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Two new workers were hired for an assembly line. Jim processed 25 units during first hour and 45 units during the second hour. Mark processed 35 units during the first hour and 50 units the second hour. Using the model of Exercise 31 and assuming that $ P(0) = 0, $ estimate the maximum number of units per hour that each worker is capable of processing.

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Calculus 2 / BC

Chapter 9

Differential Equations

Section 5

Linear Equations

Harvey Mudd College

Baylor University

Boston College

Lectures

13:37

A differential equation is a mathematical equation for an unknown function of one or several variables that relates the values of the function itself and its derivatives of various orders. An ordinary differential equation (ODE) is a differential equation containing one or more derivatives of a function and their rates of change with respect to the function itself; it can be used to model a wide variety of phenomena. Differential equations can be used to describe many phenomena in physics, including sound, heat, electrostatics, electrodynamics, fluid dynamics, elasticity, quantum mechanics, and general relativity.

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as we have derived in a quick question. Number 31 p s M plus C times it is two minus Katie, where m is maximum Learning sees an arbitrary constant that arise due to our integration. And K is positive constant. We have given the gym joints a work, and at first his learning is 25 for one hour and 45 42 wars similarly for markets 35 for one hour and 45 for to us. I'm sorry. It's 50 for to us. And we need to calculate what is the maximum learning can be for Jim and Mark. We are also provided with the condition that p of 00 Okay, so let's start with the first condition. So we have p of 00 so at equals to zero, we have P is equals to zero. Therefore, we have zero is equals to m plus three times it is to zero. Right, Kate, Times zero is zero. So we get zero is m plus C and therefore see is minus m. Okay, so now we can update our equation as a particular solution. PEs m minus m. It is two minus Katie. Now we'll solve for Jim. Okay, so Jim at equals to one we have at T equals to one. We have p as 25. So 25 is equals to what is M? We don't know. We can take em common. So this will be one minus. It is two minus k times one. And that's the equation for first. So right it is. It is two minus. Okay. Okay. So if we can, we can calculate value if it is to minus K to substitute in the next step. So we'll divide by m. So we'll have 25 by m is equals to one minus. Here is two minus K. And therefore we get We'll ship this to positions or interchange it so we'll get it is to minus K is one minus 25 by them. So let this pier first finding. Now the second condition for Jimmy's at Time t equals toe to is Peace 45. Okay, so therefore, 45 a m one minus here is two minus k into to so therefore we have 45 years and one minor. This is nothing but a minus care square. And we have already calculated this value so we can put it here. So am times one minus. I put this value on minus 25. I am squid. Okay, so if you simplify this, this becomes one a minus. Behold square is a square minus to be so that is two into 25 50. I am plus B squared. So 25 times square is 6. 25 5 M square or break it. So if you open the bracket, we get one minus one plus 50 by M minus 6 25 by M Square. And if you multiply with this m after canceling this, we get 50 minus 6. 25 by I am right, I'm square and m one and we'll get canceled. And you have You have 45 so well again, change this to position year and this will be brought here, so we'll have 6. 25 by M is 50 minus 45. So therefore we get 6. 25 I am s five and entertaining this Poggio inviting this we'll get em over 6. 25 is 1/5. So therefore and will be 6 25 by five. And that is 12 five. So maximum learning off Jim will be 1 25 on. Similarly, we can solve for Mark. So it's a condition for Mark. Market time equals to one hour is maximum working is tactic five. So let's be and t equals to two piece 50. Correct. So we have our equation s peace. M times one minus. It is two minus Katie. So for one hour, that is 35 AM one minus. It is two minus k. And what will be it is two minus care. It is two minus k will be 35 by Mm it is subtracted from one. So this is what we're going to use to substitute in the second equation. Correct. So 50 is equals to that is our second condition. We have substituting. So fifties equals two m times one minus. This is to the right. So I will first write that Well, otherwise we confusing so too. Okay, so m times one minus. This is nothing, but it is two minus k square, so we can write one minus 35 AM squared and on solving this We get one minus one square, minus two in 2. 35. I am so that ISS 70 by M plus square off this so square of 35 years. 12. 25 What? M squared. Okay, so now we'll open the bracket. So we'll get one minus one plus 75 a m minus 1225 or M square, so we can cancel one minus one. Okay, so what we have here we have. If you multiply this M inside, we'll get 70 minus 1225 or mm. Square. And we know the right hand side is how much? 50. Okay, so we again in touch in the position as we have done earlier. So this will be here, and this will go here. So we'll have 1225 or M square. Sorry. And was canceled when we multiply this in this right. So 1 25 I am 1225 a m is equals to 70 minus 50. So what we get here, So when we can invert it so on inverting. So you get em over 1 to 5 is equals to 70 minus fifties 20. So one hour, 20. Multiply one toe. Five on the right hand side. So we live m s. 122 5/20. So if we divide 10th, this will result in point. And by dividing by two will get 61 25 Okay, so that's around. So for Maxim learning off no.

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