Thursday
April 24, 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?

Answer this Question

First Name:
School Subject:
Answer:

Related Questions

CACULUS- EXPONENTIAL GROWTH - A population of animals in an ecological niche is ...
Math - It is known that if the deer population falls below a certain level, m, ...
Calculus - Assume that the rate of population growth of ants is given by the ...
Math - Assume that the rate of population growth of ants is given by the ...
Calculus - Suppose that a population develops according to the logistic ...
diffeq - Let P(t) represent the population of a non-native species introduced to...
diffeq - Suppose a species of fish in a particular lake has a population that is...
Further calculus - 1) A price p (in dollars) and demand x for a product are ...
Calculus - dP/dt = k(P-a) dP/dt = kP-ka dP/P = k-ka is this right so far? was i ...
Bio/Math - A population of glasswing butterflies exhibits logistic growth. The ...

Search
Members