Thread: Growth problem? Any help would be greatly appreciated!

thanks!

2. Re: Growth problem? Any help would be greatly appreciated!

Hello, camjenson!

A colony of people grows at a rate directly proportional to the size of the population.
The colony triples every two hours.

1. (a) Write differential equation that describes the colony's growth.
. . (b) Find the general solution.

$\displaystyle (a)\;\frac{dP}{dt} \:=\:k\!\cdot\! P$

(b) We have: .$\displaystyle \frac{dP}{P} \:=\:k\,dt$

Integrate: .$\displaystyle \ln P \:=\:kt + c \quad\Rightarrow\quad P \:=\:e^{kt+c} \:=\:e^{kt}\!\cdot\! e^c} \:=\:e^{kt}\cdot C$

Hence: .$\displaystyle P \:=\:Ce^{kt}$

When $\displaystyle t = 0,\;P = P_o$, the initial population.

So we have: .$\displaystyle P_o \:=\:Ce^{0} \quad\Rightarrow\quad C \,=\,P_o$

Therefore, the equation is: .$\displaystyle P \;=\;P_oe^{kt}$

2. What's the value of the constant of proportionality?

The population triples every two hours.
When $\displaystyle t = 2,\;P = 3P_o$

We have: .$\displaystyle P_oe^{2k} \:=\:3P_o \quad\Rightarrow\quad e^{2k} \,=\,3 \quad\Rightarrow\quad 2k \,=\,\ln3$

. . . . . . . . $\displaystyle k \:=\:\tfrac{1}{2}\ln 3 \quad\Rightarrow\quad k \:=\:\ln\left(3^{\frac{1}{2}}\right) \quad\Rightarrow\quad \boxed{k \:=\:\ln\left(\sqrt{3}\right)}$

3. At t = 9 hours, the population is 800 million, what is the initial population?

The equation is: .$\displaystyle P \;=\;P_o\,e^{\frac{1}{2}\ln(3)\cdot t} \;=\;P_o\left(e^{\ln 3}\right)^{\frac{1}{2}t}$

And we have: .$\displaystyle P \;=\;P_o\!\cdot\!3^{\frac{1}{2}t}$

When $\displaystyle t = 9,\:P = 800,\!000,\!000.$

. . $\displaystyle P_o\!\cdot\!3^{4.5} \:=\:800,\!000,\!000$

. . $\displaystyle P_o \;=\;\frac{800,\!000,\!000}{3^{4.5}} \;\approx\; 5,\!702,\!225$