a bacteria culture starts with 260 bacteria and grows at a rate proportional to
its size after 5 hours there will be 1300 bacteria express the population after t hours as a function of t
dp/dt = kp
dp/p = k dt
lnp = kt+c
p = c*e^(kt)
since p(0) = 260, c = 260 and
p = 260 e^(kt)
To find k, solve
260 e^(5k) = 1300
To express the population of the bacteria culture as a function of time, we need to determine the growth rate constant (k) first.
Let P(t) be the population at time t. According to the problem, the population grows at a rate proportional to its size.
The general formula for population growth is given by the differential equation:
dP/dt = kP
Where dP/dt represents the rate of change of the population with respect to time.
We are given that after 5 hours, there will be 1300 bacteria, so we can use this information to find the value of k.
P(5) = 1300
260 * e^(5k) = 1300
Divide both sides of the equation by 260:
e^(5k) = 1300/260
e^(5k) = 5
Taking the natural logarithm (ln) of both sides:
5k = ln(5)
Divide both sides by 5 to isolate k:
k = ln(5)/5
Now that we have the value of k, we can express the population P(t) as a function of time t:
P(t) = P₀ * e^(kt)
Where P₀ is the initial population (260 bacteria), e is the base of the natural logarithm (approximately 2.71828), k is the growth rate constant (ln(5)/5), and t is the time in hours.
Therefore, the expression for the population after t hours is:
P(t) = 260 * e^((ln(5)/5) * t)
To express the population of the bacteria culture after t hours as a function of t, we can use an exponential growth model.
Let P(t) represent the population of the bacteria culture after t hours. According to the problem, the growth rate is proportional to its size, which means that the rate of change of the population is directly proportional to the current population.
We can set up a differential equation to represent this relationship:
dP/dt = kP
Where dP/dt represents the rate of change of the population, k is the proportionality constant, and P represents the population.
To solve this differential equation, we can separate variables and integrate:
(dP/P) = kdt
∫(1/P)dP = ∫kdt
ln|P| = kt + c
P = e^(kt+c)
Since we are given the initial population as 260 bacteria, we can substitute this value into the equation:
260 = e^(k*0 + c)
260 = e^c
Now, let's use the information given that after 5 hours, there will be 1300 bacteria:
1300 = e^(k*5 + c)
We can use these two equations to solve for k and c. First, divide the second equation by the first equation:
1300/260 = e^(k*5 + c)/e^c
5 = e^(k*5)
Taking the natural logarithm of both sides:
ln(5) = ln(e^(k*5))
ln(5) = k*5
Solving for k:
k = ln(5)/5
Now, substitute this value of k back into the first equation:
260 = e^(ln(5)/5 * 0 + c)
260 = e^c
Since e^c = 260, we can find c by taking the natural logarithm of both sides:
c = ln(260)
Finally, substitute the values of k and c back into the equation P = e^(kt+c):
P(t) = e^(ln(5)/5 * t + ln(260))
Simplifying further:
P(t) = 260e^(ln(5)/5 * t)
Thus, the expression for the population after t hours as a function of t is P(t) = 260e^(ln(5)/5 * t).