Each of the following functions is an exponential function F

Each of the following functions is an exponential function. For each one find the following: i. The starting value a. ii. The growth factor b. iii. The growth/decay rate r. iv. The continuous growth/decay rate k. v. For the functions that are increasing, find the doubling time. For the functions that are decreasing, find the half-life. (a) p(t) = 5(.912)^t (b) f(t) = 3(2.1)^t/2 (c) g(t) = 2^2 + 2t (d) h(t) = 2e^-2t + 1 The quantity Q of caffeine, measured in milligrams, that remains in the blood stream t hours after consumption may be modeled by an exponential function with a half life of 8 hours. (a) Find a formula Q = Q_0b^t for the caffeine remaining t hours after a dose of Q_0 mg. (b) How many hours does it take the 100 mg dose from an 8 oz cup of coffee to decay to the 10 mg dose found in 12 oz of iced tea? (c) What is the hourly continuous decay rate, k for caffeine?

Solution

y = a(b)^t

where b is growth factor ; 1 -b or b-1 is growth/ decay rate

a) p(t) = 5(.912)^t (decay)

a = 5

r = 1- 0.912 = 0.088

k: continous decay function :Ce^-kt

5(0.912)^t = 5e^(-kt) . Plug t=1

So, 0.912 = 5e^-k

ln(0.912) = -k

k = 0.0921

Half Life:

5/2 = 5(0.912)^t

0.5 = 0.912^t

ln0.5 = t*ln0.912

t = 7.52 half life

b) f(t)= 3(2.1)^t/2 ----growth

a = 3 ; b = 2.1 - 1 = 1.1

k: continous growth function :Ce^kt

3(2.1)^t/2 = 3e^kt

Plug t =1

3(2.1)^1/2 = 3e^k

take ln on both sides: 2^1/2 = e^k

k = ln(2^1/2) = 0.346

Doubling Time

6 = 3(2.1)^t/2

ln2 = (t/2)ln(2.1)

t/2 = ln2/ln2.1 = 0.934

t = 1.87 (doubling time)

c) g(t) = 2^(2+2t) = 4(2^2t) = 4(4)^t Growth

a = 4 ; b = 4 -1 = 3

k: continous growth function :Ce^kt

4(4)^t = 4e^kt ; Plug t=1

4*4 = 4e^k

4 = e^k

ln4 = k ; k =1.386

Doubling time:

g(t) = 4(4)^t

2 = 4^t ; t= ln2/ln4 = 0.5

t = 0.5

d) h(t) = 2e^(-2t+1)

= 2e^-2t*e = 2e*e^-2t = 5.43e^-2t

h(t) = 5.43e^-2t ---- Decay

So, a = 5.43

k = -2

h(t) = 5.43(e^-2)^t = 5.43(0.135)^t

b = 1-0.135 = 0.865

Half Time:

ln0.5 = t*ln(0.135)

t =0.346