Math  /  Algebra

QuestionThe logistic growth function f(t)=116,0001+4700etf(t)=\frac{116,000}{1+4700 e^{-t}} describes the number of people, f(t)f(t), who have become ill with influenza tt weeks after its initial outbreak in a particular community. a. How many people became ill with the flu when the epidemic began? b. How many people were ill by the end of the fourth week? c. What is the limiting size of the population that becomes ill? a. The number of people initially infected is \square (Round to the nearest whole number as needed.) b. The number of people infected after 4 weeks is \square . (Round to the nearest whole number as needed.) c. The limiting size of the infected population is \square . (Round to the nearest whole number as needed.)

Studdy Solution

STEP 1

What is this asking? We're looking at how many people get sick with the flu over time, starting from the very beginning and way into the future! Watch out! Don't mix up the different times – initial, after four weeks, and the limit as time goes on forever.

STEP 2

1. Initial Sickness
2. Sickness After Four Weeks
3. Limiting Size of Sickness

STEP 3

To find out how many people were sick right at the beginning, we need to look at the function when t=0t = 0, since tt represents the number of weeks after the outbreak began.
Let's **plug in** t=0t = 0 into our function:

STEP 4

f(0)=116,0001+4700e0 f(0) = \frac{116,000}{1 + 4700 \cdot e^{-0}}

STEP 5

Remember, anything raised to the power of zero is **one**, so e0=1e^{-0} = 1.
This simplifies our equation:

STEP 6

f(0)=116,0001+47001=116,0004701 f(0) = \frac{116,000}{1 + 4700 \cdot 1} = \frac{116,000}{4701}

STEP 7

Now, we just **divide** to get our **initial value**:

STEP 8

f(0)24.6756 f(0) \approx 24.6756

STEP 9

Since we can't have parts of people, we'll **round** to the nearest whole number: approximately **25** people were sick at the start.

STEP 10

Now, we want to see how many people are sick after four weeks.
This time, we'll **plug in** t=4t = 4:

STEP 11

f(4)=116,0001+4700e4 f(4) = \frac{116,000}{1 + 4700 \cdot e^{-4}}

STEP 12

Using a calculator, we find that e40.0183e^{-4} \approx 0.0183.
Let's **substitute** that in:

STEP 13

f(4)=116,0001+47000.0183116,0001+85.97116,00086.97 f(4) = \frac{116,000}{1 + 4700 \cdot 0.0183} \approx \frac{116,000}{1 + 85.97} \approx \frac{116,000}{86.97}

STEP 14

**Dividing** gives us:

STEP 15

f(4)1333.78 f(4) \approx 1333.78

STEP 16

Again, we **round** to the nearest whole number, so about **1334** people were sick after four weeks.

STEP 17

The limiting size is what happens when time goes on forever, or as tt approaches infinity.
Think about what happens to ete^{-t} as tt gets really big.
It gets super small, approaching zero!

STEP 18

So, we can think of ete^{-t} becoming essentially zero as tt approaches infinity.
Let's see what happens to our function:

STEP 19

limtf(t)=limt116,0001+4700et \lim_{t \to \infty} f(t) = \lim_{t \to \infty} \frac{116,000}{1 + 4700 \cdot e^{-t}}

STEP 20

Since ete^{-t} approaches zero, the entire term 4700et4700 \cdot e^{-t} also approaches zero:

STEP 21

limtf(t)=116,0001+0=116,000 \lim_{t \to \infty} f(t) = \frac{116,000}{1 + 0} = 116,000

STEP 22

So, the limiting size of the infected population is **116,000** people.

STEP 23

a. The number of people initially infected is **25**. b. The number of people infected after 4 weeks is **1334**. c. The limiting size of the infected population is **116,000**.

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