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When , =what Ihe value12. In Chetokce Couny Ihe ~peeding i $17 euch mile per hour Ihe drver uaveling @e posted #Pceu limit, Cherokee Count} Kirk linted 522 spceQII_...

Question

When , =what Ihe value12. In Chetokce Couny Ihe ~peeding i $17 euch mile per hour Ihe drver uaveling @e posted #Pceu limit, Cherokee Count} Kirk linted 522 spceQII_ posled ~peed limit of 30 mph Kirk #as lined for Iravcling whale Apced miles pet tou ?0 "7 4 kIhe,standard (1 J) coordinule plane. whal I Lhe HIdnau 0f the line 20CMCIC eudpains (3.8) and_"Vhal Ihe suIn nelou"the solutions of the cuualionaX=m 2-10 = 2(--121 (-4. 44-12Muctualton 0l Waler deplh a 4 pleT shown in Ihe figur

When , = what Ihe value 12. In Chetokce Couny Ihe ~peeding i $17 euch mile per hour Ihe drver uaveling @e posted #Pceu limit, Cherokee Count} Kirk linted 522 spceQII_ posled ~peed limit of 30 mph Kirk #as lined for Iravcling whale Apced miles pet tou ? 0 "7 4 k Ihe,standard (1 J) coordinule plane. whal I Lhe HIdnau 0f the line 20CMCIC eudpains (3.8) and_" Vhal Ihe suIn nelou" the solutions of the cuualiona X=m 2-10 = 2 (--121 (-4. 4 4-12 Muctualton 0l Waler deplh a 4 pleT shown in Ihe figure neok the follox ing Vallnte$ gives the poslne difference nelween Ine Lreuie C Wile' ( depth and thc eunt Water depth shown this Eraph WUhick Vil E "7} 14. The Werage of distinct scores has the same Valc tle mledian o Ihe scotes The sum scoies 420. What i` the sum of the scores thal Are NOT the median" 315 6 4 350 1 Whal Ihe value Ihe expression belon ? Il-8 + 911 & 1 KhAALhh> 10 12 14 16 I8 20 numtrrol houn alcr 6&m 1 16. Which o the following expressions cquivalent What the slope ol the line through the standurd Lcourdina[C Dianc . I)and (2-51 }



Answers

At noon, ship $A$ was 12 nautical miles due north of ship $B$. Ship $A$ was sailing south at 12 knots (nautical miles per hour; a nautical mile is 2000 yd) and continued to do so all day. Ship $B$ was sailing east at 8 knots and continued to do so all day. a. Start counting time with $t=0$ at noon and express the distance $s$ between the ships as a function of $t$ b. How rapidly was the distance between the ships changing at noon? One hour later? c. The visibility that day was 5 nautical miles. Did the ships ever sight each other? d. Graph $s$ and $d s / d t$ together as functions of $t$ for $-1 \leq t \leq 3$ using different colors if possible. Compare the graphs and reconcile what you see with your answers in parts (b) and (c). e. The graph of $d s / d t$ looks as if it might have a horizontal asymptote in the first quadrant. This in turn suggests that ds/dt approaches a limiting value as $t \rightarrow \infty .$ What is this value? What is its relation to the ships' individual speeds?

This problem. We're looking at a bird flying from an island across the river to its nest. So let's begin with a picture. So we have an island and we'll call that island A. It is directly across the water from B, and it wants to make it all the way to its nest at D. So it's flight pattern is going to be directly across the river at some angle and then across the land over to D, and we'll call this point. See, just we have something we can call it Now we know that B and D have a has a distance of 13 factor in kilometers, so 13 kilometers. So let's call BC Let's call this X Well, if that's X, then from C to D is 13 minus X, and we know that the island is five kilometers from the shore. So we have just about all of our things marked here. Let's, uh let's also just use, um we'll call this why that distance from eight. To see that he's going to fly now what we're wanting to minimize here, we're told that the birds instinctively are going to minimize their energy that they need to expend. So let's look at the energy expended in this situation. Well, the energy expended across the water is going to be, um, energy Times distance. Now we're told that the energy to get across the water is 1.4 times more than than on the land. So if I call the energy across the land one, then the energy for why is going to be 1.4 units and however we're measuring that So the energy across the water is going to be 1.4 times why our energy across the land is going to be 13 minus x times the distance, our times, the energy expenditure on the land. Well, we said that was just gonna be one unit. So across the water is 1.4 across the land is gonna be one. So here is my energy expenditure formula. But the problem is, it's in two variables. I have why, and I have an ex. So let's put them in terms of each other. Take a look at my triangle. I can use the Pythagorean theorem to say that why squared equals five squared plus x squared, or why equals the square root of 25 plus X squared so I can plug that into my equation. So I have 1.4 times 25 plus X squared plus 13 minus x. That is my energy expenditure given in one variable instead of two. If I want to minimize this these air optimization problems, I'm going to take a derivative, set it equal to zero, and that will give me my critical points that show me where I can minimize my formula. My function here. So if I take the derivative that's going to give me well, I have a radical so that 1.4 is on the top. My denominator is going to be two times my radical and in the numerator, I'm gonna multiply by the derivative of what's under the radical sign, which is going to be two x derivative 13 is zero derivative of negative X is negative one. Now, I can do a little bit of simplification here. Not much, but a little bit. And in order to, um, find my minimum energy, I'm going to set this equal to zero. So what I get is 1.4 x over the square root of 25 plus X squared, and that equals one. Hey, now to solve this for X, let's multiply both sides by that radical that's gonna give me 1.4 X equals the square root of 25 plus X squared, comparing a little bit more room to get rid of the radical. We're going to square both sides. So the square of 1.4 is 1.96 x squared and squaring the radical gives me what's underneath Now it's just a matter of algebra combined, my like terms. So I'm going to move my X squared term and divide by 0.96 Now, when I take the square root algebraic Lee, I'm going to get a plus or a minus. But if you look at the problem X is a positive number, The smallest X could possibly be a zero, and that would be straight across the water. X is not gonna have a negative value here, so I could take the positive square root, which is going to be approximately 5.1 kilometers. Okay, so I'm gonna the bird is going to fly across the water, See, is going to be at a 0.5 point one kilometers from B and then continue the rest of the way on two D that will minimize the birds energy expenditure. Now let's tweak this problem a little bit. Let's make it more generic. Instead of saying that it's 1.4 times more energy on water than on land, let's just use I'll use blue here. Let's use W for the energy expenditure on the water an l for the expenditure on the land. So let's revamp this problem again. But using these values. So what is that going to change? Well, I am gonna scroll down in a moment and rewrite, but for the moment, let's look at what I have here. The difference in my original equation instead of 1.4. This is going to be W. That's That's, um that's my expenditure on water, and instead of one, I'm going to have an l. Everything else is going to be the same. So let me scroll down and rewrite this. I'm gonna have my energy expenditure is w times Why, that's my expenditure. Energy expenditure over the water plus 13 minus x times l. That's my energy expenditure overland. So just like before. I'm going to put this in terms of one variable. So I'm gonna have w times the square root of 25 plus x squared That peace has not changed in this problem. And I have 13 minus x times L So let's take the derivative. Well, remember, W and l are constant, so they're not going to change that. My variable is only X. So just like before, taking the derivative of a radical and the denominator is going to be two times the radical W will be in the numerator and we're going to multiply that by the derivative of what's underneath the radical sign. Okay, here l is my constant. So it's going to be l times the derivative of what's in the parentheses, which is negative. One. No, my choose cancel. And what I have is a derivative of E is W X over the square root of 25 plus X squared minus l. Then let's set that equal to zero. That's going to give me W X over my radical equals. L and I would like to find w over l. So let's do a little bit of algebra here. Cross multiply. And if I want w over l that's going to equal 25 plus X squared over X. Now, we're going to use this a little bit, but let's just talk a bit about what this means. How are w and L related? Well, the bigger W is never w is our energy expenditure over water, the more taxing it is to fly over water versus land, the less time the bird will spend flying over the water. So the bigger W is a bigger this ratio is X is gonna become smaller and smaller. We wanna have our Yes, we don't have as much time on land as possible. So a big W means that X is going to get very, very small my way to get off the water as fast as I can. The smaller this ratio is if the energy expenditure over water isn't that much bigger compared toa land, then why not take a more direct route? The smaller that ratio is, the longer my water travel will be because that's the more direct route. So X is gonna be bigger. So that's how w NL and ex are all related to each other. So let's do a couple of examples using this information first. What will the value of W over L. B if the bird goes directly to its nesting area? Well, if it goes directly to its nesting area, that means X equals that entire 13 kilometer unit that means X equals 13. So what is W over l If X is 13? Well, that gives me 25 plus 13 squared, which is 169 over 13. Plug that into my calculator and I get approximately 1.7 is my ratio. Well, what happens if my bird flies directly across the water and then across the land? That means if you look at my picture, if I go directly across the layoff across the water X is going to be zero, there is no triangle. So what happens if I let X equal zero? This was X equals 13. What effect? Zero. Well, that means I'm going toe have the square root of 25 over zero. You can't divide by zero. There is no scenario in which this ratio gets so out of whack that the best thing for the bird to do is to fly directly across the water. It's always going to go at some angle, even for a very large ratio. So that case is not gonna happen. But it is possible that it could go directly to its nesting area. Okay, last question. Ornithologists Observe that certain birds go to shore at a 0.4 kilometers from B. So other words they have found there's a certain bird that when they do this trip, X equals four. So what does that tell me about this ratio of w tell my nexus four. That tells me I have a square root of 25 plus 16/4, which is approximately 1.6. So that is the ratio of my energy expenditure on water versus land for that particular bird.

All right, So we're asked to This is a new H at noon. A ship A, which is here will say is strong nautical miles north of B. So here's B and that it says it's going at 12 knots an hour. So a prime or D A by D. T is 12 as well, and it's minus because it's going so e it's is going eight knots an hour east. So be prime is eight and it's getting positives is growing. All right. And then we have this d here. Okay, so what it wants us to do is first of all, in part a express the distance between the ships, which they've actually called. That will change this toe s. It has a function of time. So this side, he is going to be 12 minus 12 t. Hey, this side is because it's 12 right now, and it's changing at 12 nautical MPH. And then this site is going to be 80. Okay, but what we get is s is the square root of 12 minus 12 t squared, plus 80 all squared. All right. And if we expand that, we get 144 plus 144 times to actually minus not a plus because its a negative 12 t so minus 2 88 t and then plus 144 t squared and then plus a further 64 teats heard which, if we then put together, uses this 1 44 minus 280 80 plus 208 t squared. That's part a part B says. How rapidly was the distance between the ship's changing at noon? Okay, so for part B at noon, it's changing at 12. No, Excuse me, because be an a r directly on top of each other at that point. And then it says one hour later, one hour later, a is actually gonna be at point B. It's gonna B zero in the UAE. So it's gonna be eight. Not so. It's changing at that point because any become zero basically right then RT says the visibility that day was five nautical miles. The ships ever lose sight of each other. Okay, so if we start out, um, eight and 12 then they can see Oh, ever site. Sorry. Did they ever site each other, not loose. I was gonna say, um, they're not gonna see each other at the beginning, right? Because it's gonna be 12 nautical miles here, which is bigger than 10 so they could see five, and they can see five. This is an A is gonna go straight down as they go straight across when they go straight across from each other. It's eight nautical miles then. So they're not going to see each other because they could only see five. Each of them can only see five. And after that point, it's gonna be bigger than eight. So they're never going to see each other. So the answer to part C is No, because the distance between was always going to be greater then five. Okay. And then the question asked us to graph these things. So I'm gonna do that on Dez Mel's. Okay, so So, uh, I'm gonna just sketch these based on what I'm drawing from, um, Deshmukh, because I'm not actually going to show you the graphs. Uh, we're not gonna be able to, I guess because I'm not sharing my whole screen. That's okay. I will give you a sketch, and you can sort of get the idea. All right. So, um, excuse me here with me while I graft this year, so it kind of looks like this. Okay, the graph of the s, if we just and this is just a basic sketch, but it essentially comes down and then goes back up like that right now. The derivative, um, is going to look a little bit different. We're just again Bear with me while I graft that. But what's gonna happen? You have to take the derivatives, so it's gonna be 1/2. I'm in that whole 144 minus to 88 x plus 200 x squared to the negative 1/2. So we're raising to the negative 1/2 and then it's gonna be multiplied by the truth of what's inside, which is minus 288 plus 416 X. Okay. And what it looks like is this it kind of I'm gonna draw this in a different color. It kind of goes like this and then comes up and then sort of goes back down like that. Okay. And again, you can draw these on your own If you want to see a little bit better. I'm just using decimal stock come to do it. But then it says it appears that this has in Assam toad. Okay? And we want to say, What is that? What is that Assam toe? So it basically is. If you look on your graph and again, I'm just I'm just taking this from my graph. It's for re 13. Okay, you can do that on Desmond's. You can look at at that value. So this is what is the relation to the ship's individual speeds. It's basically the derivative of those, okay, and it's the idea that it's a combination of the X and Y opponents. So it's sort of the the iPod news. It's the rate at which this is changing right and saying that you, the distance between there was going to always have. It's going to sort of stabilize at four re 13. As they go to infinity, that's t goes to infinity. Okay, that's the idea in those last two

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

Okay, so we have this following. Inequality enforce its race. A boat is considered qualified if it meets this falling inequality. So let's go. That are all Volume is our total length. When are a is or a surface area and the is the displacement. Hey, so for part a, what are we given were, given that we have the length Oh, of 6 50 and then a surface ariel of 3400 It's squared and our displacement the is 6. 50 Pete squared. We're sorry. That's accused. So I'm gonna plug this into O A and the and then see what it gets. So we have 2.3 ohm times at 6. 50 and then plus 0.38 and then here, 6. 50. Actually, that sounds 6 50 or a values 3400 to the power of one. How? On the minus 83 times 6. 52 power of 1/3. So that's going to give us a That's approximately and sorry about this. Are length should be actually 60. So here we gets 14 point two is not sin or equal to 16. Well, this is true. So we see that these moderates satisfying or fulfilling conditions. And now, for part B, this time we have a length of 65 a displacements of 600 feats cute and were asked what is the largest possible sail area? So we're solving for our a value, so I'm gonna plug in what we have. So this is 1233 0.3 old times 65 close. There were 0.38 in an age. The part of what half minus three times 600 to the power of 1/3 is that they're equal to 16. Okay, so it's all for a sore point. Three times 65 is 19. Port five, and then we have three times six injured A to the power of 1/3. So this is around 25 points. Three. Okay, so we have 16 we're gonna at 25.3 and then we're going to subtract 19.5. So here we're gonna have 21.8 is less than or equal to 0.38 aged a part of one half. Let's divide. But both sides are important to me. Eight. So now we're gonna have 57 points for is less than or equal to a dip are about half. Let's take wolf science to depart to when we get that are surface area at minimum is 3281 point 14 and this is in feet cute.


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