Problem 53 - Calculus: Early Transcendentals

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 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.

Step-by-Step Solution

Answer

Fastest growth occurs at \( A(t) = \frac{1}{3} M \).

Step 1: Define the Differential Equation

The problem states that the rate of growth of the area, \( \frac{dA}{dt} \), is jointly proportional to the square root of the area \( \sqrt{A(t)} \) and the difference between the final area \( M \) and the current area \( A(t) \). This gives us the differential equation \( \frac{dA}{dt} = k \sqrt{A(t)} (M - A(t)) \), where \( k \) is the constant of proportionality.

Step 2: Express and Differentiate the Function

To find when the tissue grows fastest, we express \( \frac{dA}{dt} = k \sqrt{A(t)} (M - A(t)) \) and differentiate it with respect to \( t \). For maximal growth rate, solve \( \frac{d^2A}{dt^2} = 0 \). This involves implicit differentiation of \( \frac{dA}{dt} \).

Step 3: Set the Second Derivative to Zero

From the differentiated form \( \frac{d^2A}{dt^2} = \frac{k}{2}(M - 3A(t)) \), setting \( \frac{d^2A}{dt^2} = 0 \) gives \( M - 3A(t) = 0 \), implying \( A(t) = \frac{1}{3} M \). This shows that the growth rate is maximal at \( A(t) = \frac{1}{3} M \).

Step 4: Solve the Differential Equation for A(t)

Separating variables in \( \frac{dA}{dt} = k \sqrt{A(t)} (M - A(t)) \), we have \( \int \frac{1}{\sqrt{A(t)} (M - A(t))} dA = \int k dt \). Perform the integration using a computer algebra system. The integration yields an implicit expression in terms of \( A(t) \) and \( t \).

Step 5: Simulate the Integration with a CAS

By performing the integration using a Computer Algebra System (CAS), we get the integrated expression. Typically, this involves evaluating the integral \( \int \frac{1}{\sqrt{A} (M - A)} dA \), which is non-trivial. The solution includes inverse trigonometric functions, rendering an implicit form for \( t \) and \( A(t) \).

Key Concepts

Proportionality Constant

In the context of differential equations, a proportionality constant plays a crucial role. It is often represented by the variable \( k \). This constant serves as a scaling factor that balances the equation by relating different quantities.

For example, in this exercise regarding tissue growth, the differential equation is given as \( \frac{dA}{dt} = k \sqrt{A(t)} (M - A(t)) \). Here, \( k \) makes the equation true by determining how fast or slow the tissue grows concerning both \( \sqrt{A(t)} \) and the difference \( M - A(t) \).

• The larger the proportionality constant \( k \), the faster the tissue growth rate will be under the same conditions.
• Introducing a variable like \( k \) allows to model different scenarios and make predictions about growth rates in varying environmental conditions.

Growth Rate

The growth rate in this exercise is described as dependent on both the current size of the area, represented as \( \sqrt{A(t)} \), and the difference between the maximum possible size \( M \) and \( A(t) \). The relation is much like how a balloon expands differently when it is near full size versus nearly empty.

To find when the tissue grows fastest, it's essential to determine when \( \frac{d^2A}{dt^2} \) equals zero.

• The equation \( M - 3A(t) = 0 \) arises, which leads us to discover that maximal growth occurs when \( A(t) = \frac{1}{3} M \).
• This concept mirrors real-world scenarios where resources or certain conditions are optimal for growth.

Understanding this helps model natural phenomena where growth rates change dynamically over time.

Implicit Differentiation

Implicit differentiation is a vital technique used when dealing with differential equations that are not straightforward. It helps in finding derivatives of equations where the dependent variable is not isolated.

In solving the exercise, you need to find the second derivative of the growth rate \( \frac{d^2A}{dt^2} \). This involves using implicit differentiation to handle the complexity of combining \( \sqrt{A(t)} \) and \( M - A(t) \).

• Implicit differentiation allows us to work with terms even when they are not neatly separated on either side of the equation.
• This technique is indispensable for setting \( \frac{d^2A}{dt^2} = 0 \), revealing critical points in growth where changes are most rapid or slow.

These methods are fundamental in exploring how functions interact when direct separation is not possible, especially in biological systems where interactions are often complex and interdependent.