Problem 30 - Precalculus

Question

(A) sketch the curve represented by the parametric equations (indicate the orientation of the curve) and (b) eliminate the parameter and write the resulting rectangular equation whose graph represents the curve. Adjust the domain of the rectangular equation, if necessary. $$\begin{aligned} &x=e^{t}\\\ &y=e^{3 t} \end{aligned}$$

Step-by-Step Solution

Answer

The curve represented by the parametric equations \(x = e^{t}\) and \(y = e^{3t}\) is a rightward opening parabola. After eliminating the parameter, the resulting rectangular equation is \(y = x^{3}\) with the domain \(x > 0\).

Step 1: Identifying the Equations

Recognize that \(x = e^{t}\) and \(y = e^{3t}\). These are parametric equations depending on parameter \(t\). The curve is oriented in increasing direction of \(t\) (from left to right).

Step 2: Sketching the Curve

We sketch the curve for different values of \(t\). For \(t = -2, -1, 0, 1, 2\), we find the corresponding coordinates using the parametric equations. As you plug in these values into the parametric equations, you see that the curve is curved upward. The curve is oriented from left to right.

Step 3: Eliminating the Parameter

By manipulating the equations, we are trying to express \(y\) in terms of \(x\). Therefore, \(y\) could be written as \(x^{3}\), since from our parametric equations when we express \(t\) and substitute into \(y\), we find that \(y = (e^{t})^{3} = x^{3}\).

Step 4: Adjusting Domain of Rectangular Equation

The original parametric equation doesn't restrict \(t\) which means \(x\), that is \(e^{t}\), ranges from 0 to +∞ (excluding 0). So, for our rectangular equation \(y = x^{3}\), the domain will be \(x > 0\).

Key Concepts

Eliminating the Parameter

Eliminating the parameter means removing the variable that controls both the parametric equations. It can simplify the parametric equations into a single equation. In the given parametric equations, we have \( x = e^{t} \) and \( y = e^{3t} \). Both equations depend on the parameter \( t \).
To eliminate \( t \), we need to express one variable solely in terms of the other. By taking the natural logarithm of the equation \( x = e^{t} \), we get \( t = \ln(x) \).
We can then substitute this value back into the equation for \( y \):
\[ y = e^{3t} = e^{3 \ln(x)} = (e^{\ln(x)})^3 = x^3 \]
Thus, we rewrite the parametric equations as the single equation \( y = x^3 \), effectively merging them into one beautiful rectangular equation.

Rectangular Equation

A rectangular equation is an equation that relates \( x \) and \( y \) in a straightforward way without involving a parameter like \( t \). In our example, we found the rectangular equation by eliminating the parameter \( t \). The links between the given parametric equations \( x = e^t \) and \( y = e^{3t} \) led us to the equation \( y = x^3 \).
With \( y = x^3 \), we now have a simple relationship between \( x \) and \( y \), making it easier to analyze and graph. This rectangular form often simplifies many calculus operations, like finding tangents, derivatives, or integrations. Keep in mind that understanding how to switch between parametric and rectangular forms is a powerful tool when studying the behaviors of curves and their representations.

Domain Adjustment

Domain adjustment ensures that the rectangular equation reflects the original parametric functions' constraints. For the parametric equations \( x = e^t \) and \( y = e^{3t} \), there's an implied domain based on \( t \), since \( e^t \) is only positive. Therefore, \( x = e^t \) tells us that \( x > 0 \), because exponential functions never yield zero or negative values.
As a result, when we form the rectangular equation \( y = x^3 \), it's crucial to also state its domain. The valid domain of \( x \) must be adjusted to keep \( x \) positive. Therefore, the complete rectangular equation with its correct domain is \( y = x^3 \) with \( x > 0 \).
This adjustment ensures that our rectangular representation accurately covers all points on the original curve under the parametric description and avoids any loss of information or misinterpretation of the curve's behavior.