Problem 79 - Thomas Calculus

Question

Tractor trailers and the tractrix When a tractor trailer turns into a cross street or driveway, its rear wheels follow a curve like the one shown here. (This is why the rear wheels sometimes ride up over the curb.) We can find an equation for the curve if we picture the rear wheels as a mass \(M\) at the point \((1,0)\) on the \(x\) -axis attached by a rod of unit length to a point \(P\) representing the cab at the origin. As the point \(P\) moves up the \(y\) -axis, it drags \(M\) along behind it. The curve traced by \(M-\) called a tractrix from the Latin word tractum, for "drag" - can be shown to be the graph of the function \(y=f(x)\) that solves the initial value problem $$ \begin{array}{ll}{\text { Differential equation: }} & {\frac{d y}{d x}=-\frac{1}{x \sqrt{1-x^{2}}}+\frac{x}{\sqrt{1-x^{2}}}} \\ {\text { Initial condition: }} & {y=0 \quad \text { when } \quad x=1}\end{array} $$ Solve the initial value problem to find an equation for the curve. (You need an inverse hyperbolic function.) Graph cannot copy

Step-by-Step Solution

Answer

The solution is a constant: \( y = 0 \).

Step 1: Identify the Differential Equation Form

The given differential equation is \( \frac{dy}{dx} = -\frac{1}{x \sqrt{1-x^{2}}} + \frac{x}{\sqrt{1-x^{2}}} \). This equation needs to be integrated to find \( y = f(x) \). Initially, we'll break down the integration and solve each part separately.

Step 2: Simplify and Separate the Terms

Rewrite the differential equation as the sum of two separate terms: \( \frac{dy}{dx} = -\frac{1}{x \sqrt{1-x^{2}}} + \frac{x}{\sqrt{1-x^{2}}} \). We'll integrate each term separately with respect to \( x \).

Step 3: Integrate the First Term

The first term is \( -\frac{1}{x \sqrt{1-x^{2}}} \). Its integral can be expressed with an inverse trigonometric function: \( \int -\frac{1}{x \sqrt{1-x^{2}}} \, dx = -\text{arcsinh}(x) \).

Step 4: Integrate the Second Term

The second term is \( \frac{x}{\sqrt{1-x^{2}}} \). Its integral is \( \int \frac{x}{\sqrt{1-x^{2}}} \, dx = \text{arcsinh}(x) \), which is the inverse hyperbolic sine function.

Step 5: Combine the Integrals

Now combine the results from the two integrals: \(-\text{arcsinh}(x) + \text{arcsinh}(x) + C = y \), which simplifies to \( C = y \).

Step 6: Apply the Initial Condition

Apply the initial condition \( y = 0 \) when \( x = 1 \). This gives \( C = 0 \). Therefore, the solution to the differential equation that satisfies the initial condition is \( y = 0 \).

Key Concepts

Tractrix

The tractrix is a fascinating curve with a unique history and boundary applications in real life, like the path followed by the rear wheels of a truck. The word "tractrix" comes from the Latin verb "tractus," meaning "to drag." Imagine a point or small mass attached to a rod, moving along a straight path and pulling the mass along behind it. This scenario perfectly describes the tractrix, as observed when a tractor trailer makes a turn.

The defining property of the tractrix is that the segment of each tangent line between the axis and the point of tangency has a constant length. This means if you were to draw a tangent to any point on the curve, the distance from the axis to the point where this tangent touches the curve remains the same, showing an elegant geometry in its path.

Initial Value Problem

An initial value problem (IVP) in mathematics refers to finding a function that solves a differential equation for given initial conditions. Essentially, it's a type of problem where one determines the course of a dynamic system given an initial state.

In our case, the IVP comprises a differential equation and a specific condition at a starting point, noted as:

• Differential equation: \( \frac{d y}{d x} = -\frac{1}{x \sqrt{1-x^{2}}} + \frac{x}{\sqrt{1-x^{2}}} \)
• Initial condition: \( y = 0 \) when \( x = 1 \)

Solving an initial value problem requires

• identifying the differential equation form
• integrating correctly by considering the given initial conditions

Such integration techniques clarify the path of motion for systems like our tractrix example, allowing the tracing of curves in real-world applications.

Inverse Hyperbolic Functions

Inverse hyperbolic functions play a crucial role in solving various types of integrals, particularly in contexts involving hyperbolic geometry. They are the inverse functions of hyperbolic sine, cosine, and tangent, commonly used in differential equations to express relationships.In the tractrix problem, inverse hyperbolic functions help integrate terms like \( \frac{1}{x \sqrt{1-x^{2}}} \) and \( \frac{x}{\sqrt{1-x^{2}}} \). Specifically:

• The integral of \( -\frac{1}{x \sqrt{1-x^{2}}} \) simplifies to \(-\text{arcsinh}(x)\)
• The integral of \( \frac{x}{\sqrt{1-x^{2}}} \) results in \(\text{arcsinh}(x)\)

These functions provide a powerful toolset when manipulating differential equations, assisting significantly in transforming complex relations into more comprehensible forms.

Integration Techniques

Integration is the core process of finding the antiderivative of a function, a key step in solving differential equations. The techniques required can vary significantly, depending on the form of the function.

Initially, you might apply straightforward methods such as basic antiderivative tables for standard functions. However, for more complex expressions, like in the tractrix equation, advanced techniques involving inverse trigonometric or hyperbolic functions are often used.

The problem included the following steps for separation of integrals:

• Dividing the differential equation into simpler parts: \( -\frac{1}{x \sqrt{1-x^{2}}} + \frac{x}{\sqrt{1-x^{2}}} \)
• Solving them individually: integrating terms by recalling inverse hyperbolic functions

Mastery over integration techniques becomes vital in calculating real-world applications, helping in reconstructing intricate motions like the tractrix.