Posted by **naseba** on Friday, March 30, 2012 at 8:35am.

2. The rate of change in the number of bacteria in a culture is proportional to the number present. In a certain laboratory experiment, a culture has 10,000 bacterial initially, 20,000 bacteria at time t1 minutes, and 100,000 bacteria at (t1 + 10) minutes.

a. In terms of t only, find the number of bacteria in the culture at any time t minutes, t ≥ 0,

b. How many bacteria were there after 20 minutes?

c. How many minutes had elapsed when the 20,000 bacteria were observed?

- calculus -
**Reiny**, Friday, March 30, 2012 at 8:52am
The function must have the form

N(t) = a e^kt , where a is the intitial number and k is a constant , and t is in minutes

for t=0 , a = 10000

for t = t1

20000 = 10000 e^(kt1)

2 = e^(kt1)

for t = t1+10

100000 = 10000 e^(k(t1+10)

10 = e^(kt1 + 10k)

divide the two equations

5 = e^(10k)

ln5 = 10klne

10k = ln5

k = ln5/10

a) N(t) = 10000 e^(ln5/10 t)

b)

when t = 20

N(20) = 10000 e^(ln5/10 (20)) = 250 000

c) when N(t) = 20000

2 = e^(ln5/10 t)

ln2 = ln5/10 t

t = 10ln2/ln5 = 4.3067

so in effect we found the "doubling time" to be appr. 4.307 minutes

(we can now also check if for t = 4.307 + 10 or t=14.307 we get N = 50000

10000 e^(ln5/10 (14.307))

= 10000 e^2.302623

= 10000(10.00037..)

= 100 004 , not bad

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