A bacterial population grows at a rate proportional to the population size at time t. Let y(t) be the population size at time t. By experiment it is determined that the population at t=15 min is 16000 and at t=45 min it is 22000.
(a) What was the initial population?
(b)What will the population be at time t=65 min?
Pls help! I know how to solve this kind of question if the initial condition is given. How do I approach it this time?
780000
y(t) = a e^(kt) , where a is the intitial count, and k is a constant, and t is in minutes
You have two ordered pairs, (15,16000) and (45,22000)
for (15,16000)
16000 = a e(15k)
for (45,22000)
22000 = a e^(45k)
divide the 2nd by the 1st
1.375 =e^(30k)
ln both sides , that is : (log base e)
ln 1.375 = 30k
k = ln 1.375/30 = .010615.. (store it in your calculator )
now plut that into your first equation of
16000 = a e^(15(.0106..) )
I got a = 13645
now that you have both a and k in the equation, plug in t = 65
(here is where storing the value of k is important)
btw, jorge's answer is bogus.
I got 27203
The population of a southern city follows the exponential law. If the population doubled in size over 2929 months and the current population is 30 comma 00030,000, what will the population be 33 years from now
The population of a southern city follows the exponential law. If the population doubled in size over 29 months and the current population is 30,000, what will the population be 33 years from now
To solve this problem, we can use the concept of exponential growth and the given information about the population size at two different times.
Given that the growth rate is proportional to the population size, we can write the differential equation as:
dy/dt = k * y(t),
where k is the constant of proportionality.
To solve this differential equation, we can separate the variables and integrate both sides:
∫ (1/y) dy = ∫ k dt.
Integrating the left side gives: ln|y| = kt + C1, where C1 is the constant of integration.
Applying the initial condition y(15) = 16000, we have:
ln|16000| = k * 15 + C1.
Similarly, applying the condition y(45) = 22000, we have:
ln|22000| = k * 45 + C1.
Now you can solve these two equations simultaneously to find the values of k and C1.
(a) To find the initial population, we need to find y(0).
Substituting t=0 into the equation ln|y| = kt + C1, we get:
ln|y(0)| = k * 0 + C1,
ln|y(0)| = C1.
Therefore, the initial population is given by y(0) = e^C1.
(b) To find the population at t=65, we can use the value of k obtained from part (a) and substitute it into the equation ln|y| = kt + C1.
ln|y(65)| = k * 65 + C1.
Solving this equation will give you the population size at t=65 min.
Remember, you need to solve the two simultaneous equations from the given information first to find the values of k and C1 before proceeding to find the initial population and the population at t=65 min.