Eliminate the Parameter: Videos & Practice Problems

To eliminate the parameter in parametric equations, solve one equation for the parameter and substitute it into the other. For example, given $x=\cos \left(t\right)$ and $y=3\sin \left(t\right)$, use the Pythagorean identity $\sin \left(t\right)+\cos \left(t\right)=1$ to derive the equation of an ellipse: $x^2+(y/3)^2=1$. This process reveals the relationship between parametric and rectangular equations.

Eliminating the Parameter

In the study of parametric equations, the process of eliminating the parameter is essential for converting parametric forms into rectangular equations, which only involve the variables x and y. This transformation allows for a clearer understanding of the relationship between the two variables. To eliminate the parameter, one typically solves one of the parametric equations for the parameter (often denoted as t) and substitutes this expression into the other equation.

For instance, consider the parametric equations given by \( x(t) = \sqrt{t} \) and \( y(t) = t - 3 \). To eliminate the parameter, we can first express t in terms of x. By squaring both sides of the equation for x, we find that \( t = x^2 \). We can then substitute this expression for t into the equation for y:

\[ y = t - 3 = x^2 - 3 \]

This results in the rectangular equation \( y = x^2 - 3 \), which is a parabola. The shape of the graph reflects this relationship, demonstrating that the parametric equations describe a portion of the parabola defined by the rectangular equation.

It is important to note that the range of t values can affect the portion of the graph that is represented. In this example, since t is restricted to positive values, the resulting graph only depicts the right half of the parabola. Understanding these restrictions is crucial, as they determine the specific segment of the graph that corresponds to the parametric equations.

In summary, eliminating the parameter in parametric equations involves substituting expressions derived from one equation into another, leading to a rectangular equation that can often reveal familiar shapes, such as parabolas. This process not only simplifies the analysis of the relationship between x and y but also enhances the ability to graph these relationships accurately.

Eliminate the parameter to rewrite the following as a rectangular equation.
$x\left(t\right)=2t-1$
$y\left(t\right)=t^5-2$

• A. $y=\left(\frac{x-2}{2}\right)^5-1$
• B. $y=\left(\frac{x-1}{2}\right)^5-2$
• C. $y=\left(\frac{x+1}{2}\right)^5-2$
• D. $y=\left(\frac{x+1}{2}\right)^5-1$

Answer: (C)

Eliminating the Parameter Example 1

In this exploration of parametric equations, we begin by graphing two quadratic equations defined in terms of a parameter \( t \). The goal is to derive their equivalent rectangular equation by eliminating the parameter, resulting in a relationship solely between \( x \) and \( y \).

To graph the equations, we select values for \( t \) within the interval of all real numbers. Given that both equations are quadratic, using small integers such as \( 0, 1, 2, \) and \( 3 \) is advantageous. This choice simplifies calculations, especially since squaring negative values yields the same results as their positive counterparts. For instance, substituting these values into the equations yields coordinates: for \( t = 0 \), \( x = 0^2 - 1 = -1 \) and \( y = 0^2 - 2 = -2 \); for \( t = 1 \), \( x = 1^2 - 1 = 0 \) and \( y = 1^2 - 2 = -1 \); and so forth. The resulting points are then plotted on a graph.

Upon connecting these points, one might initially expect a parabolic shape due to the quadratic nature of the equations. However, the graph reveals a straight line, indicating that the relationship between \( x \) and \( y \) is linear. The orientation of the line can be traced by the increasing values of \( t \), confirming the direction of the graph.

To derive the rectangular equation, we eliminate the parameter \( t \) by solving one of the equations for \( t \) and substituting it into the other. Starting with the equation for \( x \), we have:

\( x = t^2 - 1 \)

Rearranging gives:

\( t^2 = x + 1 \)

Taking the square root results in:

\( t = \sqrt{x + 1} \)

Next, we substitute this expression for \( t \) into the equation for \( y \):

\( y = t^2 - 2 \)

Substituting yields:

\( y = \sqrt{x + 1}^2 - 2 \)

Since squaring the square root cancels out, we simplify to:

\( y = (x + 1) - 2 \)

Thus, the final equation is:

\( y = x - 1 \)

This result confirms that the relationship between \( x \) and \( y \) is indeed linear, with a slope of \( 1 \) and a y-intercept of \( -1 \). This example illustrates how two quadratic parametric equations can yield a linear relationship when appropriately analyzed and transformed.

Eliminate the Parameter Example 2

In this discussion, we explore parametric equations defined as \( x(t) = t - 3 \) and \( y(t) = \frac{1}{t - 5} \). The goal is to graph the corresponding plane curve and derive the equivalent rectangular equation. It's important to note that \( t \) can take any value except \( 5 \), as this would make the denominator in the \( y(t) \) equation zero, resulting in an undefined expression.

To graph the curve, we substitute specific \( t \) values into the equations to generate \( (x, y) \) coordinate pairs. For instance, when \( t = 0 \), \( x(0) = -3 \) and \( y(0) = -0.2 \). Continuing this process for other values of \( t \) such as \( 3, 4.8, 5.2, \) and \( 7 \), we obtain the following coordinates: \( (-3, -0.2), (0, -0.5), (1.8, -5), (2.2, 5), (4, 0.5) \). These points are plotted on a graph, revealing a curve that approaches a vertical asymptote at \( x = 2 \) as \( t \) approaches \( 5 \).

The next step involves eliminating the parameter \( t \) to find the rectangular equation. Starting with the equation for \( x \), we rearrange it to express \( t \): \( t = x + 3 \). Substituting this expression into the equation for \( y \), we get:

\[ y = \frac{1}{(x + 3) - 5} = \frac{1}{x - 2} \]

This results in the rectangular equation \( y = \frac{1}{x - 2} \), which is a rational function. The vertical asymptote occurs at \( x = 2 \), indicating that the function approaches infinity as \( x \) nears this value from either side, confirming the behavior observed in the graph.

In summary, we have successfully graphed the parametric equations and derived the equivalent rectangular equation, illustrating the relationship between the two forms and the characteristics of the resulting rational function.

Equations with Trigonometric Functions

In the study of parametric equations, particularly those involving trigonometric functions, a systematic approach is essential for eliminating the parameter. When given equations such as x = cos(t) and y = 3sin(t), the first step is to isolate the trigonometric functions. For instance, from x = cos(t), we can express this as cos(t) = x. Similarly, from y = 3sin(t), we can rearrange it to sin(t) = y/3.

To eliminate the parameter t, we can utilize the Pythagorean identity, which states that sin²(t) + cos²(t) = 1. This identity is versatile and can apply to any variable, including x and y. By substituting our expressions for sin(t) and cos(t) into this identity, we can derive a new equation that no longer contains the parameter.

Squaring both sides of our isolated trigonometric functions gives us cos²(t) = x² and sin²(t) = (y/3)². Plugging these into the Pythagorean identity results in:

x² + (y/3)² = 1

This equation represents an ellipse, a familiar conic section. The process of eliminating the parameter in parametric equations involving trigonometric functions thus hinges on isolating the trigonometric components and applying the Pythagorean identity to derive a new equation that encapsulates the relationship between x and y without the parameter t.

First eliminate the parameter, then graph the plane curve of the parametric equations.
$x\left(t\right)=2+\cos t$, $y\left(t\right)=-1+\sin t$; $0\le t\le2\pi$

• A. $\left(x-2\right)^2+\left(y+1\right)^2=1$
• B. $\left(x-2\right)^2+\left(y+1\right)^2=1$
• C. $\left(x+2\right)^2+\left(y-1\right)^2=1$
• D. $\left(x+2\right)^2+\left(y-1\right)^2=1$

Answer: (A)

Equations with Trigonometric Functions Example 3

In this example, we explore the process of eliminating parameters from parametric equations to derive a rectangular equation. We start with the parametric equations given by \( x(t) = 3 \cos(t) \) and \( y(t) = 2 \sin(t) \). To eliminate the parameter \( t \), we utilize the Pythagorean identity, which states that \( \cos^2(t) + \sin^2(t) = 1 \).

First, we rearrange the equation for \( x \) to express \( \cos(t) \) in terms of \( x \):

\( \cos(t) = \frac{x}{3} \)

Squaring both sides gives us:

\( \cos^2(t) = \left(\frac{x}{3}\right)^2 = \frac{x^2}{9} \)

Next, we do the same for \( y \):

\( \sin(t) = \frac{y}{2} \)

Squaring this expression results in:

\( \sin^2(t) = \left(\frac{y}{2}\right)^2 = \frac{y^2}{4} \)

Substituting these squared expressions into the Pythagorean identity yields:

\( \frac{x^2}{9} + \frac{y^2}{4} = 1 \)

This equation represents an ellipse centered at the origin. The denominators indicate the lengths of the semi-major and semi-minor axes, with 3 and 2 being the respective lengths along the x and y axes. Thus, the ellipse extends from -3 to 3 on the x-axis and from -2 to 2 on the y-axis, forming a wider shape along the x-axis.

However, it is crucial to consider the parameter restrictions. In this case, \( t \) is restricted to the interval \( \pi \leq t < 2\pi \). This means we only graph the bottom half of the ellipse, as the parameter does not cover the entire range from 0 to \( 2\pi \). Therefore, the resulting graph will only depict the lower portion of the ellipse, emphasizing the importance of parameter limits in determining the final graph.

Here’s what students ask on this topic:

How do you eliminate the parameter in parametric equations?

To eliminate the parameter in parametric equations, you solve one of the equations for the parameter, typically denoted as $t$, and substitute it into the other equation. This process results in a rectangular equation involving only $x$ and $y$. For example, if you have $x=t$ and $y=t^2$, solving for $t$ in the first equation gives $t=x$. Substituting this into the second equation results in $y=x^2$. This method is useful for converting parametric equations into more familiar forms like lines or parabolas.

What is the process for eliminating the parameter in trigonometric parametric equations?

Eliminating the parameter in trigonometric parametric equations involves a different approach due to the complexity of trigonometric functions. First, solve each equation for the trigonometric function involved. For example, if $x=\cos \left(t\right)$ and $y=3\sin \left(t\right)$, solve for $\cos \left(t\right)$ and $\sin \left(t\right)$. Then, use a Pythagorean identity, such as ${\sin }^2\left(t\right)+{\cos }^2\left(t\right)=1$, to eliminate $t$. This results in an equation like $x^2+\left(\frac{y}{3}\right)2^{}=1$, representing an ellipse.

Why is it important to understand the restrictions on $t$ values in parametric equations?

Understanding the restrictions on $t$ values in parametric equations is crucial because they determine the portion of the graph represented. The parameter $t$ often has a specific range, which affects the shape and extent of the graph. For instance, if $t$ is restricted to positive values, the graph may only represent a portion of the full curve, such as half of a parabola. These restrictions ensure that the graph accurately reflects the intended behavior of the parametric equations, and they help in understanding the relationship between the parametric and rectangular forms of the equation.

What are some common mistakes to avoid when eliminating the parameter in parametric equations?

When eliminating the parameter in parametric equations, common mistakes include incorrectly solving for the parameter $t$, not substituting correctly, and ignoring restrictions on $t$. Ensure that you solve one equation for $t$ accurately and substitute it into the other equation without errors. Additionally, be mindful of any restrictions on $t$ values, as they affect the portion of the graph represented. Another mistake is not simplifying the resulting equation properly, which can lead to incorrect interpretations of the graph. Always double-check your work to avoid these errors.

How does eliminating the parameter help in understanding the graph of parametric equations?

Eliminating the parameter in parametric equations helps in understanding the graph by converting the equations into a more familiar rectangular form involving only $x$ and $y$. This process reveals the underlying shape of the graph, such as a line, parabola, or ellipse, making it easier to analyze and interpret. By removing the parameter, you can see the relationship between $x$ and $y$ directly, which aids in visualizing the graph's behavior and understanding its properties. This conversion is particularly useful in identifying the type of conic section or curve represented by the parametric equations.