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Magazine Chapter 6: Problem 54
A falling body with friction equal to velocity squared obeys \(d v / d t=g-v^{2}\) (a) Show that \(v(t)=\sqrt{g}\) tanh \(\sqrt{g t \text { satisfies the equation. }}\) (b) Derive this \(v\) yourself, by integrating \(d v /\left(g-v^{2}\right)=d t\) (c) Integrate \(v(t)\) to find the distance \(f(t)\)
Short Answer
Expert verified
(a) Verified; (b) Derived solution is \(v(t) = \sqrt{g} \tanh(\sqrt{g}t)\); (c) Distance: \(f(t) = \frac{1}{\sqrt{g}} \ln|\cosh(\sqrt{g}t)| + C\)."
Step by step solution
01
Verify Solution for Differential Equation
To verify that \(v(t) = \sqrt{g}\tan(\sqrt{g}t)\) satisfies the given differential equation \(\frac{dv}{dt} = g - v^2\), we start by differentiating \(v(t)\) with respect to \(t\).First, calculate \(\frac{d}{dt}[\sqrt{g}\tan(\sqrt{g}t)] = \sqrt{g} \cdot \frac{d}{dt}\left(\tanh(\sqrt{g}t)\right) = \sqrt{g} \cdot \frac{1}{\cosh^2(\sqrt{g}t)} \cdot \sqrt{g}\) using the derivative of \(\tanh(x)\), which is \(\text{sech}^2(x) = \frac{1}{\cosh^2(x)}\).Thus, \(\frac{dv}{dt} = g \cdot \text{sech}^2(\sqrt{g}t)\), matching the right-hand side of the equation if we substitute \(v = \sqrt{g} \tanh(\sqrt{g}t)\). It simplifies to \(\frac{dv}{dt} = g - (\sqrt{g} \tanh(\sqrt{g}t))^2\), confirming that \(v(t)\) is a solution.
02
Integrate the Differential Equation
To derive \(v(t)\), we separate variables in the given differential equation:\[\frac{dv}{g - v^2} = dt\]We'll use partial fraction decomposition to integrate:Let \(A/(g-v) + B/(g+v)\) be the decomposition.Solving, \(A = 1/(2\sqrt{g})\) and \(B = -1/(2\sqrt{g})\) result in:\[\int \left(\frac{1}{g + v} - \frac{1}{g - v}\right) dv = \int dt\]This becomes:\[\frac{1}{2\sqrt{g}}\left(\ln|g+v| - \ln|g-v|\right) = t + C\]Solving gives:\[v(t) = \sqrt{g} \tanh(\sqrt{g}t + C)\] with \(C = 0\).
03
Integrate Velocity to Find Distance
To find the distance \(f(t)\), integrate \(v(t)\) to get:\[f(t) = \int v(t) dt = \int \sqrt{g}\tanh(\sqrt{g}t) dt\]By substitution, let \(u = \sqrt{g}t\), making \(du = \sqrt{g} dt\). Hence:\[f(t) = \frac{1}{\sqrt{g}} \int \tanh(u) du\]The integral of \(\tanh(u)\) is \(\ln|\cosh(u)|\), giving:\[f(t) = \frac{1}{\sqrt{g}} \ln|\cosh(\sqrt{g}t)| + C\]This result provides the distance \(f(t)\) travelled as a function of time \(t\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Velocity Function
In the realm of differential equations, understanding how velocity functions work is essential, especially when dealing with problems involving motion. A velocity function, like the one described by the function \( v(t) = \sqrt{g} \tanh(\sqrt{g}t) \), represents how the speed of an object changes over time.
In this particular example, the velocity function is derived from a scenario where a falling object is subject to a frictional force proportional to its velocity squared. This means that the force hindering the object's motion increases with its speed, leading to a complex motion equation.
The specific form of the velocity function, involving hyperbolic tangent, arises because it naturally satisfies the differential equation \( \frac{dv}{dt} = g - v^2 \). This form shows that the velocity approaches its limiting value, a common occurrence in physics where forces decrease the rate of change of velocity.
In this particular example, the velocity function is derived from a scenario where a falling object is subject to a frictional force proportional to its velocity squared. This means that the force hindering the object's motion increases with its speed, leading to a complex motion equation.
The specific form of the velocity function, involving hyperbolic tangent, arises because it naturally satisfies the differential equation \( \frac{dv}{dt} = g - v^2 \). This form shows that the velocity approaches its limiting value, a common occurrence in physics where forces decrease the rate of change of velocity.
Separation of Variables
Separation of variables is a powerful technique used to simplify differential equations. It involves rearranging the equation to isolate one variable on each side, allowing for easier integration. For the differential equation \( \frac{dv}{dt} = g - v^2 \), this method is employed by rewriting it as:
\[ \frac{dv}{g - v^2} = dt \]
By isolating the terms involving \( v \) on one side and the differential element \( dt \) on the other, the path is laid out for integration. This manipulation permits us to handle the equation as two integrals, which can then be solved independently.
Separation of variables is especially useful for equations that, at first glance, appear tangled. It provides a way to decompose the problem into manageable parts and is particularly common in solving first-order differential equations like this one.
\[ \frac{dv}{g - v^2} = dt \]
By isolating the terms involving \( v \) on one side and the differential element \( dt \) on the other, the path is laid out for integration. This manipulation permits us to handle the equation as two integrals, which can then be solved independently.
Separation of variables is especially useful for equations that, at first glance, appear tangled. It provides a way to decompose the problem into manageable parts and is particularly common in solving first-order differential equations like this one.
Partial Fraction Decomposition
When integrating rational functions, partial fraction decomposition is an invaluable tool. It breaks down complex fractions into simpler parts that are more manageable to integrate. For the expression \( \frac{dv}{g - v^2} \), we use partial fraction decomposition to express it in a form suitable for integration:
\[ \frac{1}{g - v^2} = \frac{A}{g-v} + \frac{B}{g+v} \]
Solving for \( A \) and \( B \), we find \( A = \frac{1}{2\sqrt{g}} \) and \( B = -\frac{1}{2\sqrt{g}} \).
This decomposition allows us to integrate each fraction more straightforwardly, reducing the problem to a sum of logarithmic functions. It's a common technique when dealing with polynomial denominators that can be factored into linear terms.
\[ \frac{1}{g - v^2} = \frac{A}{g-v} + \frac{B}{g+v} \]
Solving for \( A \) and \( B \), we find \( A = \frac{1}{2\sqrt{g}} \) and \( B = -\frac{1}{2\sqrt{g}} \).
This decomposition allows us to integrate each fraction more straightforwardly, reducing the problem to a sum of logarithmic functions. It's a common technique when dealing with polynomial denominators that can be factored into linear terms.
Integration
Integration is the mathematical tool used to accumulate values over a domain, and it plays a critical role in solving differential equations. Once we have separated variables and performed partial fraction decomposition, we integrate the resulting terms:
\[ \int \left(\frac{1}{g + v} - \frac{1}{g - v}\right) dv = \int dt \]
Through integration, we find:
\[ \frac{1}{2\sqrt{g}}(\ln|g+v| - \ln|g-v|) = t + C \]
Solving this yields the function for velocity. Additionally, we integrate the velocity function \( v(t) \) to determine the distance \( f(t) \).
Integration is not just about finding areas under curves; it also reconstructs functions representing physical quantities like distance from velocity, providing insight into the dynamics of systems.
\[ \int \left(\frac{1}{g + v} - \frac{1}{g - v}\right) dv = \int dt \]
Through integration, we find:
\[ \frac{1}{2\sqrt{g}}(\ln|g+v| - \ln|g-v|) = t + C \]
Solving this yields the function for velocity. Additionally, we integrate the velocity function \( v(t) \) to determine the distance \( f(t) \).
Integration is not just about finding areas under curves; it also reconstructs functions representing physical quantities like distance from velocity, providing insight into the dynamics of systems.
Hyperbolic Functions
Hyperbolic functions, analogous to trigonometric functions, offer a smooth transition across various domains. In calculus and, specifically, dealing with velocity functions, they provide solutions that are not apparent otherwise.
The hyperbolic tangent function, denoted \( \tanh(x) \), appears prominently in the velocity function \( v(t) = \sqrt{g} \tanh(\sqrt{g}t) \). Like its trigonometric counterpart, \( \tanh(x) \) reflects the saturated growth typical in friction-impacted dynamics.
These functions owe their usefulness to their unique properties, such as having derivatives that are simple expressions in terms of hyperbolic secant:
\( \frac{d}{dx}\tanh(x) = \text{sech}^2(x) \).
The use of hyperbolic functions thus offers solutions that are mathematically elegant and physically meaningful, capturing phenomena like decelerating forces or equilibrium states in motion systems.
The hyperbolic tangent function, denoted \( \tanh(x) \), appears prominently in the velocity function \( v(t) = \sqrt{g} \tanh(\sqrt{g}t) \). Like its trigonometric counterpart, \( \tanh(x) \) reflects the saturated growth typical in friction-impacted dynamics.
These functions owe their usefulness to their unique properties, such as having derivatives that are simple expressions in terms of hyperbolic secant:
\( \frac{d}{dx}\tanh(x) = \text{sech}^2(x) \).
The use of hyperbolic functions thus offers solutions that are mathematically elegant and physically meaningful, capturing phenomena like decelerating forces or equilibrium states in motion systems.
