Question

The time rate of change of a rabbit population PP is proportional to the square root of PP. At time t=0t=0 (months) the population numbers 100100 rabbits and is increasing at the rate of 1010 rabbits per month. Let P′=kP12P′=kP12 describe the growth of the rabbit population, where kk is a positive constant to be found. Find the formulas for kk and for the rabbit population P(t)P(t) after tt months.

Answer 1

Answer:

$\frac{dP}{\sqrt{P}} = k dt$

And if we integrate both sides we got:

$2 \sqrt{P} = kt +C$

Where C is a constant., we can rewrite the expression like this:

$\sqrt{P} = \frac{1}{2} (kt +C)$

If we square both sides we got:

$P = \frac{1}{4} (kt +C)^2$

If we use the initial condition we have that:

$P(0) = 100 = \frac{1}{4} (k*0 +C)^2$

And we can solve for C like this:

$400 = C^2$

$C = 20$

And now we can find the derivate of the function and we got:

$P'(t) = 2* \frac{1}{4} (kt + 20) * k$

Using the condition $P'(0) = 10$ we got:

$10 = \frac{1}{2} k (k*0 +20)$

$20 = 20 k$

k= 1

And then the model is defined as:

$P = \frac{1}{4} (t +20)^2$

And for t =12 months we have:

$P(12) = \frac{1}{4} (12 +20)^2 = 256$

Step-by-step explanation:

For this case we cna use the proportional model given by:

$\frac{dP}{dt} = k \sqrt{P}$

Where k is a proportional constant, P the population and the represent the number of months

For this case we know the following initial condition $P(0) =100$ and $P'(0) = 10$

we can rewrite the differential equation like this:

$\frac{dP}{\sqrt{P}} = k dt$

And if we integrate both sides we got:

$2 \sqrt{P} = kt +C$

Where C is a constant., we can rewrite the expression like this:

$\sqrt{P} = \frac{1}{2} (kt +C)$

If we square both sides we got:

$P = \frac{1}{4} (kt +C)^2$

If we use the initial condition we have that:

$P(0) = 100 = \frac{1}{4} (k*0 +C)^2$

And we can solve for C like this:

$400 = C^2$

$C = 20$

And now we can find the derivate of the function and we got:

$P'(t) = 2* \frac{1}{4} (kt + 20) * k$

Using the condition $P'(0) = 10$ we got:

$10 = \frac{1}{2} k (k*0 +20)$

$20 = 20 k$

k= 1

And then the model is defined as:

$P = \frac{1}{4} (t +20)^2$

And for t =12 months we have:

$P(12) = \frac{1}{4} (12 +20)^2 = 256$