Answer:
a. dD(t)/dt = -kD(t) b. [tex]D(t) = D_{0}e^{-kt }[/tex] c. [tex]D(t) = 90e^{-0.126t }[/tex] d. 5.5 hours
Step-by-step explanation:
a. The differential equation that models the growth of the dD(t)/dt is
-dD(t)/dt ∝ D(t)
dD(t)/dt = -kD(t)
b. The general solution of the equation in a. is gotten below
dD(t)/dt = -kD(t)
separating the variables, we have
dD(t)/D(t) = -kdt
Integrating both sides, we have
∫dD(t)/D(t) = -∫kdt
㏑D(t) = -kt + c
[tex]D(t) = e^{-kt + c}\\ D(t) = e^{-kt }e^{c}\\D(t) = D_{0}e^{-kt } (D_{0} = e^{c})[/tex]
c. Given that when t = 2 hours, D(t) = 70 grams and D₀ = initial amount of drug = 90 grams
Substituting these values into the equation, we have
[tex]D(t) = D_{0}e^{-kt } \\70 = 90e^{-2k } \\\frac{70}{90} = e^{-2k }\\\\[/tex]
㏑(7/9) = -2k
k = -[㏑(7/9)]/2
= 0.2513/2
= 0.126
[tex]D(t) = 90e^{-0.126t }[/tex]
d. The half-life of the drug is the time when D(t) = D₀/2.
So
[tex]D(t) = D_{0}e^{-0.126t }\\\frac{D_{0}}{2} = D_{0}e^{-0.126t }\\\frac{D_{0}}{2D_{0}} = e^{-0.126t }\\\frac{1}{2} = e^{-0.126t }[/tex]
taking natural logarithm of both sides, we have
t = ㏑(1/2)/-0.126 = -0.693/-0.126 = 5.5 hours
So, the half-life is 5.5 hours
