Math / Calculus

Question(1 point) Recall that one model for population growth states that a population grows at a rate proportional to its size. (a) We begin with the differential equation $\frac{d P}{d t}=\frac{1}{2} P$ . Find an equilibrium solution: $P=$ $\square$ Is this equilibrium solution stable or unstable? A. stable B. unstable
Describe the long-term behavior of the solution to $\frac{d P}{d t}=\frac{1}{2} P$ when $P(0)$ is positive. A. The value of $P$ approaches zero. B. The value of $P$ oscillates and does not approach a limit. C. The value of $P$ approaches a nonzero constant. D. The value of $P$ increases without bound. E. None of the above

Studdy Solution

STEP 1

What is this asking? We're looking at a population that grows proportionally to its size, trying to find a stable population size and see how the population changes over a long time. Watch out! Don't mix up stable and unstable equilibrium – stable means the population settles down, unstable means it runs away!

STEP 2

1. Find the Equilibrium Solution
2. Analyze Stability
3. Long-Term Behavior

STEP 3

An equilibrium solution means the population isn't changing, so the rate of change is zero.
We set $\frac{dP}{dt}$ to zero in our equation: $\frac{dP}{dt} = \frac{1}{2}P = 0$

STEP 4

To solve for $P$ , we can multiply both sides by $2$ : $2 \cdot \frac{1}{2} P = 2 \cdot 0$ $P = 0$ So our equilibrium solution is $P = 0$ .

STEP 5

Imagine the population is a tiny bit bigger than zero.
Since the growth rate $\frac{1}{2}P$ is positive, the population will increase.
If the population is slightly less than zero (which doesn't really make sense in the real world, but let's roll with it mathematically!), the growth rate will be negative, meaning the population will decrease further.

STEP 6

Since the population moves away from zero in both cases, the equilibrium is unstable.

STEP 7

We're told $P(0)$ is positive.
This means we're starting with a population greater than zero.

STEP 8

Since the rate of change $\frac{1}{2}P$ is proportional to the population size, and the population starts positive, the population will keep growing!
The bigger $P$ gets, the faster it grows.

STEP 9

This means the value of $P$ increases without bound.

STEP 10

The equilibrium solution is $P = 0$ .
This equilibrium is unstable.
If the initial population is positive, the population increases without bound.

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Studdy Solution

STEP 1

STEP 2

1. Find the Equilibrium Solution
2. Analyze Stability
3. Long-Term Behavior

STEP 3

An equilibrium solution means the population isn't changing, so the rate of change is zero.
We set $\frac{dP}{dt}$ to zero in our equation: $\frac{dP}{dt} = \frac{1}{2}P = 0$

STEP 4

To solve for $P$ , we can multiply both sides by $2$ : $2 \cdot \frac{1}{2} P = 2 \cdot 0$ $P = 0$ So our equilibrium solution is $P = 0$ .

STEP 5

STEP 6

Since the population moves away from zero in both cases, the equilibrium is unstable.

STEP 7

We're told $P(0)$ is positive.
This means we're starting with a population greater than zero.

STEP 8

Since the rate of change $\frac{1}{2}P$ is proportional to the population size, and the population starts positive, the population will keep growing!
The bigger $P$ gets, the faster it grows.

STEP 9

This means the value of $P$ increases without bound.

STEP 10

The equilibrium solution is $P = 0$ .
This equilibrium is unstable.
If the initial population is positive, the population increases without bound.

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Studdy Solution

STEP 1

STEP 2

1. Find the Equilibrium Solution2. Analyze Stability3. Long-Term Behavior

STEP 3

An *equilibrium* solution means the population isn't changing, so the rate of change is zero.We **set** dPdt\frac{dP}{dt}dtdP​ **to zero** in our equation: dPdt=12P=0 \frac{dP}{dt} = \frac{1}{2}P = 0 dtdP​=21​P=0

STEP 4

To **solve for** PPP, we can multiply both sides by 222: 2⋅12P=2⋅0 2 \cdot \frac{1}{2} P = 2 \cdot 0 2⋅21​P=2⋅0 P=0 P = 0 P=0So our **equilibrium solution** is P=0P = 0P=0.

STEP 5

STEP 6

Since the population moves *away* from zero in both cases, the equilibrium is **unstable**.

STEP 7

We're told P(0)P(0)P(0) is positive.This means we're starting with a population greater than zero.

STEP 8

Since the rate of change 12P\frac{1}{2}P21​P is proportional to the population size, and the population starts positive, the population will keep growing!The bigger PPP gets, the faster it grows.

STEP 9

This means the value of PPP **increases without bound**.

STEP 10

The equilibrium solution is P=0P = 0P=0.This equilibrium is unstable.If the initial population is positive, the population increases without bound.

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1. Find the Equilibrium Solution
2. Analyze Stability
3. Long-Term Behavior

An equilibrium solution means the population isn't changing, so the rate of change is zero.
We set $\frac{dP}{dt}$ to zero in our equation: $\frac{dP}{dt} = \frac{1}{2}P = 0$

To solve for $P$ , we can multiply both sides by $2$ : $2 \cdot \frac{1}{2} P = 2 \cdot 0$ $P = 0$ So our equilibrium solution is $P = 0$ .

Since the population moves away from zero in both cases, the equilibrium is unstable.

We're told $P(0)$ is positive.
This means we're starting with a population greater than zero.

Since the rate of change $\frac{1}{2}P$ is proportional to the population size, and the population starts positive, the population will keep growing!
The bigger $P$ gets, the faster it grows.

This means the value of $P$ increases without bound.

The equilibrium solution is $P = 0$ .
This equilibrium is unstable.
If the initial population is positive, the population increases without bound.