Related rates yeast cells problem

In summary, the conversation discusses a model for the population of yeast cells in a laboratory culture, which increases rapidly at first but levels off eventually. The model is represented by the function n=f(t)=\frac{a}{1+be^{-t}}, where t is measured in hours. The population at time t=0 is 20 cells and is increasing at a rate of 10 cells/hour. The conversation then goes on to find the values of a and b, and discusses what happens to the population in the long run. The correct solution involves taking the derivative of the function and solving for a and b using the given conditions.
  • #1
crybllrd
120
0

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

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

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



[tex]\frac{dr}{dt}=10,\ t=0,\ n=20.[/tex]

The Attempt at a Solution



(a) For a,

[tex]20=\frac{a}{1+be^{}-(0)}[/tex]

[tex]a= 20+20b[/tex]

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.
 
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  • #2
crybllrd said:

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

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

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



[tex]\frac{dr}{dt}=10,\ t=0,\ n=20.[/tex]

The Attempt at a Solution



(a) For a,

[tex]20=\frac{a}{1+be^{-0}}[/tex]

[tex]a= 20+20b[/tex]

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)?
crybllrd said:
(b) I think that after I find b, i will see a horizontal asymptote that the function will approach.
 
  • #3
OK, so I have:

[tex]n(1)= 30[/tex]

[tex]30=\frac{20+20b}{1+be^{-1}}[/tex]

[tex]30+30be^{-1}=20+20b[/tex]

[tex]30be^{-1}-20b=-10[/tex]

[tex]b(30e^{-1}-20)=-10[/tex]

[tex]b=\frac{-10}{30e^{-1}-20}[/tex]

Does that look right?
 
  • #4
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:
[tex]b=\frac{10}{20 - 30e^{-1}}[/tex]
 
  • #5
Oh, dear, I have to strongly disagree with Mark44 (and that scares me!:eek:)

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
[tex]\frac{dn}{dt}= 10[/tex]
n= 20 at t= 0.

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

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

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

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!)
 
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  • #6
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.
 
Last edited:
  • #7
HallsofIvy said:
Oh, dear, I have to strongly disagree with Mark44 (and that scares me!:eek:)
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.
 
  • #8
That's great, thanks a lot everyone!
 

What is a related rates yeast cells problem?

A related rates yeast cells problem is a type of mathematical problem that involves determining the rate of change of one variable with respect to another variable, given their relationship in a specific situation involving yeast cells. This type of problem is commonly encountered in biology and chemistry research.

What are some common variables involved in related rates yeast cells problems?

Common variables involved in related rates yeast cells problems may include the rate of growth of yeast cells, the amount of nutrients present, and the rate of diffusion of nutrients into the cells. Other variables may also be included depending on the specific problem.

How are related rates yeast cells problems solved?

Related rates yeast cells problems are typically solved using calculus, specifically the derivative and chain rule. The specific steps for solving a problem may vary depending on the given information and the problem's context.

What are some real-life applications of related rates yeast cells problems?

Related rates yeast cells problems have many real-life applications in fields such as food and beverage production, pharmaceutical research, and biofuels development. By understanding the relationship between different variables in yeast cells, scientists can better optimize production processes and develop new products.

What are some tips for solving related rates yeast cells problems?

Some tips for solving related rates yeast cells problems include carefully identifying and labeling all given information, drawing a diagram to visualize the problem, and using the chain rule to relate different variables. It is also important to understand the underlying biology and chemistry concepts to accurately interpret the problem and its solution.

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