The population of a bacteria culture increases at the rate of 3 times the square root of the present population.
A. Model the population P = P(t) of the bacteria population with a differential equation.
B. Solve the differential equation that models the population P = P(t) of the bacteria population.
C. Suppose the population at time t = 0 hours is 1000. Derive an equation for the population P as an explicit function of time t (in hours). Your equation should contain no undetermined constants.
D. What’s the population of the bacteria culture at the end of 10 hours? After 100 hours?
dp/dt = 3√p
dp/√p = 3dt
2√p = 3t+c
4p = (3t+c)^2
Now finish it off
Im confused on the function of time.
Would it be
p = (3x+20√10)^2 / 4
p = (3x^2 / 4) + 1000
A. Hmm, let me think of a funny way to model this bacteria population. How about we call the bacteria "The Mathletes"? Now, the population P of The Mathletes can be modeled using a differential equation, which can be written as dP/dt = 3√P. But remember, this mathematical equation is not as hilarious as The Mathletes themselves!
B. Ah, time to solve the differential equation! To separate the variables, we can rewrite it as 1/√P dP = 3dt. Integrating both sides will give us the lovely equation 2√P = 3t + C, where C is an arbitrary constant. Don't worry, C is a lot friendlier than The Mathletes!
C. Now, let's find the value of that arbitrary constant C using the population at t = 0 hours, which is 1000. Plugging in those values, we get 2√1000 = 3(0) + C. Simplifying this equation, we have C = 20√10. So now we have the hilarious equation 2√P = 3t + 20√10 for the population as an explicit function of time!
D. To find the population of The Mathletes at the end of 10 hours, we just need to substitute t = 10 into our equation. Plugging that in, we get 2√P = 3(10) + 20√10. Simplifying, we find that √P = 15 + 20√10/2 = 15 + 10√10. Squaring both sides, we have P = (15 + 10√10)^2. Now, that's quite a mouthful!
And for the population after 100 hours, you guessed it, we substitute t = 100 into our equation. Plugging that in, we get 2√P = 3(100) + 20√10. Simplifying further, we find √P = 300 + 20√10/2 = 300 + 10√10. Squaring both sides, we have P = (300 + 10√10)^2. That's one big, happy Mathlete family!
Remember, these are just mathematical calculations, and the actual population may vary due to various factors. But hey, at least we had some fun along the way!
A. To model the population of the bacteria culture with a differential equation, we can start by assuming that the population at any time t is denoted by P(t). The differential equation will relate the rate of change of the population, dP/dt, to the present population P.
Given that the population increases at a rate of 3 times the square root of the present population, we can express this mathematically as:
dP/dt = 3√P
B. To solve the differential equation, we want to find an expression for P(t) that satisfies the equation dP/dt = 3√P. One way to do this is by separation of variables.
Separation of variables involves moving all terms with P on one side and all terms with t on the other side. We can rewrite the equation as:
1/√P dP = 3 dt
Now, integrating both sides with respect to their respective variables gives us:
∫ 1/√P dP = ∫ 3 dt
C. To find an explicit expression for the population P as a function of time t, we can evaluate the integrals on both sides of the equation.
The integral of 1/√P with respect to P can be found using the power rule of integration:
2√P = 3t + C
Next, we solve for P by squaring both sides of the equation:
4P = (3t + C)^2
Since the problem statement mentions that the population at time t = 0 hours is 1000, we can substitute these values into the equation:
4(1000) = (3(0) + C)^2
4000 = C^2
Taking the square root of both sides and considering C = ±√4000 = ±20√10, we find:
C = ±20√10
Therefore, the equation for the population P as an explicit function of time t is:
P(t) = (3t ± 20√10)^2 / 4
D. To find the population of the bacteria culture at the end of 10 hours, substitute t = 10 into the expression for P(t):
P(10) = (3(10) ± 20√10)^2 / 4
Using the positive value of C, we get:
P(10) = (30 + 20√10)^2 / 4
Calculating this expression will give you the population at the end of 10 hours.
To find the population after 100 hours, substitute t = 100 into the expression for P(t):
P(100) = (3(100) ± 20√10)^2 / 4
Again, using the positive value of C, we get:
P(100) = (300 + 20√10)^2 / 4
Calculating this expression will give you the population after 100 hours.
p(0) = 1000, so
4000 = c^2
c = 20√10
p = (3x+20√10)^2 / 4
Odd that you should even consider the 2nd equation, by this time in your math studies.