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W(t)BCL: Workers at a local oflice building apreed work by YAM (( 60) each day: Letthetwice differentiable function y W(t) bethe amount ofworkers building at time m...

Question

W(t)BCL: Workers at a local oflice building apreed work by YAM (( 60) each day: Letthetwice differentiable function y W(t) bethe amount ofworkers building at time mnMe5 satisfying the Loxkuc differential equation 15> (1 _ 100" Where W(0) = Selected values of W (t) are given inthe table where t 0 is BAM(d) Use Euler' metnod with two steps equal size starting at t 45 minutes approxmate the number of workers inside the oflice building by 254M (t = 85 minutes)theretmesuch that 25 < â

W(t) BCL: Workers at a local oflice building apreed work by YAM (( 60) each day: Letthetwice differentiable function y W(t) bethe amount ofworkers building at time mnMe5 satisfying the Loxkuc differential equation 15> (1 _ 100" Where W(0) = Selected values of W (t) are given inthe table where t 0 is BAM (d) Use Euler' metnod with two steps equal size starting at t 45 minutes approxmate the number of workers inside the oflice building by 254M (t = 85 minutes) there tme such that 25 < € 45,when W'(c) Justity your answer: (0) Let Pz(t) be the second degree Taylor polynomial for W(t) centered att 65. Find Pz(t) and use itto approximate the number of workers inside the office building 4t9 ZSAM (( 85).



Answers

The management at a certain factory has found that the maximum number of units a worker can produce in a day is $75 .$ The rate of increase in the number of units $N$ produced with respect to time $t$ in days by a new employee is proportional to $75-N .$
(a) Determine the differential equation describing the rate of change of performance with respect to time.
(b) Solve the differential equation from part (a).
(c) Find the particular solution for a new employee who produced 20 units on the first day at the factory and 35 units on the twentieth day.

So in this problem, we have a warehouse. Um, and there's no heating or cooling, um, or active heating or cooling in that in the warehouse, and it depending on the amount of installation, the time constant for the building may range from 1 to 5 hours to illustrate the effect installation having the temperature inside the warehouse soon the outside temperature various as a sine wave with a minimum of 16 degrees C and a maximum of 32 degrees C. So minimum occurs at 2 a.m. The maximum occurs at 2 p.m. Assuming the exponential term has died off. What is the lowest temperature inside the building of the time? Constant is one hour. And if it is five hours, um and what is the highest temperature? So we have our outside temperature. I'm gonna call it t not out is an average temperature of outside. Um, plus some change. We can figure out the average is 24 degrees C. On the change up or down is eight degrees C. The period is, um, not Let's see, this is, um, 1/12 of this is ours, not seconds. Um, so pi over 12 hours, Um so basically half a day. Um, so let's see here. Now we have, um and basically, this is so when we go for when we go, 12 hours we go, we go 24 hours we go from, you know, uh, you know, go through one sine wave eso Let's see here our equation described the temperature inside the building is D T d t times k times t plus k times t equals k times t not. So here. We're looking. This is the temperature inside the building. Um, we have a homogeneous solution that looks like an exponentially decaying function. Um, and we're gonna assume that that is a die off, by the time way, reach kind of peaks temperatures. And so then we need a particular solution, and they give us the particular solution in the book. Um, one particular solution is simply, um, a constant this constant here. So obviously, if we plug a constant into here, we'll get the constant term here and then the, uh, sign or the coastline term. Here we have a particular solution that looks like like this here with an amplitude of delta t. Um, again, this is just giving derived for you in the book, so I'm not gonna derive it here. Um, what we could do, though, is to make this to get peaks. It's hard to figure out where the maximum here is because of you know, the sign, plus the coastline term. But what we can do is we can write this an amplitude phase form so we can combine these into one sine function with the same, um, time dependence, but then a phase and then the amplitude would be the square root of this down here. So again, we, um it's just basically some trig identities to turn this into an amplitude phase form on. Why we do that is because, Well, we know that whatever this phases, because we're not asked for win win the temperature hit the peak. We just asked what the peak is. Well, we know the peak of a sine function is one, and the minimum is minus one. So we know that. Then the minimum occurs when, um, sign is one. Because we have a minus sign here and we can plug in our values for omega for delta T for tea. Not. And if we have K equals one. We get 16.26 degrees Celsius. And if we use, um, K equals 1/5. So the time constant of five hours, we get 19 point for 14 degrees Celsius. So if we have ah longer time constants are minimum temperature is higher. Now for the maximum Again with the maximum is when workers when this term here is a minimum. So we have one sign when sign the sine function is minus one. So then when sine function is minus one, we get t not plus delta t all over this room and again, we can plug in values for that. And if K equals one, we get 31.74 to B C and K equals 1/5. We get 28.86 degrees c. So again we can see that the max is higher. Um, and the if the time constants lower. So when we have a local over time constant, we have a greater swing from high to low. I remember from high went, you know, we almost almost reached the outside temperatures here. Here. If we have a longer time constant, which means more insulation, we don't don't quite. You know, we get we don't reach those peaks on the outside a zoo closely, so we get a much lower swing and temperatures here, then we do from the outside. So again, the key here to do this is because they didn't ask for the time. I mean, we could solve for the time if we wanted to, but because they didn't ask for the time. We just write this in the amplitude phase for him. And then we just know that this has a peak, as a maximum sign is, the man has a maximum value of one and a minimum value of minus one. So, um, that's whatever the time is. We'll figure out what that maximum value is. That either that it's either one or it's minus one. Yeah.

Mhm. So this problem is giving us an equation that can model the number of construction workers for any given month. And so part a of this question asks us for the total number of construction workers that worked the first quarter. And so we can go ahead and evaluate that um using a graphing calculator because this problem actually asks us to use a graphing utility. And so the first thing that we are going to do is go to our graphing calculator of course turn it on and we're going to go to the top left corner where it says why equals in touch there. If you already have an equation and go ahead and hit clear and if not you should be able to enter this equation into why one. When you are entering this equation, first of all, make sure that we are in radiance. If you are unsure if you're in radiance or not go to the mode button, it's in the top left corner, next to the blue second button. The third option down should say radiant or degree. Make sure radiant is highlighted also when you are entering this equation um there is no T button on the calculator. I mean there is a letter T. But Um in terms of entering things into the calculator to graph are variable is always called X. So we will enter this value as .41 x. And that X button is right next to the green alpha. When we are ready to graph this, the next thing we have to do is set our window. Um So the window button is also in the top left corner, it's to the right of why equals so let's set window. And so in terms of our exes we are going to have things that are greater than zero because we are working with T. Which is time. And so I'm gonna set my ex minimum 20 and my ex maximum I'm going to set um 2 15 just to be a little bit Further than 12 which is 12 months. And then the X. Scale. I'm gonna go ahead and make one for each month in terms of the why um we should not be working with anything negative because the Y. Is talking about thousands of construction workers. So I'm gonna set my why minimum to be zero. And then the maximum you can see that this function is going to be circulating around our midline of 7500. Um in some chains. So almost 7600. So I'm going to set my Y max to be 10,000 just to get a great visual of this graph. And I'm going to set my Y. Scale to be 1000. Just so I can use those markers to help and let's go ahead and graph that. So on my graph it's looking something like this. It definitely has a curve to it. Like a sign graph would um it's just definitely pretty spread out, which is fine. So in order to find the total number of workers that worked the first quarter, we need to find the area under the curve Between zero and 3. And so we are going to find the area under the curve between zero and 3. And so that will be this. In order to do that, we will go to second, cowpoke An option seven is the anti derivative. It's going to ask you what your lower limit is. You're going to say zero and hit enter. It's going to ask you what your upper limit is. You're going to hit three and hit enter. And then it's going to calculate that the area under the curve between zero and three should be 2000 Excuse me? 21,640 1.91. So part B let's switch colors here is going to ask you to do something similar just between months three and six. So Part B is going to look something like this. We're finding the area under the curve between three and six. Again, we're going to use that same set of steps. So where you are right now you probably still have between zero and three highlighted on your calculator. Just go ahead and hit that clear button. Um for me that made my graph go away. Oh it didn't clear. Okay, so we should try just going to second count again So we can go all the way down to choice seven. And this time the lower limit is three. The upper limit is six And I got 22,000 950 0.54 And this should make sense because it's about the same as part A. But as that curve is curving up just a little bit we'll have just a few more people included in there. So part C. Is going to be the entire year. So this would go all the way into 12 And this was zero and we are going to find the whole year. So again we're going to do this same process. 2nd help Joyce seven are lower limit zero. Our upper limit is 12 And the total that I got was 92,000 78. Yeah 0.1. And that's it.

For this problem, we're looking at function involving construction workers which is the number of construction workers employed in the United States, it's gonna be 70,900 54 plus 455 point to kind of the sine of zero point for tea minus 1.713 These are the values right here. T. Is the time and runs and T equals one corresponds to january 1st. So we want to approximately the month in which the number of construction workers employed as a maximum. So looking at our function zooming out, we see that that's going to occur all the way up here. Um And letting our X values be much smaller, we see that going from there are 12. Thank you. 20. We end up seeing that it's going to be a maximum in um in august where it's gonna be about 8400 nine and that's in thousands. And then we see that the minimum. Um Yeah, so the maximum is here, the minimums down here, but it just asks about the maximum. So that would be august and that will be how many there are

Welcome to solve a problem. Number 30 to sense. We have p off zero equals zero. So we have be off t equal. Mm. Multiply one minus e to support negative G d. And if the off one, uh t is jumps learning curve than be one appear on one. Equal training five on B one story. Um, for fun to equal 14. What hands? 25 equal. M one, multiply one minus. It was about negative. K Andi 45 equal, ERM one multiply one minus E to the power minus two. Okay, so one minus 20 five, uh, one minus 25. Sorry. One minus 25. Over. I am one equal e to the ball. Negative K or Okay. Equal Minus. Lynn one minus 25. Um, one. So it will be equal limb M one over one. Minus 20. Why? On but 45 equal one multiple. I want mine e to the power. Negative. Okay, So 45 equal, um, one. Um, and one. I want to play one minus, um, one minus 25. Substitution. Um, over on one squared or, uh, 45 equal 51 minus 62 6 to 5 over m one. So thus, um, em one equal one to any What is a maximum number off the units per hour? Jamie's capable for off processing. Okay. Similarly, if p two if B two off t, uh is marks learning than be to off one, uh, off 21 p to one equal 35 on be to to equal 50. So so, um Okay, equal. Len, I am too over em to my naked 35. Um, and 50 equal, Um, to what? Apply one minus. I'm to minus 35 over to all squirt. So, um, to equal 61.25 And the maximum number of units per hour for mark is approximately 60 one. Another approach would be to use the midpoint off the intervals. So that be one, uh, 0.5 equal 25 and the one off 1.5 equals 45. Doing that gives us, um, one approach equal 52.6 on em. Too much equal for 51.8. Thanks for watching


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