# Calc 2 Exponential Growth(differential equation)

## Homework Statement

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.

## Homework Equations

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

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gabbagabbahey
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t
(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
Well,

$$\ln\left(\frac{9-A}{(3-\sqrt{A})^2}\right)=3kt+3C\implies \frac{9-A}{(3-\sqrt{A})^2}=Be^{3kt}$$

where $B\equiv e^{3C}$

From there, multiply both sides by $(3-\sqrt{A})^2$, expand everything out and collect terms in powers of $A$...you will be left with an equation that is quadratic in terms of $\sqrt{A}$, and I'm sure you know how to solve quadratic equations.