# Integration Question

by Shaybay92
Tags: absorption, integration, rate
 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(50-D)/dt $$\alpha$$ 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
P: 927
 Quote by Shaybay92 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 $$\alpha$$ D
This is correct. I'd only write dD/dt though.

 Quote by Shaybay92 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).
 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(50-D)/dt? Because 50-amount left = amount absorbed
P: 927

## Integration Question

That's right, but you can certainly show that d(50-D)/dt = -dD/dt. What was your question again? :p
 P: 122 How does d(50-D)/dt turn into -dD/dt???
 P: 927 Differentiation is linear, so that d(50-D)/dt = d 50 /dt - dD/dt. Differentiation of a constant gives zero.
 P: 122 Thankyou!
HW Helper
P: 3,309
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
$$\frac{dD(t)}{dt} = -kD(t)$$
for some as yet undetermined constant k

integrate, then use the given values to find the constants
assume an initial value
$$D(0) = d$$
value at time = 4hrs
$$D(4) = d/2$$

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