Math / Calculus

QuestionConsider the model for limited logistic growth given by $A=\frac{c}{1+a e^{-b t}}$ . The amount, or size, at time $t$ is represented by $\qquad$ . This value can never exceed $\qquad$ .
Consider the model for limited logistic growth given by $A=\frac{c}{1+a e^{-b t}}$ . The amount, or size, at time $t$ is represented by $\square$ This value can never exceed $\square$ A. t. $e$ . b.

Studdy Solution

STEP 1

What is this asking? This problem wants us to figure out what the different parts of the logistic growth formula mean, specifically what represents the current size and what's the biggest it can get. Watch out! Don't mix up the different parts of the formula! $a$ , $b$ , $c$ , and $t$ all play different roles.
Also, remember $e$ is just a number, like $\pi$ .

STEP 2

1. Understand the formula
2. Find the amount at time $t$
3. Find the maximum value

STEP 3

Let's break down this fancy formula: $A = \frac{c}{1 + ae^{-bt}}$ .
It describes how something grows over time, like the number of fish in a pond or the spread of a rumor.

STEP 4

Think of it like this: the pond can only hold so many fish, right?
That's our limit, and that's where $c$ comes in. $c$ represents the carrying capacity, the maximum value that $A$ can approach.

STEP 5

Now, what about $t$ ? $t$ represents time.
As time goes on, the fish population grows, and the value of $A$ changes.

STEP 6

$a$ and $b$ are constants that affect how quickly the growth happens.
Don't worry too much about them for now, just know they help shape the curve. $e$ is another constant, Euler's number, approximately equal to 2.718.
It's just a special number that pops up in growth and decay situations.

STEP 7

The formula itself tells us the amount at any given time $t$ . $A$ is the amount or size at a specific time.
So, if we plug in a value for $t$ , we get the corresponding value for $A$ .

STEP 8

As time $t$ gets bigger and bigger, what happens to $e^{-bt}$ ?
Since $b$ is a positive constant, $-bt$ becomes a very large negative number. $e$ raised to a very large negative number gets incredibly close to zero.

STEP 9

So, as $t$ approaches infinity, $e^{-bt}$ approaches zero.
This means the denominator, $1 + ae^{-bt}$ , gets closer and closer to one (because we're adding something very close to zero to one).

STEP 10

Now, look at the whole formula: $A = \frac{c}{1 + ae^{-bt}}$ .
If the denominator is approaching one, then the whole fraction is approaching $\frac{c}{1}$ , which is just $c$ .

STEP 11

Therefore, the value of $A$ can never exceed $c$ , which is the carrying capacity. $A$ gets closer and closer to $c$ as time goes on, but it can never actually reach or surpass it.

STEP 12

The amount, or size, at time $t$ is represented by $A$ .
This value can never exceed $c$ .

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Studdy Solution

STEP 1

STEP 2

1. Understand the formula
2. Find the amount at time $t$
3. Find the maximum value

STEP 3

Let's break down this fancy formula: $A = \frac{c}{1 + ae^{-bt}}$ .
It describes how something grows over time, like the number of fish in a pond or the spread of a rumor.

STEP 4

Think of it like this: the pond can only hold so many fish, right?
That's our limit, and that's where $c$ comes in. $c$ represents the carrying capacity, the maximum value that $A$ can approach.

STEP 5

Now, what about $t$ ? $t$ represents time.
As time goes on, the fish population grows, and the value of $A$ changes.

STEP 6

$a$ and $b$ are constants that affect how quickly the growth happens.
Don't worry too much about them for now, just know they help shape the curve. $e$ is another constant, Euler's number, approximately equal to 2.718.
It's just a special number that pops up in growth and decay situations.

STEP 7

The formula itself tells us the amount at any given time $t$ . $A$ is the amount or size at a specific time.
So, if we plug in a value for $t$ , we get the corresponding value for $A$ .

STEP 8

As time $t$ gets bigger and bigger, what happens to $e^{-bt}$ ?
Since $b$ is a positive constant, $-bt$ becomes a very large negative number. $e$ raised to a very large negative number gets incredibly close to zero.

STEP 9

So, as $t$ approaches infinity, $e^{-bt}$ approaches zero.
This means the denominator, $1 + ae^{-bt}$ , gets closer and closer to one (because we're adding something very close to zero to one).

STEP 10

Now, look at the whole formula: $A = \frac{c}{1 + ae^{-bt}}$ .
If the denominator is approaching one, then the whole fraction is approaching $\frac{c}{1}$ , which is just $c$ .

STEP 11

Therefore, the value of $A$ can never exceed $c$ , which is the carrying capacity. $A$ gets closer and closer to $c$ as time goes on, but it can never actually reach or surpass it.

STEP 12

The amount, or size, at time $t$ is represented by $A$ .
This value can never exceed $c$ .

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Studdy Solution

STEP 1

STEP 2

1. Understand the formula2. Find the amount at time ttt3. Find the maximum value

STEP 3

Let's break down this fancy formula: A=c1+ae−btA = \frac{c}{1 + ae^{-bt}}A=1+ae−btc​.It describes how something grows over time, like the number of fish in a pond or the spread of a rumor.

STEP 4

Think of it like this: the pond can only hold so many fish, right?That's our **limit**, and that's where ccc comes in. ccc represents the **carrying capacity**, the maximum value that AAA can approach.

STEP 5

Now, what about ttt? ttt represents **time**.As time goes on, the fish population grows, and the value of AAA changes.

STEP 6

aaa and bbb are **constants** that affect how quickly the growth happens.Don't worry too much about them for now, just know they help shape the curve. eee is another constant, Euler's number, approximately equal to **2.718**.It's just a special number that pops up in growth and decay situations.

STEP 7

The formula itself tells us the amount at any given time ttt. AAA is the **amount** or **size** at a specific time.So, if we plug in a value for ttt, we get the corresponding value for AAA.

STEP 8

As time ttt gets bigger and bigger, what happens to e−bte^{-bt}e−bt?Since bbb is a positive constant, −bt-bt−bt becomes a very large negative number. eee raised to a very large negative number gets incredibly close to **zero**.

STEP 9

So, as ttt approaches infinity, e−bte^{-bt}e−bt approaches **zero**.This means the denominator, 1+ae−bt1 + ae^{-bt}1+ae−bt, gets closer and closer to **one** (because we're adding something very close to zero to one).

STEP 10

Now, look at the whole formula: A=c1+ae−btA = \frac{c}{1 + ae^{-bt}}A=1+ae−btc​.If the denominator is approaching **one**, then the whole fraction is approaching c1\frac{c}{1}1c​, which is just ccc.

STEP 11

Therefore, the value of AAA can never exceed ccc, which is the **carrying capacity**. AAA gets closer and closer to ccc as time goes on, but it can never actually reach or surpass it.

STEP 12

The amount, or size, at time ttt is represented by AAA.This value can never exceed ccc.

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Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. Understand the formula
2. Find the amount at time $t$
3. Find the maximum value

Let's break down this fancy formula: $A = \frac{c}{1 + ae^{-bt}}$ .
It describes how something grows over time, like the number of fish in a pond or the spread of a rumor.

Think of it like this: the pond can only hold so many fish, right?
That's our limit, and that's where $c$ comes in. $c$ represents the carrying capacity, the maximum value that $A$ can approach.

Now, what about $t$ ? $t$ represents time.
As time goes on, the fish population grows, and the value of $A$ changes.

$a$ and $b$ are constants that affect how quickly the growth happens.
Don't worry too much about them for now, just know they help shape the curve. $e$ is another constant, Euler's number, approximately equal to 2.718.
It's just a special number that pops up in growth and decay situations.

The formula itself tells us the amount at any given time $t$ . $A$ is the amount or size at a specific time.
So, if we plug in a value for $t$ , we get the corresponding value for $A$ .

As time $t$ gets bigger and bigger, what happens to $e^{-bt}$ ?
Since $b$ is a positive constant, $-bt$ becomes a very large negative number. $e$ raised to a very large negative number gets incredibly close to zero.

So, as $t$ approaches infinity, $e^{-bt}$ approaches zero.
This means the denominator, $1 + ae^{-bt}$ , gets closer and closer to one (because we're adding something very close to zero to one).

Now, look at the whole formula: $A = \frac{c}{1 + ae^{-bt}}$ .
If the denominator is approaching one, then the whole fraction is approaching $\frac{c}{1}$ , which is just $c$ .

Therefore, the value of $A$ can never exceed $c$ , which is the carrying capacity. $A$ gets closer and closer to $c$ as time goes on, but it can never actually reach or surpass it.

The amount, or size, at time $t$ is represented by $A$ .
This value can never exceed $c$ .