round your answers to two decimal places.)
(a) in general, if it takes 1.3 times as much energy to fly over water as land, to what point c should the bird fly in order to minimize the total energy expended in returning to its nesting area?
km from b
(b) let w and l denote the energy (in joules) per kilometer flown over water and land, respectively. assuming the birds energy expenditure is minimized, determine a function for the ratio \\( \frac { w } { l } \\) in terms of x, the distance from b to c.
\\( \frac { w } { l } = \\)
(c) what should the value of \\( \frac { w } { l } \\) be in order for the bird to fly directly to its nesting area d?
(d) if the ornithologists observe that birds of a certain species reach the shore at a point 3 km from b, how many times more energy does it take a bird to fly over water than land?

Explanation:

Step1: Label the distance from \( B \) to \( C \) as \( x \)

Let \( BC=x \) km. Then \( CD=(11 - x) \) km. Using the Pythagorean theorem, the distance over water (from the island to \( C \)) is \( \sqrt{x^{2}+6^{2}}=\sqrt{x^{2}+36} \) km.

Step2: Set up the energy function

The energy function \( E(x) \) (assuming \( W \) is the energy per - km over water and \( L \) is the energy per - km over land) is \( E(x)=W\sqrt{x^{2}+36}+L(11 - x) \).

Step3: Differentiate the energy function

Using the chain rule, \( E^\prime(x)=\frac{Wx}{\sqrt{x^{2}+36}}-L \).

Step4: Set the derivative equal to zero for minimization

Set \( E^\prime(x) = 0 \), so \( \frac{Wx}{\sqrt{x^{2}+36}}-L=0 \). Then \( \frac{Wx}{\sqrt{x^{2}+36}}=L \), and \( \frac{W}{L}=\frac{\sqrt{x^{2}+36}}{x} \).

Step5: Solve for \( x \) in part (a)

Given \( \frac{W}{L}=1.3 \), we have \( 1.3=\frac{\sqrt{x^{2}+36}}{x} \). Squaring both sides: \( 1.69x^{2}=x^{2}+36 \). Then \( 1.69x^{2}-x^{2}=36 \), \( 0.69x^{2}=36 \), \( x^{2}=\frac{36}{0.69}\approx52.17 \), \( x=\sqrt{52.17}\approx7.22 \) km.

Answer:

(a) \( 7.22 \) km