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Chapter 2: Problem 17

(a) Solve the Gompertz equation $$ d y / d t=r y \ln (K / y) $$ subject to the initial condition \(y(0)=y_{0}\) (b) For the data given in Example 1 in the text \([ \leftr=0.71 \text { per year, } K=80.5 \times 10^{6} \mathrm{kg}\), \right. \(\left.y_{0} / K=0.25\right]\), use the Gompertz model to find the predicted value of \(y(2) .\) (c) For the same data as in part (b), use the Gompertz model to find the time \(\tau\) at which \(y(\tau)=0.75 K .\) Hint: You may wish to let \(u=\ln (y / K)\).

Short Answer

Expert verified
The expression for the Gompertz growth model is: $$y(t) = K e^{\ln(\frac{y_0}{K}) e^{-rt}}.$$

Step by step solution

01

Solve the Gompertz equation

Let \(u=\ln(y/K).\) Then, \(y=Ke^u\) and differentiating both sides with respect to \(t\), we have $$ \frac{dy}{dt}=Ke^u\frac{du}{dt} $$ Substitute this expression for \(\frac{dy}{dt}\) and \(y=Ke^u\) into the original Gompertz equation: $$ Ke^u\frac{du}{dt}=ry\ln\left(\frac{K}{y}\right) \Rightarrow \frac{du}{dt}=r\ln\left(\frac{K}{Ke^u}\right) $$ We now have a separable equation: $$ \frac{du}{dt}=r\ln\left(\frac{1}{e^u}\right) =-r\ln\left(e^{u}\right) =-ru $$ Now, we can solve this equation by separating the variables and integrate: $$ \int \frac{du}{u}=-r\int dt $$
02

Apply the initial condition

Using integration, we have: $$ \ln |u| = -rt+C $$ Or $$ u=e^{-rt+C}=e^{-rt}e^C $$ Recall that \(u=\ln(y/K)\), thus, we have: $$ \ln\left(\frac{y}{K}\right)=e^C e^{-rt} $$ Applying the initial condition \(y(0)=y_0:\) $$ \ln\left(\frac{y_0}{K}\right)=e^C $$ So, the Gompertz equation can be written as: $$ y(t) = K e^{\ln(\frac{y_0}{K}) e^{-rt}} $$
03

Find the predicted value of \(y(2)\)

Using the values given in Example 1: \(r=0.71\text{ per year}, K=80.5\times 10^6\mathrm{kg},\) and \(y_0/K=0.25, \) so \(y_0 = 0.25K\). We can find the predicted value of y at t=2: $$ y(2)=K e^{\ln\left(\frac{0.25K}{K}\right) e^{-0.71\times 2}} $$ Now, we can calculate the value of \(y(2)\).
04

Find the time \(\tau\) at which \(y(\tau)=0.75 K\)

We have the equation: $$ y(\tau)=0.75K \Rightarrow 0.75K=K e^{\ln\left(\frac{0.25K}{K}\right)e^{-0.71\tau}} $$ This simplifies to: $$ 0.75=e^{\ln(0.25)e^{-0.71\tau}} $$ Taking the natural logarithm on both sides and isolating \(\tau\), we get: $$ \tau =-\frac{\ln(\ln(0.75)/\ln(0.25))}{0.71} $$ Finally, we can calculate the value of \(\tau\).

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Key Concepts

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

Differential Equations
At the heart of modeling real-world phenomena like population growth, chemical reactions, and disease spread are differential equations. They describe how a variable, typically representing physical quantities such as concentration, velocity, or, in our case, population size, changes over time or space. A differential equation sets up a relationship between a function and its derivatives, giving rise to complex dynamics of change.

Particularly, the Gompertz equation is a specific type of differential equation used to model growth, which is characterized by a slowing increase as the variable approaches a certain limit. It can be used not only in biology to model tumor growth but also in economics and other fields where a saturation point might occur. By solving this kind of equation, predictions on certain variables like population at a future time can be made, providing valuable insights for planning and analysis.
Boundary Value Problems
Boundary Value Problems (BVPs) are special cases within differential equations where the solution is specified by conditions at different points, often at the extremes of the interval over which the equation is defined. For example, in a rod with varying temperature along its length, the temperatures at both ends of the rod are the boundary values.

In the exercise dealing with the Gompertz equation, the initial condition, given as the value at time zero, acts similarly to a boundary condition, enabling the precise determination of the solution. BVPs are crucial because they give us the necessary information to produce a unique solution from a sea of possibilities, tailoring the general solution of a differential equation to a specific scenario.
Separable Variables
Separable variables come into play when we encounter a differential equation where the two types of variables, such as those representing time and growth, can be separated on different sides of an equation. It simplifies solving as each variable can be dealt with independently.

In other words, a 'separable' differential equation is one in which the two variables can be written as a product of two functions, each depending exclusively on one of the variables. This property is what makes it possible to integrate both sides with respect to their respective variables, which is exactly what we did with the Gompertz equation by setting a substitution for a more workable form and then integrating to find a general solution. For students, recognizing a differential equation as separable is crucial for applying the right method to solve it efficiently.
Exponential Growth
The Gompertz equation models a form of exponential growth where the rate of growth decreases over time. Traditional exponential growth is characterized by a rate of change that is proportional to the current value of a variable, resulting in a rapid increase of the variable over time.

However, unlike unrestricted exponential growth, the Gompertz model introduces a boundary, which represents a real-world limiting factor, such as carrying capacity in population dynamics. This ensures that growth slows down as this upper limit is approached, which is frequently how populations and economic systems work. Its representation as a slightly modified exponential function allows us to forecast based on initial conditions and rates, which is a valuable tool in scientific and financial forecasting.

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Most popular questions from this chapter

Consider the initial value problem $$ y^{\prime}=t^{2}+y^{2}, \quad y(0)=1 $$ Use Euler's method with \(h=0.1,0.05,0.025,\) and 0.01 to explore the solution of this problem for \(0 \leq t \leq 1 .\) What is your best estimate of the value of the solution at \(t=0.8 ?\) At \(t=1 ?\) Are your results consistent with the direction field in Problem \(9 ?\)

Radiocarbon Dating, An important tool in archeological research is radiocarbon dating. This is a means of determining the age of certain wood and plant remains, hence of animal or human bones or artifacts found buried at the same levels. The procedure was developed by the American chemist Willard Libby ( \(1908-1980\) ) in the early 1950 s and resulted in his winning the Nobel prize for chemistry in 1960 . Radiocarbon dating is based on the fact that some wood or plant remains contain residual amounts of carbon- \(14,\) a radioactive isotope of carbon. This isotope is accumulated during the lifetime of the plant and begins to decay at its death. Since the half-life of carbon- 14 is long (approximately 5730 years'), measurable amounts of carbon- 14 remain after many thousands of years. Libby showed that if even a tiny fraction of the original amount of carbon- 14 is still present, then by appropriate laboratory measurements the proportion of the original amount of carbon- 14 that remains can be accurately determined. In other words, if \(Q(t)\) is the amount of carbon- 14 at time \(t\) and \(Q_{0}\) is the original amount, then the ratio \(Q(t) / Q_{0}\) can be determined, at least if this quantity is not too small. Present measurement techniques permit the use of this method for time periods up to about \(50,000\) years, after which the amount of carbon- 14 remaining is only about 0.00236 of the original amount. $$ \begin{array}{l}{\text { (a) Assuming that } Q \text { satisfies the differential equation } Q^{\prime}=-r Q, \text { determine the decay }} \\\ {\text { constant } r \text { for carbon-14. }} \\ {\text { (b) Find an expression for } Q(t) \text { at any time } t, \text { if } Q(0)=Q_{0}}\end{array} $$ $$ \begin{array}{l}{\text { (c) Suppose that certain remains are discovered in which the current residual amount of }} \\ {\text { carbon- } 14 \text { is } 20 \% \text { of the original amount. Determine the age of these remains. }}\end{array} $$

A recent college graduate borrows 100,000 dollar at an interest rate of \(9 \%\)to purchase a condominium. Anticipating steady salary increases, the buyer expects to make payments at a monthly rate of 800 dollar (1+t / 120), where t is the number of months since the loan was made. $$ \begin{array}{l}{\text { (a) Assuming that this payment schedule can be maintained, when will the loan be fully }} \\ {\text { paid? }} \\ {\text { (b) Assuming the same payment show large a loan could be paid off in exactly }} \\ {\text { 20 years? }}\end{array} $$

The method outlined in Problem 30 can be used for any homogeneous equation. That is, the substitution \(y=x v(x)\) transforms a homogeneous equation into a separable equation. The latter equation can be solved by direct integration, and then replacing \(v\) by \(y / x\) gives the solution to the original equation. In each of Problems 31 through 38 : $$ \begin{array}{l}{\text { (a) Show that the given equation is homogeneous. }} \\\ {\text { (b) Solve the differential equation. }} \\ {\text { (c) Draw a direction field and some integral curves. Are they symmetric with respect to }} \\ {\text { the origin? }}\end{array} $$ $$ \frac{d y}{d x}=\frac{x^{2}+x y+y^{2}}{x^{2}} $$

Show that any separable equation, $$ M(x)+N(y) y^{\prime}=0 $$ is also exact.
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