# Homework Help: Problem Solving Question - Polynomial/Exponential Functions

1. May 23, 2010

### Liparulo

1. The problem statement, all variables and given/known data

When the kidneys eliminate a chemical from the blood, they tend to eliminate a certain proportion each time period. For example, the average person eliminates about 13% of the caffeine in his or her body each hour. In doses larger than 20mg, caffeine can act as a mild stimulant (it can be toxic in large doses). Assume that a 375mL can of Cola has about 45mg of caffeine.
Suppose that you have a can of Cola on the hour for three hours. If you had your last Cola at 10 p.m, when can you expect the caffeine that you have ingested to stop having its effect?

2. Relevant equations

All the information is supposedly above. The teacher did say that Microsoft Excel could be used.

3. The attempt at a solution

I'm really stuck. I asked someone else and they said that the equation they would use would be dC/dt = -.13C. Could someone explain what they meant by this? I don't understand why it's a negative. I don't want to answer, just some assistance as how to start. It's been years since I've used Excel and I really don't know how to integrate the problem. I'm sorry that there isn't much of an attempt at solution, but I'm just stuck as how to start. Thank you.

2. May 23, 2010

### The Chaz

1. (As you know...) you need to express mg of caffeine as a function (of time), and then you need to find when this function dips below 20.
2. It's negative because the amount of caffeine is decreasing.
3. When the rate of change of "f" is proportional to the value of "f" (as in this case), 1. ends up being the exponential function. If you want to derive this yourself...
4. "Separate and integrate". Get all of the differentials "on top" (i.e. in the numerators), and all of the "C"s on one side... "t"s on the other. you'll have something like C/C. Do you remember which function has this derivative?

3. May 23, 2010

### Tedjn

A stepwise way to work on this problem is to make a table:

Code (Text):

Hour     Caffeine (mg)
0        45
1        45 - 0.13*45 + 45
2        ...
3        ...
The only thing to remember is that you add 45 mg of caffeine at hours 0, 1, and 2. This is probably where you would use Excel, giving each cell a formula based on the previous cell. You look for when the caffeine in the body dips below the stimulant threshold, which seems to be 20 mg.

To solve this problem exactly, you would use a differential equation, for which you wouldn't need Excel. That your teacher suggested Excel seems to indicate that he/she didn't expect you to solve a differential equation.

4. May 24, 2010

### vaibhav1803

try looking at it like this,
time rate of fall(caffiene conc.)= 13% of present conc.
{-d/dt}[C]= .13C...
{-d/dt}=rate of fall....if u haven't guesed by now.

this approach is although wrong let me tell you since you mentioned that the caffiene elimination occurs on an hourly basis.
it's a discrete function of "n",n being no. of hours passed since the caffiene was consumed.
try work it out that way purely in symbols(even for the 13% thats given) without excel and by yourself to get a taste of problem solving.
find the discrete function youre looking for and post it for verification, we're all here to help you any further only you should give it a fair amount of tries to know the problem well.

Last edited: May 24, 2010
5. May 24, 2010

### Liparulo

Okay. I've walked away from it for a little while, but I'm still not sure whether I'm on the right track. Here's what I have so far.

The amount of caffeine in your body after a single can may be represented as a function:

C(t)=-0.13c
Where C is the caffeine in your body t hours after ingestion – 13% is expressed as a negative value due to the caffeine in the blood decreasing.
We know that when the can is first drunk at 0 hours (t), we can assume from the above information that the body will have received 45mg of caffeine. Label the first hour as:

C1(0)=45
Then solve dC1/dt= -0.13Cwith respect to C1(0)=45.
dC1/dt= 45-0.13(45)
dC1/dt= 39.15
dC1/dt= 39.15-0.13(39.15)
dC1/dt= 34.06

Anywhere near solved/on the right track?

Last edited: May 24, 2010
6. May 24, 2010

### vaibhav1803

is there also an hourly consumption of caffiene.?..i apparently seem to have missed that
C(t) is positive , its the rate of change thats negative.

7. May 24, 2010

### vaibhav1803

ohk its like this c(n) represent caffiene amt. in ur body,assuming u are taking cola for the first time in your life and consumption starts @n=0, n number of hrs passed since u drunk cola(all values in mg, unless mentioned as something else)
"a" is caffiene supplied per drink, k is the fraction excreted an the hourly basis.
c(0)=a
c(1)=a(1-k)+a=(c(0))(1-k)+a
c(2)=(c(1))(1-k)+a={expand c(1)}=a(1-k)^2 +a(1-k)+a..notice some pattern..?
c(3)=(c(2))(1-k)+a={expand c(1)}=.....this ones for you.....=you will notice a pattern.
find c(n), post the formula(we'll verify,yet again)

8. May 24, 2010

### vaibhav1803

to end it you notice that, its the sum of a GP, probably your teacher didnt expect you to know that...
c(3) is your answer, then find out after how much time c(3) will decaty below 20mg
hence solve by sayin tha after c(3) is achieved you lose k th part (0.13 here) of the present caffiene per hour,
c'(1)=c(3)(1-k)...c'(n) represents caffiene amt. n hours after consumption stopped
c'(2)=c(4)(1-k)=c(3)(1-k)^2=c(3)(0.87)^2...c'(n)=c(3)(1-k)^n
we know that effect must stop once caffiene drops below 20 mg
thus we must solve....c'(n)<20 mg in order to determine the time(n) at which the effect stops after having the last cola
(remember you will not get an integral n for this this you must report the answer choosing thw value wisely)
if you calculate you get n=14.36 hours thus, i will thus choose "14 hrs" as my answer if it requires to be an integer, otherwise its on the platter for you.

9. May 24, 2010

### Liparulo

Ah, thank you! I think I understand now. I'll let you know once I've finished.

10. May 24, 2010

### Liparulo

Okay, here is what I have so far:

The function C(t)rrepresents the amount of caffeine present in your body, t symbolizing the number of hours passed since the first cola, assuming that there is no other cola or caffeine present from previous days. Let a represent the caffeine supplied per drink and k the percentage excreted on an hourly basis. Therefore:
C(0)=a or C(0)=45
With C(0) representing the level of caffeine present in the body at 8:00PM. 13% of this value is then cleansed from the body and we end up with the following formula:
=a(1-k)
The person then drinks another cola adding a further 45mg of caffeine to the body, shown as follows:
C(1)=a(1-k)+a
Or:
C(1)=(C(0))(1-k)+a
The values for a and k can then be substituted into the formula.
C(1)=45(1-13/100)+45
C(1)=84 3/20 mg of caffeine
In other words, after ingesting the second cola at 9:00PM, there is 84 3/20 mg of caffeine in the body. Repeat this formula again, except this time for C(2).
C(2)=(C(1))(1-k)+a(1-k)
Or:
C(2)=45(1-13/100)+45(1-13/100)+45(1-13/100)
C(2)=117 9/20 mg of caffeine

Any good?

Last edited: May 24, 2010
11. May 24, 2010

### The Chaz

(edit... I just now reread the actual question, with the "drinking a cola on the hour" stuff... the following solution is for solving the differential equation dC/dt = -.13t, which I now believe does NOT model our situation)

Lip, I don't mean to be pompous, but there are some wrong answers on this thread. Here's the method. (And I have "mis"quoted you so that I could edit a few of your statements. The grey-ed out stuff ain't right)

solve dC/dt = -.13C (we are going to integrate, and I am going to use "P" for the constant of integration since "C" is the name of our function!)
dC/C = -.13dt
ln|C| = -.13t + P (integrating u/u gives natural log, and integrating a constant gives a linear function)
exponentiate...
C = e^(-.13t + P) = e^(-.13t) * e^(P) = e^(P) * e^(-.13t).
Since e^(P) is a constant, replace it with just P

C(t) = Pe^(-.13)

General equation.
Using the initial value C(0) = 45, we solve for P, which turns out to just be 45.

C(t) = 45e^(-.13)

Specific equation.

To see when this = 20, solve
20 = 45e^(-.13t)
20/45 = e^(-.13t)
ln(20/45) = -.13t
ln(20/45)/-.13 = t.

12. May 24, 2010

### The Chaz

Ok, here goes nothing!

"you have a can of Cola on the hour for three hours" -

I assume that this means you have a can at 8pm, 9pm, and 10pm. This is THREE cans, but only TWO hours actually pass between your first and last can.

We want to find when the amount of caffeine in your body ( = C(t) )is less than 20mg.

We can measure the impact of each can (i.e. the amount of caffeine that each can contributes individually), and - since there are three cans - call these C8(t), C9(t), and C10(t). (the subscripts are the time that they were consumed. You could also just put c1, c2, and c3)

C(t) = c8(t) + c9(t) + c10(t).

Now let's find these!

C10(t).... well, we know that the rate of change of caffeine is proportional to the amount (A, but I'll switch back to "C" in a minute) of caffeine in the body, which is modeled by
dA/dt = kA.
Solving this using the methods in my previous post yields
A(t) = Pe^(kt).

"body eliminates 13% of the caffeine per hour" -

this does NOT mean that k = .13 (or -.13). This means that A(t+1) = .87*A(t).
(That last line is pretty dense. The .87 is how much is LEFT after losing 13%.)

For example, if you start with 100mg in your system, one hour later you should have 87mg.

Solve A(t+1) = .87*A(t) using t=0 for simplicity.
A(0+1) = .87*A(0)
Pe^(k*1) = .87*Pe^(k*0) ........cancel the P's and simplify...
e^k = .87e^0 = .87*1 = .87
ln(e^k) = ln(.87)
k = ln(.87)

So each of the lower-case "c" functions will be (something like)
c(t) = 45e^(k(t+h)), where k = ln(.87) and "h" is the amount of time since ingestion.

You'll get three pretty similar equations, add them up, then set it = 20 and solve.

13. May 24, 2010

### Tedjn

In this case, our job is simplified by the fact that you drink a cola every hour for only 3 hours, so you can calculate the amount of caffeine left hour by hour (adding 45 mg each time) until the last cola. After that, the original differential equation would work out. Of course, you guys seem to have a more general solution.

14. May 24, 2010

### vaibhav1803

you forgot the square in c(2)
c(2)=45(.87)^2+45(.87)+45..remember c(r+1)=(1-.13)c(r)...

15. May 24, 2010

### vaibhav1803

the point is that the guy who having the cola is taking it on an hourly basis, so we must calculate it as a discrete-input function....

now we'd require a DE to find the solution only if the guy has a cola dispenser installed in his bathroom, from which he continuosly drinks while excreting caffiene(please excuse the language..couldn't find a more polite way to say it) ..you need a DE for that sort of a case...where the cola consumption doesnt stop ..and so does the elimination.

16. May 24, 2010

### The Chaz

This is insanity. Have you read my posts? The body does NOT eliminate caffeine on an HOURLY BASIS!!!!!!!!!!!!!!!!!!!!!! It is a CONTINUOUS process, so we model it with an exponential function. This function is actually the sum of the functions pertaining to each of the three cans of cola consumed.

To be direct: you are wrong OR I am wrong. I think it is the former, and have given sufficient evidence towards that conclusion. Please respond to this or...QUIT POSTING

Last edited by a moderator: May 24, 2010
17. May 24, 2010

### vaibhav1803

ok, i have 1 question.
so when after a coffee/any drink do you immediately, go to the bathroom and spend the rest of your life pissing..? or do you visit the loo after a certain period of time..
look kid, i took in a sane assumption that the person is human..and no caffiene seperating machine..besides I am allowed that by common sense.

18. May 24, 2010

### vaibhav1803

@Liparulo,
please make the problem statement more clear to me, and i think you must have arrived at the solution by now,(there is no effect after 14.36 hours)..and its quite the time the thread be closed so it won't be spammed, for any further details on the solution you may contact me by messages(on my PF account)
thanks.

Last edited: May 25, 2010
19. May 24, 2010

### Tedjn

I would just like to post one last time to clear up any confusion with my response. Like The Chaz, I interpreted the problem as a continuous one, although it is possible to discretize in the first 3 hours for simplicity. I feel that this is the usual interpretation for such problems.

One can also assume that this person only gets rid of caffeine on an hourly basis (or on some discrete scale). This doesn't make the differential equation solution invalid; merely, you would round up, just as vaibhav1803 did in his solution.

Physically, the interpretation depends on how the kidneys work in removing caffeine. If they break it down into an intermediary compound, then one could think of caffeine removal as continuous. Mathematically, the continuous interpretation is more likely.

In any case, more arguing will probably not help the OP. We're all trying to help out, after all. Liparulo, if you have any more questions, feel free to ask and we'll try to help out best we can.

20. May 25, 2010

### vaibhav1803

I agree Tedjn to a certain extent though, i still stick to my discrete interpretation though, as the removal of caffiene from the body(urination) occurs at the period of one hour, that i consider as "caffiene eleimaination complete", but you have clarified the second model to me which "the Chaz" apparently assumed to be the explanatory one so thank you, buti'd like to ask
what would be considered better as caffiene removal from the body, the stage after filtering from kidneys, or the removal from body through the urinary tract.?
its this interpretation that decides the modl we choose, it didn't occur to me before, till Tedjn pointed out.
so..we need to ask Liparulo, what he'd consider as a complete removal of caffiene from the body, the kidney stage or the emptying the bladder?..the answer must be by his needs right..?..hence the problem.