Drug Absorption: Proportional to Amount Present at Time t

[Shaybay92]

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 \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

 

[clamtrox]

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.

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).

 

[Shaybay92]

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

 

[clamtrox]

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

 

[Shaybay92]

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

 

[clamtrox]

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

 

[Shaybay92]

Thankyou!

 

[lanedance]

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