Problem 50 - Calculus: Early Transcendentals

Question

An object of mass \( m \) is moving horizontally through a medium which resists the motion with a force that is a function of the velocity; that is, \( m \frac {d^2s}{dt^2} = m \frac {dv}{dt} = f(v) \) where \( v = v(t) \) and \( s = s(t) \) represent the velocity and position of the object at time \( t, \) respectively. For example, think of a boat moving through the water. (a) Suppose that the resisting force is proportional to the velocity, that is \( f(v) = -kv, k \) a positive constant. (This model is appropriate for small values of \( v. \)) Let \( v(0) = v_0 \) and \( s \) at any time \( t. \) What is the total distance that the object travels from time \( t = 0? \) (b) For larger values of \( v \) a better model is obtained by supposing that the resisting force is proportional to the square of the velocity, that is, \( f(v) = kv^2, k > 0\. \) (This model was first proposed by Newton.) Let \( v_0 \) and \( s_0 \) be the initial values of \( v \) and \( s. \) Determine \( v \) and \( s \) at any time \( t. \) What is the total distance that the object travels in this case?

Step-by-Step Solution

Answer

(a) The total distance is \( \frac{mv_0}{k} \). (b) The distance is infinite.

Step 1: Set up the differential equation for Case (a)

In this first case, the resisting force is proportional to velocity, i.e., \( f(v) = -kv \). The equation \( m \frac{dv}{dt} = f(v) \) becomes \( m \frac{dv}{dt} = -kv \). This simplifies to \( \frac{dv}{dt} = -\frac{k}{m}v \).

Step 2: Solve the differential equation for velocity (Case a)

This is a separable differential equation. We separate variables and integrate: \[ \int \frac{1}{v} dv = -\frac{k}{m} \int dt \]After integration, we get:\[ \ln |v| = -\frac{k}{m}t + C \]Exponentiating both sides gives \( v(t) = C' e^{-\frac{k}{m}t} \). Using the initial condition \( v(0) = v_0 \), we find \( C' = v_0 \). Thus, \( v(t) = v_0 e^{-\frac{k}{m}t} \).

Step 3: Determine position \( s(t) \) for Case (a)

We integrate the velocity function to find position function \( s(t) \):\[ s(t) = \int v(t) \, dt = \int v_0 e^{-\frac{k}{m}t} \, dt \]This results in:\[ s(t) = -\frac{mv_0}{k} e^{-\frac{k}{m}t} + C'' \]Using the initial condition \( s(0) = s_0 \), we solve for \( C'' \) and find that \( C'' = s_0 + \frac{mv_0}{k} \). Thus, \( s(t) = -\frac{mv_0}{k} e^{-\frac{k}{m}t} + s_0 + \frac{mv_0}{k} \).

Step 4: Determine total distance traveled for Case (a)

As \( t \to \infty \), \( v(t) \to 0 \), meaning the motion stops. The total distance is the limit as \( t \to \infty \) of \( s(t) - s(0) \). Thus, the total distance traveled is:\[ \frac{mv_0}{k} \].

Step 5: Set up the differential equation for Case (b)

In the second case, the resisting force is proportional to the square of the velocity: \( f(v) = -kv^2 \). The equation \( m \frac{dv}{dt} = f(v) \) becomes \( m \frac{dv}{dt} = -kv^2 \), or \( \frac{dv}{dt} = -\frac{k}{m}v^2 \).

Step 6: Solve the differential equation for velocity (Case b)

This is also a separable differential equation. We separate variables and integrate:\[ \int \frac{1}{v^2} dv = -\frac{k}{m} \int dt \]After integration, we get:\[ -\frac{1}{v} = -\frac{k}{m}t + C \]Solving for \( v(t) \), we have \( v(t) = \frac{1}{C + \frac{k}{m}t} \). Using \( v(0) = v_0 \), we find \( C = \frac{1}{v_0} \). So, \( v(t) = \frac{v_0}{1 + \frac{kv_0}{m}t} \).

Step 7: Determine position \( s(t) \) for Case (b)

To find \( s(t) \), integrate \( v(t) \):\[ s(t) = \int v(t) \, dt = \int \frac{v_0}{1 + \frac{kv_0}{m}t} \, dt \]The result is:\[ s(t) = \frac{m}{k} \ln\left(1 + \frac{kv_0}{m}t\right) + s_0 \].

Step 8: Determine total distance traveled for Case (b)

As \( t \to \infty \), \( v(t) \to 0 \), meaning the object stops. The total distance is the limit of \( s(t) - s(0) \), which approaches infinity. The object continues to travel infinitely as time increases, never fully coming to a stop.

Key Concepts

Resisting Force

In the study of motion, a resisting force is a force that works against the motion of an object. It is often caused by friction or drag from the medium the object is moving through, such as air or water. In our exercise, we consider two specific cases:

• When the resisting force is proportional to velocity: This means the force depends linearly on the velocity of the object. For small speeds, this is a practical assumption, as seen in equations where the force \( f(v) = -kv \).
• When the resisting force is proportional to the square of velocity: This case is more realistic at higher speeds. The resisting force in this scenario is described by \( f(v) = -kv^2 \), a more complex relationship first proposed by Newton.

Understanding the resisting force is essential in physics because it helps predict how an object's motion will change over time. By quantifying it, we calculate the actual path and stopping distance of moving objects like boats or cars.

Velocity

Velocity refers to the speed of an object in a specific direction. In differential equations, velocity is often a function of time \( v(t) \), illustrating how an object speeds up or slows down.

For our exercise, two scenarios are given where resisting force affects velocity differently:

• Case (a): With a linear resisting force, the equation simplifies to help find velocity as an exponential decay process: \( v(t) = v_0 e^{-\frac{k}{m}t} \). This equation tells us that as time goes on, velocity decreases towards zero.
• Case (b): With a quadratic resisting force, velocity decreases more swiftly, described by \( v(t) = \frac{v_0}{1 + \frac{kv_0}{m}t} \). Here, velocity decreases with a hyperbolic function, signifying a faster slowdown with time.

Precise velocity calculations are crucial in engineering applications, enabling accurate designs and safety assessments for moving systems.

Separable Differential Equations

Separable differential equations are a type of equation that can be rewritten so that each of the two variables occurs on a separate side of the equation. This separation allows for straightforward integration to solve the equation. In our problem:

• For Case (a) with resistive force proportional to velocity, we have \( \frac{dv}{dt} = -\frac{k}{m}v \), which can be rearranged to \( \frac{1}{v} dv = -\frac{k}{m} dt \). After integrating both sides, we find the velocity as a function of time.
• For Case (b) with force proportional to the square of velocity, the differential equation \( \frac{dv}{dt} = -\frac{k}{m}v^2 \) becomes \( \frac{1}{v^2} dv = -\frac{k}{m} dt \), simplifying integration.

Separable differential equations are common in physics, making them valuable tools for solving real-world problems involving rate changes over time. They offer clear solutions once variables are efficiently separated and integrated.

Integration by Parts

Integration by Parts is a method used to integrate products of functions. It is particularly useful when functions are multiplied together, and simple integration doesn't suffice. The formula for integration by parts is given by:\[ \int u \, dv = uv - \int v \, du \]While our exercise does not explicitly require integration by parts, understanding it can be key when dealing with more complex integrated functions. Here, both problem scenarios utilize simple integration due to the separability of the differential equations, using different methods than integration by parts.

However, in advanced settings, integration by parts might be necessary when confronted with non-linear systems or more sophisticated equations involving products where separating variables is not possible. This expands the toolkit of methods one may use to solve various types of integrals in differential equations.