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The growth of cancer tumors may be modeled by the Gompertz growth equation. Let \(M(t)\) be the mass of a tumor, for \(t \geq 0 .\) The relevant initial value problem is \(\frac{d M}{d t}=-r M(t) \ln \left(\frac{M(t)}{K}\right), \quad M(0)=M_{0}\), where \(r\) and \(K\) are positive constants and \(0

Short Answer

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Question: Estimate the limiting size of the tumor as \(t\) approaches infinity, given the Gompertz growth equation \(M(t) = K\left(\frac{M_0}{K}\right)^{\exp(-rt)}\) with \(M_0 = 100\), \(r = 0.05\), and an example value \(K = 1000\). Answer: As \(t\) approaches infinity, the limiting size of the tumor is approximately equal to the given value of \(K\), which for this example is \(1000\).

Step by step solution

01

Write down the given equation and solution.

First, let's write down the given differential equation and the solution we need to verify. Initial value problem: \(\frac{dM}{dt} = -r M(t) \ln\left(\frac{M(t)}{K}\right)\), with \(M(0)=M_0\). Solution to verify: \(M(t) = K\left(\frac{M_0}{K}\right)^{\exp(-rt)}\).
02

Calculate the derivative of the given solution.

We need to find \(\frac{dM}{dt}\) for the solution \(M(t) = K\left(\frac{M_0}{K}\right)^{\exp(-rt)}\). Using the chain rule, we get: \(\frac{dM}{dt} = \frac{d}{dt}\left[K\left(\frac{M_0}{K}\right)^{\exp(-rt)}\right] = -K\left(\frac{M_0}{K}\right)^{\exp(-rt)}\cdot \exp(-rt) \cdot r\)
03

Substitute the derivative into the equation.

Now, we'll substitute the derivative we found in Step 2 into the original equation: \(-rM(t)\ln\left(\frac{M(t)}{K}\right)=-r\left[K\left(\frac{M_0}{K}\right)^{\exp(-rt)}\right]\ln\left(\frac{K\left(\frac{M_0}{K}\right)^{\exp(-rt)}}{K}\right)\) This simplifies to: \(-rM(t)\ln\left(\frac{M_0}{K}\right)^{\exp(-rt)} = -rM(t)\ln\left(\frac{M_0}{K}\right)^{\exp(-rt)}\) Since the left side of the equation is equal to the right side after substituting the derivative and the function, the given solution is correct. #b. Graph the solution#
04

Set given values for constants.

We are given the values for \(M_0\) and \(r\) as: \(M_0 = 100\) \(r = 0.05\)
05

Plot the function.

With given values, the function can be written as: \(M(t) = K \left(\frac{100}{K}\right)^{\exp(-0.05t)}\) Let's assume a value for \(K\), for example \(K = 1000\) (with \(M_0 < K\) as given). Now, plot this function for \(t\) ranging from \(0\) to a suitable value, say \(t = 12\). (The student will have to use a graphing calculator or an online graphing tool to create the graph for this function.) #c. Estimate the limiting size#
06

Analyze the graph.

From the graph created in part (b), we can observe the behavior of the function as \(t\) approaches infinity.
07

Estimate the limiting size of the tumor.

By analyzing the graph, we can see that the function approaches a constant limit as \(t\) increases. Thus, we can estimate the value of \(\lim_{t\rightarrow \infty}M(t)\), which represents the limiting size of the tumor. Since we used \(K = 1000\) as an example, the limiting size of the tumor would be close to \(K\) when \(t \rightarrow \infty\). Different values of \(K\) could yield different results, but the method to estimate the limiting size remains the same.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gompertz Growth Model
The Gompertz Growth Model is a mathematical model used to describe the growth of biological organisms, such as cells and tumors over time. It is particularly useful for modeling situations where growth slows down as it approaches a limiting size. This characteristic is often observed in biological systems, where resources become limited as growth nears a maximum possible size.

The model is expressed with the differential equation: \( \frac{dM}{dt} = -r M(t) \ln \left( \frac{M(t)}{K} \right) \). Here, \( M(t) \) represents the size or mass of the tumor at time \( t \); \( r \) is a growth rate constant; and \( K \) is the carrying capacity or the limiting size that the tumor tends to approach.

This logarithmic growth model shows that as \( M(t) \) approaches \( K \), the growth rate slows significantly, providing a realistic prediction for various biological growth processes.
Initial Value Problem
An Initial Value Problem (IVP) is a type of differential equation accompanied by specific conditions that specify the state of the system at a starting point. This is crucial in solving differential equations as it helps to find a unique solution that fits the given conditions.

For the Gompertz Growth Model, the initial value problem is defined as \( \frac{dM}{dt} = -r M(t) \ln \left( \frac{M(t)}{K} \right) \), with the condition \( M(0) = M_0 \). This means that at time \( t = 0 \), the tumor mass is \( M_0 \).

The solution to an IVP describes how the system evolves over time, starting from this initial condition. Solving the given IVP for the Gompertz Growth Model, we find the function \( M(t) = K \left( \frac{M_0}{K} \right)^{\exp(-rt)} \), which is valid for all \( t \geq 0 \).
Tumor Growth Modeling
Tumor Growth Modeling using mathematical equations like the Gompertz model helps researchers and medical professionals understand and predict how a tumor might grow over time. This modeling provides insight into the potential progression of the disease, enabling better strategic planning for treatments and interventions.

In the context of the Gompertz Growth Model, it involves parameters such as:
  • \( r \), the growth rate constant: Determines how quickly the tumor grows initially.
  • \( K \), the carrying capacity: The tumor size at which growth virtually stops.
The model's logarithmic nature means that tumors will grow rapidly at first and then slow down as they approach the limiting size \( K \). This characteristic suits real-world scenarios where resources and space limit further growth.
Limit of a Function
The concept of the limit of a function is fundamental in calculus, indicating the value that a function \( f(x) \) approaches as \( x \) tends toward a certain point, often infinity. In the study of tumor growth, finding the limit as \( t \) approaches infinity gives us an idea of the maximum size the tumor will reach.

In the exercise, estimating the limit \( \lim_{t \to \infty} M(t) \) allows us to understand the ultimate growth ceasing at size \( K \). This is because as time increases, the exponential decay term \( \exp(-rt) \) approaches zero, making \( M(t) \) approach \( K \).

In essence, the limit highlights the constraints imposed by environmental factors and resource availability, providing a mathematical framework to predict stable long-term behavior in growth processes.

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Most popular questions from this chapter

A fish hatchery has 500 fish at \(t=0\), when harvesting begins at a rate of \(b>0\) fish/year. The fish population is modeled by the initial value problem \(y^{\prime}(t)=0.01 y-b, y(0)=500,\) where \(t\) is measured in years. a. Find the fish population, for \(t \geq 0\), in terms of the harvesting rate \(b\) b. Graph the solution in the case that \(b=40\) fish/year. Describe the solution. c. Graph the solution in the case that \(b=60\) fish/year. Describe the solution.

Solve the following initial value problems. When possible, give the solution as an explicit function of \(t\) $$y^{\prime}(t)=\frac{y+3}{5 t+6}, y(2)=0$$

The following models were discussed in Section 1 and reappear in later sections of this chapter. In each case carry out the indicated analysis using direction fields. The delivery of a drug (such as an antibiotic) through an intravenous line may be modeled by the differential equation \(m^{\prime}(t)+k m(t)=I,\) where \(m(t)\) is the mass of the drug in the blood at time \(t \geq 0, k\) is a constant that describes the rate at which the drug is absorbed, and \(I\) is the infusion rate. Let \(I=10 \mathrm{mg} / \mathrm{hr}\) and \(k=0.05 \mathrm{hr}^{-1}\). a. Draw the direction field, for \(0 \leq t \leq 100,0 \leq y \leq 600\). b. What is the equilibrium solution? c. For what initial values \(m(0)=A\) are solutions increasing? Decreasing?

A special class of first-order linear equations have the form \(a(t) y^{\prime}(t)+a^{\prime}(t) y(t)=f(t),\) where \(a\) and \(f\) are given functions of \(t.\) Notice that the left side of this equation can be written as the derivative of a product, so the equation has the form $$a(t) y^{\prime}(t)+a^{\prime}(t) y(t)=\frac{d}{d t}(a(t) y(t))=f(t)$$ Therefore, the equation can be solved by integrating both sides with respect to \(t.\) Use this idea to solve the following initial value problems. $$\left(t^{2}+1\right) y^{\prime}(t)+2 t y=3 t^{2}, y(2)=8$$

Two curves are orthogonal to each other if their tangent lines are perpendicular at each point of intersection. A family of curves forms orthogonal trajectories with another family of curves if each curve in one family is orthogonal to each curve in the other family. Use the following steps to find the orthogonal trajectories of the family of ellipses \(2 x^{2}+y^{2}=a^{2}\) a. Apply implicit differentiation to \(2 x^{2}+y^{2}=a^{2}\) to show that $$ \frac{d y}{d x}=\frac{-2 x}{y} $$ b. The family of trajectories orthogonal to \(2 x^{2}+y^{2}=a^{2}\) satisfies the differential equation \(\frac{d y}{d x}=\frac{y}{2 x} .\) Why? c. Solve the differential equation in part (b) to verify that \(y^{2}=e^{C}|x|\) and then explain why it follows that \(y^{2}=k x\) Therefore, the family of parabolas \(y^{2}=k x\) forms the orthogonal trajectories of the family of ellipses \(2 x^{2}+y^{2}=a^{2}\)

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