At any time t=>0 , in days, the rate of growth of a bacteria population is given by y' = ky, where y is the number of bacteria present and k is a constant.
The initial population is 1500 and the population quadrupled during the first two days. By what factor will the population have increased during the first three days?
Thanks!!
To find out how the population will have increased during the first three days, we need to solve the given differential equation and use the information about the population quadrupling during the first two days.
The differential equation y' = ky represents the rate of growth of the bacteria population, where y is the number of bacteria present and k is a constant. We can solve this equation by separating the variables and integrating both sides.
Let's solve the differential equation step by step:
Step 1: Start with the differential equation y' = ky.
Step 2: Separate the variables by dividing both sides by y. This gives us:
y' / y = k
Step 3: Integrate both sides with respect to time (t). This gives us:
∫ (y' / y) dt = ∫ k dt
Step 4: Simplify the integrals:
ln|y| = kt + C1 (where C1 is the constant of integration)
ln|y| = ln|Ce^kt| (using properties of logarithms)
|y| = Ce^kt
Step 5: Apply the initial condition y(0) = 1500. This means that when t = 0, the population is 1500.
Since y can never be negative, we can drop the absolute value signs:
y = Ce^kt, where C is a positive constant.
Step 6: Use the information that the population quadrupled during the first two days.
At t = 2, the population is 4 times the initial population.
y(2) = 4 * y(0)
C * e^(2k) = 4 * 1500
Ce^2k = 6000
Step 7: Use this information to determine the value of C.
Divide both sides by e^2k:
C = 6000 / e^2k
Now we have the equation for the bacteria population:
y = (6000 / e^2k) * e^kt
Step 8: Calculate the population at t = 3 days.
y(3) = (6000 / e^2k) * e^3k
To find the factor by which the population has increased, we can divide the population at t = 3 days by the initial population:
Factor = y(3) / y(0)
Substituting the equations for y(3) and y(0):
Factor = [(6000 / e^2k) * e^3k] / 1500
= (4 * 6000) / 1500 (since e^3k / e^2k = e^k = 4)
= 16
Therefore, the population will have increased by a factor of 16 during the first three days.
then y = c e^(kt)
when t = 4
4c = c e^(4k)
4 = e^ 4k
4k = ln4
k = ln4/4
y = 1500 e^((ln4/4)t)
so when t = 3
y = 1500 (e^(3ln4/4) = appr 4243
check:
at t=4
y = 1500 e^(4ln4/4) = 6000
quadrupled is doubled twice.
In two doubles, takes 4 days
doubles in 2 days.
What is 2^(3/2) ?