Related rates yeast cells problem

[crybllrd]

Homework Statement

The number of yeast cells in a laboratory culture increases rapidly at first but levels off eventually. The population is modeled by the function below

n=f(t)=\frac{a}{1+be^{-t}}

where t is measured in hours. at time t=0 the population is 20 cells and is increasing at a rate of 10 cells/hour.
(a) Find a and b.
(b) According to the model, what happens to the population in the long run?

Homework Equations

\frac{dr}{dt}=10,\ t=0,\ n=20.

The Attempt at a Solution

(a) For a,

20=\frac{a}{1+be^{}-(0)}

a= 20+20b

I am not sure how to find b. I am assuming I will need to take the derivative of something, but I am not sure what or why.

(b) I think that after I find b, i will see a horizontal asymptote that the function will approach.

 

[Mark44]

Homework Statement

The number of yeast cells in a laboratory culture increases rapidly at first but levels off eventually. The population is modeled by the function below

n=f(t)=\frac{a}{1+be^{-t}}

where t is measured in hours. at time t=0 the population is 20 cells and is increasing at a rate of 10 cells/hour.
(a) Find a and b.
(b) According to the model, what happens to the population in the long run?

Homework Equations

\frac{dr}{dt}=10,\ t=0,\ n=20.

The Attempt at a Solution

(a) For a,

20=\frac{a}{1+be^{-0}}

a= 20+20b

I am not sure how to find b. I am assuming I will need to take the derivative of something, but I am not sure what or why.

No, just use the fact that the population is increasing at a rate of 10 cells/hour. You know that n(0) = 20, so what is n(1)?

(b) I think that after I find b, i will see a horizontal asymptote that the function will approach.

 

[crybllrd]

OK, so I have:

n(1)= 30

30=\frac{20+20b}{1+be^{-1}}

30+30be^{-1}=20+20b

30be^{-1}-20b=-10

b(30e^{-1}-20)=-10

b=\frac{-10}{30e^{-1}-20}

Does that look right?

 

[Mark44]

I don't have my work in front of me, but that looks like what I came up with. You can also write b this way:
b=\frac{10}{20 - 30e^{-1}}

 

[HallsofIvy]

Oh, dear, I have to strongly disagree with Mark44 (and that scares me!)

For a continuous function, saying that it is "is increasing at a rate of 10 cells/hour" does NOT mean that rate continues for the entire hour and does NOT mean that the number of cells at t= 1 is n(0)+ 10.

crybllrd, you were right in your first post when you said that the conditions were
\frac{dn}{dt}= 10
n= 20 at t= 0.

n(0)= \frac{a}{1+ b}= 20
so, as you said, a= 20(1+ b).

Also
\frac{dn}{dt}(0)= \frac{ab}{(1+ b)^2}

From a= 20(1+ b) that becomes
\frac{20b(1+ b)}{(1+ b)^2}= \frac{20b}{1+ b}= 10

Solve that for b and then use a= 20(1+ b) to find a.

(But, of course, the "number of cells" is NOT continuous! My solution gives a much simpler formula but gives a number between 29 and 30 for the number of cells when t= 1! I rather suspect that Mark44 is right but that my solution is the one that was intended by whoever posed this problem!)

 

[epenguin]

I also get 29.24 for number of cells after 1h, increase of not quite 10, think HoI's is intended approach. If they had meant 10 in an hour they would have said so clearly I think.

Not actually very different.

With other parameters the difference between the two could increase a lot with time before they again reconverge and get to the same place in the end. With these parameters the difference is quite small.

For check, I get a = 40, b = 1.

I think you are supposed to by calculus derive the expression for dn/dt which I think no one has yet stated explicitly, and use that.

 

[Mark44]

Oh, dear, I have to strongly disagree with Mark44 (and that scares me!)
For a continuous function, saying that it is "is increasing at a rate of 10 cells/hour" does NOT mean that rate continues for the entire hour and does NOT mean that the number of cells at t= 1 is n(0)+ 10.

Mea culpa. My approach was an oversimplification. HallsOfIvy's approach is the more reasonable one.

 

[crybllrd]

That's great, thanks a lot everyone!