Rate-Of-Change Models in Microbiology part 2 of 2?

Question

Thanks for the answer to the previous question.

Part 2 Question: Like Humans, many bacteria also need oxygen to survive. The oxygen demand for bacteria is
D(of O2)=10(mc)((dmb) / dt) [litres per hour]
*c and b and bases of m

What is the oxygen demand after five days??

The link to the previous question with the rest of the information is here:

http://answers.yahoo.com/question/index;_ylt=AqnKeOOv1wqAWaPFQEadNbPzy6IX?qid=20070104071802AAVf8gs

****By ((dmb) / dt), I mean the derivative of mb with respect to t?

****By ((dmb) / dt), I mean the derivative of mb with respect to t.

Answer 1

I found a mistake in my last solution...just so you know...

I made a mistake here:

1.7333 = 1 + (2.4/0.2)e^(-3.9k)
0.7333 = 1.2e^(-3.9k)

2.4/0.2 = 12 not 1.2

Everything after that needs to be recalculated.... so all this work is wrong too...

--------

mb(t) = 2.6/(1 + 1.2e^(-8.860t))
mb'(t) = -2.6(1.2(-8.860)e^(-8.860t))/(1 + 1.2e^(-8.860t))²
mb'(t) = 5.74e^(-8.860t))/(1 + 1.2e^(-8.860t))²
mb'(5) = 2.398 x 10^-34 (seems really small)

Then

mc(5) = -log3 (√5 + 1) + 2.5
= -log (1 + √5)/log 3 + 2.5
= 1.431

Then

D(5) = 10(1.431)(2.398 x 10^-34)
= 3.431538 x 10^-33

Again, seems kind of low to me.