/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 30 A tumor may be regarded as a pop... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A tumor may be regarded as a population of multiplying cells. It is found empirically that the "birth rate" of the cells in a tumor decreases exponentially with time, so that \(\beta(t)=\beta_{0} e^{-\alpha t}\) (where \(\alpha\) and \(\beta_{0}\) are positive constants), and hence $$ \frac{d P}{d t}=\beta_{0} e^{-\alpha t} P, \quad P(0)=P_{0} $$ Solve this initial value problem for $$ P(t)=P_{0} \exp \left(\frac{\beta_{0}}{\alpha}\left(1-e^{-\alpha t}\right)\right) $$ Observe that \(P(t)\) approaches the finite limiting population \(P_{0} \exp \left(\beta_{0} / \alpha\right)\) as \(t \rightarrow+\infty\).

Short Answer

Expert verified
The population solution is \( P(t) = P_0 \exp \left(\frac{\beta_0}{\alpha}(1-e^{-\alpha t})\right) \) and it approaches \( P_0 \exp \left(\frac{\beta_0}{\alpha}\right) \) as \( t \to \infty \).

Step by step solution

01

Write Down the Differential Equation

Starting with the given differential equation for the population of cells, we have: \( \frac{dP}{dt} = \beta_{0} e^{-\alpha t} P \). This equation describes how the population \( P \) changes with time \( t \), given the exponentially decreasing birth rate \( \beta(t) \).
02

Apply Separation of Variables

To solve the differential equation, use the method of separation of variables. Write the equation as: \( \frac{dP}{P} = \beta_{0} e^{-\alpha t} dt \). This separates the variables, with \( P \) terms on one side and \( t \) terms on the other.
03

Integrate Both Sides

Integrate both sides of the equation. The left side becomes \( \int \frac{dP}{P} = \ln |P| + C_1 \). The right side becomes \( \int \beta_0 e^{-\alpha t} dt = -\frac{\beta_{0}}{\alpha} e^{-\alpha t} + C_2 \).
04

Exponentiate to Solve for P(t)

Simplify the integrated equation: \( \ln |P| = -\frac{\beta_0}{\alpha} e^{-\alpha t} + C \). To solve for \( P(t) \), exponentiate both sides to get: \( P = e^{C} e^{-\frac{\beta_0}{\alpha} e^{-\alpha t}} \).
05

Use Initial Condition to Solve for Constant

Apply the initial condition \( P(0) = P_0 \) to find the constant \( e^C \). At \( t = 0 \), \( P_0 = e^C e^{-\frac{\beta_0}{\alpha}} \), giving \( e^C = P_0 e^{\frac{\beta_0}{\alpha}} \). Substitute back: \( P(t) = P_0 e^{\frac{\beta_0}{\alpha}(1 - e^{-\alpha t})} \).
06

Analyze the Limiting Behavior

Evaluate the behavior of \( P(t) \) as \( t \) approaches infinity. As \( t \to \infty \), \( e^{-\alpha t} \to 0 \), so \( P(t) \to P_0 e^{\frac{\beta_0}{\alpha}} \), confirming the limiting population. Thus, the tumor population reaches a finite limit.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Initial Value Problem
In differential equations, an initial value problem is a situation where we need to find a function that satisfies a given differential equation and also meets a specified value at an initial point. When solving these problems, the initial condition provides crucial information allowing us to determine a particular solution from a family of solutions.
In the exercise above, the problem starts with the differential equation \( \frac{dP}{dt} = \beta_{0} e^{-\alpha t} P \), with the initial condition \( P(0) = P_{0} \). This means that at time \( t = 0 \), the population of the tumor cells is \( P_{0} \). Particularly, the initial condition helps us solve for the constant of integration when we integrate our differential equation solution.
To resolve this, once we have integrated the equation using separation of variables, we substitute the initial values to find any unknown constants, uniquely determining the function \( P(t) \) that describes our system.
Separation of Variables
Separation of variables is a core technique for solving differential equations. It involves rearranging the equation such that each variable occurs on a distinct side of the equation. This transformation allows for effective integration since each side can be integrated with respect to its own variable.
For our differential equation \( \frac{dP}{dt} = \beta_{0} e^{-\alpha t} P \), variables \( P \) and \( t \) are separated by rearranging into \( \frac{dP}{P} = \beta_{0} e^{-\alpha t} dt \). This layout means we can integrate \( P \) with respect to \( P \) on the left, and \( t \) with respect to time on the right.
Upon integration, we find expressions for both variables, involving constant integration terms. These are then manipulated and solved to provide expressions carrying functional dependencies on initial conditions.
Exponential Growth and Decay
Exponential growth and decay are phenomena where a quantity grows or decreases at a rate proportional to its current value. In our case, the population of cells in the tumor decreases according to the exponential function \( \beta(t)=\beta_{0} e^{-\alpha t} \). This indicates the growth rate itself is declining over time exponentially, slowing the growth of the population.
This is described by the solution to the equation \( P(t) = P_{0} \exp \left(\frac{\beta_{0}}{\alpha} \left(1 - e^{-\alpha t}\right)\right) \) which combines exponential decay in the birth rate with an exponential growth model of the population.
  • When \( t \) is small, \( e^{-\alpha t} \approx 1 \) and the growth rate is near its initial maximum.
  • As \( t \) increases, \( e^{-\alpha t} \to 0 \), substantially slowing growth.
This interplay importantly shapes the tumor's population dynamics over time.
Limiting Behavior
Understanding the behavior of a function as time approaches infinity is essential in grasping the long-term outcome of processes described by differential equations. Knowing the limiting behavior helps predict the eventual stabilization of systems like our tumor cell population.
In this exercise, as \( t \rightarrow +\infty \), the term \( e^{-\alpha t} \) tends to zero. Therefore, the expression for \( P(t) \) simplifies to \( P_{0} \exp \left( \frac{\beta_{0}}{\alpha} \right) \), indicating that the population levels off to a constant value as time goes on.
This behavior demonstrates that despite any initial exponential growth, the population will eventually cease to grow indefinitely, stabilizing when further growth is balanced by the reduced birth rate. Understanding this limiting behavior ensures that predictions about population size remain grounded in realistic dynamics.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An initial value problem and its exact solution \(y(x)\) are given. Apply Euler's method twice to approximate to this solution on the interval \(\left[0, \frac{1}{2}\right]\), first with step size \(h=0.25\), then with step size \(h=0.1 .\) Compare the threedecimal-place values of the two approximations at \(x=\frac{1}{2}\) with the value \(y\left(\frac{1}{2}\right)\) of the actual solution. $$ y^{\prime}=-3 x^{2} y, v(0)=3 ; y(x)=3 e^{-x^{3}} $$

In Problems first solve the equation \(f(x)=0\) to find the critical points of the given autonomous differential equation \(d x / d t=f(x) .\) Then analyze the sign of \(f(x)\) to determine whether each crifical point is stable or unstable, and construct the corresponding phase diagrant for the differential equarion. Next, solve the differential equarion explicitly for \(x(t)\) in terms of t. Finally use either the exact solution or a computer-generated slope field to sketch typical solution curves for the given differential equation, and verify visually the stability of each critical point. $$ \frac{d x}{d t}=7 x-x^{2}-10 $$

In Jules Verne's original problem, the projectile launched from the surface of the earth is attracted by both the earth and the moon, so its distance \(r(t)\) from the center of the earth satisfies the initial value problem $$ \frac{d^{2} r}{d t^{2}}=-\frac{G M_{e}}{r^{2}}+\frac{G M_{m}}{(S-r)^{2}} ; \quad r(0)=R, \quad r^{\prime}(0)=v_{0} $$ where \(M_{e}\) and \(M_{m}\) denote the masses of the earth and the moon, respectively; \(R\) is the radius of the earth and \(S=384,400 \mathrm{~km}\) is the distance between the centers of the earth and the moon. To reach the moon, the projectile must only just pass the point between the moon and earth where its net acceleration vanishes. Thereafter it is "under the control" of the moon, and falls from there to the lunar surface. Find the minimal launch velocity \(v_{0}\) that suffices for the projectile to make it "From the Earth to the Moon."

(a) Show that if a projectile is launched straight upward from the surface of the earth with initial velocity \(v_{0}\) less than escape velocity \(\sqrt{2 G M / R}\), then the maximum distance from the center of the earth attained by the projectile is $$ r_{\max }=\frac{2 G M R}{2 G M-R v_{0}^{2}} $$ where \(M\) and \(R\) are the mass and radius of the earth, respectively. (b) With what initial velocity \(v_{0}\) must such a projectile be launched to yield a maximum altitude of 100 kilometers above the surface of the earth? (c) Find the maximum distance from the center of the earth, expressed in terms of earth radii, attained by a projectile launched from the surface of the earth with \(90 \%\) of escape velocity.

Separate variables and use partial fractions to solve the initial value problems in Problems \(1-8 .\) Use either the exact solution or a computer- generated slope field to sketch the graphs of several solutions of the given differential equation, and highlight the indicated particular solution. $$ \frac{d x}{d t}=10 x-x^{2}, x(0)=1 $$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.