Problem 69 - Calculus for Biology and Medicine

Question

The Gompertz equation is used to model the growth of a tumor. We will study it in Chapter \(8 .\) In this model the number of cells \(N(t)\) in a tumorgrows over time at a rate that depends on \(N\), that is, tumors of different sizes grow at different rates, producing a differential equation: $$ \frac{d N}{d t}=a N \ln (b / N) $$ where a and b are positive constants that depend on the type of tumor, whether the tumor is being treated, and on the kind of treatment. In Chapter 8 we will see that the solution to this equation is given by evaluating the integral $$ t=\int \frac{d N}{a N \ln (b / N)} $$ Assume \(a=b=1\); then evaluate the integral \(t=\int \frac{d N}{N \ln (1 / N)}\). Your answer will contain an unknown constant of integration.

Step-by-Step Solution

Answer

The evaluated integral is \( t = -\ln|\ln(N)| + C \).

Step 1: Simplify the Integral

Given that \( a = b = 1 \), the integral simplifies to:\[t = \int \frac{dN}{N \ln(1/N)}\]Since \( \ln(1/N) = -\ln(N) \), substitute to get:\[t = \int \frac{-dN}{N \ln(N)}\]This is equivalent to:\[t = -\int \frac{dN}{N \ln(N)}\]

Step 2: Use a Substitution Method

Let \( u = \ln(N) \), therefore \( du = \frac{1}{N} dN \). The integral becomes:\[t = -\int \frac{1}{u} du\]

Step 3: Solve the Integral

The integral \(-\int \frac{1}{u} du\) can be solved as follows:\[\int \frac{1}{u} du = \ln|u| + C\]Therefore, the solution is:\[t = -\ln|u| + C\]Substituting back \( u = \ln(N) \), we get:\[t = -\ln(\ln(N)) + C\]

Step 4: Finalize the Solution

From substituting back, the original integral evaluates to:\[t = -\ln|\ln(N)| + C\]This includes the constant of integration \( C \).

Key Concepts

Tumor Growth Modeling

The Gompertz equation serves as a powerful tool for modeling tumor growth. Unlike simpler models that assume constant growth rates, the Gompertz equation considers that the growth rate of a tumor changes over time. This aligns better with biological observations, where small tumors might grow relatively quickly compared to larger tumors whose growth decelerates.
In the Gompertz model, the rate of tumor growth is a function of the number of cells present, modeled by the equation \[\frac{d N}{d t}=a N \ln (b / N)\] where:

• \(N(t)\) is the number of tumor cells at time \(t\).
• \(a\) and \(b\) are constants influencing growth characteristics, often impacted by treatment methods.
• The logarithmic term \(\ln (b / N)\) reflects the decreasing growth rate as the tumor expands.

The complexity in the equation allows us to capture how tumors respond to variables like treatment, making it invaluable for predictive modeling in oncology.

Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives. In the context of tumor growth modeling, they help describe the dynamics of how the tumor grows over time.
The differential equation given is \[\frac{d N}{d t}=a N \ln (b / N)\] This is a first-order equation because it involves the first derivative of \(N\). Such equations are vital in modeling the complex interplay of factors affecting tumor growth because they provide a deep understanding of how each component influences rate changes.
Solving a differential equation often involves finding an expression for \(N\) as a function of \(t\). The solution offers insights into how the cell population evolves, and the mathematical methods used in solving them, such as substitution, can simplify this process, rendering it accessible and interpretable.

Integrals in Calculus

Integrals are essential in solving differential equations, and they help in determining the growth function from its rate of change. The original equation in the exercise requires evaluating an integral to solve for time \(t\). The integral involved is:\[t=\int \frac{d N}{a N \ln (b / N)} \]Given that \(a\) and \(b\) are 1, the integral simplifies to: \[t = \int \frac{d N}{N \ln (1 / N)}\]This integral is solved using substitution, converting a complex integral into a manageable form. In this problem, substitution helps by setting \(u = \ln(N)\), turning the integral into a form where the antiderivative is known, \[-\int \frac{1}{u} du\] The solution involves natural logarithms, resulting in \[t = -\ln(\ln(N)) + C\]where \(C\) is the integration constant. Integrals are about finding areas under curves, but in differential equations, they help solve for original functions based on given rates. This approach transforms the abstract concept of change into concrete solutions.