Problem 32 - A Graphical Approach to Precalculus with Limits

Question

Find a rectangular equation. State the appropriate interval for \(x\) or \(y .\) $$x=\sqrt{t}, y=t^{2}-1, \text { for } t \text { in }[0, \infty)$$

Step-by-Step Solution

Answer

Rectangular equation: \( y = x^4 - 1 \) for \( x \geq 0 \).

Step 1: Identify Relations

The given parametric equations are: \( x = \sqrt{t} \) and \( y = t^2 - 1 \). We need to express \( y \) as a function of \( x \) to find the rectangular equation.

Step 2: Express \( t \) in Terms of \( x \)

From \( x = \sqrt{t} \), we can square both sides to remove the square root: \( x^2 = t \).

Step 3: Substitute \( t \) in the Equation for \( y \)

Substitute \( x^2 = t \) into \( y = t^2 - 1 \): \[ y = (x^2)^2 - 1 \] Simplifying gives: \[ y = x^4 - 1 \].

Step 4: Determine the Interval for \( x \)

Since \( t = x^2 \) and \( t \) is defined for \([0, \infty)\), \( x \) must satisfy \( \sqrt{0} \leq x < \infty \). Therefore, \( x \geq 0 \).

Key Concepts

Parametric Equations

When working with parametric equations, each variable is expressed as a function of one or more independent parameters. In our given exercise, the equations are:

• \( x = \sqrt{t} \)
• \( y = t^2 - 1 \)

Here, both \( x \) and \( y \) are functions of the parameter \( t \). Often in problems like this, the goal is to remove the parameter to find a direct relationship between \( x \) and \( y \), known as a rectangular equation.
This involves expressing \( y \) directly as a function of \( x \) by eliminating the parameter \( t \), leading to a more straightforward equation that uses just the Cartesian coordinates \( x \) and \( y \).
This conversion provides a clearer view of the path or curve described by the equations without explicitly involving the parameter.

Interval Notation

Interval notation is a crucial part of understanding the range of values a variable can take. In our exercise, the parameter \( t \) is defined in the interval \([0, \infty)\), meaning \( t \) can be any non-negative value but not infinity itself.
This translates into understanding the permissible values of \( x \) once \( t \) is expressed in terms of \( x \).
After substituting \( t \) with \( x^2 \) to form the rectangular equation, we determine that \( x \) also follows the interval \([0, \infty)\), as the square root operation requires non-negative inputs.

• The notation \([0, \infty)\) represents all real numbers starting from 0 up to, but not including, infinity.
• This helps define the domain of functions when solving mathematical expressions.

Understanding such notations is essential for accurately describing the range over which a function or equation is valid.

Function of a Variable

A function of a variable expresses one quantity directly in terms of another. In the context of this exercise, this means rewriting equations to eliminate the parameter \( t \) and showing \( y \) as a function of \( x \).
Initially, we had:

• \( x = \sqrt{t} \)
• \( y = t^2 - 1 \)

By identifying \( t = x^2 \), we substitutely formed the rectangular equation:
\( y = x^4 - 1 \).
This expression shows \( y \) directly as a function of \( x \), offering a simpler representation of the relationship between \( x \) and \( y \) without the intermediate parameter.

• Such expressions help to analyze how output variables depend on input variables.
• This helps us understand the behavior of equations in a more intuitive manner.

Recognizing functions in variables is key to solving real-world problems where relationships between variables are involved.