dialysis-treatment-removes-urea-and-other-waste-products-from-a-patients-blood-by-diverting-some-o-3

Question

Dialysis treatment removes urea and other waste products from a patient's blood by diverting some of the bloodflow externally through a machine called a dialyzer. The rate at which urea is removed from the blood (in mg/min) is often well described by the equation $$c(t)=\frac{K}{V} c_{0} e^{-K_{I} / V}$$ where $K$ is the rate of flow of blood through the dialyzer $(\mathrm{in}$$\mathrm{mL} /\mathrm{min} ), V$ is the volume of the patient's blood (in mL) and $c_{0}$ is the amount of urea in the blood (in mg at time $t=0 .$ Evaluate the integral $\int_{0}^{30} c(t) d t$ and interpret it.

Fuzail Shakir · Accepted Answer

So this is our integral. We end up with K divided by V. C zero multiplied by e to the power Negative. Katie divided. Be divided by t over V and we&#x27;re gonna evaluate this from 0 0 to 30 and that&#x27;s your final answer.

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.

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

Patrick Vaughan · Answer

Hello. So this is a nice problem. So let&#x27;s read it. Dialysis removes Yuria from a patient&#x27;s blood by diverting some blood flow externally through a duct. Violet dialyzers The rate at which your he is removed from the blood and milligrams permanent leader is modeled by this formula right here where K is the rate of blood flow and Mel leaders per minute T is the time in minutes and V is the volume of the patient&#x27;s blood and mill leaders. And that would be in milliliters cute and then see not is the amount of Yuria present in the blood in milligrams at time. Zero. So this problem asked us first to you calculate the partial derivative of tea. Then it&#x27;s asks us, Teoh, calculate the partial derivative with respect to um Oops, Sorry. The first one have a typo. So the first one is the partial derivative with respect. Okay, So with respect, Teoh, the rate of the blood flow then the second one is with respect to TV. So with respect to time and then the third one is the concentration on Syria with respect to the patient&#x27;s blood volume. So I&#x27;m gonna do these in a color coded fashion, starting with the partial derivative of that function with respect. Okay, so the partial derivative of sea with respect to Okay, well, we have two K&#x27;s in this. And first, let&#x27;s just rewrite the function so that all of the V&#x27;s are, uh, actually, no, let&#x27;s just let&#x27;s just go straight at it. Let&#x27;s separate the constants. So see, not be from the variable of interest, which is K So we have K and we have e, um to the negative TV Overbey. That&#x27;s also a constant times. Okay, So as a reminder, when we&#x27;re confronted with a variable times the variable again, we have to do the derivative by parks. So this is what I like to do is, say, the left side times the derivative of the right side, plus the derivative of the left side times the right side retained. And I think that&#x27;s the chain rule product rule of differentiation. So when you use this method, then you mean that constant out front and then so then this becomes K. So the lefts entertained times. So first will treat this as something like either the acts. So what&#x27;s the derivative of the X Well, I just needed the X. So eat to the negative T for the times K and then times the derivative of that quote unquote X, which will It&#x27;s a constant give to Ruby Times. So the variable of interest cake so than that negative t Overbey comes down and then we add it. So the derivative of K, that&#x27;s just one times that right side retained. So then we can supply that out into our final form. So we have that partial derivative of R. C of our concentration of Yuria with respect to the blood flow is can be written as c not, um, Overbey times you to the negative, Katie. Over. B. So I&#x27;m gonna pull that out because we have that here and here and then times Negative, Katie Overbey plus one that would be my most simplified version of this partial derivative. So then it asks us to interpret what this is. So this is the change in concentration of, uh, Yuria. So the solid of urine with respect to you, the patient&#x27;s blood flow Oops, too, on the nose in there. And that&#x27;s what this new expression is telling us. Then we can move on to the second one. We&#x27;re taking me partial derivative. Come on. The partial derivative of this formula with respect to t So looking at this, we only have one team, the equation, So okay, the sea, not her all constants that could be pulled out. And then we basically air left with e to the quote unquote x, where t is the variable interest. So this becomes e to the negative k Overbey, and I&#x27;ll just isolate that times t. So it makes it really straightforward. So then what&#x27;s the derivative of X? Because the derivative even the access tv x times the derivative of the X So are quote unquote X here is going to be our t. So then that brings us down a negative K Overbey. So the final form here is going to be negative. K squared over b squared with that initial concentration times e to the negative k t over b. And that&#x27;s gonna be the partial derivative with respect to T. And what this is saying is the change in a concentration of Yuria with respect to time, and then finally will take that final partial derivative. So we have the partial derivative of our function with respect to, uh, be. And I&#x27;m gonna rewrite it so that we have all of our bees. So first we pull out of Constance R k and R C. Not. That&#x27;s a constant in the first part of the expression, and then we have one were V so I&#x27;ll just regret that is e to the negative one times e to the negative k t. That&#x27;s the constant times e to the negative one again because we have that one of her V. So again, we&#x27;re going to use that product. Real differentiation. So we have our the negative one times the derivative of our quote you. Yes. So we have e to the negative k t over b times the derivative of that quote unquote x. So that&#x27;s going to be our negative, Katie. And then we&#x27;re gonna add Well, what&#x27;s the derivative of you tonight? If one well at spey to the negative two times negative one and then we have our right side retained so e to the negative k t Overbey again. So let&#x27;s rewrite this. Make it a little more straightforward. So we&#x27;ve got so I&#x27;m gonna back up a step, so I actually made a mistake right here. So this should be the derivative of negative Katie beat in the negative one. As we just saw, The derivative of even the negative one is redid the negative two times. Negative one. So this becomes a positive Katie times. I&#x27;ll just add the negative, too. Right there. So since we have a veto, the negative too, in both parts of this expression. And I&#x27;ll just put there that we have our case, you know? Then you can pull out that needed the negative to you. So we have kay, not Kate. Times the initial concentration divided by V squared on. Then we also have this in both expressions, so e to the negative k t Overbey, and then this will be multiplied by Katie over B, which is what&#x27;s left in that expression. You pull out this components and that component and then minus, uh, one. So that would be the partial derivative of sea with respect to be and well, what does this mean? So this again means the change and the concentration of the area in patients blood with respect to the patients. What volume and zooming out so you can see all the different partial derivatives. We don&#x27;t care. So the reason why we would be interested in these is we might be interested in how the if the patients in critical condition, how that you&#x27;re in the area and their patients blood is changing with time. So if it&#x27;s changing quickly with time, that&#x27;s good. We&#x27;re pulling it out of the system. Maybe we&#x27;re interested in the changing concentration of that Yuria with respect to the patient&#x27;s blood volume. So maybe the patient has a large maybe a small volume of blood. So having even a low concentration of Yuria would be severely impactful to that particular patient and then this final one. So we wanted to look at the changing concentration of your re ever with respect some patients blood flow, so is the is a high concentration of Yuria, causing the patient&#x27;s blood to slow down. What a lower concentration cause they&#x27;re paid their blood speed up. That&#x27;s kind of thing you can look at by using this type of equation here. So that&#x27;s why we would be interested in taking all these partial derivatives, and this is sort of what each one implies

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.

Problem

Gompertz tumor growth In Chapter 7 we will explor…

Problem 60 Medium Difficulty

Answer

$C_{0}\left(1-e^{-30 r / V}\right)$

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James Stewart

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Integration Review - Intro

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$C_{0}\left(1-e^{-30 r / V}\right)$

Biocalculus Calculus for the Life Sciences

Heather Z.

Kayleah T.

Kristen K.

Samuel H.

Integration Review - Intro

Definite vs Indefinite

James StewartBiocalculus Calculus for the Life Sciences

Heather Z.

Kayleah T.

Kristen K.

Samuel H.

James Stewart

Biocalculus Calculus for the Life Sciences