Please answer #8 and #9. Thank You! The concentration of a drug in grams/cubic c

Question

Please answer #8 and #9. Thank You!

The concentration of a drug in grams/cubic centimeter (g/cm3) t min after it has been injected into the bloodstream is given by the following where a, b, and k are positive constants, with b > a. C(t) = k/b - a(e-at - e-bt) At what time is the concentration of the drug the greatest? What will be the concentration of the drug in the long run? A radioactive substance decays according to the formula below where Q(t) denotes the amount of the substance present at time t (measured in years), Q0 denotes the amount of the substance present initially, and k (a positive constant) is the decay constant. Q(t) = Q0e-kt Find the half-life of the substance in terms of k. Suppose that a radioactive substance decays according to the formula Q(t) = 21e-0.0001591t How long will it take for the substance to decay to half the original amount? (Round your answer to the nearest whole number.) yr

Explanation / Answer

1.)

a.) c'(t) = k(-ae^(-at) + be^(-bt)) / (b-a) = 0

be^(-bt) = ae^(-at)

b/a = e^(b-a)t

t= ln(b/a) / (b-a) min

b.)

In the long run the concentration c(t) will be zero

c(infinity) = k(e^(-a(infinity)) - e^(-b(infinity))) / (b-a) = k(0-0)/b-a = 0

2.)

a.)Q=Q0(e^-kt)

Let the half life time be x

Thus, Q0(e^-kx) = Q0/2

(e^-kx) = 1/2

e^kx = 2

kx = ln(2)

x = ln(2) / k years

b.)

Q(t) = 21(e^-0.0001591t)

By the above formula

time t =  ln(2)/k = ln(2)/0.0001591 = 4356.676 years

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.