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!

## Answer this Question

## Related Questions

- Math (Calculus II) - Rate of change proportional to size: P(t)= P(i)e^kt, where ...
- Calculs I - A bacterial population grows at a rate proportional to the ...
- Math - The size of a bacteria population is given by P=C*e^(kt) Where C is the ...
- maths --plse help me.. - A bacteria culture has an initial population of 600. ...
- Algebra - Solve. The population of a particular city is increasing at a rate ...
- Calculus AB - The rate of growth dP/dt of a population of bacteria is ...
- Calculus - The rate of growth of a particular population is given by dP/dt=50t^2...
- Pleaaaaaase help with differential graph question - Suppose further that the ...
- Calculus - Suppose that you follow the size of a population over time. When you ...
- precal - You can find the size of a population after t years using the formula N...

More Related Questions