Solved on Feb 12, 2024

Find the carrying capacity of a population that grows according to the logistic equation $\frac{dP}{dt} = 0.084P - 0.00168P^2$ , where $t$ is measured in weeks.

STEP 1

Assumptions
1. The population $P(t)$ develops according to the logistic differential equation given by $\frac{d P}{d t}=0.084 P-0.00168 P^{2}$ .
2. The variable $t$ represents time measured in weeks.
3. The carrying capacity is the population size $P$ at which the growth rate $\frac{d P}{d t}$ becomes zero.

STEP 2

The logistic differential equation can be written in the form:
$\frac{d P}{d t}=rP\left(1-\frac{P}{K}\right)$
where $r$ is the intrinsic growth rate and $K$ is the carrying capacity.

STEP 3

Comparing the given logistic equation with the standard form, we identify the coefficients corresponding to $r$ and $K$ :
$0.084 P = rP$ $-0.00168 P^{2} = -r\frac{P^{2}}{K}$

STEP 4

From the first comparison, we can see that $r = 0.084$ .

STEP 5

From the second comparison, we solve for $K$ :
$-0.00168 P^{2} = -0.084\frac{P^{2}}{K}$

STEP 6

Divide both sides by $-P^{2}$ , assuming $P \neq 0$ (since we are interested in the non-trivial carrying capacity where the population is not zero):
$0.00168 = 0.084 / K$

STEP 7

Solve for $K$ by multiplying both sides by $K$ and then dividing by $0.00168$ :
$K = 0.084 / 0.00168$

STEP 8

Calculate the value of $K$ :
$K = \frac{0.084}{0.00168}$

STEP 9

Perform the division to find the carrying capacity:
$K = 50$
The carrying capacity of the population is 50.

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Find the carrying capacity of a population that grows according to the logistic equation $\frac{dP}{dt} = 0.084P - 0.00168P^2$ , where $t$ is measured in weeks.

STEP 1

STEP 2

The logistic differential equation can be written in the form:
$\frac{d P}{d t}=rP\left(1-\frac{P}{K}\right)$
where $r$ is the intrinsic growth rate and $K$ is the carrying capacity.

STEP 3

Comparing the given logistic equation with the standard form, we identify the coefficients corresponding to $r$ and $K$ :
$0.084 P = rP$ $-0.00168 P^{2} = -r\frac{P^{2}}{K}$

STEP 4

From the first comparison, we can see that $r = 0.084$ .

STEP 5

From the second comparison, we solve for $K$ :
$-0.00168 P^{2} = -0.084\frac{P^{2}}{K}$

STEP 6

Divide both sides by $-P^{2}$ , assuming $P \neq 0$ (since we are interested in the non-trivial carrying capacity where the population is not zero):
$0.00168 = 0.084 / K$

STEP 7

Solve for $K$ by multiplying both sides by $K$ and then dividing by $0.00168$ :
$K = 0.084 / 0.00168$

STEP 8

Calculate the value of $K$ :
$K = \frac{0.084}{0.00168}$

STEP 9

Perform the division to find the carrying capacity:
$K = 50$
The carrying capacity of the population is 50.

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Find the carrying capacity of a population that grows according to the logistic equation dPdt=0.084P−0.00168P2\frac{dP}{dt} = 0.084P - 0.00168P^2dtdP​=0.084P−0.00168P2, where ttt is measured in weeks.

STEP 1

STEP 2

The logistic differential equation can be written in the form:dPdt=rP(1−PK)\frac{d P}{d t}=rP\left(1-\frac{P}{K}\right)dtdP​=rP(1−KP​)where rrr is the intrinsic growth rate and KKK is the carrying capacity.

STEP 3

Comparing the given logistic equation with the standard form, we identify the coefficients corresponding to rrr and KKK:0.084P=rP0.084 P = rP0.084P=rP −0.00168P2=−rP2K-0.00168 P^{2} = -r\frac{P^{2}}{K}−0.00168P2=−rKP2​

STEP 4

From the first comparison, we can see that r=0.084r = 0.084r=0.084.

STEP 5

From the second comparison, we solve for KKK:−0.00168P2=−0.084P2K-0.00168 P^{2} = -0.084\frac{P^{2}}{K}−0.00168P2=−0.084KP2​

STEP 6

Divide both sides by −P2-P^{2}−P2, assuming P≠0P \neq 0P=0 (since we are interested in the non-trivial carrying capacity where the population is not zero):0.00168=0.084/K0.00168 = 0.084 / K0.00168=0.084/K

STEP 7

Solve for KKK by multiplying both sides by KKK and then dividing by 0.001680.001680.00168:K=0.084/0.00168K = 0.084 / 0.00168K=0.084/0.00168

STEP 8

Calculate the value of KKK:K=0.0840.00168K = \frac{0.084}{0.00168}K=0.001680.084​

STEP 9

Perform the division to find the carrying capacity:K=50K = 50K=50The carrying capacity of the population is 50.

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Find the carrying capacity of a population that grows according to the logistic equation $\frac{dP}{dt} = 0.084P - 0.00168P^2$ , where $t$ is measured in weeks.

The logistic differential equation can be written in the form:
$\frac{d P}{d t}=rP\left(1-\frac{P}{K}\right)$
where $r$ is the intrinsic growth rate and $K$ is the carrying capacity.

Comparing the given logistic equation with the standard form, we identify the coefficients corresponding to $r$ and $K$ :
$0.084 P = rP$ $-0.00168 P^{2} = -r\frac{P^{2}}{K}$

From the first comparison, we can see that $r = 0.084$ .

From the second comparison, we solve for $K$ :
$-0.00168 P^{2} = -0.084\frac{P^{2}}{K}$

Divide both sides by $-P^{2}$ , assuming $P \neq 0$ (since we are interested in the non-trivial carrying capacity where the population is not zero):
$0.00168 = 0.084 / K$

Solve for $K$ by multiplying both sides by $K$ and then dividing by $0.00168$ :
$K = 0.084 / 0.00168$

Calculate the value of $K$ :
$K = \frac{0.084}{0.00168}$

Perform the division to find the carrying capacity:
$K = 50$
The carrying capacity of the population is 50.