Integration Question

  • Thread starter Shaybay92
  • Start date
  • #1
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Homework Statement


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.

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(50-D)/dt [tex]\alpha[/tex] D

Is this correct, in that the rate of absorption is Amount absorbed/time, so 50mL-D = amount absorbed?

By doing this, I integrate and just get Ae^kt = D
 

Answers and Replies

  • #2
938
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I'm confused as we have to make D=the amount present (not yet absorbed)

This means that it would be:

d(50-D)/dt [tex]\alpha[/tex] D

This is correct. I'd only write dD/dt though.


By doing this, I integrate and just get Ae^kt = D

Correct as well, now you just have two constants and two constraints on them: D(0) and D(4 hours).
 
  • #3
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But the rate of absorption is the amount absorbed over time? D represents the amount left, so it would have to be d(50-D)/dt? Because 50-amount left = amount absorbed
 
  • #4
938
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That's right, but you can certainly show that d(50-D)/dt = -dD/dt. What was your question again? :p
 
  • #5
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How does d(50-D)/dt turn into -dD/dt???
 
  • #6
938
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Differentiation is linear, so that d(50-D)/dt = d 50 /dt - dD/dt. Differentiation of a constant gives zero.
 
  • #7
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Thankyou!
 
  • #8
lanedance
Homework Helper
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i don't think you need the 50 at all to formulate the differntial equation

the statement
The rate at which a particular drug is absorbed by the body is proportional to the amount of the drug present (D)

is enough to give you the differential equation
[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]
 

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