Problem 34 - A Graphical Approach to Precalculus with Limits

Question

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

Step-by-Step Solution

Answer

The rectangular equation is \(x = y^2\) for \(x > 0\).

Step 1: Understand the given parametric equations

We are given the parametric equations for a curve, \( x = e^{2t} \) and \( y = e^{t} \). These equations express \(x\) and \(y\) in terms of the parameter \(t\), and our goal is to eliminate \(t\) to find an equation solely in terms of \(x\) and \(y\).

Step 2: Express t in terms of y

From the equation \(y = e^{t}\), we need to solve for \(t\). Take the natural logarithm of both sides: \( t = \ln y \). This allows \(t\) to be expressed in terms of \(y\).

Step 3: Substitute t in the equation for x

Substitute \(t = \ln y\) from Step 2 into the equation \(x = e^{2t}\). This gives us \(x = e^{2(\ln y)}\). Using the property of exponents, \(e^{2\ln y} = (e^{\ln y})^2 = y^2\). Hence, we obtain the equation \(x = y^2\).

Step 4: Determine the appropriate interval

Since \(t\) can range from \(-\infty\) to \(\infty\), and \(y = e^t\) implies \(y > 0\), the appropriate interval for \(y\) corresponds to \(y > 0\). Substituting \(y\) in \(x = y^2\), we see that \(x\) can take any positive value as \(y \to \infty\). Hence the interval for \(x\) is \((0, \infty)\).

Key Concepts

Rectangular Equation

In mathematics, rectangular equations refer to expressions that relate the coordinates \(x\) and \(y\) directly, without involving any parameters like \(t\). These are typically the type of equations used in Cartesian coordinate systems.
To convert a parametric equation into a rectangular one, you must eliminate the parameter. Generally, this involves solving one of the parametric equations for the parameter and substituting it back into the other equation.
In this case, our parametric equations are \(x = e^{2t}\) and \(y = e^{t}\). By expressing \(t\) from one equation, such as \(y = e^{t}\), and substituting it into \(x = e^{2t}\), we find \(x = y^2\).
The resulting rectangular equation \(x = y^2\) describes a parabola that opens to the right in the Cartesian plane. This conversion from parametric to rectangular form is crucial for a deeper analysis and understanding of the curve's properties.

Natural Logarithm

The natural logarithm, denoted \(\ln\), is a mathematical function closely related to the exponential function. It is the inverse of the exponential function \(e^x\).
In simpler terms, if \(y = e^t\), then \(t = \ln(y)\). The natural logarithm helps convert expressions involving exponents into simpler additive forms.
In our exercise, we used \(\ln\) to express \(t\) in terms of \(y\). By taking the natural logarithm of both sides of \(y = e^t\), we derive \(t = \ln y\). This step is essential to solve the parametric equations by removing the parameter and finding a rectangular form.
The ability to invert growth offered by exponentials is incredibly useful in solving equations and simplifying complex mathematical problems.

Exponents

Exponents are mathematical notations indicating how many times a number, known as the base, is used as a factor. For instance, \(e^{2t}\) in our parametric equations indicates that the base \(e\) is multiplied by itself repeatedly according to the exponent's value.
Exponential expressions are often involved in growth and decay processes, capturing the essence of increasing or decreasing quantities over time.
In the given problem, understanding the exponent rules allows us to simplify the expression \(x = e^{2(\ln y)}\). Using the property \(e^{2\ln y} = (e^{\ln y})^2 = y^2\), we transformed the expression into \(x = y^2\).
This transformation highlights how exponent properties like \(e^{a\ln b} = b^a\) can be applied to manage and rearrange expressions effectively.

Interval of a Function

An interval in mathematics defines a specific range of values for which an expression or function is valid. In the context of parametric equations, the interval typically corresponds to the parameter that generates the curve.
For the original parametric function \(y = e^t\), the interval of \(t\) is \((-\infty, \infty)\), which affects the interval of \(y\). Since \(e^t > 0\) for all real numbers, \(y > 0\). This gives the interval for \(y\) as \((0, \infty)\).
Consequently, as \(x = y^2\) for \(y > 0\), the interval for \(x\) is \((0, \infty)\) because squaring a positive \(y\) yields positive \(x\).
Understanding the intervals helps us determine the valid range of the function and restrict graph plotting to applicable regions, ensuring realistic mathematical interpretation.