Question

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?

Answer 1

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