Problem 8 - Thomas Calculus

Question

Give parametric equations and parameter intervals for the motion of a particle in the \(x y\) -plane. Identify the particle's path by finding a Cartesian equation for it. Graph the Cartesian equation. (The graphs will vary with the equation used.) Indicate the portion of the graph traced by the particle and the direction of motion. \(x=\csc t, \quad y=\cot t ; \quad 0< t<\pi\)

Step-by-Step Solution

Answer

The Cartesian equation is \(x^2 - y^2 = 1\), a hyperbola.

Step 1: Determine Parametric Equations

The given parametric equations for the motion of the particle in the \(xy\)-plane are: \[ x(t) = \csc(t) \] \[ y(t) = \cot(t) \] The parameter interval is \(0 < t < \pi\).

Step 2: Express \(x\) and \(y\) in terms of \(\sin(t)\) and \(\cos(t)\)

Recall that \(\csc(t) = \frac{1}{\sin(t)}\) and \(\cot(t) = \frac{\cos(t)}{\sin(t)}\). Thus, we have: \[ x = \frac{1}{\sin(t)} \] \[ y = \frac{\cos(t)}{\sin(t)} \]

Step 3: Derive the Cartesian Equation

Replace \(\sin(t)\) and \(\cos(t)\) in terms of \(x\) and \(y\). Since \(x = \frac{1}{\sin(t)}\), we can say \(\sin(t) = \frac{1}{x}\). Substituting this in \(y = \frac{\cos(t)}{\sin(t)}\): \[ y = \cos(t) \cdot x \] From \(\sin^2(t) + \cos^2(t) = 1\), replace \(\sin(t)\) as \(\frac{1}{x}\). \[ \left(\frac{1}{x}\right)^2 + \cos^2(t) = 1 \Rightarrow \cos^2(t) = 1 - \left(\frac{1}{x}\right)^2 \]Thus, \(\cos(t) = \sqrt{1 - \frac{1}{x^2}}\) leading to \[ y = x \cdot \sqrt{1 - \frac{1}{x^2}} \]

Step 4: Simplify the Cartesian Equation

To simplify, note that \(y^2 = x^2(1 - \frac{1}{x^2}) = x^2 - 1\). Thus, \[ y^2 = x^2 - 1 \] The Cartesian equation of the path is \(x^2 - y^2 = 1\), which is a hyperbola.

Step 5: Graph the Cartesian Equation and Indicate Direction

Graph the hyperbola given by \(x^2 - y^2 = 1\). Note that this equation represents a hyperbola centered at the origin with the transverse axis along the x-axis. Since \(x = \csc(t)\) and challenges with \(t\) expand from \(0\) to \(\pi\), the particle will trace the right branch from \(x = +\infty\) to \(x = 1\), and then it moves downwards and leftwards, following the block from \(x = 1\) to \(x = -1\).

Key Concepts

Cartesian Equation

Converting parametric equations to a Cartesian equation helps in visualizing the path of a particle on the xy-plane. In our exercise, we start with the given parametric equations:

• \[ x = \csc(t) \]
• \[ y = \cot(t) \]

To find the Cartesian equation, we express these equations in terms of sine and cosine:

• \(\csc(t) = \frac{1}{\sin(t)}\)
• \(\cot(t) = \frac{\cos(t)}{\sin(t)}\)

Simplifying further, using the trigonometric identity, \(\sin^2(t) + \cos^2(t) = 1\), helps us to derive:\[ y^2 = x^2 - 1 \]This equation, \(x^2 - y^2 = 1\), is a standard form of a hyperbola. This Cartesian equation elegantly illustrates the particle's path.

Particle Motion

Understanding the motion of a particle is essential in connecting the parametric and Cartesian representations of its path. The particle in our problem moves in the xy-plane, dictated by the parametric equations:

• \(x = \csc(t)\)
• \(y = \cot(t)\)

The interval \(0 < t < \pi\) allows the particle to traverse along a particular path. Here, each value of \(t\) defines a unique point \((x,y)\) on this path. As \(t\) changes, the sine and cosine components reflect the specific coordinates of the particle's position. Comprehending parametric equations aids in predicting where the particle is at any time \(t\), based on its dependencies on trigonometric functions.

Hyperbola

A hyperbola is a type of conic section represented by the Cartesian equation \(x^2 - y^2 = 1\). In our case, the hyperbola describes the path traced by the particle as it moves.This particular hyperbola is notable for its transverse axis along the x-axis, differently from ellipses which are oval in shape. The hyperbola has two branches, commonly understood as mirror images about the center.Our equation indicates the hyperbola is centered at the origin, and this unique path allows for significant insights into the symmetry and shape of the particle's motion. Hyperbolas are not only seen in geometry but also appear in physics and engineering scenarios.

Direction of Motion

The direction of motion of a particle along a path can determine critical information about its behavior. Here, the parametric time interval \(0 < t < \pi\) specifies how the particle traces the hyperbola.Initially, as \(t\) approaches zero, the value of \(x\) starts from positive infinity and moves toward 1. Simultaneously, \(y\) decreases as \(t\) increases. After reaching \(x = 1\), the particle direction shifts: \(x\) then decreases, and \(y\) continues decreasing, taking the particle leftwards and downwards.Thus, the trace goes from positive to negative along the x-axis, providing evidence of the dynamic movement from one end of the hyperbola to the other.