Question

Sketch the parametric equations by eliminating the parameter. Indicate any asymptotes of the graph. $$ x=e^{t}, \quad y=e^{2 t}+1 $$

Short answer

The graph is a parabola \(y = x^2 + 1\), with no asymptotes.

Step-by-step solution

Express x in terms of t

Starting with the parametric equation for \(x\): \(x = e^t\). To eliminate the parameter \(t\), we need to express \(t\) in terms of \(x\). By taking the natural logarithm on both sides, we obtain \(t = \ln x\).

Substitute t in terms of x into y equation

With \(t\) expressed as \(\ln x\), substitute it into the equation for \(y\): \(y = e^{2t} + 1\). This becomes \(y = e^{2\ln x} + 1 = (e^{\ln x})^2 + 1 = x^2 + 1\).

Write the Cartesian form of the equation

The previous step gives us the Cartesian form of the parametric equations: \(y = x^2 + 1\).

Identify asymptotes of the equation

The equation \(y = x^2 + 1\) is a parabola opening upwards with its vertex at (0, 1). This graph does not have any asymptotes, as it extends indefinitely in both the positive and negative directions of the x-axis and always has positive values.

Key concepts

Eliminating the Parameter

When dealing with parametric equations, one of the common tasks is to convert them into a Cartesian form, which is a familiar format using only the variables \(x\) and \(y\). This process involves eliminating the parameter, which is often represented by \(t\) or another variable. Let’s break down the steps using the provided parametric equations:

• First, identify the equations: \(x = e^t\) and \(y = e^{2t} + 1\).
• The goal is to remove \(t\) and express \(y\) directly in terms of \(x\).
• To start, solve the first equation for \(t\) by applying the natural logarithm: \(t = \ln x\).
• Substitute \(t = \ln x\) back into the second equation for \(y\): \(y = e^{2t} + 1\).
• This becomes \(y = (e^{\ln x})^2 + 1 = x^2 + 1\), eliminating the parameter completely and providing the Cartesian form.

The result is a single equation expressing \(y\) in terms of \(x\), known as the Cartesian form.

Asymptotes

Asymptotes are lines that a graph approaches but never actually touches. They are common in rational and exponential graph equations but not all graphs have them. When analyzing the provided equation after parameter elimination, such as the Cartesian form \(y = x^2 + 1\), it's important to identify if asymptotes are present.For this particular equation:

• It's a parabola, and parabolas have a distinctive 'U' shape.
• They extend indefinitely and do not approach a straight line specifically, even at infinity.
• This means that \(y = x^2 + 1\) does not possess any asymptotes.
• The graph is open upwards and shifts one unit up in the \(y\)-axis, emerging from its vertex at (0, 1).

Thus, when determining asymptotes, understanding the overall behavior and equation type is crucial.

Cartesian Form

Cartesian form refers to an equation expressed in terms of \(x\) and \(y\), without any other parameters. It’s often more convenient for graphing and analysis due to its simplicity and familiarity. Here’s why converting parametric equations to Cartesian form is so valuable:

• Allows for easier visualization on a standard Cartesian coordinate plane.
• With the given example, \(y = x^2 + 1\) directly gives insights into the graph structure: a parabola.
• Helps in analyzing and predicting geometric features of the graph like intercepts, vertices, and axes of symmetry.
• Makes identifying features like vertex positions much clearer; here, it sits at (0, 1) with \(x = 0\) and \(y = 1\).

The Cartesian form simplifies the comprehension of the graph, providing a complete picture of the relationship between \(x\) and \(y\) without any extra steps.