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I cant solve this problem. Can you solve it for me?If the injection rate of a drug (R) is 7mg/min and the maximumsolubility of the drug Sm is 0,5mg/cm3, how much bl...

Question

I cant solve this problem. Can you solve it for me?If the injection rate of a drug (R) is 7mg/min and the maximumsolubility of the drug Sm is 0,5mg/cm3, how much blood flow Qis required to prevent the drug from settling? Be precise with theanswer.

I cant solve this problem. Can you solve it for me? If the injection rate of a drug (R) is 7mg/min and the maximum solubility of the drug Sm is 0,5mg/cm3, how much blood flow Q is required to prevent the drug from settling? Be precise with the answer.



Answers

When a drug is given intravenously, the concentration of the drug in the blood is $C_{i}(t)=250 e^{-0.08 t},$ for $t \geq 0$ When the same drug is given orally, the concentration of the drug in the blood is $C_{o}(t)=200\left(e^{-0.08 t}-e^{-1.8 t}\right),$ for $t \geq 0 .$ Compute the bioavailability of the drug.

In this problem were given to functions that represent the concentration of a drug. When administered intravenously, C I and then administered orally CEO were asked to calculate the bio availability. Example. Six of the section poses a very similar problem and we're told that when calculating by availability for each method of administration Shin, it's important to calculate the area under the curve. Therefore, we need to take the integral of the function of C I and CEO and those bounds. We're going to go from zero to infinity and then we need toe divide the mount for aural over the amount of intravenously which will give us the final answer by availability. So let's start by integrating the C I function suspect one well, just to ensure obviously so you know, we're calculating, so we have to take the integral of C I from zero to infinity. Already we see that 200 fifty's a constant so we can pulled outside the integral and we have eaten negative 0.0 80 d t. So now before we start integrating, we also realize that we have to apply limits because you have infinite so we can do 250 times the limit as a a purchase affinity of zero to a E to the negative 0.8 t do you cheap? And now we can take the integral. So first, that just right the rest of the expression, which will give us negative E to negative 0.8 t over a 0.8 and that will be evaluated from zero to a. Now let's plug in those bounds of a and zero who got 215 times the limit as a approaches infinity. So I'm plugging in a Remember, he has a negative power so we can write them the denominator. So negative one over E to the 0.8 a time 0.8 minus. But since we have a negative sign here will be plus one over. And then, since we'll have each of zero, that will just be one, so we'll have 0.8 So now me to evaluate the limit of each of the terms. The second term is really easy because it's a constant. The limit will just remain itself, So that will be one over 0.8 and this could be rewritten, actually, as 100 over eight. Or then you can simplify that to be 25/2 12.5, which is just easier to write. And then for the first term, the denominator will go to infinity of Approaches infinity, meaning that this whole fraction will get infinite Lee small finally approaching zero as its limit. So now we can, right? The expression has 250 times. Well, our first from zero. So you can ignore that 12.5, which is a limit the second term and then multiplying. Now we get 3125. But when you put it into your calculator, Okay, so this is our intravenous value salts. But intra now what we need to do is calculate the oral administration value so similarly Now we need to integrate our function from zero to infinity. Since we see there's a 200 we can already pull that out of the integral. And now let's right the rest of the function inside the integral. So even negative 0.8 e minus e to the negative 1.8 t DT similar to other function Now we can apply limits so 200 times the limit as a purchase, infinity from zero to a and then we'll just have the function. Now let's take the derivative of them. Well, first, I just read out the rest of the expression, so the integral of E to the negative 0.8 will be negative e to the negative 0.8 t over 0.8 which is what we did previously. And then the second term will be plus because we have a negative here you tonight of 1.8 t over 1.8 and that will be valued from 80 So now when you do is plug in those values. So 200 times the limit is a approaches infinity. So for first term, remember, he has a negative power for both of these, so I can write them in the denominator. I could have 1/0 0.8 times beat to the 0.8 a Puts one over E to 1.88 times 1.8. Now let's plug in zero. Since you have a negative sign in the first term, this walks to give us plus one over 0.8 because I will just be e to the zero, which is one then minus because this is a positive term, 1/1 0.8. Now all we need do is find a limit of each term. Well, the last part, we already know that the limit of the first term will go to zero, which is what we did over here. Similarly, since the denominator ghost infinity as a purchase infinity of the second term this term will also go to zero. Since the last two terms or constant, they will just remain the same. So now let's rewrite the expression 200 times 12.5 minus. And this would be, ah 100 over 1 80 which can also be simplified down to 10/18 and then finally five over nine. Okay, so now you're gonna you can plug that into your calculator and you're going to get something around 2388.8 repeating. Great. So now all we need to do and this is for or administration. All we need to do now is talking at the final bio availability, which will be the aural administration over the intravenous administration. So we're gonna get to 1003 188 when eight. Repeating over 3125 and that will give you a number around 0.7644 which will give us your final answer. Weaken rounds it up of 0.76

So for the first part of this, we're trying to find out when this equation here achieves a maximum. So we need to first take the first right of here. You get sea of negative B into negative bt. Thank you to the part plus rather so plus A. It's in the negative. Okay, now, since we want this to set equal to zero, I want to get the same denominators here, what will happen is that we can further see here, we have negative B. E. To the 80. That would be plus A. E. To the B. T. If you want to get like the same denominators because of the negatives they go down. And so we end up with this being equal to zero. Right? So I'm going to go ahead and isolate get the bring the A over to the other side here and I get negative A. You to the Bt here equals negative B. Into the 18th. And so that when we do from here is I'm going to divide by negative view on both sides. I have you to the A. T equals A over B into the Bt. And so then from here, I'm gonna divide by E to the beats you to both sides. And so what that simplifies down to is A over B equals E to the A minus B types team. Right, divide you subtract the exponents. And so taking the natural log of both sides now, so, alan of a over B is equal to a minus B. I'm steve. And so then I have T is equal to and then a overpay divided by you might just be okay. And that's the answer to, you know, the first part since seconds. All right. So now, let's take a look at the second derivative. So, second rate of here. So that's when it's going to be decreasing the fastest. Right? So, we want second derivative, which is just see of b squared, it's the negative VT minus a squared, heats the negative A T. Here. Now this well look very familiar because it's going to just be C b squared E to the 80 may be minus A squared E to the BT calling zero. And so because we're going to end up with a similar notion to what we had from the first part, this will just be the same thing as doing Ln of a squared B squared over a minus B equals T. Okay. And since they're both squared, we could actually simplify that down to two allen of ap for a minus B. So this is what our solution would be for that.

Hi. So now we are going to see an application of multi variable functions. So in this case we have this value Z. That will model the concentration of milligrams bleacher of some drug on blood. Okay. So what happened is that this concentration is given as a function of two variables X. The amount of drugs but we we inject and the the time signs the injection. Okay so this is the fine esti exponential -65 miles x. And the values of X is between zero and four. Anti is just any positive value. Okay so the first thing is to calculate f of free two. So if we replace the values we have here two exponential -2 times five minus three. And we can observe that this will be to e to the minus walk. So there's the value. But what is the meaning? What is going to be the units of this function? Well, this represent as a mentioned before the concentration of milligrams of the drug per litre on blood. So this is milligrams per liter of blood the patient in this case. And what is the interpretation of this? What is the meaning? 3? And to hear the function? Well let's remember the T. Is the time in hours. Science the science injection and next is the amount of drugs and and measured in milligrams. So basically this means that we're measuring the concentration of the drug of the period of the drug in the patient. If we yeah object Here I'm going to buy like three milligrams of the drug And then we wait two hours concentration of the drug in the patient. If we inject after Yeah two worse. And that concentration we can see is too. So that concentration is equal to eat the -4 which is around 0.0 36

Welcome to this lesson in this lesson we've been given why that is the concentration of drug in a blast room after you turn it into the Margo and we relaxed to find the the time I watched the concentration its maximum. So this would required us to take the Differential of why equated to zero then find the tea from there. So A the N. C. Regular Constance. Okay. Yes, speaking, why prime would be called to the sea is out, then this is negative B. Don't last A. Yeah. Uh huh. Okay. Oh so if this is equal to zero, we would have, they get to, so for months From one point we can buy through mhm. By sea. Oh so that we have zero. Let's go to negative B. Yeah. Uh huh. Yeah. Okay. What so this becomes B. Then E. Let you be T score too. Okay. Yeah, negative 18 at this point can take the land of both sides. Oh and this becomes. Yeah. Yeah. Mhm. Yeah. Oh, so this is land be minus meaty, mm hmm. Where does it go to Len A -80? So you have to be minus late eh? That is a call through negative 18. Bless Bt. So that is lend B minus line A. And I can fuck the out A sort of B minus A. So T. S accord to land b minus name A B minus a old len A minus lindy All over A minus B. Okay. Mhm. So in more simplified way we can make it land A on the all over a minus B. All right. So this is the time that the the constant question will be my maximum. All right. This step by the next part, we are looking at the time at which the concentration of the drug Decreasing most rapidly. So 4: one. We'll take the second differential of high. We had the first differential as all of this. Mhm. Oh yeah. Oh yeah. Okay. Yeah. So the second friend shell would become a CLS still that is negative. Yeah that is B squared E. Yeah. You get two big squared yeah. Oh okay. So if we quit this to zero again Mhm. Then it goes through the core process. So but we have Squired, eat the poor negative 80 Mexico to be squared a. Of the squid. Eat the bar negative B. T. We now take the land on both sides. Mhm. Uh huh. Yeah. Yeah. Mhm. And we have Mhm. Land is quite minus 80 score to land B squared minus B. T. So that we can have yeah learning squared minus Land being square. That is a quote negative bt last Katie. Okay so land is squared on the squared sick or to tease out the A minus B. So the time is lane is squared all over P squared. Then that is a minus and which is simplified to two. We stripped the squared from them. Yeah. So this is twice the time of the first one, so at this point received it the last two, the concentration in the drug is decreasing most rapidly. Yeah. Yeah. Mhm. Okay, so thanks for time this is the end of the lesson.


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