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#1
Nov 3 2013 02:44am
1. The function C(t) = K(e^-at - e^-bt) where a,b, and K are positive constants and b > a, is used to model the concentration at time t of a drug injected into the bloodstream. (Different drugs have different values for the constants, but the model itself doesnt change.)

d.) For what values of t is C'(t) = 0? What does this mean in this context?


Work:
C'(t) = [-ae^-at + be^ -bt]K <--------------------- this i know is correct
set C'(t) = 0 and solve for t


K[-ae^-at + be^ -bt] = 0

ln(-ae^-at) + ln(be^ -bt) = ln 0 <-------------- this line of work is what mainly concerns me

-at ln(-ae) - bt ln(be) = 1

t [-a ln(-ae) - b ln(be)] = 1

t = 1 / [-a ln(-ae) - b ln(be)]

This post was edited by CamelFinger on Nov 3 2013 03:11am
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#2
Nov 3 2013 03:32am
Quote (CamelFinger @ Nov 3 2013 01:44am)
1. The function C(t) = K(e^-at - e^-bt) where a,b, and K are positive constants and b > a, is used to model the concentration at time t of a drug injected into the bloodstream. (Different drugs have different values for the constants, but the model itself doesnt change.)

d.) For what values of t is C'(t) = 0? What does this mean in this context?


Work:
C'(t) = [-ae^-at + be^ -bt]K
set C'(t) = 0 and solve for t


K[-ae^-at + be^ -bt] = 0

ln(-ae^-at) + ln(be^ -bt) =ln 0
That is where I stopped reading what you wrote. What is the value of ln(0)? Yeah that is a problem...

Here is what I did:
[-a*e^(-at) + b*e^(-bt)]K = 0

b*e^(-bt) = a*e^(-at)

ln{b*e^(-bt)} = ln{a*e^(-at)}

ln{b} - bt = ln{a} - at

at - bt = ln{a} - ln{b}

t = ln{a/b}/a-b

/e
oh as far as the meaning... hmmm well a-b is going to be negative.... uhh and ln{a/b} will be negative... so the right hand side is positive.... hmmm all i can think of is when time = ln{a/b}/a-b the concentration of the drug in your blood stream will stop changing... Which might mean that is when the drug will no longer be present, but I'm not sure about that part...

This post was edited by Azrad on Nov 3 2013 03:41am
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#3
Nov 3 2013 04:59am
thinking about it some more, perhaps my interpretation isn't right. Maybe that just represents a point in time when the concentration reaches a local min/max... But you would think the concentration would be maxed at t=0 and just decrease from there...Well clearly I don't know shit about drugs and bloodstreams... :(
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#4
Nov 3 2013 08:14am
t_0 = ln{a/b} / (a-b)
indeed.
You can also compute : t_0 = ( lnb - lna ) / ( b-a )

C'(t) > 0 <=> t < t_0
C'(t) < 0 <=> t > t_0
C'(t) = 0 <=> t = t_0

C is increasing on [0 ; t_0]
C reaches a maximum at t = t_0
C is decreasing on [t_0 ; +infinity [

Notice that C(0) = 0 (meaning that t=0 must be the moment when drug is injected by some way, a begins to access the bloodstream), and limit C (as t -> +infinity) = 0, meaning that drug will disappear after an infinite time.
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#5
Nov 3 2013 08:42am
Quote (feanur @ Nov 3 2013 07:14am)
Notice that C(0) = 0 (meaning that t=0 must be the moment when drug is injected by some way, a begins to access the bloodstream), and limit C (as t -> +infinity) = 0, meaning that drug will disappear after an infinite time.
ah i get it now, heh thanks.

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#6
Nov 3 2013 10:55am
Is this work correct?

1. The function C(t) = K(e^-at - e^-bt) where a,b, and K are positive constants and b > a, is used to model the concentration at time t of a drug injected into the bloodstream. (Different drugs have different values for the constants, but the model itself doesnt change.)

d.) For what values of t is C'(t) = 0? What does this mean in this context?


Work:

[-a*e^(-at) + b*e^(-bt)]K = 0

b*e^(-bt) = a*e^(-at)

ln{b*e^(-bt)} = ln{a*e^(-at)}

ln{b} - bt = ln{a} - at

at - bt = ln{a} - ln{b}

t = ln{a/b}/a-b<--------------- Final Answer

This just means that the rate of change of the concentration of the drug in the blood is 0 (the conc is not changing) when t = ln{a/b}/a-b<---------------------What this means in context.


Btw feanur, You are correct in the limit approachin infin =0, and also when t=0 we get 0 as well (those two questions were also on my hw and i got the same answer).

This post was edited by CamelFinger on Nov 3 2013 10:57am
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#7
Nov 3 2013 11:46am
Azrad's calculations are correct of course.

Yes, the rate of change of the concentration of the drug is 0 when t=ln (a/b) / (a-b).
"The concentration is not changing when t=..." has no meaning. At a precise moment, of course nothing changes (it's like taking a picture of a race car : the car is not moving on the picture).

I'm pretty sure that the expected answer is : concentration of drug reaches a maximum at t=ln (a/b) / (a-b).

Actually, in most problems, the rate of change (the derivative) is not important in itself. It's only a tool to study the changes of the variable, its maximum or minimum.
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#8
Nov 3 2013 01:15pm
Quote (feanur @ Nov 3 2013 12:46pm)
Azrad's calculations are correct of course.

Yes, the rate of change of the concentration of the drug is 0 when t=ln (a/b) / (a-b).
"The concentration is not changing when t=..." has no meaning. At a precise moment, of course nothing changes (it's like taking a picture of a race car : the car is not moving on the picture).

I'm pretty sure that the expected answer is : concentration of drug reaches a maximum at t=ln (a/b) / (a-b).

Actually, in most problems, the rate of change (the derivative) is not important in itself. It's only a tool to study the changes of the variable, its maximum or minimum.


since the derivative means the rate of change, wouldnt it mean that when C'(t) = 0, the rate of change of the concentration of the drug in the bloodstream would be 0 AKA the same and therefore mean the concentration would be changing at the same rate. It just means that the rate of change of the concentration is either at a constant rate or the concentration of the blood is not changing at all because it is completely out of the system...


In other words if the rate of change is 0 aka the same rate of change, that would mean the concentration of the drug is diminishing at the same rate. Or the drug is already completely out of the bloodstream.

This post was edited by CamelFinger on Nov 3 2013 01:22pm
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#9
Nov 3 2013 03:25pm
Quote (CamelFinger @ Nov 3 2013 08:15pm)
since the derivative means the rate of change, wouldnt it mean that when C'(t) = 0, the rate of change of the concentration of the drug in the bloodstream would be 0 AKA the same and therefore mean the concentration would be changing at the same rate. It just means that the rate of change of the concentration is either at a constant rate or the concentration of the blood is not changing at all because it is completely out of the system...


In other words if the rate of change is 0 aka the same rate of change, that would mean the concentration of the drug is diminishing at the same rate. Or the drug is already completely out of the bloodstream.


According to the mathematical model ( C(t) = K * exp(...) ), the drug is never off the bloodstream. Some quantity always remains.
I won't discuss the validity of the model, of course it cannot be true if we suppose that the quantity of drug can be counted with molecules (whole numbers), whereas exponantiels reaches "infinite" small values.

Just notice that lim C' (t) = 0 as t tends to +infinity, so the rate of change (as well as the concentration itself) tends to zero according to this model.
Your problem is not refeering to this in my opinion.

Via pm :

Quote
Your question is the very fundamental question about derivative. For many centuries, people didn't understand that a quantity could change over a period of time even if it doesn't change at any specific moment.

The derivative is indeed the rate of change. Think about it as a speed (a velocity).

When you throw an object in the air, its speed is always decreasing (according to the gravity field).
If the initial speed is positive (you throw the object upwards), at a moment the speed is zero then it becomes negative.
As long as the speed is positive, the object goes up, and when the speed is negative, it goes down (it falls).

When the speed is zero, the object reaches its highest position. But this doesn't mean that its position is not changing anymore.

Same goes for your concentration : when the derivative is zero, you're at the turning point of the concentration :
- before that moment, concentration was increasing,
- after that moment, concentration will decrease,
- at that precise moment, concentration is at its maximum value.

But it has no meaning to say that concentration doesn't change at that moment (at any precise moment, concentration never changes, it requires some period of time for the concentration to change).


This post was edited by feanur on Nov 3 2013 03:25pm
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#10
Nov 3 2013 03:44pm
- before that moment, concentration was increasing,
- after that moment, concentration will decrease,
- at that precise moment, concentration is at its maximum value.

When C'(t) = 0 , we are at the highest point of concentration in the bloodsteam!!!!

THANK YOU!!!
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