Bacterial Growth A colony of bacteria increases according to the law of uninhibited

Question

Bacterial Growth

A colony of bacteria increases according to the law of uninhibited growth.

(a) If N is the number of cells and t is the time in hours, express N as a function of t.

(b) If the number of bacteria doubles in 3 hours, find the function that gives the number of cells in the culture.

(c) How long will it take for the size of the colony to triple?

(d) How long will it take for the population to double a second time (that is, increase four times)?

Accepted Answer

(a) Using formula (2), the number N of cells at time t is

[latex]\mathbf{N}(t)=N_{\mathrm{0}}e^{k t}[/latex] k>0 (2)

where [latex]N_{0}[/latex] is the initial number of bacteria present and k is a positive number.

(b) To find the growth rate k. note that the number of cells doubles in 3 hours Hence

[latex]\mathbf{N}(3)=2N_{\mathrm{0}}[/latex]

But [latex]N(3)\,=\,N_{0}e^{k(3)},[/latex] so

[latex]N_{0}e^{k(3)}\ =\ 2N_{0}[/latex]

[latex]e^{k(3)} = 2[/latex] Divide both sides by [latex]N_{0}[/latex]

[latex] 3 k = \ln 2 [/latex] Write the exponential equation as a logarithm.

[latex]k={\frac{1}{3}}\ln2\ \approx0.23105[/latex]

The function that models this growth process is therefore

[latex]N(t)\,=\,N_{0}e^{0.23105t}[/latex]

(c) The time t needed for the size of the colony to triple requires that N = 3[latex]N_{0}[/latex] Substitute [latex]3N_0[/latex] for N to get

[latex]3N_{0}=N_{0}e^{0.23105t}[/latex]

[latex]3=e^{0.23105t}[/latex] Divide both sides by [latex]N_{0}[/latex]

[latex]0.23105t=\ln3[/latex] Write the exponential equation as a logarithm.

[latex]t={\frac{\ln3}{0.23105}}\approx4.755\,\mathrm{hours}[/latex]

It will take about 4.755 hours or 4 hours, 45 minutes for the size of the colony to triple.

(d) If a population doubles in 3 hours, it will double a second time in 3 more hours, for a total time of 6 hours.

Question 4.4.8.2: Bacterial Growth A colony of bacteria increases according to......

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