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Mathematics 31 Online
OpenStudy (lgbasallote):

APPLICATION OF FIRST ORDER DIFFERENTIAL EQUATIONS: blood carries a drug into an organ at the rate of 3 cm^3 / sec and leaves at the same rate. the organ has a liquid volume of 125 cm^3. if the concentration of the drug in the blood entering the organ is 0.2 g/cm^3, what is the concentration of the drug in the organ at time t? after how many seconds will the concentration of the drug in the organ reach 0.1 g/cm^3?

OpenStudy (lgbasallote):

WAITTTTTTT

OpenStudy (lgbasallote):

go slow! :/

OpenStudy (lgbasallote):

dont spam me with formulas lol i cant keep up..please explain step by step :(

OpenStudy (lgbasallote):

what's the first step?

OpenStudy (valpey):

Okay sorry, I just had to get it all out there. First I am saying that the change in concentration is equal to the weighted average of 125-3=122 cc's of current concentration and 3 cc's of drugged concentration.

OpenStudy (valpey):

Let y = the current drug concentration. y' = change in drug concentration per second? (I'm already sure I've done it wrong, so don't take anything to the bank just yet)

OpenStudy (lgbasallote):

"the change in concentration is equal to the weighted average of 125-3=122 cc's of current concentration and 3 cc's of drugged concentration. " what does this mean?

OpenStudy (valpey):

Ah wait, at t= infinity, the concentration in the organ must equal 0.2 g/cc. Now I am thinking: \[y'+\frac{3y}{125}=3*(0.2)\] I'm also looking at questions 39-44 from: http://college.cengage.com/mathematics/larson/calculus_early/3e/shared/chapter15/clc7eap1502.pdf

OpenStudy (lgbasallote):

what's cc??

OpenStudy (valpey):

cubic centimeters. Same as cm^3

OpenStudy (valpey):

You know, "I need 100 cc's of morphine, STAT!"

OpenStudy (lgbasallote):

lol familiar

OpenStudy (lgbasallote):

uhmm let's start this explanation wiht your first post...what did you do there? what formula?

OpenStudy (valpey):

The idea is that the concentration at t+1 will be 122/125 times the concentration at t plus 3/125 times the concentration of the drug.

OpenStudy (lgbasallote):

did you use \[\frac{dQ}{dt} = r_1 c_1 - r_2 (\frac{Q}{V + (r_1 - r_2)t})\]

OpenStudy (valpey):

Yes, I think that's where we get to. \[C_{t+1}=\frac{122}{125}C_t+\frac{3}{125}C_d\] The change in concentration should resemble: \[\frac{C_{t+1}-C_t}{(t+1)-t}\]

OpenStudy (lgbasallote):

lol what's this o.O

OpenStudy (lgbasallote):

can you explain your notations? and also why?

OpenStudy (valpey):

But I am worried they have a typo, because I get: \[y'=\frac{-3}{125}y+\frac{3}{125}(0.2)\]

OpenStudy (valpey):

the capital C is for Concentration, the sub-t+1 is concentration at time t+1. the sub-d is Concentration of the drug.

OpenStudy (lgbasallote):

okay im still not getting what you're doing -_-

OpenStudy (lgbasallote):

i'll try to solve this...

OpenStudy (valpey):

So like, the way I think their equation should read is: \[\frac{dQ}{dt}=\frac{r_1}{V}c_1−\frac{r_2Q}{V+(r1−r2)t}\]

OpenStudy (lgbasallote):

why over V? there shouldnt be an over V for ric1

OpenStudy (valpey):

I think r1/V because the dQ/dt approaches 0 as t goes to infinity so r1/V -> r2/V If I am right, the ODE becomes: \[y'-\frac{3}{125}y=\frac{3}{125}0.2\] This consists of a complimentary function which solves: \[y_c'-\frac{3}{125}y_c=0\] and a particular integral which solves: \[y_p'-\frac{3}{125}y_p=\frac{3}{125}0.2\]

OpenStudy (valpey):

Also as t goes to infinity Q approaches c1.

OpenStudy (valpey):

Again, if I am right we can see that \[y_c = Ae^{\frac{3}{125}t}\] solves the complimentary function for any constant A. For the particular integral, we can just look for constants.

OpenStudy (lgbasallote):

ugh i cant do it :/

OpenStudy (valpey):

The constant that solves the particular integral is -0.2 because \[0-\frac{3}{125}(-0.2)=\frac{3}{125}0.2\] Hence: \[y=Ae^{\frac{3}{125}t}-0.2\]

OpenStudy (valpey):

Okay, let's go back to this equation: \[\frac{dQ}{dt}=\frac{r_1}{V}c1−\frac{r_2Q}{V+(r_1−r_2)t}\] Do you feel like you know what each of these variables means?

OpenStudy (lgbasallote):

i was able to arrive with \[-\ln (25-Q) = \frac{3}{125}t + c\]

OpenStudy (lgbasallote):

yes but there isnt supposed to be a V why is there a V there? below r1c1 i mean

OpenStudy (valpey):

Oh, I see, for this equation Q isn't a concentration, but actually a quantity.

OpenStudy (valpey):

So do you want to model concentration or quantity?

OpenStudy (lgbasallote):

Q is the amt of drug already in the organ i guess

OpenStudy (valpey):

Let y(t) = the quantity of drug (in grams) in the organ at time t. \[\frac{dQ}{dt}\frac{g}{s}+\frac{r_2\frac{cc}{s}Qg}{Vcc+(r_1\frac{cc}{s}-r_2\frac{cc}{s})t}=q_1\frac{g}{cc}r_1\frac{cc}{s}\]

OpenStudy (valpey):

I just threw the units in there so we are clear that this equation is all about grams per second.

OpenStudy (lgbasallote):

okay you're plugging in a lot of thingies...please explain

OpenStudy (valpey):

s=seconds, cc = cubic centimeters, g = grams we will ignore units from now on. This is the same as: \[\frac{dQ}{dt}+\frac{r_2Q}{V+(r_1-r_2)t}=q_1r_1\] V is the total volume of the organ. r1 and r2 are the rates we are adding and subtracting fluid. q1 is the drug concentration. t is time.

OpenStudy (lgbasallote):

q1 is like c1 right

OpenStudy (lgbasallote):

okay go on

OpenStudy (valpey):

Yes we might have had c1 before. Fortunately r1=r2=3cc/s so they cancel out in the denominator.

OpenStudy (lgbasallote):

yup

OpenStudy (valpey):

so replacing Q and dQ/dt with y and y' we have: \[y'+\frac{r_2}{V}y=q_1r_1\] This is an ordinary differential equation.

OpenStudy (lgbasallote):

ohhh yes

OpenStudy (lgbasallote):

\[y' + \frac{3}{125} y = 0.6\]

OpenStudy (valpey):

Yes, and \[y=y_c+y_p\] \[y_c'+\frac{3}{125}y_c=0\] \[y_p'+\frac{3}{125}y_p=0.6\]

OpenStudy (lgbasallote):

go on..im following

OpenStudy (valpey):

The complimentary function will need to take the form: \[y_c=Ae^{\frac{-3}{125}t}\]

OpenStudy (valpey):

Do you see why that is?

OpenStudy (lgbasallote):

complimentaru function? what;'s that?

OpenStudy (valpey):

The possible functions which solve this equation so far are infinite. They can be thought of as a linear combination of complimentary functions who resolve to zero on the left plus a particular one which resolves to the expression on the right. The particular integral (I think that's what they call it) y_p can just be a constant since its derivative will be zero. \[y_p = 125*0.2=25\]

OpenStudy (valpey):

I think y_c is also called the characteristic equation. Anyway, now that we have the form of y_c and y_p we can combine them: \[y=Ae^{\frac{-3}{125}t}+25\]

OpenStudy (valpey):

But we have an important initial condition. When t = 0 there is no drug in the organ. So what is A?

OpenStudy (lgbasallote):

wahhh what's happening

OpenStudy (lgbasallote):

what's a complimentary function first?

OpenStudy (valpey):

There are two parts to the function y(t). Let's call them g(t) and h(t). y(t) = g(t)+h(t). For the purposes of our ODE which looks like y'(t)+Xy(t) = C, we could rewrite this as: g'(t)+h'(t)+X(g(t)+h(t)) = C

OpenStudy (lgbasallote):

what's X?

OpenStudy (valpey):

X is just some constant. Saying our unknown function y is really the sum of two unknown functions g and h isn't all that special.

OpenStudy (lgbasallote):

is it the thingy in linear de?

OpenStudy (valpey):

Yes, the thingy in linear de. What is special is that we distinguish g(t) and h(t) like this: g'(t)+h'(t)+X(g(t)+h(t)) = C g'(t) + X g(t) = 0 h'(t) + X h(t) = C

OpenStudy (lgbasallote):

mmm

OpenStudy (valpey):

We choose g(t) and h(t) this way because it makes it easy to solve.

OpenStudy (valpey):

So what I was calling y_c is g(t) and what I was calling y_p is h(t).

OpenStudy (lgbasallote):

so what's yc and yp in my formula?

OpenStudy (valpey):

\[y=y_c+y_p\] \[y_c'+\frac{3}{125}y_c=0\] \[y_p'+\frac{3}{125}y_p=0.6\]

OpenStudy (lgbasallote):

so after this what? that's when i got lost

OpenStudy (valpey):

Because the derivative of e^Bt =Be^Bt we know that is a place to start for our solution to u_c.

OpenStudy (lgbasallote):

what? e???

OpenStudy (lgbasallote):

why is there e??

OpenStudy (valpey):

Yeah, Euler's constant, the base of the natural logarithm. The button on your calculator. \[y_c=Ae^{\frac{-3}{125}t}\]

OpenStudy (lgbasallote):

i kknow but why????

OpenStudy (valpey):

Because \[\frac{d(e^{Bx})}{dx}=Be^Bx\]

OpenStudy (lgbasallote):

i know...but why e????

OpenStudy (valpey):

e^x is a special class of functions whose derivative is itself.

OpenStudy (lgbasallote):

what the heck does e have to do with this?

OpenStudy (valpey):

We need a base so that when we add up y' with By we get zero

OpenStudy (valpey):

functions with e^x in them are perfect for that job.

OpenStudy (lgbasallote):

wait..ill try to get this -_-

OpenStudy (lgbasallote):

okay... 1) what's By 2)why do we need to add to By 3) why must the sum be 0

OpenStudy (valpey):

B is just a constant. We are solving for y_c. And above I meant: \[\frac{d(e^{Bx})}{dx}=Be^{Bx}\]

OpenStudy (lgbasallote):

i really dont get your method :/

OpenStudy (valpey):

Do you see that?: \[\frac{d(7e^{4x})}{dx}=28e^{4x}\]

OpenStudy (lgbasallote):

i know derivatives ..it's your method i dont understand

OpenStudy (valpey):

Now what is y(x) if: \[y'(x)-4y(x) = 0\]

OpenStudy (lgbasallote):

4?

OpenStudy (valpey):

I'm looking for a function like: \[y(x) = 7e^{4x}\]

OpenStudy (lgbasallote):

why???

OpenStudy (valpey):

Right, so the whole fun thing about ODE's is that we are doing algebra to solve for unknowns, but the unknowns we are solving for are not numbers, the unknowns are entire functions.

OpenStudy (lgbasallote):

your method is so vague o.O what method is this?

OpenStudy (valpey):

I agree that it seems vague because we don't know where we are going. I am picking things apparently out of thin air, but because I know what we need to resolve to.

OpenStudy (valpey):

But you agree that if \[y(x)=7e^{4x}\] then \[y'(x)−4y(x)=0\]

OpenStudy (lgbasallote):

yes

OpenStudy (valpey):

So what about \[y(x) = 11e^{4x}\]

OpenStudy (lgbasallote):

where are you getting these numbers?

OpenStudy (valpey):

Right, so the 11 and the 7 are arbitrary. The point is that we don't know A yet, but we know that \[y(x)=Ae^{4x}\] is a class of solutions.

OpenStudy (lgbasallote):

i dont think this is necessary...i think linear de or variable separable would suffice...we didnt even discuss tmethod..

OpenStudy (lgbasallote):

this method*

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