Integral of the reciprocal Function

• Jan 18th 2010, 09:35 PM
sacsha1
Integral of the reciprocal Function
Hi guys, its my first post here...hoping you guy can help me solve this problem...

Population Problem: Population of a community is growing at a rate of 5% per year.

DP/DT = .05P

Suppose the population, P, is 1000 at time t=0, and grows to P=N at t=10, where you must fine the value of N. By separating the variables and then integrating you get,

1/p Dp = .05dt

N 10
1 DP = .05dt
p
1000 0

Evaluate the integral numerically for N= 1000, 1500, 2000, 2500, 500, and 100.
• Jan 18th 2010, 09:41 PM
pickslides
Quote:

Originally Posted by sacsha1

Suppose the population, P, is 1000 at time t=0, and grows to P=N at t=10, where you must fine the value of N. By separating the variables and then integrating you get,

1/p Dp = .05dt

N 10
1 DP = .05dt
p

sacsha1, you did not finish the integration

$\int \frac{dp}{p} = \int \frac{dt}{2}$

gives

$\ln(p) = \frac{t}{2}+c$

$p = e^{\frac{t}{2}+c}$

$p = Ae^{\frac{t}{2}}$

Now use $p(0)=1000$ to find A