Given a population growth $P$ milligrams at time $t$ hours, such that $\frac{dP}{dt} = 0.005P(120-P)$ If the initial population is $10$mg find $P$ as a function of $t$ in order to express how many hours it will take for the population to reach $110$mg. State the value of $lim P(t)$ as $t→∞$
Now, I believe I can arrange $P$ as a function of $t$ by separating variables
$\frac{dP}{P(120-P)} = 0.005dt$
$\frac{1}{P(120-P)} = \frac{A}{P} + \frac{B}{120-P}$
$1 = A(120-P) + BP = 120A + (B-A)P$ when $B-A = 0$, $120A = 1$ so $A = B = \frac{1}{120}$
$\frac{1}{P(120-P)} = \frac{1}{120}[\frac{1}{P} + \frac{1}{120-P}]$
$∫\frac{dP}{P(120-P)} = \frac{1}{120}∫[\frac{1}{P} + \frac{1}{120-P}]dP = ∫0.005dt$
$\frac{1}{120}[ln|P|-ln|120-P|] = 0.005t + c$
$ln|120-P| - ln|P| = -0.6t - 120c$
using log laws $ln|\frac{120-P}{P}| = -0.6t - 120c$
$|\frac{120-P}{P}| = e^{(-0.6t-120c)} = e^{-0.6t}e^{-120c}$
$\frac{120-P}{P} = Ae^{-0.6t}$ where $A = e^{-120c}$
$\frac{120-P}{P} = \frac{120}{P} - 1 = Ae^{-0.6t}$
$\frac{120-P}{P} = 1 + Ae^{-0.6t}$
$P = \frac{120}{1 + Ae^{-0.6t}}$
For the initial condition, $A = \frac{120-10}{10} = \frac{110}{10} = 11$
$P = \frac{120}{1 + 11e^{-0.6t}}$
Is my working out correct? If so, my assumption is that to find when the population is $110$ we find $t$ when
$110 = \frac{120}{1 + 11e^{-0.6t}}$
As for the limit, I'm not totally sure how to calculate that. I assumed that $t→∞$ would be when $P = 120$ but there must be some method that can calculate it correctly? Thanks.
All your work is correct.
When you arrive at $$P = \frac{120}{1 + 11e^{-0.6t}}$$ when $t\to \infty$, $e^{-0.6t}\to 0^+$ and $P_\infty=120^-=120$.
What you missed is the solution of $t$ such that $$110 = \frac{120}{1 + 11e^{-0.6t}}\implies 11e^{-0.6t}=\frac 1{11}\implies e^{-0.6t}=\frac 1{121}$$ that is to say $$e^{0.6t}=121\implies 0.6t=\log(121)\implies t=\frac{\log(121)}{0.6} \approx 8$$