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A differential equation governing the velocity \(v\) of a falling mass \(m\) subjected to air resistance proportional to the square of the instantaneous velocity is $$ m \frac{d v}{d t}=m g-k v^{2} $$ where \(k\) is a positive constant of proportionality. (a) Solve this equation subject to the initial condition \(v(0)=v_{0}\). (b) Determine the limiting, or terminal, velocity of the mass. (c) If distance \(s\) is related to velocity by \(d s / d t=v\), find an explicit expression for \(s\) if it is further known that \(s(0)=s_{0}\).

Short Answer

Expert verified
(a) Solve by separation of variables and integration. (b) Terminal velocity is \( \sqrt{\frac{mg}{k}} \). (c) Integrate velocity function to find distance \( s(t) \).

Step by step solution

01

Separate Variables

To solve \( m \frac{dv}{dt} = mg - kv^2 \), begin by separating the variables. This gives us\[ m \frac{dv}{mg - kv^2} = dt. \]
02

Integrate Both Sides

Integrate both sides of the separated equation. The left side becomes \( \int \frac{m}{mg - kv^2} \) dv and the right side becomes \( \int dt \). Upon simplifying, we find \[ \frac{m}{k} \int \frac{1}{\frac{mg}{k} - v^2} \] for the left side.
03

Solve the Integral

Recognize the left side integral as a standard form, \( \int \frac{1}{a^2 - x^2} \) dx = \( \frac{1}{2a} \ln \left| \frac{a+x}{a-x} \right| \) + C. This gives \[ \frac{m}{2k} \ln \left| \frac{\sqrt{mg/k} + v}{\sqrt{mg/k} - v} \right| = t + C. \]
04

Solve for Constant and Velocity

Use the initial condition \( v(0) = v_0 \) to solve for the constant \( C \). Substitute \( v = v_0 \) and \( t = 0 \) in the equation to find \( C \), then solve for \( v \) as a function of \( t \).
05

Find the Terminal Velocity

The terminal velocity \( v_t \) is the steady state solution when \( \frac{dv}{dt} = 0 \). Setting \( mg - kv^2 = 0 \) gives \( v_t = \sqrt{\frac{mg}{k}} \).
06

Express Distance in Terms of Time

Since \( \frac{ds}{dt} = v \), integrate \( v(t) \) with respect to time to find an expression for \( s(t) \). Use \( s(0) = s_0 \) as the initial condition to solve completely for \( s(t) \).

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Key Concepts

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

Initial Conditions
In the context of differential equations, initial conditions are crucial. They provide the necessary values at the starting point of the system under examination. Initial conditions help us determine specific solutions from a family of solutions that a differential equation may possess.
For example, given the differential equation\[ m \frac{dv}{dt} = mg - kv^2 \]we have an initial condition given as \( v(0) = v_0 \). This condition specifies that the velocity \( v \) of the falling object is \( v_0 \) when time \( t \) is zero.
Applying this initial condition is critical to finding the constant of integration that arises after solving the differential equation. Without it, we would only have a general solution, leaving ambiguity about which specific path the system actually follows.
Terminal Velocity
Terminal velocity refers to the constant velocity that a falling object reaches when the force of gravity is balanced by the drag force due to air resistance. In effect, the object stops accelerating and continues to fall at this constant speed.
The differential equation \[ m \frac{dv}{dt} = mg - kv^2 \]can show us terminal velocity. At terminal velocity, the acceleration (\( \frac{dv}{dt} \)) becomes zero because forces balance out.
Setting\[ mg - kv^2 = 0 \]we can solve for the velocity \( v \), resulting in\[ v_t = \sqrt{\frac{mg}{k}} \].
This equation allows us to calculate the terminal velocity \( v_t \) using known values such as the mass \( m \), gravitational constant \( g \), and drag coefficient \( k \). When your object reaches this velocity, it will no longer accelerate.
Separation of Variables
Separation of variables is a powerful technique for solving differential equations. It involves rearranging the equation so that each variable and its derivatives are on separate sides.
Consider the equation:\[ m \frac{dv}{dt} = mg - kv^2 \] We start by moving all terms involving \( v \) to one side and all terms involving \( t \) to the other:\[ m \frac{dv}{mg - kv^2} = dt \]This allows us to address each side independently by integrating. This method transforms the differential equation into an equation we can solve more easily.
Effective use of separation of variables often simplifies complex equations, making them more approachable and solvable through standard integration.
Integration Methods
Once variables are separated in differential equations, we need to integrate both sides to proceed towards a solution. This requires applying appropriate integration techniques.
In our exercise, we encountered an integral of the form:\[ \int \frac{1}{\frac{mg}{k} - v^2} dv \]Recognizing this as a standard integral, we can use the formula:\[ \int \frac{1}{a^2 - x^2} \, dx = \frac{1}{2a} \ln \left| \frac{a+x}{a-x} \right| + C \].
Using this formula, we substitute the values to compute the integral and solve for velocity \( v \) as a function of time \( t \). Integration methods are essential tools that help solve differential equations, especially when they align with known integral forms.
Mastering these methods simplifies solving complex problems and reaching specific solutions tied to initial conditions.

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Most popular questions from this chapter

A differential equation governing the velocity \(v\) of a falling mass \(m\) subjected to air resistance proportional to the instantaneous velocityis $$ m \frac{d v}{d t}=m g-k v_{n} $$ where \(k\) is a positive constant of proportionality. (a) Solve the equation subject to the initial condition \(v(0)=v_{0}\). (b) Determine the limiting, or terminal, velocity of the mass. (c) If distance \(s\) is related to velocity \(d s / d t=v\), find an explicit expression for \(s\) if it is further known that \(s(0)=s_{0}\).

(a) If a constant number \(h\) of animals is removed or harvested per unit time, then a model for the population \(P(t)\) of animals at any time \(t\) is given by $$ \frac{d P}{d t}=P(a-b P)-h, \quad P(0)=P_{0}, $$ where \(a, b, h\), and \(P_{0}\) are positive constants. Solve the problem when \(a=5, b=1\), and \(h=4\). (b) Use an ODE solver to determine the long-term behavior of the population in part (a) in the cases \(P_{0}>4,1

Air containing \(0.06 \%\) carbon dioxide is pumped into a room whose volume is \(8000 \mathrm{ft}^{3}\). The air is pumped in at a rate of \(2000 \mathrm{ft}^{3} / \mathrm{min}\), and the circulated air is then pumped out at the same rate. If there is an initial concentration of \(0.2 \%\) carbon dioxide, determine the subsequent amount in the room at any time. What is the concentration at 10 minutes? What is the steady-state, or equilibrium, concentration of carbon dioxide?

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A 100 -volt electromotive force is applied to an \(R C\) series circuit in which the resistance is 200 ohms and the capacitance is \(10^{-4}\) farad. Find the charge \(q(t)\) on the capacitor if \(q(0)=0\). Find the current \(i(t)\).

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