Question

Let C(t) be the concentration of a drug in the bloodstream. As the body eliminates the drug, C(t) decreases at a rate that is proportional to the amount of the drug that is present at the time. Thus C 0 (t) = −2kC(t), where 2k is a positive number. Here is it is half of the elimination constant of the drug. (a) If Co is the concentration at time t = 0, find the concentration at time . (b) If the body eliminates half the drug in 30 hours, how long does it take to eliminate 80% of the drug?

Answer 1

See explanation below

Explanation:

In this case, let's answer this by parts:

a) Concentration at time t when Co is concentration at t = 0:

In this case, we will use the initial expression but without the 2, so:

C(t) = -kC

In this case, we want to know the concentration at time t so:

C'(t) = -kC(t) derivating:

dC/dt = -kC

dC/C = -kdt From here, we can do integrals so:

lnC = -kt + C₁ (1)

Now, it's time to replace t = 0 and C = C₀:

lnC₀ = -k(0) + C₁

lnC₀ = C₁ (2)

Replacing (2) in (1) we have:

lnC = -kt + lnC₀

lnC - lnC₀ = -kt

ln(C/C₀) = -kt

C/C₀ = e^(-kt)

C(t) = C₀ e^(-kt) (3)

This is the expression for C at given time t.

2. time to eliminate 80% of the drug:

With the first data, we need to calculate the value of k, which will be constant at any given time so:

C(t) = C₀ e^(-kt)

0.5C₀ = C₀ e^(-30k)

0.5 = e^(-30k)

ln(0.5) = -30k

k = 0.02310

Now with this value we can calculate the time to eliminate 80% of the drug or simply in other words, that we just have a remaining of 0.2C₀:

0.2C₀ = C₀ e^(-0.0231t)

ln(0.2) = -0.0231t

-ln(0.2) / 0.0231 = t

t = 69.67 h

This is the time to eliminate 80% of the drug