The rate of growth of a population of bacteria is proportional to the square root of t, where P is the population size and t is the time in days initial size of the population is After one day the population has grown to Estimate the
Population after ays.Round your answer to the nearest integer.
A) bacteria
B) bacteria
C) bacteria
D) bacteria
E)bacteria

Question

The rate of growth  of a population of bacteria is proportional to the square root of t, where P is the population size and t is the time in days  initial size of the population is  After one day the population has grown to  Estimate the
Population after  ays.Round your answer to the nearest integer.
A) bacteria 
B) bacteria 
C) bacteria 
D) bacteria 
E)bacteria

Quizplus · Accepted Answer

We can use the formula for exponential growth, where P(t) is the population at time t, P0 is the initial population, and k is a constant of proportionality:

P(t) = P0*e^(kt)

We know that the rate of growth is proportional to the square root of t, so we can set up the equation:

P'(t) = k*sqrt(t)

Taking the integral of both sides with respect to t, we get:

P(t) = (2/3)*k*t^(3/2) + C

Using the initial condition, we can find the value of C:

110 = (2/3)*k*1.732 + C

C = 110 - (2/3)*k*1.732

Now we can use this equation to estimate the population after 11 days:

P(11) = (2/3)*k*11^(3/2) + (110 - (2/3)*k*1.732)

We don't know the value of k, but we can use the fact that the population has doubled after one day to find it:

2*110 = P(1) = (2/3)*k*sqrt(1) + C
220 = (2/3)*k*1.732 + C
220 = (2/3)*k*1.732 + (110 - (2/3)*k*1.732)
110 = (1/3)*k*1.732
k = 223.6

Now we can plug in k and t = 11 to find the population:

P(11) = (2/3)*223.6*11^(3/2) + (110 - (2/3)*223.6*1.732) = 1048

Rounding to the nearest integer, we get 1048, which corresponds to choice C. Let's first find the constant of proportionality. We know that the rate of growth is proportional to the square root of t, so we can write this as dP/dt = k*sqrt(t), where k is the constant of proportionality. We also know that P(0) = 100. To find k, we can use the fact that the population has grown to 121 after one day, so we have:

dP/dt = k*sqrt(t)
dP/dt = k*sqrt(1)
dP/dt = k

So, we know that k = (dP/dt)|t=1 = 21/2. Now we can solve the differential equation:

dP/dt = k*sqrt(t)
dP = k*sqrt(t)*dt
∫dP = ∫k*sqrt(t)*dt
P = 2/3*k*t^(3/2) + C

Using the initial condition P(0) = 100, we find that C = 100. So the general solution to the differential equation is:

P = 2/3*k*t^(3/2) + 100

To find the population after 11 days, we plug in t = 11:

P = 2/3*(21/2)*11^(3/2) + 100
P ≈ 437

Rounded to the nearest integer, we get P ≈ 437, so the answer is choice C.

The Rate of Growth of a Population of Bacteria