{Use of Tech} Free fall Using th e background given in Exercise 47, assume the resistance is given by f(v)=−Rv, for t≥0, where R\u003e0 is a drag coefficient (an assumption often made for a heavy medium such as water or oil).c. Find the solution of this separable equation assuming v(0)=0 and 0\u003cv\u003cg/b .

Ch. 9 - Differential Equations

Briggs - Calculus: Early Transcendentals 3rd Edition

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Briggs 3rd Edition

Ch. 9 - Differential Equations

Problem 9.3.48c

Chapter 9, Problem 9.3.48c

{Use of Tech} Free fall Using th e background given in Exercise 47, assume the resistance is given by f(v)=−Rv, for t≥0, where R>0 is a drag coefficient (an assumption often made for a heavy medium such as water or oil).

c. Find the solution of this separable equation assuming v(0)=0 and 0<v<g/b.

Verified step by step guidance

Start with the given differential equation for velocity under resistance: \[\frac{dv}{dt} = g - bv - Rv,\] where \(g\) is the acceleration due to gravity, \(b\) is a constant, and \(R > 0\) is the drag coefficient. Combine the terms involving \(v\) to rewrite the equation as \[\frac{dv}{dt} = g - (b + R)v.\]

Recognize that this is a first-order linear ordinary differential equation in \(v(t)\), which can be rearranged as \[\frac{dv}{dt} + (b + R)v = g.\]

To solve this linear ODE, find the integrating factor \(\mu(t)\), which is given by \[\mu(t) = e^{\int (b + R) dt} = e^{(b + R)t}.\]

Multiply both sides of the differential equation by the integrating factor to get \[e^{(b + R)t} \frac{dv}{dt} + (b + R) e^{(b + R)t} v = g e^{(b + R)t},\] which simplifies to \[\frac{d}{dt} \left( e^{(b + R)t} v \right) = g e^{(b + R)t}.\]

Integrate both sides with respect to \(t\( to find \[e^{(b + R)t} v = \int g e^{(b + R)t} dt + C,\] where \)C\) is the constant of integration. Then solve for \(v(t)\) by dividing both sides by \(e^{(b + R)t}\). Use the initial condition \(v(0) = 0\) to determine \(C\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Separable Differential Equations

A separable differential equation can be written as a product of a function of the dependent variable and a function of the independent variable. This allows the variables to be separated on opposite sides of the equation and integrated independently, simplifying the solution process.

Modeling Drag Force as Proportional to Velocity

In fluid resistance problems, drag force is often modeled as proportional to velocity, expressed as f(v) = -Rv, where R is the drag coefficient. This linear resistance opposes motion and affects the acceleration and velocity of the falling object.

Initial Conditions in Differential Equations

Initial conditions specify the value of the solution at a particular point, such as v(0) = 0. They are essential for determining the unique solution to a differential equation from the family of possible solutions obtained after integration.

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Related Practice

Textbook Question

38–43. Equilibrium solutions A differential equation of the form y′(t)=f(y) is said to be autonomous (the function f depends only on y). The constant function y=y0 is an equilibrium solution of the equation provided f(y0)=0 (because then y'(t)=0 and the solution remains constant for all t). Note that equilibrium solutions correspond to horizontal lines in the direction field. Note also that for autonomous equations, the direction field is independent of t. Carry out the following analysis on the given equations.

c. Sketch the solution curve that corresponds to the initial condition y0=1.

y′(t) = 2y + 4

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Textbook Question

{Use of Tech} Logistic equation for an epidemic When an infected person is introduced into a closed and otherwise healthy community, the number of people who contract the disease (in the absence of any intervention) may be modeled by the logistic equation

dP/dt=kP(1−P/A),P0=P_0,

where K is a positive infection rate, A is the number of people in the community, and P0 is the number of infected people at t=0. The model also assumes no recovery.

c. For a fixed value of K and A, describe the long-term behavior of the solutions, for any P0 with 0<P0<A.

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Textbook Question

46–48. Analyzing models The following models were discussed in Section 9.1 and reappear in later sections of this chapter. In each case, carry out the indicated analysis using direction fields.

Drug infusion The delivery of a drug (such as an antibiotic) through an intravenous line may be modeled by the differential equation m′(t)+km(t)=I, where m(t) is the mass of the drug in the blood at time t≥0, K is a constant that describes the rate at which the drug is absorbed, and I is the infusion rate. Let I=10mg/hr and k=0.05 hr^−1.

c. What is the equilibrium solution?

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Textbook Question

29–32. {Use of Tech} Errors in Euler’s method Consider the following initial value problems.

c. Which time step results in the more accurate approximation? Explain your observations.

y′(t) = 4−y, y(0) = 3; y(t) = 4−e⁻ᵗ

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Textbook Question

27–30. Predator-prey models Consider the following pairs of differential equations that model a predator-prey system with populations x and y. In each case, carry out the following steps.

c. Find the equilibrium points for the system.

x′(t) = −3x + xy, y′(t) = 2y − xy

Textbook Question

27–30. Predator-prey models Consider the following pairs of differential equations that model a predator-prey system with populations x and y. In each case, carry out the following steps.

c. Find the equilibrium points for the system.

x′(t) = −3x + 6xy, y′(t) = y − 4xy