Problem 40 - Essential Calculus

Question

Find two different sets of parametric equations for the rectangular equation. $$ y=\frac{2}{x-1} $$

Step-by-Step Solution

Answer

The two sets of parametric equations for the given rectangular equation are (1) \(x = 1 + t\), \(y = \frac{2}{t}\) and (2) \(x = p + 1\), \(y = \frac{2}{p}\).

Step 1: Choose the First Parameter

Consider \(t\) as \(x-1\). This choice comes from the denominator of the equation \(y=\frac{2}{x-1}\). Then, \(x=1+t\). Substitute for \(x\) into the equation to get \(y = \frac{2}{t}\). So, the first parametric set of equations is \(x = 1 + t\) and \(y = \frac{2}{t}\).

Step 2: Choose the Second Parameter

Let's choose another parameter, for instance \(p = x-1\). Then, \(x = p + 1\). Substituting for \(x\) into the equation gives \(y = \frac{2}{p}\). Consequently, the second parametric set of equations is \(x = p + 1\) and \(y = \frac{2}{p}\).

Key Concepts

Rectangular Equation

In algebra and coordinate geometry, a rectangular equation is the term given to an equation representing a curve in which the coordinates of the points on the curve are given by a single equation in two variables. For instance, in the given problem, the rectangular equation is \( y=\frac{2}{x-1} \). This type of equation is known as rectangular because it uses the conventional Cartesian coordinates, which create a rectangle-like grid pattern.

To convert from a rectangular equation to parametric equations, which express the coordinates as functions of a third variable, typically known as a parameter, we can manipulate the equation algebraically to isolate the independent variable (in this case \(x\)) and represent \(y\)) in terms of this new parameter. The reason for creating parametric equations is to simplify certain calculus operations, like taking derivatives or integrals, as they often make these procedures more straightforward.

Calculus

Calculus is a field of mathematics that studies rates of change (differential calculus) and accumulation of quantities (integral calculus). It's an essential tool in various scientific disciplines. When we deal with a rectangular equation like \(y=\frac{2}{x-1}\), calculus helps in finding slopes, areas, and volumes related to the curve defined by the equation.

In the context of the given exercise, calculus might be employed to find the derivative or the integral of \(y\)) with respect to \(x\) for the curve. However, the process can be more intuitive if the equation is expressed in parametric form, because we can work with the derivatives and integrals of single-variable functions that describe the curve, rather than implicitly defined multi-variable functions in the rectangular form. This direct approach can simplify the task immensely and is a key reason for converting equations to parametric form.

Algebraic Manipulation

Algebraic manipulation involves rearranging and transforming equations to simplify them or to put them in a more useful form. In the problem provided, the step-by-step solution illustrates algebraic manipulation by introducing a parameter to transform the rectangular equation into a system of parametric equations. This process starts by selecting an appropriate substitution that will make subsequent calculations simpler.

For example, in the solution we define a new variable \(t\) as \(x-1\), which simplifies the equation to \(y = \frac{2}{t}\). Then, expressing \(x\) in terms of \(t\) as \(x=1+t\) completes the transformation to parametric form. The choice of parameter is somewhat arbitrary, but wise choices, as shown in the solution, can simplify the expression of the curve. Thus, algebraic manipulation is pivotal in the transition from one equation form to another, enabling deeper insights into the behavior of the equations and their corresponding graphs.