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adelina 88 [10]
3 years ago
12

The E. coli bacteria has a volume of 6 µm3 . In optimal conditions, an E. coli bacteria will double about every 30 minutes. Unde

r these conditions, how long will it take for a single bacterium to grow to fill a thimble with volume 1 cm3 ? How long will it take for the volume to fill the entire earth 1.08 × 108 km3 ?
Mathematics
1 answer:
Zinaida [17]3 years ago
4 0

Answer:

It is going to take around 19 hours and 15 minutes for a single bacteria to grow to fill a thimble with volume 1 cm3.

It will take 57.6 hours to for the volume to fill the entire earth

Step-by-step explanation:

We can say that the volume of the bacteria is a geometric sequence, and each time moment is an arithmetic sequence.

Each geometric sequence has the following format:

{a, ar, ar^{2}, ar^{3},...}

In which:

a is the first term

r is the common ratio.

We can find any term of the sequence by the following equation:

x_{n} = ar^{(n-1)}

Each arithmetic sequence has the following format:

{a, a+d, a+2d,...}

In which:

a is the first term

d is the difference between the terms.

We can find any term of the sequence by the following equation:

x_{n} = a + d(n)

How i am going to solve this problem.

We have the sequence that is the volume of the bacteria:

Obs: 1(um)^{3} = 10^{-18}m^{3}

V_{n} = {6*10^{-18}, 12*10^{-18},...}

a = 6*10^{-18}

r = 2

And the following arithmetic sequence that are the time(in hours).

T_{n} = {0,0.5,1,...}

a = 0

d = 0.5

T_{n} = 0.5n

How long will it take for a single bacterium to grow to fill a thimble with volume 1 cm3 ?

I am going to find the value of n for which V_{n} = 1cm^{3}, then i find the value at this position in the arithmetic sequence. So

1cm^{3} = 10^{-6}m^{3}

V_{n} = 6*10^{-18}*(2^{(n-1)})

10^{-6} = 6*10^{-18}*(2^{(n-1)})

\frac{10^{-6}}{6*10^{-18}} = 2^{(n-1)})

Obs: a^{b-c} = \frac{a^{b}}{a^{c}}

\frac{10^{12}}{6} = \frac{2^{n}}{2}

2^{n} = \frac{10^{12}}{3}

Now, we have to apply these following logarithim proprierties to find the value of n:

log a^{n} = n log a

log(\frac{a}{b}) = log a - log b

log 10^{n} = n

log 2 = 0.30

log 3 = 0.48

log 2^{n} = log(\frac{10^{12}}{3})

n log 2 = log(10^{12}) - log 3

0.30n = 12 - 0.48

0.30n = 11.52

n = 38.4

Lets find T_{38} and T_{39} in the arithmetic sequence.

T_{n} = 0.5n

T_{38} = 0.5*38 = 19

T_{39} = 0.5*39 = 19.5

It is going to take around 19 hours and 15 minutes for a single bacteria to grow to fill a thimble with volume 1 cm3.

How long will it take for the volume to fill the entire earth 1.08 × 108 km3 ?

1 (km)^{3} = 10^{9}m^{3}

So

1.08*10^{8} km^{3} = 1.08*10^{17} m^{3} = 108*10^{15}m^{3}

Lets solve the same way as the first question.

V_{n} = 6*10^{-18}*(2^{(n-1)})

108*10^{15} = 6*10^{-18}*(2^{(n-1)})

\frac{108*10^{15}}{6*10^{-18}} = 2^{(n-1)})

18*10^{33} = \frac{2^{n}}{2}

2^{n} = 36 * 10^{33}

Aside from the proprierties seen in the first exercise, we also have that

log a*b = log a + log b

log (2^{n}) = log(36 * 10^{33})

n log 2 = log 36 + log 10^{33}

0.30n = 1.56 + 33

0.30n = 34.56

n = \frac{34.56}{0.30}

n = 115.2

T_{115} = 0.5*115.2 = 57.6

It will take 57.6 hours to for the volume to fill the entire earth

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