Logistic Growth
Exponential Decay
Exponential Growth
100
The relative growth rate of Lussierville is 2.5%. Write a differential equation for the population.
dP/dt = .025P
100
The decay equation for radon-222 gas is known to be y=(y0)e^(-.18t), with t in days, and y0 as the initial amount. About how long will it take the amount of radon in a sealed sample of air to decay to 90% of its original value?
t=.585 days
100
Suppose that the cholera bacteria in a colony grows unchecked according to the Law of Exponential Change. The colony starts with 1 bacterium and doubles in number every half hour. How many bacteria will the colony contain at the end of 24 hours?
2.815x10^14
200
Write the differential equation for the logistic growth model with k=.05, M=200.
dP/dt=.00025P(200-P)
200
Suppose the amount of oil pumped from one of the canyon wells in Chisholm, CO, decreases at the continuous rate of 10% per year. When will the well's output fall to 1/5 of its present level?
t=16.09 years
200
A colony of bacteria is grown under ideal conditions in a laboratory so that the population increases exponentially with time. At the end of 3 hours there are 10,000 bacteria. At the end of 5 hours there are 40,000 bacteria. How many bacteria were present initially?
1251
300
A national park is known to be capable of supporting no more than 100 adult lions. 10 lions are in the park at time t=0. The population can be modeled by the solution to the following differential equation: dP/dt=.001P(100-P). Find the logistic model P(t)
P(t)=100/(1+9e^(-.1t))
300
Physicists using the radioactive decay equation y=(y0)e^(-kt) call the number 1/k the mean life of a radioactive nucleus. The mean life of a radon-222 nucleus is about 1/.18=5.6 days. The mean life of a carbon-14 nucleus is more than 8000 years. Show that 95% of the radioactive nuclei originally present in any sample will disintegrate within three mean lifetimes, that is, by time t=3/k. Thus, the mean life of a nucleus gives a quick way to estimate how long the radioactivity of a sample will last.
y=(y0)e^(-kt) .05(y0)=(y0)e^(-kt) .05=e^(-kt) ln|.05|=-kt -2.9957=-kt t=3/k
300
see p. 338, #15
y=2e^(.458t)
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