Problem 35 - A Graphical Approach to Precalculus with Limits

Question

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

Step-by-Step Solution

Answer

The rectangular equation is \(y = 1 - 2x^2\); \(x\) is in \((-\infty, 0) \cup (0, \infty)\).

Step 1: Express t in terms of x

Start by addressing the expression given for \(x\): \[x = \frac{1}{\sqrt{t+2}}\]. By isolating \(\sqrt{t+2}\), we have \(\sqrt{t+2} = \frac{1}{x}\). Squaring both sides, \(t+2 = \frac{1}{x^2}\). Finally, solve for \(t\) to get \(t = \frac{1}{x^2} - 2\).

Step 2: Substitute t into y's expression

Now that we have \(t\) in terms of \(x\), we substitute it into the expression for \(y\). The expression for \(y\) is \(y = \frac{t}{t+2}\). Substitute \(t = \frac{1}{x^2} - 2\) into this expression, giving us:\[y = \frac{\frac{1}{x^2} - 2}{\left(\frac{1}{x^2} - 2\right) + 2}\].

Step 3: Simplify the expression for y

Simplify the expression found during substitution. The denominator simplifies as follows:\[t + 2 = \left(\frac{1}{x^2} - 2\right) + 2 = \frac{1}{x^2}\].Now substitute this into the expression for \(y\):\[y = \frac{\frac{1}{x^2} - 2}{\frac{1}{x^2}}\]. Expand and simplify the equation:\[y = \frac{1}{x^2} \cdot \frac{1}{x^2} - 2 \cdot \frac{1}{x^2}\].After simplifying, we obtain:\[y = 1 - 2x^2\].

Step 4: Determine the interval for x

Recall that \(t\) is in \((-2, \infty)\). Given \(t = \frac{1}{x^2} - 2\), the inequality \(-2 < \frac{1}{x^2} - 2\) must hold. Solve \(\frac{1}{x^2} - 2 > -2\) to find \(\frac{1}{x^2} > 0\), which implies that \(x\) cannot be 0. Therefore, \(x\) is in the intervals \((-rac{1}{\sqrt{0}}, 0) \cup (0, \frac{1}{\sqrt{0}})\), simplifying to \((-\infty, 0) \cup (0, \infty)\).

Key Concepts

Interval Notation

Interval notation is a mathematical method of denoting a range of values, typically for numbers like the variable in an equation. It employs the use of brackets and parentheses to indicate whether endpoints are included or excluded from the interval.

• Parentheses "( )" are used to exclude an endpoint.
• Brackets "[ ]" are used to include an endpoint.

A crucial part of solving problems with variable constraints is determining the correct interval. When dealing with the parametric equations, interval notation helps specify valid ranges for which the equations apply. In our exercise, for example, the interval for the variable \(x\) derived from the parametric equations must not include zero, leading to a solution in the intervals:

• \(x \in (-\infty, 0) \cup (0, \infty)\)

This means \(x\) can take any real number value except zero, providing clear boundaries for the solution.

Parametric Equations

Parametric equations involve expressing variables using a third parameter, usually \(t\). In these equations, rather than having a single function-type relationship between variables \(x\) and \(y\), both \(x\) and \(y\) are expressed in terms of \(t\).
This provides flexibility and power in describing curves that aren't functions of \(x\) or \(y\) alone.
In the problem example, our parametric equations were given as:

• \(x = \frac{1}{\sqrt{t+2}}\)
• \(y = \frac{t}{t+2}\)

These expressions are tied to the parameter \(t\), inviting a comprehensive way to analyze relationships between \(x\), \(y\), and \(t\). When these parametric equations are used, we can explore the behavior of a point \((x, y)\) as \(t\) varies within a defined interval.

Algebraic Manipulation

Algebraic manipulation refers to the process of rewriting mathematical expressions in different forms to reveal simpler interpretations or to solve them. It's a crucial skill for mathematics and involves operations like factoring, expanding, and simplifying.
In this exercise, you leverage algebraic manipulation to express \(t\) in terms of \(x\) from the parametric form to solve for \(y\). The key manipulation involved isolating \(\sqrt{t+2}\) before squaring both sides to achieve:
\[ \sqrt{t+2} = \frac{1}{x} \quad \Rightarrow \quad t + 2 = \frac{1}{x^2} \quad \Rightarrow \quad t = \frac{1}{x^2} - 2 \]This result is then substituted back into the expression for \(y\) to facilitate solving the rectangular equation, showing how transformations and algebraic strategies simplify complex parametric forms.

Simplification Steps

Simplifying expressions involves reducing them to their most compact and understandable form. Here’s how simplification helps in our exercise:
Initially, the parametric form of \(y\) needed substitution:
\[ y = \frac{\frac{1}{x^2} - 2}{\frac{1}{x^2}} \]
This was simplified through performing numerator and denominator operations:

• The denominator \(t+2\) simplifies as \(\frac{1}{x^2}\).
• The equation then becomes: \[ y = \frac{\frac{1}{x^2} - 2}{\frac{1}{x^2}} \]
• Expanding and simplifying to: \[ y = 1 - 2x^2 \]

Every part of the simplification process brings us to a clearer picture of how \(x\) and \(y\) relate, ensuring the mathematical description is as direct and concise as possible.