The number of a certain type of bacteria increases continuously at a rate proportional to the number present. There are 100 present initially, and 200 present 7 hours later. How many will there be 20 hours after the initial time? Round your answer to the nearest integer.
A)14 bacteria
B)620 bacteria
C)725 bacteria
D)27 bacteria
E)22 bacteria

Question

Quizplus · Accepted Answer

Let's denote the number of bacteria present after time t as N(t). According to the problem, the rate of increase is proportional to the number present, which means we can write:

dN/dt = k*N

where k is the proportionality constant. This is a separable differential equation, which we can solve as follows:

dN/N = k*dt
∫(1/N)dN = ∫k*dt
ln|N| = kt + C
N = e^(kt+C) = Ce^(kt)

where C is the constant of integration. We can find C using the initial condition N(0) = 100:

100 = Ce^(k*0)
C = 100

So the solution to the differential equation is:

N(t) = 100*e^(kt)

We can find k using the second initial condition N(7) = 200:

200 = 100*e^(7k)
e^(7k) = 2
7k = ln(2)
k = ln(2)/7

Now we can use this value of k to find N(20):

N(20) = 100*e^(ln(2)/7 * 20) ≈ 725

Rounding to the nearest integer, we get:

Answer: C

The Number of a Certain Type of Bacteria Increases Continuously