Register to reply 
Integration Question 
Share this thread: 
#1
Nov909, 12:24 AM

P: 122

1. The problem statement, all variables and given/known data
The rate at which a particular drug is absorbed by the body is proportional to the amount of the drug present (D) at any time t. If 50mL is initially administered to a patient and 50% remains after 4 hours, find A) D as a function of t. 3. The attempt at a solution I'm confused as we have to make D=the amount present (not yet absorbed) This means that it would be: d(50D)/dt [tex]\alpha[/tex] D Is this correct, in that the rate of absorption is Amount absorbed/time, so 50mLD = amount absorbed? By doing this, I integrate and just get Ae^kt = D 


#2
Nov909, 12:39 AM

P: 939




#3
Nov909, 12:40 AM

P: 122

But the rate of absorption is the amount absorbed over time? D represents the amount left, so it would have to be d(50D)/dt? Because 50amount left = amount absorbed



#4
Nov909, 01:09 AM

P: 939

Integration Question
That's right, but you can certainly show that d(50D)/dt = dD/dt. What was your question again? :p



#5
Nov909, 01:11 AM

P: 122

How does d(50D)/dt turn into dD/dt???



#6
Nov909, 01:18 AM

P: 939

Differentiation is linear, so that d(50D)/dt = d 50 /dt  dD/dt. Differentiation of a constant gives zero.



#7
Nov909, 01:28 AM

P: 122

Thankyou!



#8
Nov909, 01:34 AM

HW Helper
P: 3,307

i don't think you need the 50 at all to formulate the differntial equation
the statement [tex]\frac{dD(t)}{dt} = kD(t) [/tex] for some as yet undetermined constant k integrate, then use the given values to find the constants assume an initial value [tex]D(0) = d [/tex] value at time = 4hrs [tex]D(4) = d/2 [/tex] 


Register to reply 
Related Discussions  
Integration Question  Calculus & Beyond Homework  2  
Integration Question  Calculus  17  
Integration Question  Calculus  13  
Question about Integration  Calculus  21  
Question arrising from integration homework (advanced integration i guess?)  Calculus  9 