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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 the urea is removed from the blood (in mg/min) is often well described by the equation

$$ u(t) = \frac{r}{V} C_0 e^{-rt/V} $$

where $ r $ is the rate of flow of blood through the dialyzer (in mL/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 $ \displaystyle \int^{30}_0 u(t) \, dt $ and interpret it.

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

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Luke W.

April 14, 2020

what happened to the r/v from the original equation? is that somewhere cancelled out? if so where?

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in this exercise we're asked to consider a situation of dialysis. And we are given a function. You represents the amount of your area in a patient'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'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's blood. So what percentage of the patient's blood or what ratio of the patient'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'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'm evaluating the furniture of 2 30. The definite it'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'm gonna change my color to black. It's so the problem is I'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'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'm going to put to you here. I'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'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'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'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's blood. The initial amount of area and the patient's blood minus the amount of your area that has been removed. Which is based on the rate Of the patient'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.

University of Southern California