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))
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.
y(t) = b + e^kt ?
The Attempt at a Solution
For part A, I took the derivative and set that equal to 0, which gave me back 3.
For part B, i took it as a separable differential equation and got
(1/3)(ln((9-A)/(3-sqrt(A))^2)) = kt + C
but i have no idea how to solve that for A
Thanks for any help