Question 4.4.8.5: fruit fly population Fruit flies are placed in a half-pint m......
fruit fly population
Fruit flies are placed in a half-pint milk bottle with a banana (for food) and yeast plants (for food and to provide a stimulus to lay eggs). Suppose that the fruit fly population after t days is given by
P(t)\,=\,{\frac{230}{1\,+\,56.5e^{-0.37t}}}(a) State the carrying capacity and the growth rate.
(b) Determine the initial population.
(c) What is the population after 5 days?
(d) How long does it take for the population to reach 180?
(e) Use a graphing utility to determine how long it takes for the population to reach one-half of the carrying capacity.
(a) As t\rightarrow\infty.\;e^{-0.37t}\rightarrow0 and P(t)\to{\frac{230}{1}}. The carrying capacity of the half-pint bottle is 230 fruit flies. The growth rate is |b|\,=\,|0.37|\,=\,37% , per day.
(b) To find the initial number of fruit flies in the half-pint bottle, evaluate P(0)
\begin{array}{c}{{P(0)=\frac{230}{1+56.5e^{-0.37(0)}}}}\\ {{=\frac{230}{1+56.5}}}\end{array}= 4
Thus, initially, there were 4 fruit flies in the half-pint bottle.
(c) To find the number of fruit flies in the half-pint bottle after 5 days, evaluate P(5) .
P(5)\,=\,{\frac{230}{1\,+\,56.5e^{-0.37(5)}}}\approx23 fruit flies
After 5 days, there are approximately 23 fruit flies in the bottle.
(d) To determine when the population of fruit flies will be 180, solve the equation P(t) = 180.
{\frac{230}{1+56.5e^{-0.37t}}}=180 \\ 230=180(1+56.5e^{-0.37t})1.2778=1\,+^{'}56.5e^{-0.37t} Divide both sides by 180.
0.2778=56.5e^{-0.37t} Subtract 1 from both sides.
0.0049=e^{-0.37t} Divide both sides by 56.5.
\ln\left(0.0049\right)\;=\;-0.37t Rewrite as a logarithmic expression.
t\approx{\mathrm{14.4~days}} Divide both sides by – 0.37.
It will take approximately 14.4 days (14 days, 10 hours) for the population to reach 180 fruit flies.
(e) One-half carrying capacity is 115 fruit flies. Solve P (t) = 115 by graphing Y_{1}={\frac{230}{1+56.5e^{-0.37x}}}\,\mathrm{a} and Y_{2}=115 and using INTERSECT. See Figure 46. The population will reach one-half of the carrying capacity in about 10.9 days (10 days, 22 hours).
