# Application of differentiation, modeling bacteria

https://www.dropbox.com/s/defhs0u02yuqtyw/differentiation%20bacteria.jpg

I basically don't understand the first sentence of the question. A good explanation and a a complete solution would be appreciated

Edit: Ok since people asked for my work here it is a) innitial population is when t=0 hence P=11 and I guess this would translate to 11 000 bacteria .Bacteria will reach 14 mil when 10+e^t-3t=14 Ok so I don't understand why 14 and not 14 000 it gives t=2.42 hours b) e^t-3 I don't know how to approach this part because I am confused from the previous one c) e^t the rate of change of the growth rate of the bacteria

• Please keep the question self contained. – Tom Nov 20 '13 at 20:11
• A complete solution is not what people do (Or at least should do ) on this website. Tell us what you think, show us work. – LASV Nov 20 '13 at 20:12
• Thousands here is just a unit, like meters, or pounds. But when you solve actual equations you don't use units, unless you have some quantities that have different units so you need to convert them. For example, if they ask you at what time population will be 15 million, then millions is a different unit, not thousands, so you need to convert it - $15\ mil = 15000\ thousand$ and put that value into equation. – Kaster Nov 20 '13 at 20:26
• Damn, I misread the question, they actually ask when it's 14 million, not 14 thousand :) – Kaster Nov 20 '13 at 20:31
• Thanks it turned out there was a mistake in the answers in my textbook. I was correct the first time – Rinik Nov 20 '13 at 22:25

Your thoughts on part a are correct. For when the population will reach 14 million, set $P=14,000$ since our units are in thousands. For part b, you differentiated correctly, and you do a very similar thing as in part a, considering that $\frac{dP}{dt}$ is the rate of growth. Your work for part c is also correct. For the second part of c, recall that a min/max of a function is found where its derivative is $0$, and you can justify that the extremum is a minimum using the second derivative test.