A population is modeled by the differential equation. $\frac { d P } { d t } = 1.4 P \left( 1 - \frac { P } { 4560 } \right)$ For what values of P is the population increasing? A) $P 1.4$ B) $P 4560$ C) $P

Question

A population is modeled by the differential equation. $\frac { d P } { d t } = 1.4 P \left( 1 - \frac { P } { 4560 } \right)$ For what values of P is the population increasing? A) $P > 1.4$ B) $P > 4560$ C) $P < 4580$ D) $0 < P < 4560$ E) $0 < P < 1.4$

Quizplus · Accepted Answer

The population is increasing if $\frac{dP}{dt}>0$. Given the differential equation, we have $$\frac{dP}{dt} = 1.4P\left(1-\frac{P}{4560} ight)$$ To determine when the population is increasing, we need to find the values of $P$ where $\frac{dP}{dt}>0$. Notice that $\frac{dP}{dt}=0$ at $P=0$, $P=4560$, and $P=9120$. These values divide the $P$-axis into four intervals: $(-\infty,0)$, $(0,4560)$, $(4560,9120)$, and $(9120,\infty)$. We can choose one test value in each of these intervals to determine the sign of $\frac{dP}{dt}$ in each interval. Let's choose $P=-1000$, $P=2000$, $P=6000$, and $P=10000$ as our test values. If $P=-1000$, then we have $\frac{dP}{dt} = 1.4(-1000)\left(1-\frac{-1000}{4560} ight)= 2.73\cdot10^3 >0$. Therefore, $\frac{dP}{dt}>0$ on the interval $(-\infty,0)$. If $P=2000$, then we have $\frac{dP}{dt} = 1.4(2000)\left(1-\frac{2000}{4560} ight)= 709.09 >0$. Therefore, $\frac{dP}{dt}>0$ on the interval $(0,4560)$. If $P=6000$, then we have $\frac{dP}{dt} = 1.4(6000)\left(1-\frac{6000}{4560} ight)= -7560 <0$. Therefore, $\frac{dP}{dt}<0$ on the interval $(4560,9120)$. If $P=10000$, then we have $\frac{dP}{dt} = 1.4(10000)\left(1-\frac{10000}{4560} ight)= -2.73\cdot10^4 <0$. Therefore, $\frac{dP}{dt}<0$ on the interval $(9120,\infty)$. Putting all of this together, we see that $\frac{dP}{dt}>0$ on the interval $(0,4560)$, and $\frac{dP}{dt}<0$ on the intervals $(-\infty,0)$, $(4560,9120)$, and $(9120,\infty)$. Therefore, the population is increasing when $0

A Population Is Modeled by the Differential Equation dPdt=1.4P(1−P4560)\frac { d P } { d t } = 1.4 P \left( 1 - \frac { P } { 4560 } \right)dtdP​=1.4P(1−4560P​)

A Population Is Modeled by the Differential Equation $\frac { d P } { d t } = 1.4 P \left( 1 - \frac { P } { 4560 } \right)$