Question

Let \(A(t)\) be the area of a tissue culture at time \(t\) and let \(M\) be the final area of the tissue when growth is complete. Most cell divisions occur on the periphery of the tissue and the number of cells on the periphery is proportional to \(\sqrt {A(t)} \). So, a reasonable model for the growth of tissue is obtained by assuming that the rate of growth of the area is jointly proportional to \(\sqrt {A(t)} \) and \(M - A(t)\).

(a) Formulate a differential equation and use it to show that the tissue grows fastest when \(A(t) = \frac{1}{3}M\).

(b) Solve the differential equation to find an expression for \(A(t)\). Use a computer algebra system to perform the integration.

Short answer

a)

Third derivative of\({\rm{A}}\)is positive at\(A = \frac{A}{3}\)which tells\(\frac{{dA}}{{dt}}\)is maximum.

b)

\(A = M \cdot {\tanh ^2}\left( {\frac{{\sqrt M (kt + C)}}{2}} \right)\) where\(C\) is the constant of integration.

Step-by-step solution

Concept used

Differential equation:

it is defined as the equation that contains derivatives of one or more dependent variables with respect to one or more independent variables.

The differential equation for some constant k

a)

The equation is

\(\frac{{dA}}{{dt}} = k\sqrt A (M - A)\)

Take derivative of\(\frac{{dA}}{{dt}}\) with respect to A.

\(\begin{aligned}{l}\frac{d}{{dA}}\left( {\frac{{dA}}{{dt}}} \right) = k\sqrt A ( - 1) + k(M - A)\left( {\frac{1}{{2\sqrt A }}} \right)\\ = k\left( {\frac{{ - 2A + M - A}}{{2\sqrt A }}} \right)\\ = k\left( {\frac{{M - 3A}}{{2\sqrt A }}} \right)\end{aligned}\)

Set the expression equal to 0 to find the max.

\(\frac{d}{{dA}}\left( {\frac{{dA}}{{dt}}} \right) = 0\)when\(A = \frac{1}{3}M\)

Therefore, that is the value of\(A\)that makes the largest growth rate.

Third derivative of \({\rm{A}}\) is positive at \(A = \frac{A}{3}\) which tells \(\frac{{dA}}{{dt}}\) is maximum.

The differential equation from part (a)

b)

In part (a) the differential equation is

\(\frac{{dA}}{{dt}} = k\sqrt A (M - A)\)

Rearrange the terms

\(\frac{{dA}}{{\sqrt A (M - A)}} = kdt\)

Integrate both sides

\(\int {\frac{{dA}}{{\sqrt A (M - A)}}} = \int k dt\)

\(\frac{2}{{\sqrt M }}{\tanh ^{ - 1}}\sqrt {\frac{A}{M}} = kt + C\)

Multiply both sides by\(\frac{{\sqrt M }}{2}\)

\({\tanh ^{ - 1}}\sqrt {\frac{A}{M}} = \frac{{\sqrt M (kt + C)}}{2}\)

Apply the hyperbolic tangent function

Apply the hyperbolic tangent function to both sides

\(\sqrt {\frac{A}{M}} = \tanh \left( {\frac{{\sqrt M (kt + C)}}{2}} \right)\)

Square both sides

\(\frac{A}{M} = {\tanh ^2}\left( {\frac{{\sqrt M (kt + C)}}{2}} \right)\)

Multiply both sides by\(M\)

\(A = M \cdot {\tanh ^2}\left( {\frac{{\sqrt M (kt + C)}}{2}} \right)\) where\(C\) is the constant of integration.