Problem 24 - Precalculus with Limits

Question

In Exercises 7-26, (a) sketch the curve represented by the parametric equations (indicate the orientation of the curve) and (b) eliminate the parameter and write the corresponding rectangular equation whose graph represents the curve. Adjust the domain of the resulting rectangular equation if necessary. \(x=e^{2t}\) \(y=e^{t}\)

Step-by-Step Solution

Answer

The curve represented by the parametric equations is sketched by plotting points for different values of \(t\). The parameter is eliminated by substituting \(t = ln(y)\) from the equation \(y=e^{t}\) into the equation \(x=e^{2t}\) to get \(x=y^{2}\). The domain is adjusted to \(y>0\).

Step 1: Sketch the Curve

To sketch out the curve represented by \(x=e^{2t}\) and \(y=e^{t}\), plot some points for various values of \(t\). For instance, for \(t=0\), \(x=e^{2*0}=1\) and \(y=e^{0}=1\), giving the point (1,1). Try out several more values to get more points and draw the curve through them. The positive direction of \(t\) would give the orientation of the curve.

Step 2: Eliminate the Parameter

To eliminate the parameter \(t\), solve one of the equations for \(t\) and substitute in the other equation. In this case, take the natural logarithm of both sides of the equation \(y=e^{t}\) to get \(t = ln(y)\) . Substitute \(t\) into the equation \(x=e^{2t}\) to get \(x=e^{2ln(y)}\). Simplify to get \(x=y^{2}\).

Step 3: Adjust the Domain

The original parametric equations suggest the domain of \(x\) and \(y\) is \((0, \infty)\) due to the properties of an exponential function. So, in rectangular form, this implies the same domain for \(y\), i.e. \(y>0\). Adjusted rectangular equation becomes \(x=y^{2}\) , where \(y>0\).

Key Concepts

Rectangular Equation

Transforming parametric equations into rectangular equations helps to simplify the graphing process. The given parametric equations are:

• \(x = e^{2t}\)
• \(y = e^{t}\)

To find a rectangular equation, we need to eliminate the parameter, \(t\). First, solve the equation \(y = e^{t}\) for \(t\):\[ t = \ln(y) \] Now, substitute \(t\) back into the first equation: \(x = e^{2t}\). Replacing \(t\) gives us:

• \(x = e^{2\ln(y)}\)

Using the property \(e^{2\ln(y)} = y^{2}\), the equation becomes:

• \(x = y^{2}\)

This equation is now free of the parameter \(t\) and is expressed as an equation in terms of \(x\) and \(y\). It represents a parabola that opens to the right, which is the rectangular form of the given parametric equations.

Curve Sketching

Curve sketching from parametric equations involves finding points that lie on the curve and understanding its behavior. To do this, we must plot multiple points by substituting various values of \(t\). For instance:

• When \(t = 0\), \(x = e^{2 \times 0} = 1\) and \(y = e^{0} = 1\). Hence, point \((1, 1)\).
• When \(t = 1\), \(x = e^{2 \times 1} = e^{2}\) and \(y = e^{1} = e\). Thus, \((e^{2}, e)\).
• When \(t = -1\), \(x = e^{-2}\) and \(y = e^{-1}\). Thus, \((e^{-2}, e^{-1})\).

Plot these points and connect them smoothly to represent the curve. The orientation, indicated by the positive direction of \(t\), shows the curve moving from left to right as \(t\) increases.

Parameter Elimination

Eliminating the parameter in parametric equations is a crucial step in the graphing and analysis process. It simplifies the relationship between \(x\) and \(y\) by removing \(t\). For the given parametric set:\[\begin{align*}x &= e^{2t} \y &= e^{t}\end{align*}\]We solved for \(t\) using the equation \(y = e^{t}\) by taking the natural logarithm:\[t = \ln(y)\]Substituting into \(x = e^{2t}\) yields:\[x = e^{2\ln(y)}\]Simplifying this using properties of exponents results in \(x = y^{2}\). Eliminating the parameter successfully converts the parametric form to a rectangular equation. This step is foundational for understanding the curve algebraically.

Domain Adjustment

Domain adjustment is crucial when converting parametric equations to rectangular equations. It ensures that the new equation faithfully represents the original parametric version. Exponential functions like \(e^{t}\) inherently imply certain domain restrictions. For the parametric equations

• \(x = e^{2t}\)
• \(y = e^{t}\)

the domain of \(t\) is all real numbers, but both \(x\) and \(y\) must be positive because exponential functions are always positive. This means, when converting to the rectangular form \(x = y^{2}\), we must adjust so that \(y > 0\), maintaining consistency with the original parametric model. Ensuring this restriction keeps the graph accurate and true to the nature of the exponential function.