fruit fly population Fruit flies are placed in a half-pint milk bottle with a banana

Question

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

[latex]P(t)\,=\,{\frac{230}{1\,+\,56.5e^{-0.37t}}}[/latex]

(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.

Accepted Answer

(a) As [latex]t ightarrow\infty.\;e^{-0.37t} ightarrow0[/latex] and [latex]P(t) o{\frac{230}{1}}.[/latex] The carrying capacity of the half-pint bottle is 230 fruit flies. The growth rate is [latex]|b|\,=\,|0.37|\,=\,37[/latex]% , per day.

(b) To find the initial number of fruit flies in the half-pint bottle, evaluate P(0)

[latex]\begin{array}{c}{{P(0)=\frac{230}{1+56.5e^{-0.37(0)}}}}\ {{=\frac{230}{1+56.5}}}\end{array}[/latex]

= 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) .

[latex]P(5)\,=\,{\frac{230}{1\,+\,56.5e^{-0.37(5)}}}\approx23[/latex] 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.

[latex]{\frac{230}{1+56.5e^{-0.37t}}}=180 \ 230=180(1+56.5e^{-0.37t})[/latex]

[latex]1.2778=1\,+^{'}56.5e^{-0.37t}[/latex] Divide both sides by 180.

[latex]0.2778=56.5e^{-0.37t}[/latex] Subtract 1 from both sides.

[latex]0.0049=e^{-0.37t}[/latex] Divide both sides by 56.5.

[latex]\ln\left(0.0049 ight)\;=\;-0.37t[/latex] Rewrite as a logarithmic expression.

[latex]t\approx{\mathrm{14.4~days}}[/latex] 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 [latex]Y_{1}={\frac{230}{1+56.5e^{-0.37x}}}\,\mathrm{a}[/latex] and [latex]Y_{2}=115[/latex] 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).

Question 4.4.8.5: fruit fly population Fruit flies are placed in a half-pint m......

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