Assume that the bird instinctively chooses a path that will minimize its energy expenditure. Points B and D are 14 km apart.
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 5 km from the nearest point B on the 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 14 km apart. (Round your answers to two decimal places.)
(a) In general, if it takes 1.1 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?
(b) Let W and L denote the energy (in joules) per kilometer flown over water and land, respectively. Assuming the bird's energy expenditure is minimized, determine a function for the ratio W/L in terms of x, the distance from B to C.
(c) What should the value of 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 2 km from B, how many times more energy does it take a bird to fly over water than land?
First let's set up the problem
Let x be the distance from C to B. Then 14-x is the distance from C to D.
So the bird flies sqrt(5^2+x^2) to reach C, and then 14-x from C to D along the shoreline.
a) E = 1.1sqrt(25+x^2)+14-x
(dE)/(dx) = 1.1(2x)/(2sqrt(25+x^2))-x = (1.1x)/(sqrt(25+x^2))-1
At a critical point (dE)/(dx) = 0 so (1.1x)/(sqrt(25+x^2))-x =0
We need to solve for x.
`1.1x=sqrt(25+x^2)` , `1.21x^2=25+x^2` ,` 0.21x^2 = 25` , `x^2=25/0.21` , `x ~~10.911"km"`
Plugging this into E we get E=16.291
We should also check x=0 and x=14, x=0 we get E=19.5, and x=14 E=16.352 so the answer to a) is C should be 10.911km.
b) `E = Wsqrt(25+x^2) + L(14-x)`
So `(dE)/(dx) = (Wx)/(sqrt(25+x^2))-L`
To minimize we find the critical points `(dE)/(dx)= 0` and solve for x.
`Wx/(sqrt(25+x^2))=L` , rearanging we get `Wx=Lsqrt(25+x^2)` , squareing we get
`W/L = sqrt(25+x^2)/x` ``
We also need the end points, x=0 and x=14. Substituting we get
`E=Wsqrt(25)+L(14) = 5W+14L` , at x=0 and `E=Wsqrt(25+14^2)+0=Wsqrt(221)` , at x=14
c) `W/L=sqrt(25+14^2)/14=sqrt(221)/14=1.062` . We would obviously need W/L to be less than or equal to that.
d) at `W/L=sqrt(25+2^2)/2 = sqrt(29)/2 ~~ 2.693`
So it takes 2.693 times as much energy for this bird to fly over water than land.