Homework Help: Math (Calculus II)

Posted by John on Sunday, September 5, 2010 at 1:29am.

Rate of change proportional to size:

P(t)= P(i)e^kt, where P is population, t is time, k is a constant and P(i) is the initial p or "p naught"; sorry, I don't know the typed notation for subscript. By the way, p(i) is given as 100.

The problems are

a.) "Find the expression after t hours"

b.) "Find the number of bacteria after t hours."

The answer to a is(via back of the book) is: 100(4.2)^t

I'm not sure exactly how to get there though I believe it has to do with taking natural logarithms and solving for e^k but since p is a function of t I'm not sure how to differentiate "k" (treat it as a constant or differentiate implicitly). If you could show the step by step process I'd appreciate it.

I got to: p(t)= 100(e^t+e^k) and stalled out.

Thanks in advance!

Math (Calculus II) - drwls, Sunday, September 5, 2010 at 2:06am
Without some specific numerical information on growth rate or the value of k, I don't see how that back-of-the-book equation can be derived. Is t in seconds in the original equation?

I also don't see the difference between "the population" and "the number of bacteria"

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