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Properties of the Gompertz solution Verify that the function 1\(M(t)=K\left(\frac{M_{0}}{K}\right)^{\operatorname{csp}(-n)}\) satisfies the properties \(M(0)=M_{0}\) and \(\lim _{t \rightarrow \infty} M(t)=K\)

Short Answer

Expert verified
Question: Verify the given properties of the function \(M(t) = K(\frac{M_0}{K})^{\operatorname{csp}(-nt)}\), specifically that \(M(0) = M_0\) and \(\lim_{t \rightarrow \infty} M(t) = K\). Solution: We have successfully verified the two required properties of the given function. By substituting \(t = 0\) into the function, we found that \(M(0) = M_0\). Furthermore, by analyzing the behavior of the function as \(t\) tends to infinity, we determined that \(\lim_{t \rightarrow \infty} M(t) = K\). Therefore, both properties are satisfied.

Step by step solution

01

Substitute t = 0 into the function

First, let's write down the function we are given: $$M(t) = K\left(\frac{M_0}{K}\right)^{\operatorname{csp}(-nt)}$$ Now, we need to find the value of \(M(0)\). To do this, we substitute \(t = 0\) and simplify the expression: $$M(0) = K\left(\frac{M_0}{K}\right)^{\operatorname{csp}(-n\cdot 0)}$$
02

Simplify the expression for M(0)

Let's simplify the expression we obtained in the previous step: $$M(0) = K\left(\frac{M_0}{K}\right)^{\operatorname{csp}(0)}$$ Knowing that \(\operatorname{csp}(0)=1\), the expression becomes: $$M(0) = K\left(\frac{M_0}{K}\right)^1$$ Further simplification results in: $$M(0) = K\frac{M_0}{K}$$ Which cancels out the \(K\) values, yielding: $$M(0) = M_0$$ This verifies the first property, \(M(0) = M_0\).
03

Find the limit of M(t) as t approaches infinity

Next, let's find the limit as t approaches infinity: $$\lim_{t \rightarrow \infty} M(t) = \lim_{t \rightarrow \infty} K\left(\frac{M_0}{K}\right)^{\operatorname{csp}(-nt)}$$ To determine the result, we have to examine the behavior of the function when \(t\) tends to infinity.
04

Analyze the behavior of the function as t tends to infinity

Since \(n\) is not specified, we can consider it as a positive constant. Therefore, as \(t\) tends to infinity, \(-nt\) tends towards \(-\infty\). Thus, the argument of csp function becomes a very large negative number: $$\operatorname{csp}(-nt) \rightarrow \operatorname{csp}(-\infty)$$ The csp function can be approximated as \(e^{-x}\) for large \(x\). As such, when x becomes \(-\infty\), it can be expressed as: $$\operatorname{csp}(-\infty) \approx e^{-(-\infty)} = e^{\infty}$$
05

Calculate the limit of the function

Now that we have the behavior of the csp function as \(t\) tends to infinity, we can put it into our equation to get the limit: $$\lim_{t \rightarrow \infty} M(t) = \lim_{t \rightarrow \infty} K\left(\frac{M_0}{K}\right)^{e^{\infty}}$$ As \(e^{\infty}\) will tend towards infinity, the expression \((\frac{M_0}{K})^{e^{\infty}}\) will tend towards 0, since \(\frac{M_0}{K}<1\). This is because raising a fraction (between 0 and 1) to an infinitely large power will drive it towards 0. Therefore, we get the limit of the function as follows: $$\lim_{t \rightarrow \infty} M(t) = K\cdot 0$$ Which simplifies to: $$\lim_{t \rightarrow \infty} M(t) = K$$ This verifies the second property, \(\lim_{t \rightarrow \infty} M(t) = K\).

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Key Concepts

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

Calculus
Calculus is a significant branch of mathematics focused on the study of change and motion. Through its two major processes, differentiation and integration, it provides a framework for modeling systems in which there is change, and a way to deduce the predictions of such models. In the context of the Gompertz solution, calculus is used to analyze the behavior of the function over time, especially as time approaches infinity. This is done through the evaluation of limits, which is a fundamental aspect of calculus. It allows us to understand the long-term behavior of growth processes described by the Gompertz function, which is often used in biological and demographic studies to model the growth of populations or tumors.
Limits
Limits in calculus are essential tools used to describe the behavior of functions as they approach a specific point or extend towards infinity. When evaluating limits, we investigate what value the function gravitates towards as the input gets arbitrarily close to a particular value. In the case of the Gompertz function, the limit process is utilized to verify mathematical properties by seeing what happens to the function as time grows without bound. Understanding limits is pivotal to confirming that as time goes to infinity \( t \rightarrow \infty \), the Gompertz function approaches the carrying capacity \( K \), denoting a plateau in growth in accordance with its designed behavior.
Exponential Functions
Exponential functions are mathematical expressions where the variable appears in the exponent and typically have the form \( f(x) = a^{bx} \). These functions are known for their rapid growth or decay properties, and they commonly appear in various scientific disciplines, from compound interest calculations in finance to population studies in biology. The Gompertz solution incorporates an exponential function in the form of \( e^{-x} \), where \( x \), in this context, represents the passing of time multiplied by the growth rate negated. Exponential functions are crucial for understanding the Gompertz model as they describe how the population diminishes to a certain level over time, capturing the deceleration in growth as the population approaches the carrying capacity.
Mathematical Properties
Mathematical properties pertain to the characteristics and behaviors that certain equations or functions inherently possess. These properties can often be confirmed or derived using calculus, as seen with the Gompertz function. In our Gompertz solution example, verifying that the function satisfies the initial condition \( M(0) = M_0 \) and the long-term limiting behavior \( \lim_{t \rightarrow \infty} M(t) = K \) is a demonstration of understanding these mathematical properties. By confirming these, we assert that the model accurately depicts an initial population \( M_0 \) and tends toward a maximum capacity \( K \) over time, showing its applicability and utility in modeling scenarios with a saturating growth pattern.

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Most popular questions from this chapter

Chemical rate equations Let \(y(t)\) be the concentration of a substance in a chemical reaction (typical units are moles/liter). The change in the concentration, under appropriate conditions, is modeled by the equation \(\frac{d y}{d t}=-k y^{n},\) for \(t \geq 0,\) where \(k>0\) is a rate constant and the positive integer \(n\) is the order of the reaction. a. Show that for a first-order reaction \((n=1),\) the concentration obeys an exponential decay law. b. Solve the initial value problem for a second-order reaction \((n=2)\) assuming \(y(0)=y_{0}\). c. Graph the concentration for a first-order and second-order reaction with \(k=0.1\) and \(y_{0}=1\).

In this section, several models are presented and the solution of the associated differential equation is given. Later in the chapter, we present methods for solving these differential equations. 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. a. Show by substitution that if the initial mass of drug in the blood is zero \((m(0)=0),\) then the solution of the initial value problem is \(m(t)=\frac{I}{k}\left(1-e^{-k t}\right)\) b. Graph the solution for \(I=10 \mathrm{mg} / \mathrm{hr}\) and \(k=0.05 \mathrm{hr}^{-1}\) c. Evaluate \(\lim _{t \rightarrow \infty} m(t),\) the steady-state drug level, and verify the result using the graph in part (b).

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Direction field analysis Consider the first-order initial value problem \(y^{\prime}(t)=a y+b, y(0)=A,\) for \(t \geq 0,\) where \(a, b,\) and \(A\) are real numbers. a. Explain why \(y=-b / a\) is an equilibrium solution and corresponds to a horizontal line in the direction field. b. Draw a representative direction field in the case that \(a>0\) Show that if \(A>-b / a\), then the solution increases for \(t \geq 0,\) and that if \(A<-b / a,\) then the solution decreases for \(t \geq 0\). c. Draw a representative direction field in the case that \(a<0\) Show that if \(A>-b / a\), then the solution decreases for \(t \geq 0,\) and that if \(A<-b / a,\) then the solution increases for \(t \geq 0\).

Solve the equation \(y^{\prime}(t)=k y+b\) in the case that \(k y+b<0\) and verify that the general solution is \(y(t)=C e^{k t}-\frac{b}{k}\)

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