The rate of change of the population of a small town is dP/dt=kP, where P is the population, t is time in years and k is the growth rate.
If P=20000 when t=3 and P=30000 when t=5, what is the population when t=10? Round your answer to the nearest integer.
To find the population when t=10, we can start by finding the value of the constant k.
Given the rate of change equation dP/dt = kP, we can solve for k using the two given data points.
When t=3, P=20000:
dP/dt = kP
k = (dP/dt)/P = (30000 - 20000)/(5 - 3) = 10000/2 = 5000
Now we have the value of k, which is 5000.
Using the rate of change equation, we can now find the population when t=10:
dP/dt = kP
dP = kP dt
1/P dP = k dt
Integrating both sides:
∫1/P dP = ∫k dt
ln|P| = kt + C
P = e^(kt+C)
Since P=20000 when t=3, we can substitute these values into the equation:
20000 = e^(5000*3+C)
Dividing both sides by e^(5000*3):
e^-C = 20000/e^(5000*3)
e^-C = 20000/e^15000
Taking the natural logarithm of both sides:
ln(e^-C) = ln(20000/e^15000)
-C = ln(20000) - 15000*ln(e)
-C = ln(20000) - 15000
C = -ln(20000) + 15000
Now we have the value of C, which is approximately -2.71319 + 15000.
Substituting the value of C into the equation:
P = e^(kt+C)
P = e^(5000t-2.71319+15000)
Finally, we can find the population when t=10:
P = e^(5000*10-2.71319+15000)
P ≈ 986073
Therefore, when t=10, the population of the town is approximately 986073.
Just to get you started,
dP/dt=kP
dP/P = k dt
ln P = kt + ln c
P = c*e^(kt)