2) Suppose a certain type of bacteria reproduces continuously at a rate of 25% per minute. If the initial
amount of bacteria is 27.
a) Write an equation for the growth of bacteria over time.
b) How money bacteria will there be in 5 minutes? 1 hour?
c) How long will it take for the bacteria to reach a population of 10,000?

Respuesta :

A) The equation for the growth of bacteria over time is : n(t) = 27(   [tex]e^{25t}[/tex])

B) The number of bacteria the would be in 5 minutes 1 hour is :

n(5mins) = 215.055 ,  n(1) = 1944132282109.422

C) The time it will take for the bacteria to reach a population of 10,000 is : t = 0.237 hours

Determine the Equation, number of bacteria and time taken

Applying the mathematical growth and decay modelling

a. n(t) = [tex]n_{0}[/tex] [tex]e^{rt}[/tex]

[tex]n_{0}[/tex] = initial amount of bacterial

[tex]e^{rt}[/tex] = exponent

r = rate

t = time

n(t) = 27(   [tex]e^{25t}[/tex])

b. for 5 minutes

converting 5 minutes to hours; we divide by 60 minutes

[tex]\frac{5}{60}[/tex] = 0.083 hour

n(5 mins) = 27(   [tex]e^{25t}[/tex])

n(5 mins) = 27(   [tex]e^{25 * 0.083}[/tex])

n(5 mins) = 27(   [tex]e^{2.075}[/tex])

n(5mins) = 27(   7.965)

n(5mins) = 215.055

For 1 hour we have

n(1) = 27(   [tex]e^{25t}[/tex])

n(1) = 27(   [tex]e^{25 * 1}[/tex])

n(1) = 27(   [tex]e^{25}[/tex])

n(1) = 27 * (72004899337.386)

n(1) = 1944132282109.422

c. How long will it take for the bacteria to reach a population of 10,000?

10000 = 27(   [tex]e^{25t}[/tex])

[tex]\frac{10000}{27}[/tex] = (   [tex]e^{25t}[/tex])

370.370 = (   [tex]e^{25t}[/tex])

㏑ (370.370 ) = ㏑ (   [tex]e^{25t}[/tex])

5.915 = 25t

[tex]\frac{5.915}{25}[/tex] = t

t = 0.237 hours

In conclusion this type of question can be done by mathematical modelling

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