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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.(a) 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?(b) Let $ W $ and $ L $ denote the energy (in joules) per kilometer flown over water and land, respectively. What would a large value of the ratio $ W/L $ mean in terms of the bird's flight? What would a small value mean? Determine the ratio $ W/L $ corresponding to the minimum expenditure of energy.(c) What should the value of $ W/L $ be in order for the bird to fly directly to tis nesting area $ D $? What should the value of $ W/L $ be for the bird to fly to $ B $ and then along the shore to $ D $.(d) If the ornithologists observe that birds of a certain species reach the shore at a point $ 4 $ km from $ B $, how many times more energy does it take a bird to fly over water than over land?

a) the bird should fly to the point about 5.1 km from $B$b) The above ratio will minimize the energy if the birds aims for the point that is $x$ kilometers from $\mathrm{B}$ .c) There is no value of $\frac{W}{L}$ for which the bird should go directly to B.d) We have $1.4 k=c, x=4,$ and $k=1$ so $c \cdot 4=1 \cdot \sqrt{4^{2}+25} \Rightarrow c \approx 1.6$

08:49

Wen Z.

Calculus 1 / AB

Calculus 2 / BC

Chapter 4

Applications of Differentiation

Section 7

Optimization Problems

Derivatives

Differentiation

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So first we want X to be the distance from B to C so the distance from the island to see is going to be the square root of X squared plus 25 because five squared is 25 and the distance from C to D. Um well, right. Like this, C D is going to be 13 minus X, so the total distance travel will be the square root of X squared plus 25 plus 13 minus x. So if K is the energy per kilometer it takes to fly over the land, then our total energy, which will be modeled by the function F X, is going to be 1.4 k times X squared plus 25 plus K times 13 minus x. Then we find e prime of X. And keep in mind that K is a constant and what we end up getting as a result is K Times 1.4 x minus square root of X squared minus 25 over the square root of X squared plus 25 as plus 25 in both cases. And then we want to set that derivative equal zero. Now we always remember that the bottom doesn't really matter. We're more focused on the top because when the top equals zero, the numerator is equal zero. Um, the whole fraction equals zero. So we have since this constant could just be divided out. What we really have is 1.4 x minus the square root of X squared plus 25. We solve for X, and we end up getting that X is equal to plus or minus the square root of 25 over, um 196 but an extra 4.96 section. But in actuality, since it's obviously just going to be the plus answer, what we have is 51 kilometers. Then we see that a prime of one gives us a negative 07 k. So that's obviously less than zero, and then the prime of 10. So we're picking a point above and below this 51 we see that this one right here is 0 to 5 k, which is greater than zero. So what we have is we are decreasing in slope, and then we reach this 5.1 kilometers and then we're increasing in slope. So what that tells us. Is that 5.1? Um, this is a minimum. Then we have our next part, part B. So we know that if w over l is large, then the bird would fly to a point C that is closer to be than two D in order to minimize the energy used flying over the water. However, if it's small, then the bird would fly to point C that is closer to D than to be in order to minimize the distance of the flight. So what we have is the e equals w times X squared plus 25 plus L Times 13 minus x Then a prime of X is going to give us X w fax over squared of X square plus 25 minus l vax. When we isolate, we set this equal zero And when we isolate W over l, we end up getting that. This is X squared plus 25 over X so the above ratio will minimize the energy. If the bird aims for the point that is X kilometers away from beat. For part C, going directly to D means that X will equal 13, so w over l is going to equal the square root of 13 squared, which is 169 plus 25 over 13. And that's gonna give us 107 Um, so there is no value of w over l for which the birds should go directly to be, and then our last problem, the if the birds choose a path path that minimizes energy And that's gonna be e D X, which equals 14 K x over Route 25 plus X squared minus K. We set that equal to zero, and what we end up finding is that 14 k x equals K times X, the square root of X squared plus 25 case cancel. We solve for X. We get that, um 14 k. If we let this equal see and then x equal four and K equal one we get The sea is approximately 1.6

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