The rate of change of the population of a small town is dP/dt=kP
Where P is the population, t is the 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.
Well, you know what they say about small towns. They might not be big, but they can be quite fruitful! Now, let's put on our mathematic clown glasses and solve this problem.
We have the differential equation dP/dt = kP, where P is the population and t is the time.
To solve this, we'll use the fact that dP/dt = kP. Rearranging the equation, we get dP/P = kdt.
Integrating both sides, we have ∫(1/P)dP = ∫kdt.
Taking the integral, we get ln(P) = kt + C.
Now let's use the given information. When t = 3, P = 20000. Substituting these values into our equation, we get ln(20000) = 3k + C.
Next, when t = 5, P = 30000. Substituting these values, we get ln(30000) = 5k + C.
Now let's subtract the first equation from the second equation to eliminate C. We have ln(30000) - ln(20000) = 5k - 3k.
Using the properties of logarithms, we can simplify this to ln(30000/20000) = 2k.
Now, let's solve for k: 2k = ln(3/2), which means k = (1/2)ln(3/2).
Finally, we can plug this value of k back into our original equation to find the population when t = 10.
Using P(t) = P(0)e^(kt), where P(0) is the initial population at t = 0, we get P(t) = 20000e^((1/2)ln(3/2)t).
Simplifying further, we have P(t) = 20000(3/2)^(t/2).
Now, let's calculate the population when t = 10, rounding to the nearest integer: P(10) ≈ 20000(3/2)^(10/2).
Drumroll, please! After calculating, we find that P(10) is approximately 34825.
So, the population of the small town is around 34825 when t = 10. Keep growing, little town!
To find the population when t = 10, we need to solve the differential equation dP/dt = kP. Since we are given two points where P and t values are known, we can use those points to find the value of k.
First, let's use the given information to find the value of k. We know that when t = 3, P = 20000. Substituting these values into the equation, we have:
dP/dt = kP
k = (dP/dt)/P = (30000 - 20000)/(5 - 3) = 10000/2 = 5000
Now we have the value of k, we can solve the differential equation to find the population when t = 10.
dP/dt = kP
Separating variables, we get:
dP/P = k dt
Integrating both sides:
∫dP/P = ∫k dt
ln|P| = kt + C
Applying the initial condition P = 20000 at t = 3:
ln|20000| = 5000*3 + C
ln|20000| = 15000 + C
C = ln|20000| - 15000
So, the equation becomes:
ln|P| = 5000t + ln|20000| - 15000
Now, let's solve for P when t = 10:
ln|P| = 5000*10 + ln|20000| - 15000
ln|P| = 50000 + ln|20000| - 15000
ln|P| = 35000 + ln|20000|
|P| = e^(35000 + ln|20000|)
P = e^(35000) * e^(ln|20000|)
P ≈ 3.49 * 10^15
Rounding to the nearest integer, the population when t = 10 is approximately 3,490,000,000,000.
To find the population when t=10, we can use the given information and the differential equation dP/dt=kP. We can solve this differential equation using separation of variables.
First, let's integrate both sides of the equation:
∫(dP/P) = ∫k dt
Integrating, we get:
ln|P| = kt + C1
Next, we can exponentiate both sides to eliminate the natural logarithm:
|P| = e^(kt + C1)
Since we are interested in the population, we can drop the absolute value sign and rewrite the equation as:
P = ±e^(kt + C1)
To determine the correct sign of the exponential expression, we can substitute the given values of P and t when t=3 and t=5:
P = ±e^(k(3) + C1) = ±20000
P = ±e^(k(5) + C1) = ±30000
From these two equations, it is clear that P is increasing, so we can drop the negative signs. Therefore, we have:
P = e^(3k + C1) = 20000
P = e^(5k + C1) = 30000
Now, let's solve for C1 using the first equation when t=3:
e^(3k + C1) = 20000
C1 = ln(20000) - 3k
Substituting this expression for C1 into the second equation, we have:
e^(5k + ln(20000) - 3k) = 30000
Simplifying the exponent, we get:
e^(2k + ln(20000)) = 30000
Now, let's solve for k. Taking the natural logarithm of both sides, we have:
2k + ln(20000) = ln(30000)
2k = ln(30000) - ln(20000)
k = (ln(30000) - ln(20000))/2
With the value of k, we can substitute it back into either of the previous equations to find P when t=10. Let's use the first equation:
P = e^(3k + C1) = e^(3((ln(30000) - ln(20000))/2) + ln(20000) - 3k)
Calculating this expression, we get:
P ≈ 48158.88
Rounding to the nearest integer, the population when t=10 is approximately 48159.
Oh, well, when the rate is proportional to the amount, we have an e^kx kind of a thing
P = Pi e^kt
at t = 3
20,000 = Pi e^3k
at t = 5
30,000 = Pi e^5t
so
3/2 = e^5k/e^3k = e^2k
so
ln (1.5) = 2 k
got it?