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# The dye dilution method is used to measure cardiac output with 6 mg of dye. The dye concentrations, in mg/L, are modeled by $c(t) = 20 te^{-0.6t} , 0 \le t \le 10$, where $t$ is measured in seconds. Find the cardiac output.

## Cardiac output is 6.6 $\mathrm{L} / \mathrm{min}$

#### Topics

Applications of Integration

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##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

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### Video Transcript

All right. So in this problem, uh, we're going to be using the dye dilution method to measure cardiac output. Now, we're going to be starting with six milligrams of died, and we're going to have a concentration function. Are concentration that's modeled by the function twenty e to the negative zero point six t dt. Right. And we're gonna let this run for ten seconds, and I write that down like this. This is going to be the demand of our concentration function. Zero from zero to ten. Now, we want the cardiac out. Now, the cart, the cardiac output is given by this function, you let a be the initial value. You know, our initial amount, which is six. And we let and we let that be over the inner girl over the domain of the concentration and so plugging everything we have in we get this in a girlfriend zero to ten. Twenty e to the negatives. You're a point sixteen dt, right? No. And so we're gonna want to solve this inner grow because that's the hardest part of this problem. Now, solving this inner girl, uh, basically boils down to missing a t hear t boils down to recognizing that this is a product of two functions. But one of the factors of product becomes very simple once I take its derivative the tea, right, Um, in the other products are the other factor of the product Doesn't get any more difficult. When I integrated er the e to the negative zero point sixty. And so this should be telling you that you should look at integration by parts his integration by parts for its really well, when you have a product of two functions and one of them gets easier when you take its derivative and the other one doesn't get any harder when you take its integral Okay, so we are going to want to let Well, let's just go to another page. So I'm going to write the twenty on the outside, right? The inner girl. We've got t e to the negative zero point six t d t. And I'm going to do integration by parts. I'm gonna let you equal t Will it do you equal DT That's what it is, right? So Devi is gonna be Our other factor in this product is going to eat to the negative zero point six t and the is just going to be the integral of that. So v is just going to be e to the negative zero point six all over negative zero point six, right? This is what happens when you take the inner girl of an exponential police for you. Just get the same thing, but times the divided by the derivative of the experiment. Okay, let's go ahead and write what this ends up being. So we want to take you times, Fi. You know, what's the integration by parts function formula? It's gonna be you times of you witches. T times need to the negatives. You're a point sixty all over and negative zero point six evaluated from zero to ten and then we're going to minus the integral of zero to ten. And now we have dear times, me. But do you times be is just be right. And so we have me to the native zero point six t all over negatives. You're a point six dt. And let's go ahead and close our privacy's. So this is going to be the answer that we want so we can start by either evaluating the left hand side or we can start by finishing the right hand side. That's going to start by finishing the right hand side. So we get twenty times. Okay, so let's leave this the same. Right. So now let's evaluate this righthand side. So when I evaluate the right hand side, I'm going to be doing exactly what I did before when I looked at the integral of each of the negatives. You're quite sixty. You just divide by negative zero point six again. So what do we get? We get minus E to the negative zero point sixty all over. And now we get night of zero point six squared, and this is going to be evaluated from zero ten. Okay, go ahead and close our parentheses here. So the next thing we need to do is just evaluate all this. So let's go ahead and rewrite this The next bridge. So we still have our twenty out here way have tee times e to the negative zero point six all over negative zero point six, and then we have We're now minus ng uh, e to the negative. Zero point sixty all over. Negative zero point six squared now noticed that negative zero point six squared is the same thing, is just taking positive. Zero point six squared. So let's go ahead and erase this negative here, and we're going to be evaluating from zero to ten. Go ahead and put the wine the evaluation line here, although I'm sure you know what I mean. So, one, we plug in ten. What do we get? Wow, we get twenty. Always have the twenty five year. And so we have ten times. There's supposed to be a team here, ten times e to the negative, six all over negatives. You're a point six and then we have minus E to the negative. Six all over, zero point six squared. Okay, so now we have to subtract What happens when we play gin zero while when we plug in zero, this left hand side becomes heroes. We don't have to worry about that. And this righthand side here is going to become, uh, he. So it's just going to become one over's your point six square. So we're going to be adding one over's. You're a point six words. All right, so this is our final answer for the integral and you can suck you, Khun, multiply in this twenty blah, blah, blah. But the key here is that this is an applications problem. So you're going to want to plug this into your calculator to get actual number. The actual number that you're going to get when you plug this in is going to be right around fifty four point fifty nine. So the next thing we need to do is we need to finish out the problem by figuring out the actual cardiac output. And so the cardiac output is going to be six over. It's gonna be roughly six over fifty four point fifty nine and said, this is going to be roughly zero point eleven and this is going to be in leaders for seconds. Uh, you, Khun multiply by sixty two also rewrite this as six point six leaders per minute, right? And so this is going to be our final, our final answer with in two different units that leaders per second and leaders from minute uh And so that's the That's the gist of the problem. So the main point of this problem is that it's an integration by parts problem, and you can recognise that by looking at the integral and realizing that it's a product of two functions where one of the factors get simpler when you take the derivative and the other factor doesn't stay hurt doesn't get any more difficult when you take its in all right.

AH
University of Miami

#### Topics

Applications of Integration

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

Lectures

Join Bootcamp