Problem 28 - Contemporary Precalculus

Question

The given curve is part of the graph of an equation in \(x\) and \(y .\) Find the equation by eliminating the parameter. $$x=2 e^{t}, \quad y=1-e^{t}, \quad t \geq 0$$

Step-by-Step Solution

Answer

Question: Find the equation of the curve given the parametric equations \(x = 2e^t\) and \(y = 1 - e^t\). Answer: The equation of the curve is \(y = 1 - \frac{x}{2}\).

Step 1: Solve one equation for the parameter t

We have the equations \(x = 2e^t\) and \(y = 1 - e^t\). Let's solve the first equation for t: $$t = \ln(\frac{x}{2})$$

Step 2: Substitute the value of t in second equation

Now, substitute the value of t we found in step 1 into the second equation: $$y = 1 - e^{\ln(\frac{x}{2})}$$

Step 3: Simplify the equation

Using the property \(e^{\ln a}=a\), we can simplify the equation to: $$y = 1 - \frac{x}{2}$$

Step 4: Write the final equation

We have successfully eliminated the parameter t and found the relationship between x and y: $$y = 1 - \frac{x}{2}$$ This is the equation of the given curve.

Key Concepts

Eliminating the Parameter

When dealing with parametric equations like \(x = 2e^t\) and \(y = 1 - e^t\), the goal of eliminating the parameter is to express \(y\) directly in terms of \(x\), removing \(t\). This process simplifies the relationship between \(x\) and \(y\) from a parametric to a Cartesian form.

To commence, solve one of the equations for the parameter \(t\). Here, we solve for \(t\) in terms of \(x\) by rearranging \(x = 2e^t\) as \(t = \ln(\frac{x}{2})\). By substituting this expression for \(t\) into the second equation, \(y = 1 - e^t\), we can eliminate \(t\).

This approach is useful because it allows you to go from describing a curve based on a specific parameter to obtaining a more standard equation form, which is often easier to analyze and graph. It's crucial in calculus, plotting, and understanding the contour of the curves more clearly.

Exponential Functions

Exponential functions are a core component of parametric equations like the ones given: \(x = 2e^t\) and \(y = 1 - e^t\). These functions have the general form \(a^x\) or \(e^x\), where \(e\) is the base of natural logarithms (approximately equal to 2.71828).

In this context, \(e^t\) grows rapidly as \(t\) increases, illustrating how these functions model real-world phenomena, such as growth and decay processes. The parameter \(t\), often represents time, demonstrating how variables change across moments.

Understanding how exponential functions work is crucial. For example, the property \(e^{\ln a} = a\) helps tremendously when simplifying expressions after eliminating the parameter. Recognizing this property allows one to simplify expressions like \(e^{\ln \left(\frac{x}{2}\right)}\) directly to \(\frac{x}{2}\), yielding more manageable equations.

Logarithmic Functions

Logarithmic functions are the inverse of exponential functions and are vital tools for solving equations involving exponential growth. The natural logarithm, \(\ln x\), is especially important because it relates directly to the base \(e\) used in the exponential functions.

To eliminate the parameter in our exercise, we used the natural logarithm to solve for \(t\) in terms of \(x\), reformulating \(x = 2e^t\) to \(t = \ln\left(\frac{x}{2}\right)\). This crucial step employs the understanding that the logarithm and exponential functions are inverses, simplifying the procedure significantly.

By calling upon logarithmic properties, we make solving otherwise complex exponential equations more straightforward. Logarithms provide a way to handle exponential functions' rapid growth, bringing the expressions into a manageable scale and allowing equations between variables to be expressed directly and clearly.