2-hemodialysis-is-a-process-by-which-a-machine-is-used-to-filter-urea-and-other-waste-products-from-

Question

2. Hemodialysis is a process by which a machine is used to
filter urea and other waste products from a patient's blood if
the kidneys fail. The amount of urea within a patient during
dialysis is sometimes modeled by supposing there are two
compartments within the patient: the blood, which is
directly filtered by the dialysis machine, and another com-
partment that cannot be directly filtered but that is con-
nected to the blood. A system of two differential equations
describing this is
$$\frac{d c}{d t}=-\frac{K}{V} c+a p-b c \quad \frac{d p}{d t}=-a p+b c$$
where $c$ and $p$ are the urea concentrations in the blood and
the inaccessible pool (in $\mathrm{mg} / \mathrm{mL} )$ and all constants are
positive (see also Exercise 14 in the Review Section of this
chapter). Suppose that $K / V=1, a=b=\frac{1}{2},$ and the initial
urea concentration is $c(0)=c_{0}$ and $p(0)=c_{0} \mathrm{mg} / \mathrm{mL}$
$$\begin{array}{l}{	ext { (a) Classify the equilibrium of this system. }} \ {	ext { (b) Solve this initial-value problem. }}\end{array}$$

Ashley Boni · Accepted Answer

So we have the following system and first we want to classify the equilibrium. So in order to do so, we need to rewrite this stone in matrix form. So we have, um e c e d T and dp dt which we can just call the ex d. T as a vector. And this is equal to some matrix times. Uh, c p because those are labels. And so it&#x27;s filling this matrix. Um, so let&#x27;s rewrite D c e d t. Since we&#x27;re given that A and B, you&#x27;re 1/2 and k over V is one. Ah, this is just equal to negative C plus 1/2 p minus 1/2 see, or 1/2 p minus 3/2. See, So you can feel that in our matrix here s O. P. Needs to be multiplied by 1/2 and see he needs to be multiplied by negative. We could be right. Our second equation also as negative 1/2 p plus 1/2 seat soapy needs to be multiplied by negative 1/2 and see needs to be multiplied by positive 1/2. So the sir matrix. So we need to find I convey I use and I can factors. Um, so we&#x27;ve done this a lot by this point, so we just find the determinant of Ah, this matrix a minus lambda I and we get the Lambda. Wanted to are equal to negative one closer, minus square to two over too. And then we saw the system. Um, a man is ling chi times the equal zero to get our wagon vectors, which come out to be, um, one and one plus or minus square to two. So to classify the equilibrium, which is what this question is all about, um r Agon values rewrite those reminder we have that they are the same sign which tells us that we have a node and this What is their sign? Their signs Negative. So this tells us that we have a stable equilibrium So we have a stable node now for party. We need to solve the initial body problem. So we know our general solution is equal to some constant times Eat of the land that won t So we have negative one plus squared a two over to tea times. Corresponding. I am vector one and one poor square to two, plus another constant heat of the Lambda to tea. So we have minus and corresponding Aiken vector with a minus sign. Don&#x27;t order us all for the C one and C two. We need to use the initial condition. So using the initial condition, um, X zero is equal. Thio, uh, CEO zero times p 00 which are both given to be, uh, these value. See? Not so if the actually plug zero into our general solution. We see that exponential go away. So we get, um, c one times the first Aiken vector plus C two times the second in vector. So this tells me that, um, c one plus c two equals this sea, not value. Those is just, uh, see nauseous. The number do you want to see? Two are actually constants that we&#x27;re solving for. So that&#x27;s the first equation. And second equation tells me that one full square to two. So you want close? One minus squared. A juicy too asked equal zero, do you not? So now let&#x27;s, um, substitute C two equals c not minus seat one into this equation. So we&#x27;re gonna leave the one postcard to to see one plus one minus square to two instead of C two. I&#x27;m gonna put see, not mine issue on. And this is equal to see Not so let&#x27;s combine our coefficients off. See one. So I have one plus square to two. And then I also have, um, a minus this. So we have the opposite of that. So we have negative one plus square to two, and this is all multiplied by C one. And we have, um, C zero on the other side. So if we move the sea, not to the other side, we have the opposite of this again. So negative one plus squared too. Plus the one we already have on the right hand side. So one in minus one. Cancel one minus one. Cancel here. So we get to square roots or to see one equals square to to see. Not so solving. For C one, we could see one equals square root of two over two square roots or 20 square. It&#x27;s cancelled. So we get C one equals 1/2 seen hot. So we have this now we&#x27;re gonna plug it in over here to get seats, so it&#x27;s see, not minus C one, which is 1/2. See? Not so we get 1/2. See, Not again. So C one and C two are both equal to each other. So now all we need to do is substitute those in for our general solution. So I&#x27;m gonna get rid of, uh, you wanted to see to hear. And we can substitute in our constants we found So see, one is 1/2 see, not and see to is the same value. So this is our final solution to the initial value problem.

Mike Gaerlan · Answer

all right. So part a asked us to explain each term in the equilibrium or in the system of differential equations here. So if we notice here, these two terms are the same in both equations. Because, um, here, see, is in the blood represents the amount in the blood. So if you look at a P first, this is the our rate of transfer rate of Yuria. Uh, this is from the blood. Sorry. Not from the blood from the pool to blood. Okay. And then BC is gonna be the rate of Yuria. Um, be going the other direction from blood, uh, to the pool. Now, the first term here. Now, this is the hemodialysis part. Okay, the K over VC, this is the rate of hemodialysis. You know, part be asked us to construct the face plot, so let&#x27;s do so by first looking at the, um no clients. So for the sea, no client, we have negative K v. Um, plus a P minus specie is equal to zero. So we have that Pete is going to be able to be over a time C plus que over V also time, see, Or we could back out the sea and it becomes, um, k over V all that time. See, you know, for the wine no client we just have here. P is going to be equal to, um or sorry, the negative AP first plus B c is equal to zero. Or we get that. Um uh, he is gonna be be over a time. See here this slope, we&#x27;ll always be greater than this slope. So when we construct that face plain and everything should always be positive. So we have. See here, pee here. Okay, we have this, um, and then this should be slightly less slow. Okay, um, now, for let&#x27;s say that we have the 0.1 For example, if we plug in 01 as in Ah, see, zero. What? The test point here. Now we should get, um Let&#x27;s put that into our c prime. So we get negative K over V time. Zero plus a times one minus B time. Zero. That&#x27;s equal to a which is greater than zero. So we&#x27;ll be going to the right over here. And since we&#x27;re linear, we&#x27;ll also be going to the left in this region. Now let&#x27;s play the inn for our next one. So if you have negative eight times one plus B times zero isn&#x27;t me negative A which is less than zero. So in this region, we&#x27;re gonna be going down like so and then up in this region here, our equilibrium point is just going to be at the origin. 00 Okay, so then we&#x27;ll have this. So if you see here that all of our, um, all of our solutions, they will eventually go towards the, um 00 here. So for a part, be what happened to the Yuria concentrations as he goes to infinity. Ah, they the Yuria concentrations, they both go to zero concentrations, uh, go towards the equilibrium. So go to zero. Right, and then you&#x27;re done.

Fuzail Shakir · Answer

Okay, so we know the population&#x27;s blood volumes vi milliliters on that the blood is diverted through the allies dialyzers at a rate of K milliliters per minute. So therefore, we can say that at the start of the treatment, the patient&#x27;s blood contained CF zero is equal to see zero milligrams per milliliter Furia. So if we formulate the process of dialysis as an initial value probable, we thesis the initial value problem. And if we do so for the concentration, we can take the integral of both sides that plug for C a time tease equal to zero. We can plug and see not to sell for a final last year.

Fuzail Shakir · Answer

Okay, so we know that the rate at which Yuria is removed from from the blood is describing the equation of Cft, where Kay is the rate of flow through the diary to the dialyzers and me, the volume of the patient zeros, the amount of Yuria and the blood that Time t Z zero. So we had to evaluate the that status. The limit as be approaches infinity of the integral from zero to be off K over v times C zero Eat the power of negative Katie divided by V, which is equal to K over V. C. Zero times that limit. As he approaches infinity off Beat the power of negative Katy over over V divided by native Katy over v So this is equal to K me. C zero divided by Katia Feast Looked like that. Be divided by dated Katy Times the limit A And then we&#x27;ll have to evaluate e to the power of naked and Katy over V from zero to be so simplify things down. This castle&#x27;s with this, uh, we and hey, we can&#x27;t love the T, can we? No, it&#x27;s lovely tea stuck in there. All right, this cake is against love with this cake. So we end up negative C zero times eat the power of negative G&#x27;s. That&#x27;s one over won over E to the power of Katy over V multiplied by t. Evaluated from zero to be. This is equal to negative. C zero times one over E to the power off K V over V. Times be minus one over zero. It turns out, yeah, minus one over zero toe one over. Real small numbers. A really big number like infinity. So this is infinity. So therefore, this whole thing is equal to infinity. So it&#x27;s time Urgent.

Eduard Sanchez · Answer

Hi there. So for this problem we have some kind of bashing U. Of T. When she is following our older B is zero. The exponential of minus error ties the over the volume. So we know that art is the rate of law of glass through the dialyzers B is the volume and center to is the amount of urea in the blood at time T equals to zero. Dad when we need to calculate first is the integral from zero to positive infinity of this function. So we&#x27;re going to do that. You&#x27;re the infinity or the function UFT which is simply the discussion here and now we need to determine what of this. What are the of these quantities? Depends on the time Banks County. And we can see that the only turn to depends on T. Is the exponential function which has in its argument depends on T. So we can just we&#x27;re all the rest outside the integral in which is simply need to integrate this afforded. We are integrated this in the body old Bt which is the only valuable in here D. T. To make easier the ism integral. We can&#x27;t that that w we can we can do what we call a change memorials and then call w minus R. T over B. So that if we degree initiate this w we will obtain minus are over because that is a constant. The of the differential heat. No, we are going to put this information home so we can see from here. Uh and deal teams is my news the over our nine don&#x27;t live you. So putting that we can see that that factor that will came from these differential goes away where they want outside injured role. Um they create this way we can see that this goes with this and this with this we all just off this and we will have the exponential of value. This integral of course is not defining these intervals. We could do that but which is simply going to solve the interval and later on um return to to the national volleyball team. So we know that um the integral of the exponential is just the exponential function. So it&#x27;s that easy. So we will obtain the U. S. Financial on W. And we need to know who is W. So w is this in here? So we just put that and we need to evaluate in the initial Interpol&#x27;s which is zero to positive infinity. So we&#x27;re going to do that uh w is minus hard times the over B. And this is from zero to cool ass. Um infinity from zero. We know that the exponential function is just want so you know there&#x27;s we need to, you know when you have all played an integral, you first emulated in the upper limit which is infinity. But when you put black community in here you can see that it is now the financial, it is the exponential of minus ability. And when you see the graph of the exponential function which is this when we go to minus infinity, Biggs function tends to be zero. So yes, that&#x27;s zero and that is minus. We know that if the revelation in the upper limit minus the evolution in the and lower in it, so we&#x27;ll take minus Heat of zero because that&#x27;s the the lower limit and this science council and we will attain C. O. Um the o the exponential of zero. And you know the the exponential zero is want because in here it is one and will obtain the constant. These is zero. So where we are to obtain it is that the interval of this function give us the amount of urea in the blood at time zero. So that&#x27;s they&#x27;re not seeming amount of syria you can structure for block. So that is the interpretation of this office in the room, mounds of yuri, um our time, The Evil zero. So that&#x27;s the amount of period, the initial period. So we have extracted all of that amount of furia when we enhance the time to infinity because that&#x27;s the maximum we can strive from block. Thanks

Michael Cooper · Answer

in this exercise we&#x27;re asked to consider a situation of dialysis. And we are given a function. You represents the amount of your area in a patient&#x27;s blood in milligrams at a given time t where time is minute. Sorry. It took me a lot to verify. And the equation or in the problem but it says that. Uh huh. The rate is milligrams per minute. So time is in minutes. R. Is the rate of flow of blood through the dialogue. This machine. Mhm. Yeah. Oh mm B. Is volume of the patient&#x27;s blood in milliliters and Cease of zero is equal to the initial concentration of Yuria and blood. So looking at this R. Is the rate of flow divided by the volume. So that gives me the amount of blood that can how much of the blood flows through the ratio of the rate to the volume of the patient&#x27;s blood. So what percentage of the patient&#x27;s blood or what ratio of the patient&#x27;s blood is flowing through the machine each minute times the initial concentration times he raised to the power of negative are times time divided by volume. So um the more time that passes, the smaller this exponential. Um Yes exponential expression is the less the higher percentage or the less of the initial concentration remains. And then we are asked to evaluate Yeah you can&#x27;t a girl. Mhm. From zero. Yeah. Uh huh. Yeah. Yeah. 30. I believe this big away the integral from time equals 02 tiny those 30 minutes of this dialysis dialysis from shen with respect to T. Yeah and explain what I mean. So this integral written in terms of the actual equation actually gets simplified a little bit because our and the and see subzero which is the initial concentration are all constants in this situation. The only thing that is changing is time, time is my variable of integration. And so all of that can fall outside. Uh huh. The actual integral. And then I&#x27;m evaluating the furniture of 2 30. The definite it&#x27;s a girl E. To the power of negative R. T over B. With respect to T. Yeah. Yeah. Well and when I do that that is going to require us to do some substitution. So everybody to switch to just being able to draw wrong. Yeah. I&#x27;m gonna change my color to black. It&#x27;s so the problem is I&#x27;ve got this weird funky exponents in my exponential situation. I know how to find the integral of E. To the X power dx is just itself. Right? The exponent here has more than just the variables. I am going to do a substitution. I&#x27;m going to call function you that whole exponent negative. RT divided by V. And in order to evaluate an integral I need to know what the derivative of that is. So do you is just going to be negative are over being DT to make the substitution a little easier because I have a D. T. Here. I am going to find out what D. T. Equals by rewriting that. Mhm. So if I divide both sides by negative are over B. I get the DT equals negative B. Over our. Do you? And now I am going to substitute negative Rt over be for you. And I am going to substitute. No. Yeah T. T. With negative V. Over R. D. Use put the DT here and change that. I&#x27;m going to put to you here. I&#x27;m gonna rewrite all of this. Mm. Yeah. So I still have the R. I still have to be I still need the initial concentration. I am still evaluating an integral. Yeah I&#x27;m going to come back to the limits of integration in just a second and I still have but instead of raising it to the power of negative RT over mean I&#x27;m going to raise it to the U. Power. And instead of D. T. I now have negative. This is multiplied by negative being over R. D. T. Yeah simplifying that a little bit negative. The over our is still a constant negative B. Over our times are over. B cancels each other out. So. Uh huh. Look what happens to all of that. So this are going to cancel how this are because our divided by R. Is one. This baby is going to cancel how the Mississippi because a divided by B. Is one the negative is a constant. So I can go outside and can rewrite all this as. Uh huh negative. I still have my initial concentration of the definite and to grow you know where that came from. Get rid of it. The definite integral of he to new power. Okay that should be D. You because I rewrote D. T. As negative V. Over R. D. Use this was negatively over hard to do instead of D. T. That makes this. Yeah. But then I&#x27;ve got these limits. I have changed my variable of integration from T. T. You. So I need to change the limits from T. T. You as well. If T equals zero you is going to equal negative. Are times T divided by a Z. Are divided by the negative are times zero is zero divided by via zero. So when T equals zero U. Equals zero. When T equals 30 you is going to equal negative are times 30 divided by the so my upper limit of integration is going negative. 30? Mm hmm. Over. Okay. Got Okay give us me definitely integral that I can that I can find the anti derivative for I know that the anti derivative of each of the U. D. U. Is just each of the U. Power. So this all becomes negative. Initial concentration keeps Sorry about that initial concentration times. Yeah. Uh huh. Yeah. To the youth power Evaluated over the integral from zero To -34. Right? Yes. And then evaluating that I still have initial consultation tom. Uh huh. Okay. Very quickly being as one else today, times Plugging the negative 30 are over be into. Each of the power gives me me to the negative. Uh huh. There are over B. Yeah Plugging zero into the eat the U. power anything to visit all. Power is one. I can multiply the initial concentration through. And given the initial concentration of syria and the blood minus. These are the power of the amount of time. 30 minutes times are divided by the total volume of blood. And this will give me the total milligrams of syria that remain in the patient&#x27;s blood. The initial amount of area and the patient&#x27;s blood minus the amount of your area that has been removed. Which is based on the rate Of the patient&#x27;s blood flowing through. As a ratio to the total amount of patients blood. Multiplied by the amount of time that has passed 30 minutes. Oh okay. Uh huh.

Problem

3. Prostate cancer treatment During the treatment…

Problem 2 Medium Difficulty

Answer

$\frac{1}{2} c_{0} e^{\left(-1+\frac{\sqrt{2}}{2}\right) t} \left[ \begin{array}{c}{1} \\ {1+\sqrt{2}}\end{array}\right]+\frac{1}{2} c_{0} e^{\left(-1-\frac{\sqrt{2}}{2}\right)} t \left[ \begin{array}{c}{1} \\ {1-\sqrt{2}}\end{array}\right]$

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$\frac{1}{2} c_{0} e^{\left(-1+\frac{\sqrt{2}}{2}\right) t} \left[ \begin{array}{c}{1} \\ {1+\sqrt{2}}\end{array}\right]+\frac{1}{2} c_{0} e^{\left(-1-\frac{\sqrt{2}}{2}\right)} t \left[ \begin{array}{c}{1} \\ {1-\sqrt{2}}\end{array}\right]$

Biocalculus Calculus for the Life Sciences

Catherine R.

Heather Z.

Samuel H.

Michael J.

Vectors Intro

Vector Basics - Example 1

James StewartBiocalculus Calculus for the Life Sciences

Catherine R.

Heather Z.

Samuel H.

Michael J.

James Stewart

Biocalculus Calculus for the Life Sciences