Problem 24 - Calculus Early Transcendentals: Pearson New International Edition

Question

The differential equation for a falling body near the earth's surface with air resistance proportional to the velocity \(v\) is \(d v / d t=-g-a v\), where \(g=32\) feet per second per second is the acceleration of gravity and \(a>0\) is the drag coefficient. Show each of the following: (a) \(v(t)=\left(v_{0}-v_{\infty}\right) e^{-a t}+v_{\infty}\), where \(v_{0}=v(0)\), and $$ v_{\infty}=-g / a=\lim _{t \rightarrow \infty} v(t) $$ is the so-called terminal velocity. (b) If \(y(t)\) denotes the altitude, then $$ y(t)=y_{0}+t v_{\infty}+(1 / a)\left(v_{0}-v_{\infty}\right)\left(1-e^{-a t}\right) $$

Step-by-Step Solution

Answer

The solution verifies the expressions for both velocity and altitude functions.

Step 1: Verify the Terminal Velocity

The terminal velocity is defined as the steady-state velocity when the acceleration is zero. This occurs when the net force is zero, i.e., when \(\frac{dv}{dt} = 0\). Thus, setting the original differential equation to zero, we have:\[-g - av = 0\]Solving this for \(v\) gives:\[v = -\frac{g}{a} = v_{\infty}\]This confirms that the terminal velocity \(v_{\infty}\) is indeed \(-\frac{g}{a}\).

Step 2: Solve the Differential Equation for Velocity

Start by rewriting the differential equation:\[\frac{dv}{dt} = -g - av\]This is a first-order linear differential equation. To solve it, we can use an integrating factor. The integrating factor \(\mu(t)\) is given by:\[\mu(t) = e^{\int a \, dt} = e^{at}\]Multiplying both sides of the differential equation by this integrating factor:\[e^{at} \frac{dv}{dt} + e^{at} av = -ge^{at}\]The left side can be rewritten as the derivative of a product:\[\frac{d}{dt}(e^{at} v) = -ge^{at}\]Integrate both sides with respect to \(t\):\[e^{at} v = -\frac{g}{a} e^{at} + C\]Solving for \(v\) gives:\[v = -\frac{g}{a} + Ce^{-at}\]Since \(v_{\infty} = -\frac{g}{a}\), we substitute and simplify:\[v = v_{\infty} + Ce^{-at}\]By letting \(v(0) = v_0\), determine \(C\):\[v_0 = v_{\infty} + C\]Solving for \(C\) gives:\[C = v_0 - v_{\infty}\]Thus, the velocity as a function of time is:\[v(t) = (v_0 - v_{\infty}) e^{-at} + v_{\infty}\]

Step 3: Find the Expression for Altitude

Given the velocity function, we can find the position \(y(t)\) by integrating the velocity:\[v(t) = (v_0 - v_{\infty}) e^{-at} + v_{\infty}\]The altitude is given by:\[y(t) = \int v(t) \, dt = \int ((v_0 - v_{\infty}) e^{-at} + v_{\infty}) \, dt\]Separate the integral:\[y(t) = (v_0 - v_{\infty}) \int e^{-at} \, dt + \int v_{\infty} \, dt\]Compute the integrals:\[\int e^{-at} \, dt = -\frac{1}{a} e^{-at}\]\[\int v_{\infty} \, dt = v_{\infty}t\]Therefore, substituting back:\[y(t) = (v_0 - v_{\infty}) \left(-\frac{1}{a} e^{-at}\right) + v_{\infty}t + y_0\]Simplifying gives:\[y(t) = y_0 + v_{\infty}t + \frac{(v_0 - v_{\infty})}{a}(1 - e^{-at})\]

Step 4: Conclusion

Both parts have been shown: (a) The velocity function \(v(t)\) is \((v_0 - v_{\infty}) e^{-at} + v_{\infty}\), with \(v_{\infty} = -\frac{g}{a}\).(b) The altitude \(y(t)\) is given by \(y_0 + v_{\infty}t + \frac{(v_0 - v_{\infty})}{a}(1 - e^{-at})\).

Key Concepts

Terminal Velocity

Terminal velocity is a fascinating concept in physics, particularly when studying the motion of objects through a fluid such as air. In the context of our differential equation scenario, terminal velocity, denoted as \(v_{\infty}\), is the velocity at which the force of gravity pulling a falling object down is exactly balanced by the force of air resistance pushing it up. This results in a net force of zero, allowing the object to reach a constant velocity.
In mathematical terms, terminal velocity can be found by setting the acceleration to zero in the differential equation describing the motion. This is described by:

• Equation: \(\displaystyle\frac{dv}{dt} = 0\)
• Solving for \(v\) in \(-g - av = 0\)

This yields the expression \(v_{\infty} = -\frac{g}{a}\).
Thus, terminal velocity depends on both the gravitational force \(g\) and the drag coefficient \(a\).

Air Resistance

Air resistance, also known as drag, is a crucial force that acts opposite to the relative motion of an object moving through the air. It's proportional to the velocity, which means the faster an object moves, the greater the air resistance it encounters.
In the exercise's differential equation, air resistance is represented by the term \(-av\), where \(a\) is the drag coefficient. This coefficient quantifies how much resistance an object experiences as it moves through the air.
The impact of air resistance is significant in reaching terminal velocity. Without it, an object would continue to accelerate under gravity, but due to air resistance:

• The velocity of the object gradually approaches a maximum limit.
• At this point, the object no longer accelerates.

This interaction between gravitational pull and air resistance is vital to understanding the motion of falling objects.

First-Order Linear Differential Equation

A first-order linear differential equation is a type of equation that involves the rate of change of a variable. These equations are fundamental in modeling dynamic systems in calculus and physics. In the context of the problem, the given differential equation is \(\frac{dv}{dt} = -g - av\).
This equation models how velocity \(v\) changes with respect to time \(t\). To solve such an equation, the form of the equation must be recognized as linear, which means it can be expressed as:

• General form: \(\frac{dy}{dt} + p(t)y = g(t)\)
• In our case: \(p(t) = a\) and \(g(t) = -g\).

Solving the equation involves finding a function \(v(t)\) that satisfies this relationship over time, incorporating initial conditions like \(v_0 = v(0)\). Understanding these equations is essential for modeling various real-world scenarios.

Integrating Factor

The integrating factor is a powerful mathematical tool used to solve first-order linear differential equations. It simplifies the process to find solutions by transforming the equation into one that is straightforward to integrate.
For the equation \(\frac{dv}{dt} + av = -g\), the integrating factor \(\mu(t)\) is calculated as:

• \(\mu(t) = e^{\int a \, dt} = e^{at}\)

Multiplying the entire differential equation by this factor yields an expression whose left-hand side can be rewritten as the derivative of a product:

• \(\frac{d}{dt}(e^{at}v) = -ge^{at}\)

This allows us to integrate both sides easily to find the function \(v(t)\), which represents velocity over time considering air resistance. The use of integrating factors is advantageous for simplifying complex equations, making them accessible and solvable.