28. Table 1.21 shows the concentration, C of creatinine in the bloodstream of a dog: (a) Including units, find the average rate at which the concentration is chang...

Table 1.21 shows the concentration, $c,$ of creatinine in the bloodstream of a dog. $^{41}$ (a) Including units, find the average rate at which the concentration is changing between the (i) $6^{\mathrm{th}}$ and $8^{\mathrm{th}}$ minutes. (ii) $\quad 8^{\mathrm{th}}$ and $10^{\mathrm{th}}$ minutes. (b) Explain the sign and relative magnitudes of your results in terms of creatinine. $$\begin{array}{c|ccccc} \hline t \text { (minutes) } & 2 & 4 & 6 & 8 & 10 \\ \hline c(\mathrm{mg} / \mathrm{ml}) & 0.439 & 0.383 & 0.336 & 0.298 & 0.266 \\ \hline \end{array}$$

Today we'll be finding the average rate of change of a few time intervals based on the table, and we will use the equation. The average rate of change equals F A B minus f eb over B minus a. The first interval. We have one comma two and, uh, for this one A is one. And these two so ffb will be 1018 And if the vehicle will be 0.33 divided by two minus one, and that will give us negative 0.1 5/1, which is negative, 0.15 and the unit is grams for a desolate ER over hours. For the second one, our intervals 1.5 comma two and at 1.5 it's 0.24 and again at 2.18 7.18 minus point 0 to 4 over two, minus 1.5. And that will be negative 0.6 over 0.5, which will give us negative 0.12 grams. Her desolate er over ours. The third interval. We have two commas 2.5, and at 2.5 it's 0.12 and as to its again 0.18 So we will have 0.1 to minus 0.18 divided by 2.5 minus two. I'll give us negative 0.6 over 0.5, which will be negative 0.12 grams for a desolate ER over ours. Our last interval is two comma three at 3.7 and that, too, is 0.18 So we'll have 0.7 minus 0.18 over three minus two, and I will give us negative 0.1 1/1, which is just 0.11 grams her. That's the leader over ours and that finishes off part A for part B. It asked us to find the incident rate of instantaneous rate of change. AT T equals two, if you look back in the chart were given two equals two. However, this is not what we want to be looking at. This number is not the instantaneous sort of change. What we need to do is find two numbers. They give us an average of two. For example, 1.5 and 2.5 and find the average rate of change between those two. And we will use again this equation, so the interval will be 1.52 point five. That's a bracket, and at 2.5 it's 0.12 and at 1.5 it's 0.0 to 4. So we'll have 0.1 to minus 0.0 to 4, divided by 2.5 minus 1.5. That will give us negative 0.12 over one, which is 10.12 grams for a desolate ER over ours. And what this means is, after two hours, the uh, the amount is decreasing by £0.12 per desolate as per hour.

Okay, so we have a function here, see? Of tea. And she is, what, zero point six, then times your point nine six Explain it is two ten t divided by fifteen hundred minus one. Of course. Three kinds Team over one. Twenty six T minus nine hundred. And then this factor one minus Syria point nine six. And then the same exponent two hundred ten t over fifteen hundred minus one. All right then, sir, Part a simply asks us to find Sea of one eighty. Come on, eight. Okay, so let me first say that losses sees the concentration of Yuria and the dialysis patient and T is the number of minutes of the session jesu rating Which body generates Yuria and milligrams per minute. Okay, so here we have one hundred eighty in a section session, generating a milligrams per minute of Korea. What did you get? So just flinging the calculator about your point two six or nine milligrams OK, and so good being there's a little bit crazy. So it asks us to approximate approximate approximate sea of one seventy nine. So what we want to do is there that this is sea of one eighty common aid, plus some change and see, it's given by Okay. DT is negative. Ten kichi is one and sir D C. It's gonna be It would be approximately equal to Okay, so we're using this differential. Took rocks beneath the change with titties may attend. Each is once a PC will be the partial of C with respect to t of one eighty eight times two teams, plus the partial with respected G one eighty nate Times the gene. And what I did here is, I just use a computer out of her system. This is one of these derivatives that I mean, if you really want to, you can you can to my hand But, you know, with thes systems, I hear the drilling on point. I'm doing this by hand on practice. You would use the computer to do this. So you do it. You take the partial derivatives, you get the computer to evaluated those points on you get that. The CT one eighty coming eight evaluated. Okay, is to your point zero zero zero nine five four tee tee's native ten c g Evaluated is so your point zero zero five one eight seven times one and then. So when you plug this in, you get that C one seventy nine approximately zero point two six zero. It's the concentration of Yuria.

In this problem we are given the blood alcohol content and units of milligrams similar leader after two minutes of consumption that E two equals 20.0 to 252.46 17 patients information. We want to answer any which requires us to use differentiation via the chain rule that is a derivative of a function F of X with respect to X. Is F prime G n x n G. Products which is a function inside about seven A. We use the general to find the rate of change of BF 10 minutes first, we find the prime th in the chain rule as 100.2 to 5.46 17 plus. By the product rule 170.467 times 0.0 to +25 to 0.46 17. Thus our solution is for plugging into equals tend to be primed for a. Be prime 10 equals 7.5, 10 to the negative 30 mg per milliliters per minute and b we find the rate of change is 30 minutes. For this we simply forget 30 giving be a third time of 30 equals three times and 93rd milligram milligrams per milliliters per minute.

Right. Were asked to answer a question about dialysis using derivatives of the log arrhythmic and inverse tangent function. Were given a model for the removal of the area from the bloodstream via dialysis we're told, given the initial Maria concentration measured in milligrams, millimeter is C. Where C is going to be greater than one. The duration of dialysis required for certain conditions is given by the equation. Time T equals the natural laws of three times c plus square roots of nine times C squared minus eight C over to mm breast catholic, the derivative of tea with respect to see and to interpret it well. You have that to crime to see as the independent variable well. But the chain rule, this is going to be two, two over three C plus the square root of nine C squared minus eight C times the derivative of the inside your perspective. See this is rehabs was 1/2 times one half times nine C square's minus eight seats in the negative. One half times 18 C -8. Police. Sir, this simplifies to 3/3 C plus the square root of nine C squared minus eight C. That's the yes, This is 18 c minus fate over. And then we have is three C. Then what? Plus the square root of nine C squared minus eight C. So times the square root of nine C squared minus eight season. You guys kiss? Do you have? Yeah. Look, that's guy who's Actually, I should divide through by two But this is 18 seed But that too is 9C -4 but right away, this is our answer in circus. Sorry, wearing all great, this is called. Now we want to interpret this this well, as a real change rid of a T. With respect to see this is the rate of change. Yes. Um should yeah, duration of dialysis? She has. Yeah. Yes. Just with respect to the initial you're really a concentration if you don't answer correct in the bloodstream, what do you think you're also yes. Yeah. Snake.