Math / Calculus

Question2 Multiple Choice 1 point
The spread of a virus through a community can be modeled with the logistic equation $\mathrm{P}(\mathrm{t})$ , where $t$ is time in weeks and $\mathrm{P}(\mathrm{t})$ represents the number of people infected with the virus. Suppose that 10 people originally have the virus, and the number of people infected is increasing approximately exponentially, with a continuous growth rate of 1.45 . It is estimated that, in the long run, approximately 4600 people will be infected. What is the logistic equation that could model this data? $P(t)=\frac{10}{\left(1+459 e^{-1.45 t}\right)}$ $P(t)=\frac{4600}{\left(1+1.45 e^{-10 t}\right)}$ $P(t)=\frac{4600}{1+10 e^{-1.45 t}}$ $P(t)=\frac{4600}{\left(1+459 e^{-1.45 t}\right)}$

Studdy Solution

STEP 1

What is this asking? We need to find the equation that describes how a virus spreads, knowing the initial number of infected people, how quickly the infection spreads initially, and the maximum number of people who could get infected. Watch out! Don't mix up the initial number of infected people with the maximum possible number of infected people!
Also, make sure to use the correct growth rate in the equation.

STEP 2

1. Set up the logistic equation
2. Plug in the given values

STEP 3

Alright, let's start with the general form of the logistic equation.
This equation is super handy for modeling things that grow quickly at first, but then slow down as they approach a limit.
The equation looks like this: $P(t) = \frac{C}{1 + A e^{-kt}}$ Here, $P(t)$ is the number of infected people at time $t$ . $C$ is the carrying capacity, which is the maximum number of people that can get infected. $k$ is the growth rate of the infection, and $A$ is a constant we'll figure out shortly!

STEP 4

We're given that the carrying capacity, $C$ , is 4600 people.
Let's plug that into our equation: $P(t) = \frac{4600}{1 + A e^{-kt}}$

STEP 5

We also know the initial growth rate, $k$ , is 1.45.
Let's plug that in too: $P(t) = \frac{4600}{1 + A e^{-1.45t}}$

STEP 6

Now, we're told that initially, 10 people have the virus.
That means when $t = 0$ , $P(t) = 10$ .
Let's substitute these values: $10 = \frac{4600}{1 + A e^{-1.45 \cdot 0}}$

STEP 7

Since anything raised to the power of zero is one, $e^{-1.45 \cdot 0} = e^0 = 1$ .
So, our equation simplifies to: $10 = \frac{4600}{1 + A}$

STEP 8

To solve for $A$ , we can multiply both sides by $(1 + A)$ and then divide both sides by 10: $10 \cdot (1 + A) = 4600$ $1 + A = \frac{4600}{10}$ $1 + A = 460$

STEP 9

Finally, subtract one from both sides to find $A$ : $A = 460 - 1$ $A = 459$

STEP 10

Now we have all the pieces!
Let's plug $A = \textbf{459}$ back into our equation: $P(t) = \frac{4600}{1 + 459 e^{-1.45t}}$

STEP 11

The logistic equation that models this data is $P(t) = \frac{4600}{1 + 459 e^{-1.45t}}$ .

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Studdy Solution

STEP 1

STEP 2

1. Set up the logistic equation
2. Plug in the given values

STEP 3

STEP 4

We're given that the carrying capacity, $C$ , is 4600 people.
Let's plug that into our equation: $P(t) = \frac{4600}{1 + A e^{-kt}}$

STEP 5

We also know the initial growth rate, $k$ , is 1.45.
Let's plug that in too: $P(t) = \frac{4600}{1 + A e^{-1.45t}}$

STEP 6

Now, we're told that initially, 10 people have the virus.
That means when $t = 0$ , $P(t) = 10$ .
Let's substitute these values: $10 = \frac{4600}{1 + A e^{-1.45 \cdot 0}}$

STEP 7

Since anything raised to the power of zero is one, $e^{-1.45 \cdot 0} = e^0 = 1$ .
So, our equation simplifies to: $10 = \frac{4600}{1 + A}$

STEP 8

To solve for $A$ , we can multiply both sides by $(1 + A)$ and then divide both sides by 10: $10 \cdot (1 + A) = 4600$ $1 + A = \frac{4600}{10}$ $1 + A = 460$

STEP 9

Finally, subtract one from both sides to find $A$ : $A = 460 - 1$ $A = 459$

STEP 10

Now we have all the pieces!
Let's plug $A = \textbf{459}$ back into our equation: $P(t) = \frac{4600}{1 + 459 e^{-1.45t}}$

STEP 11

The logistic equation that models this data is $P(t) = \frac{4600}{1 + 459 e^{-1.45t}}$ .

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Studdy Solution

STEP 1

STEP 2

1. Set up the logistic equation2. Plug in the given values

STEP 3

STEP 4

We're given that the **carrying capacity**, CCC, is **4600** people.Let's plug that into our equation: P(t)=46001+Ae−kt P(t) = \frac{4600}{1 + A e^{-kt}} P(t)=1+Ae−kt4600​

STEP 5

We also know the **initial growth rate**, kkk, is **1.45**.Let's plug that in too: P(t)=46001+Ae−1.45t P(t) = \frac{4600}{1 + A e^{-1.45t}} P(t)=1+Ae−1.45t4600​

STEP 6

Now, we're told that **initially**, 10 people have the virus.That means when t=0t = 0t=0, P(t)=10P(t) = 10P(t)=10.Let's substitute these values: 10=46001+Ae−1.45⋅0 10 = \frac{4600}{1 + A e^{-1.45 \cdot 0}} 10=1+Ae−1.45⋅04600​

STEP 7

Since anything raised to the power of zero is one, e−1.45⋅0=e0=1e^{-1.45 \cdot 0} = e^0 = 1e−1.45⋅0=e0=1.So, our equation simplifies to: 10=46001+A 10 = \frac{4600}{1 + A} 10=1+A4600​

STEP 8

To solve for AAA, we can multiply both sides by (1+A)(1 + A)(1+A) and then divide both sides by 10: 10⋅(1+A)=4600 10 \cdot (1 + A) = 4600 10⋅(1+A)=4600 1+A=460010 1 + A = \frac{4600}{10} 1+A=104600​1+A=460 1 + A = 460 1+A=460

STEP 9

Finally, subtract one from both sides to find AAA: A=460−1 A = 460 - 1 A=460−1 A=459 A = 459 A=459

STEP 10

Now we have all the pieces!Let's plug A=459A = \textbf{459}A=459 back into our equation: P(t)=46001+459e−1.45t P(t) = \frac{4600}{1 + 459 e^{-1.45t}} P(t)=1+459e−1.45t4600​

STEP 11

The logistic equation that models this data is P(t)=46001+459e−1.45tP(t) = \frac{4600}{1 + 459 e^{-1.45t}}P(t)=1+459e−1.45t4600​.

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1. Set up the logistic equation
2. Plug in the given values

We're given that the carrying capacity, $C$ , is 4600 people.
Let's plug that into our equation: $P(t) = \frac{4600}{1 + A e^{-kt}}$

We also know the initial growth rate, $k$ , is 1.45.
Let's plug that in too: $P(t) = \frac{4600}{1 + A e^{-1.45t}}$

Now, we're told that initially, 10 people have the virus.
That means when $t = 0$ , $P(t) = 10$ .
Let's substitute these values: $10 = \frac{4600}{1 + A e^{-1.45 \cdot 0}}$

Since anything raised to the power of zero is one, $e^{-1.45 \cdot 0} = e^0 = 1$ .
So, our equation simplifies to: $10 = \frac{4600}{1 + A}$

To solve for $A$ , we can multiply both sides by $(1 + A)$ and then divide both sides by 10: $10 \cdot (1 + A) = 4600$ $1 + A = \frac{4600}{10}$ $1 + A = 460$

Finally, subtract one from both sides to find $A$ : $A = 460 - 1$ $A = 459$

Now we have all the pieces!
Let's plug $A = \textbf{459}$ back into our equation: $P(t) = \frac{4600}{1 + 459 e^{-1.45t}}$

The logistic equation that models this data is $P(t) = \frac{4600}{1 + 459 e^{-1.45t}}$ .