# calc extra credit wokr, i need it!!!

• December 17th 2007, 06:57 PM
Mr.Obson
calc extra credit wokr, i need it!!!
i'm having trouble with this last problem:

The rate of growth dP/dt of a population of bacteria is proportional to the square root of t, where P is the population size and t is the time in days. (0<=t<=10).

That is:
dP/dt = ksqrt(t) sqrt(t) = square root of t, dunno how to type that out with the keyboard.

The initial size of the population is 500. After 1 day the population has grown to 600. Estimate the population after 7 days.

any help would be greatly appreciated!!!!!
• December 17th 2007, 07:47 PM
colby2152
$\frac{dP}{dt}=k\sqrt{t}$

$P(0) = 500$

$P(1) = 600$

$P(7) = ???$

There are a variety of ways to solve this problem...

$\frac{dP}{dt}=k\sqrt{t}$

$dP=k\sqrt{t}dt$

Integrate both sides..
$P(t)=\frac{2k*t^{3/2}}{3}+C$

$600=\frac{2k}{3}+C$

$500=C$

$100=\frac{2k}{3}$

$k = 150$

$P(7)=\frac{150*2*(7)^{3/2}}{3}$

$P(7)=2352$
• December 17th 2007, 08:00 PM
Soroban
Hello, Mr.Obson!

Quote:

The rate of growth $\frac{dP}{dt}$ of a population of bacteria
is proportional to the square root of $t$, where $P$ is the population size
and $t$ is the time in days. . $(0 \leq t \leq 10)$

The initial size of the population is 500.
After 1 day, the population has grown to 600.
Estimate the population after 7 days.

We are told: . $\frac{dP}{dt} \:=\:k\!\cdot\!\sqrt{t}\quad\Rightarrow\quad dP \:=\:k\!\cdot\!t^{\frac{1}{2}}$

Integrate: . $P \;=\;\frac{2}{3}k\!\cdot\!t^{\frac{3}{2}} + C$ .[1]

When $t=0,\:P = 500\!:\;\;500 \:=\:\frac{2}{3}k\!\cdot\!0^{\frac{3}{2}} + C\quad\Rightarrow\quad C \:=\:500$

When $t = 1,\:P = 600\!:\;\;600 \:=\:\frac{2}{3}k\!\cdot\!1^{\frac{3}{2}} + 500\quad\Rightarrow\quad k = 150$

Substitute into [1]: . $P \:=\:\frac{2}{3}(150)t^{\frac{3}{2}} + 500\quad\Rightarrow\quad P \:=\:100t^{\frac{3}{2}} + 500$

When $t = 7\!:\;\;P \;=\;100\!\cdot\!7^{\frac{3}{2}} + 500 \;\approx\;2352$