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Vi Nguyen · May 21, 2009, 08:38 AM

If I am culturing an organism and it is growing at a rate proportional to the number, N, of organism present so that:

dN/dt= kN

t is the time elapsed in hours and k is a constant, initially the culture contains 3000 organisms, after 3 hours of uninterrupted growth it contains 9000 bacteria.

"From this info I have found k= (1/3)loge(3) so equation is t=(3/loge(3))loge(N/3000)"

*However, if there after (when, t>3) the organisms are sprayed with a solution that kills them at a steady rate of 75 organisms per min. (thus the rate of change of the number of organism is as described above but diminished at a rate of 4500 organisms per hour.

How would I find the number of bacteria in the culture for any time t>3

also

how long after spraying (in hours and minutes) do the bacteria die out?

galactus · May 21, 2009, 02:53 PM

From the original DE, we have $N=3000e^{\frac{ln(3)t}{3}}$ , for $t\leq 3$

Now, at 3 hours, there are 9000 bacteria. Let t=3 be t=0 in the new time frame.

Therefore, initially, we have 9000 bacteria at t=0.

So, assuming they grow at the same rate. $N=9000e^{\frac{ln(3)t}{3}}$

But, they are also decreasing at a steady rate of 4500/hr.

$9000e^{\frac{ln(3)t}{3}}-4500t, \;\ t\geq 0$

The thing is, the bacteria are not dying out. As time passes they grow exponentially and the steady rate of decrease of 4500/hr does not do much after a long time. Now, if they decreased proportional to their growth, then that is a different matter.

Vi Nguyen · May 23, 2009, 02:00 AM

I see... one of those tricky questions eh? Are you sure? Because in 3 hours there is 9000 and in one hour of poison 4500 are knocked out, so in two hours shouldn't that wipe out 9000, taking in consideration that they they are still reproducing but wouldn't they be decreasing?

Originally Posted by galactus

From the original DE, we have $N=3000e^{\frac{ln(3)t}{3}}$ , for $t\leq 3$

The thing is, the bacteria are not dying out. As time passes they grow exponentially and the steady rate of decrease of 4500/hr does not do much after a long time. Now, if they decreased proportional to their growth, then that is a different matter.

Vi Nguyen · May 24, 2009, 08:02 AM

Was wondering how you worked out that they aren't dying out.

galactus · May 24, 2009, 08:56 AM

Perhaps they are.I may have made a mistake, but it seems logical.

Let's start at t=3 being t=0. So, as before, we have 9000 microbes.

Then, in the next hour, they reproduce at the same rate but 4500 are dead.

$9000e^{\frac{ln(3)(1)}{3}}-4500(1)=8480$

2 hours: $9000e^{\frac{ln(3)(2)}{3}}-4500(2)=20,717$

10 hours: $9000e^{\frac{ln(3)(10)}{3}}-4500(10)=755,012,498$

They are dying at 4500 STEADY each hour. So, we subtract off 4500 times however many hours they have been multiplying. Unless I misunderstood.