## 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 $$\alpha$$ 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

 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).
 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

## Integration Question

That's right, but you can certainly show that d(50-D)/dt = -dD/dt. What was your question again? :p
 How does d(50-D)/dt turn into -dD/dt???
 Differentiation is linear, so that d(50-D)/dt = d 50 /dt - dD/dt. Differentiation of a constant gives zero.
 Thankyou!

Recognitions:
Homework Help
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
$$D(4) = d/2$$

 Tags absorption, integration, rate