Question

Medical Science Soon after an injection, the concentration D (in milligrams per milliliter) of a drug in a patient's bloodstream is 500 milligrams per milliliter. After 6 hours, 50 milligrams per milliliter of the drug remains in the bloodstream.

(a) Find an exponential model for the concentration D after t hours.
(b) What is the concentration of the drug after 4 hours?

Answer 1

a)
`D(t) = 500 e^(-0.3837641822t)`

b) For this case we just need to replace t=4 and see what we got:

`D(t=4) =500 e^(-0.3837641822*4)= 107.721 \approx 102`

Explanation:

For this case the proportional model is given by the following differential equation:

`(dD)/(dt) =kD`

Where D is the concentration, t the time and k a proportional constant.

We can rewrite the differential equation like this:

`(dD)/(D) =kdt`

And if we integrate both sides we got:

`ln |D| = kt +c`

Part a

And if we apply exponentiation on both sides we got:

`D(t) = e^(kt) e^c = e^(kt) D_o = D_o e^(kt)`

So then our model is given by:

`D(t) = D_o e^(kt)`

Where
`D_o` is the initial amount for t =0.

For this case we have the initial condition assumed:

`D(0) = 500`

So then our model is given by:

`D(t) = 500 e^(kt)`

We have another condition given:

`D=50 , t = 6`

`50 = 500 e^(6k)`

If we divide both sides by 500 we got:

`(1)/(10) =e^(6k)`

And if we apply natutral log on both sides we got:

`ln((1)/(10)) = 6k`

`k = ((1)/(10))/(6)= -0.3837641822`

And our model then is given by:

`D(t) = 500 e^(-0.3837641822t)`

Part b

For this case we just need to replace t=4 and see what we got:

`D(t=4) =500 e^(-0.3837641822*4)= 107.721 \approx 102`