# Math Help - Need Help with Logarithmic function problem???

1. ## Need Help with Logarithmic function problem???

Logarithmic Function Problem

The public health service monitors the spread of an epidemic of a particularly long-lasting strain of the flu in a city of 500,000 people using the logistic function. At the begining of the first week (time zero), 200 cases had been reported. During the first week 300 new cases were reported.

a. Determine the logistic function.
b. Estimate the number of individuals infected after 5 weeks.
c. When will the epidemic spread at the greatest rate?
d. At what rate will the epidemic spread when 40% of the population has been infected?
e. Graph the logistic function for the first 20 weeks of the epidemic's spread.

2. Originally Posted by CalculusChallenge
Logarithmic Function Problem

The public health service monitors the spread of an epidemic of a particularly long-lasting strain of the flu in a city of 500,000 people using the logistic function. At the begining of the first week (time zero), 200 cases had been reported. During the first week 300 new cases were reported.

a. Determine the logistic function.
b. Estimate the number of individuals infected after 5 weeks.
c. When will the epidemic spread at the greatest rate?
d. At what rate will the epidemic spread when 40% of the population has been infected?
e. Graph the logistic function for the first 20 weeks of the epidemic's spread.
What is the logistic function?

CB

3. Your first difficulty is thinking that it involves a logarithm when it doesn't! Your problem says the spread of the disease is modeled by a logistic function, not a logarithm. So I repeat Captain Black's question, "What is a logistic function"?

4. Originally Posted by CaptainBlack
What is the logistic function?

CB

Calculus Challenge:

Ahh. thank you. logistic function is y = a / [1 + be^ -kt]
But still How do I figure out a, b, and k????

5. Originally Posted by CalculusChallenge
Calculus Challenge:

Ahh. thank you. logistic function is y = a / [1 + be^ -kt]
But still How do I figure out a, b, and k????
you were given information about points on the curve ...

in a city of 500,000 people
... what constant in the function represents this value?

At the begining of the first week (time zero), 200 cases had been reported.
... (0,200) correct?

During the first week 300 new cases were reported.
... for t measured in weeks, this would be (1,300) , correct?

use this info to determine each constant (a, b, and k) in the function.

6. Originally Posted by skeeter
you were given information about points on the curve ...

... what constant in the function represents this value?

... (0,200) correct?

... for t measured in weeks, this would be (1,300) , correct?

use this info to determine each constant (a, b, and k) in the function.

CalculusChallenge

Thank you for the enlightenment:

So I used to two point to find the slope to be (300-200)/(1-0)=100. using y - mx+b to find 200= 100(0) +B, therefore b = 200.

So the constant a = 500000
constant b = 200
and constant k = 100
Is this correct???

7. Originally Posted by CalculusChallenge
CalculusChallenge

Thank you for the enlightenment:

So I used to two point to find the slope to be (300-200)/(1-0)=100. using y - mx+b to find 200= 100(0) +B, therefore b = 200.

So the constant a = 500000
constant b = 200
and constant k = 100
Is this correct???
sorry ... not even close.

$b$ is not a y-intercept ... you're dealing with a logistic growth curve, not a linear function.

8. Can you give me some formulas and hints on how to find constant a, b and k cause I haven't got a clue. thanks..

9. $y = \frac{a}{1 + be^{-kt}}$

$a$ is the limiting value; the maximum possible (also called the carrying capacity) value for $y$.

$y = \frac{500000}{1 + be^{-kt}}$

at $t = 0$ , $y = 200$

$200 = \frac{500000}{1 + be^{0}} = \frac{500000}{1 + b}$

solve for $b$ ...

$b = 2499$

$y = \frac{500000}{1 + 2499e^{-kt}}$

now use the point $(1, 300)$ and solve for $k$

10. Originally Posted by skeeter
$y = \frac{a}{1 + be^{-kt}}$

$a$ is the limiting value; the maximum possible (also called the carrying capacity) value for $y$.

$y = \frac{500000}{1 + be^{-kt}}$

at $t = 0$ , $y = 200$

$200 = \frac{500000}{1 + be^{0}} = \frac{500000}{1 + b}$

solve for $b$ ...

$b = 2499$

$y = \frac{500000}{1 + 2499e^{-kt}}$

now use the point $(1, 300)$ and solve for $k$

Oddly my prof. gave me the exact same problem. The only thing is, is that I don't know how to solve the equation in terms of k.

I got:
dy/dt=Ky(1+y/L) for the growth model y=L/1+be^-kt

So when solving for the point 1, 500 (I believe it's 500 and not 300 because they said 300 new cases, not 300 total), I got that k=500.5. That doesn't seem to make sense though because everything different number of weeks I put in I always get 500,000.

Any help is much appreciated!

11. Originally Posted by duriliim
Oddly my prof. gave me the exact same problem. The only thing is, is that I don't know how to solve the equation in terms of k.

I got:
dy/dt=Ky(1+y/L) for the growth model y=L/1+be^-kt

So when solving for the point 1, 500 (I believe it's 500 and not 300 because they said 300 new cases, not 300 total), I got that k=500.5. That doesn't seem to make sense though because everything different number of weeks I put in I always get 500,000.

Any help is much appreciated!
I did more thinking and I messed up how I got 500.5 big time. I did it again and I found that it does come to k=.9168. I tested it and it came out right. So for part b I got 18,850 infections.

The only trouble I'm having is parts c and d. How do I find the greatest rate and how do I find the rate when 40% of the population is infected?

Thanks!