Question 4.4.8.1: Bacterial Growth A colony of bacteria grows according to the......
Bacterial Growth
A colony of bacteria grows according to the law of uninhibited growth according to the function , where N(t) = 100e^{0.045(t)} is measured in grams and t is measured in days.
(a) Determine the initial amount of bacteria.
(b) What is the growth rate of the bacteria?
(c) What is the population after 5 days?
(d) How long will it take for the population to reach 140 grams?
(e) What is the doubling time for the population?
(a) The intial amount of bacteria, N_{0}, is obtained when t = 0, so
N_{0}=N(0){=}100e^{0.045(0)}=100\,\mathrm{grams}(b) Compare N(t) = 100e^{0.045(t)} to N(t) = N_{0}e^{kt}. The value of k, 0.045, indicates a growth rate of 4.5%.
(c) The population after 5 days is N(5) = 100e^{0.045(5)} \approx 125.2 grams.
(d) To find how long it takes for the population to reach 140 grams, solve the equation N(t) = 140
100e^{0.045(t)} = 140e^{0.045(5)} = 1.4 Divide both sides of the equation by 100.
0.045t = \ln 1.4 Rewrite as a logarithm.
t = \frac{\ln 1.4}{0.045} Divide both sides of the equation by 0.045.
≈ 7.5 days
The population reaches 140 grams in about 7.5 days.
(e) The population doubles when N(t) = 200 gram, so the doubling time can be found by solving the equation 200 = 100e^{0.045t} for t.
200 = 100e^{0.045t}
2 = e^{0.045t} Divide both sides of the equation by 100.
\ln 2 = 0.045t Rewrite as a logarithm
t = \frac{\ln 2 }{0.045} Divide both sides of the equation by 0.045
≈ 15.4 days.
The population doubles approximately every 15.4 days.