Respuesta :
Answer:
Step-by-step explanation:
THe logistic equation has the following form
[tex]\frac{dP}{dt} = rP(1-\frac{P}{K})[/tex] where, in this case, P i s the mass of the Bacteria. The constant K is the maximum capacity (K = 140) and r is the growth rate (r=1) we have an initial condition of [tex]P_0 = 2[/tex]
a) The logistic equation is
[tex]\frac{dP}{dt} = P(1-\frac{P}{140})[/tex]
In order for us to solve both questions, we must solve the differential equation, which turns out to be a separable equation. Hence
[tex]\int \frac{dP}{P(1-\frac{P}{140})} =\int dt = t + C[/tex]
The integral on the left is solved by the method of partial fractions,
we have the following equation
[tex]\frac{1}{P(1-\frac{P}{140})} = \frac{A}{P} + \frac{B}{(1-\frac{P}{140}}[/tex]
or equivalently
[tex] 1 = A(1-\frac{P}{140})+B(P)[/tex]
Note that this equation most hold for all values of P. If we choose P =0, we have that A=1. If P=140, then B=1/140.
Then,
[tex] \int \frac{dP}{P(1-\frac{P}{140})} = \int \frac{dP}{P} + \frac{1}{140} \int \frac{dP}{(1-\frac{P}{140})}[/tex]
Using the substitution u = 1-P/140, we get -140du = dP, this leads to the solution
[tex]\ln(|p|)-\ln(1-\frac{P}{140}) = t+C = \ln (\frac{P}{1-\frac{P}{140}})[/tex]
Then, the solution is given by
[tex] \frac{P}{1-\frac{P}{140}} = ke^{t}[/tex], where k is a constant. Note that when t=0, P=2, then
[tex]\frac{2}{1-\frac{2}{140}} = k=\frac{140}{69} [/tex]
b) For solving part b, we will use the derived equation, letting P=60 and solving for t, then
[tex]= 105=\frac{60}{1-\frac{60}{140}}=\frac{140}{69} e^t[/tex]. Then
[tex]t = \ln(105*69/140) = 3.9464[/tex]. Then after 4 days the bacteria will be really close to 60 mg.
c) from the equation
[tex] \frac{P}{1-\frac{P}{140}} = \frac{140}{69}e^{t}[/tex] with some algebra we can solve for P. This gives us
[tex]P = \frac{\frac{140}{69}e^{t}}{(1+\frac{1}{69}e^{t})}[/tex]
We want to know first, the amount in mg for t=10. Then
[tex]P = \frac{\frac{140}{69}e^{10}}{(1+\frac{1}{69}e^{10})}=139.56[/tex]
Then, P = 0.13956 Grams after 10 days.
