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Integration Question |
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| Nov9-09, 12:24 AM | #1 |
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Integration Question
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(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 |
| Nov9-09, 12:39 AM | #2 |
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| Nov9-09, 12:40 AM | #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
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| Nov9-09, 01:09 AM | #4 |
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Integration Question
That's right, but you can certainly show that d(50-D)/dt = -dD/dt. What was your question again? :p
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| Nov9-09, 01:11 AM | #5 |
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How does d(50-D)/dt turn into -dD/dt???
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| Nov9-09, 01:18 AM | #6 |
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Differentiation is linear, so that d(50-D)/dt = d 50 /dt - dD/dt. Differentiation of a constant gives zero.
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| Nov9-09, 01:28 AM | #7 |
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Thankyou!
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| Nov9-09, 01:34 AM | #8 |
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Recognitions:
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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] |
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| absorption, integration, rate |
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