Expert: Paul Klarreich - 3/6/2007

Question
Hey again Mr. Klarreich, I'm having some trouble (again :() trying to figure out how this problem models up and how to solve it. It is exponential growth in calculus, here is the problem:

A tissue culture grows until it has an area of 9 cm^2. Let A(t) be the area of the tissue at time t. A model of the growth rate is that:  A'(t) = k*(sqrt(A(t))*(9-A(t)) [as in, k times (the square root of A(t)) * (9-A(t))] for some constant k.

a. Without solving the equation, show that the maximum rate of growth occurs at any time when A(t) = 3 cm^2.

b. Assume that k = 6. Find the solution corresponding to A(0) = 1 and sketch its graph.

c. Do the same for A(0) = 4.

Thanks so very much again sir!

Answer
Questioner:   Aliza
Category:  Calculus
Private:  No
 
Subject:  Exponential growth and decay
Question:  Hey again Mr. Klarreich, I'm having some trouble (again :() trying to figure out how this problem models up and how to solve it. It is exponential growth in calculus, here is the problem:

A tissue culture grows until it has an area of 9 cm^2. Let A(t) be the area of the tissue at time t. A model of the growth rate is that:  A'(t) = k*(sqrt(A(t))*(9-A(t)) [as in, k times (the square root of A(t)) * (9-A(t))] for some constant k.

a. Without solving the equation, show that the maximum rate of growth occurs at any time when A(t) = 3 cm^2.

b. Assume that k = 6. Find the solution corresponding to A(0) = 1 and sketch its graph.

c. Do the same for A(0) = 4.

Thanks so very much again sir!
.......................................
Hi, Aliza,

Sorry, I don't have a complete solution for you.  I'm sending as much as I was able to work out and if I come with some more, I'll send it along.
.........................
If the growth is given by: [I'm just writing A instead of A(t) to save typing.]

A' = k sqrt(A)(9-A)  

Let R = sqrt(A)(9 - A)
To maximize, try the usual suspects, er.. methods:
                      1
R' = sqrt(A)(-1) + --------(9 - A)
                  2sqrt(A)  

                 9 - A
R' = - sqrt(A) + --------
                2sqrt(A)  

    - 2A + 9 - A
R' = -------------
       2sqrt(A)  

Set that (just the top, actually) equal to zero and you have your  A = 3
...................................
Now the differential equation.

dA/dt = k sqrt(A)(9-A)
Separate:
   dA
-------------- = kt
sqrt(A)(9 - A)
We have to integrate:

{      dA
| --------------
} sqrt(A)(9 - A)

So we have to resort to our usual methods of integration. In this case, I like a rationalizing substitution:  

Let  y = sqrt(A)
A = y^2
dA = 2y dy

{     2y dy
| ------------
}  y (9 - y^2)

{   2 dy
| --------
}  9 - y^2

There is a standard form (look in the table of integrals in the back of your book):

{    du        1       u + a
| --------- = --- ln | ----- |
} a^2 - u^2   2a       u - a

Which we can use, with  a = 3 and u = y


  1       y + 3
= --- ln | ------ |
  3       y - 3

  1       sqrt(A) + 3
= --- ln | ----------- |
  3       sqrt(A) - 3

I'm not sure how to proceed from here.  If I can come up with something clever I'll let you know.