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September 30, 2014

Homework Help: Calculus AB

Posted by Savannah on Friday, December 16, 2011 at 4:51pm.

The rate of growth dP/dt of a population of bacteria is proportional to the square root of t, where p is the population size and t is the time in days (0<equal t <equal 10). That is, dp/dt = ksqrt(t). The initial size of the population is 500. After 1 day the population has grown to 600. Estimate the population after 7 days.

I know that f(0)=500 and f(1)=600.

The first thing I would do would be to find the integral of ksqrt(t)dt, right? I don't understand what that k means. Do I just leave it there?

This is what I have so far, but I think it is wrong.

dp/dt = ksqrt(t)
dp=integral(ksqrt(t)dt)
P= 2/3k(t^3/2)

Now I'm stuck. Can someone help point me in the right direction?

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