# Thread: Quick calc worded problem

1. ## Quick calc worded problem

im stuck on ii
3) A culture of bacteria is growing at a rate proportional to the number,
$\displaystyle N$, of bacteria present, so that: $\displaystyle \frac{dN}{dt} = kN$
where $\displaystyle t$ is the time elapsed in hours and $\displaystyle k$ is a constant.

Initially, the culture contains 3000 bacteria,
and after three hours, it contains 9000 bacteria.
(i) Find k.

Thereafter (that is, for t > 3), the culture is washed with a solution
which is harmful to the bacteria, and the bacteria are killed off
at a constant rate of 4500 per hour.

(ii) Find the number of bacteria in the culture for any time t > 3.

(iii) After how much washing time (hours/minutes) do the bacteria die out?
(i) find k

I got $\displaystyle k \:=\: \frac{\ln(3)}{3}$

Which gives:$\displaystyle N(t) \;=\;3000e^{(\frac{ln(3)}{3})t}$...............................$\displaystyle (=\;3000\cdot3^{\frac{t}{3}})$

ii) find the population after 3 hrs (i.e. t>3)

Ok. so first I thought it was $\displaystyle N(t) = 3000e^{\frac{ln3}{3}t} - 4500t$

then i thought it this: $\displaystyle N(t) = (9000-4500)e^{\frac{ln3}{3}t}$

Now im thinking $\displaystyle N(t) = 9000e^{\frac{ln3}{3}t} - 4500t$

yeh so any tips...is it the last one?

2. Originally Posted by foreg0ne
im stuck on ii

(i) find k

I got $\displaystyle k \:=\: \frac{\ln(3)}{3}$

Which gives:$\displaystyle N(t) \;=\;3000e^{(\frac{ln(3)}{3})t}$...............................$\displaystyle (=\;3000\cdot3^{\frac{t}{3}})$

ii) find the population after 3 hrs (i.e. t>3)

Ok. so first I thought it was $\displaystyle N(t) = 3000e^{\frac{ln3}{3}t} - 4500t$

then i thought it this: $\displaystyle N(t) = (9000-4500)e^{\frac{ln3}{3}t}$

Now im thinking $\displaystyle N(t) = 9000e^{\frac{ln3}{3}t} - 4500t$

yeh so any tips...is it the last one?
Measuring time from the instant that the solution is added, yes it is the last one, but $\displaystyle t$ is not measured from that time so it is:

$\displaystyle N(t) = 9000e^{\frac{ln3}{3}(t-3)} - 4500(t-3)$

for $\displaystyle t>3$

CB

3. I don't understand...

4. Originally Posted by foreg0ne
I don't understand...
What is it exactly that you don't understand? You have it right if you were measuring time from when the solution was added, but you are not you are adding it at t=3 hours.

CB