Question

The Gompertz growth equation is often used to model the growth of tumors. Let \(M(t)\) be the mass of a tumor at time \(t \geq 0 .\) The relevant initial value problem is $$ \frac{d M}{d t}=-r M \ln \left(\frac{M}{K}\right), M(0)=M_{0} $$ a. Graph the growth rate function \(R(M)=-r M \ln \left(\frac{M}{K}\right)\) (which equals \(M^{\prime}(t)\) ) assuming \(r=1\) and \(K=4 .\) For what values of \(M\) is the growth rate positive? For what value of \(M\) is the growth rate a maximum? b. Solve the initial value problem and graph the solution for \(r=1, K=4,\) and \(M_{0}=1 .\) Describe the growth pattern of the tumor. Is the growth unbounded? If not, what is the limiting size of the tumor? c. In the general solution, what is the meaning of \(K ?\) where \(r\) and \(K\) are positive constants and \(0 < M_{0} < K\)

Short answer

In summary: a. The growth rate is positive for values \(0 < M < 4\). The maximum growth rate occurs when \(M=4e\). b. The growth pattern of the tumor is initially rapid but slows down as M approaches K. The growth is not unbounded, and the limiting size of the tumor is K. c. K represents the limiting size of the tumor, also called the carrying capacity, and it indicates the maximum mass that the tumor can reach given the specific environment and growth parameters.

Step-by-step solution

Part a: Graph the growth rate function and find the positive growth rate and maximum value.

To graph the growth rate function \(R(M) = -rM\ln(\frac{M}{K})\), with \(r=1\) and \(K=4\), we can plug in the values of r and K: $$ R(M) = -M\ln(\frac{M}{4}). $$ We will use software such as Desmos or Geogebra to graph this function. From the graph, we can see that the growth rate is positive for values \(0 < M < 4\). To find the maximum value of the growth rate, we need to find the critical points: $$ \frac{d}{dM}R(M) = \frac{d}{dM}(-M\ln(\frac{M}{4})) = 0. $$ We can use the product rule and the chain rule for this differentiation: $$ \frac{d}{dM}(-M\ln(\frac{M}{4})) = -\ln(\frac{M}{4}) - 1. $$ Now, we set the derivative equal to 0 and solve for M: $$ -\ln(\frac{M}{4}) - 1 = 0 $$ Adding 1 and exponentiating, $$ \frac{M}{4} = e^1 $$ So, \(M=4e\). The maximum growth rate occurs when \(M=4e\).

Part b: Solving the initial value problem and graphing the solution.

To solve the initial value problem, let's rewrite the Gompertz growth equation: $$ \frac{dM}{dt}=-M\ln\left(\frac{M}{K}\right), M(0)=M_{0} $$ To solve this differential equation, we will use separation of variables. Divide by M and multiply by dt: $$ \frac{dM}{M\ln(\frac{M}{K})} = -dt $$ Now, integrate both sides: $$ \int \frac{dM}{M\ln(\frac{M}{K})} = -\int dt. $$ The right-hand side integration is straightforward, yielding \(-t + C\) (C is the constant of integration). For the left-hand side, the integration requires a substitution, let \(u = \ln(\frac{M}{K})\). Now, \(du = \frac{dM}{M}\), and the integral becomes: $$ \int \frac{du}{u} = -t + C. $$ Integrating, we obtain $$ \ln |u| = -t + C. $$ Substituting back \(u = \ln(\frac{M}{K})\), $$ \ln(\ln(\frac{M}{K})) = -t + C. $$ To solve for M, we can exponentiate both sides, and then multiply both sides by K: $$ M(t) = K\left(e^{\ln(\frac{M_{0}}{K})e^{-t}}\right) $$ Now, we can substitute the given values \(r=1, K=4\), and \(M_0=1\): $$ M(t) = 4\left(e^{\ln(\frac{1}{4})e^{-t}}\right) $$ We will use a graphing tool such as Desmos or Geogebra to plot this function and observe the growth pattern of the tumor. From the graph, we can see that the growth of the tumor is initially rapid but slows down as M approaches K (which is 4 in this case). As time goes to infinity, the tumor size converges to K, indicating that the growth is not unbounded. The limiting size of the tumor is K.

Part c: Meaning of K in the general solution.

In the general solution, K represents the limiting size of the tumor (also called carrying capacity). It indicates the maximum mass that the tumor can reach given the specific environment and growth parameters. In biological terms, K represents the limit or saturation point in which the tumor growth may be constrained due to limited resources or carrying capacity in the host.

Key concepts

Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives. They are essential in modeling various natural phenomena where rates of change are involved. In the context of the Gompertz growth model, the differential equation is used to describe how a tumor grows over time.
The specific formula used here is \[\frac{dM}{dt} = -rM\ln\left(\frac{M}{K}\right)\]where \(M(t)\) represents the tumor mass at time \(t\), \(r\) is the growth rate constant, and \(K\) is the carrying capacity. These equations help predict future states based on current conditions. By solving these equations, we can understand the dynamics of tumors and make predictions about their growth over time. Integrating differential equations involves techniques like separation of variables and substitution, making them crucial for finding analytical solutions in growth modeling.

Tumor Growth Modeling

Tumor growth modeling is a method used to understand and predict how a tumor increases in size over time. Using mathematical frameworks like the Gompertz model helps capture the essential dynamics of tumor progression.
In practice, the Gompertz model captures the initial exponential growth phase that tumors typically exhibit, followed by a plateau as growth slows down. This model is realistic because it accounts for biological constraints, such as nutrient supply and space, which limit unlimited growth. The model provides a way to represent complex biological processes with relatively simple mathematical expressions.
The equation for tumor growth used here adapts to the changing growth rate, showcasing how tumors rapidly grow initially and then stabilize as they encounter biological limitations in their environment. This kind of modeling is critical for designing effective treatments and understanding the growth limits of tumors.

Carrying Capacity

Carrying capacity, in the context of the Gompertz growth model, refers to the maximum size a tumor can attain in its environment. It is denoted by \(K\) in the growth equation. The concept originates from ecology, where it's used to define the maximum population size that an environment can sustain indefinitely.
In tumor growth terms, carrying capacity suggests the tumor's ability to grow is finite due to limited resources, such as nutrients and oxygen. When a tumor reaches its carrying capacity, its growth rate declines, eventually flattening out. Understanding the carrying capacity helps anticipate when a tumor will stop growing and aids in planning treatment strategies.
The parameter \(K\) shapes the model's predictions by setting a limit to growth, offering a realistic view of a tumor's potential size under given biological conditions.

Maximum Growth Rate

The maximum growth rate describes the highest speed at which a tumor can grow, occurring at a certain tumor size. For the Gompertz model, this happens before the size reaches the carrying capacity. Finding this maximum involves deriving the growth rate function, identified here as \(-rM\ln\left(\frac{M}{K}\right)\).
To determine this point, we evaluate where the derivative of the growth rate function equals zero, revealing critical points. This calculation helps identify optimal conditions for growth, typically, when the tumor is at an intermediate size.
Understanding the maximum growth rate is important for clinical purposes, as it gives insights into the phase when the tumor is most aggressive. Clinicians can use this information to target therapies during peak growth periods, improving treatment outcomes by slowing down the growth rate when it is at its fastest.