Question

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?

Answer 1

WAITTTTTTT

Answer 2

go slow! :/

Answer 3

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

Answer 4

what's the first step?

Answer 5

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.

Answer 6

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)

Answer 7

"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?

Answer 8

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

Answer 9

what's cc??

Answer 10

cubic centimeters. Same as cm^3

Answer 11

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

Answer 12

lol familiar

Answer 13

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

Answer 14

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.

Answer 15

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

Answer 16

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:
\[\frac{C_{t+1}-C_t}{(t+1)-t}\]

Answer 17

lol what's this o.O

Answer 18

can you explain your notations? and also why?

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

i'll try to solve this...

Answer 23

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

Answer 24

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

Answer 25

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:
\[y_p'-\frac{3}{125}y_p=\frac{3}{125}0.2\]

Answer 26

Also as t goes to infinity Q approaches c1.

Answer 27

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.

Answer 28

ugh i cant do it :/

Answer 29

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

Answer 30

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?

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

So do you want to model concentration or quantity?

Answer 35

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

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

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.

Answer 40

q1 is like c1 right

Answer 41

okay go on

Answer 42

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

Answer 43

yup

Answer 44

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.

Answer 45

ohhh yes

Answer 46

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

Answer 47

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

Answer 48

go on..im following

Answer 49

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

Answer 50

Do you see why that is?

Answer 51

complimentaru function? what;'s that?

Answer 52

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.
\[y_p = 125*0.2=25\]

Answer 53

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:
\[y=Ae^{\frac{-3}{125}t}+25\]

Answer 54

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

Answer 55

wahhh what's happening

Answer 56

what's a complimentary function first?

Answer 57

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

Answer 58

what's X?

Answer 59

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.

Answer 60

is it the thingy in linear de?

Answer 61

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

Answer 62

mmm

Answer 63

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

Answer 64

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

Answer 65

so what's yc and yp in my formula?

Answer 66

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

Answer 67

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

Answer 68

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

Answer 69

what? e???

Answer 70

why is there e??

Answer 71

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

Answer 72

i kknow but why????

Answer 73

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

Answer 74

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

Answer 75

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

Answer 76

what the heck does e have to do with this?

Answer 77

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

Answer 78

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

Answer 79

wait..ill try to get this -_-

Answer 80

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

Answer 81

B is just a constant. We are solving for y_c.

And above I meant:
\[\frac{d(e^{Bx})}{dx}=Be^{Bx}\]

Answer 82

i really dont get your method :/

Answer 83

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

Answer 84

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

Answer 85

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

Answer 86

4?

Answer 87

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

Answer 88

why???

Answer 89

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.

Answer 90

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

Answer 91

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.

Answer 92

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

Answer 93

yes

Answer 94

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

Answer 95

where are you getting these numbers?

Answer 96

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.

Answer 97

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

Answer 98

this method*