Problem 8
Question
\begin{array}{l}{1-22 \text { a pair of parametric equations is given. }} \\\ {\text { (a) Sketch the curve represented by the parametric equations. }} \\\ {\text { (b) Find a rectangular-coordinate equation for the curve by }} \\\ {\text { eliminating the parameter. }}\end{array} $$ x=t+1, \quad y=\frac{t}{t+1} $$
Step-by-Step Solution
Verified Answer
The curve is a hyperbola, given by the equation \( y = 1 - \frac{1}{x} \).
1Step 1: Understand the Parametric Equations
We are given parametric equations with \( x = t + 1 \) and \( y = \frac{t}{t+1} \). The variable \( t \) is the parameter, which describes a curve in the Cartesian plane when \( t \) varies over all real numbers.
2Step 2: Sketch the Curve
To sketch the curve, compute several points by substituting values of \( t \) into the parametric equations to get the corresponding \( (x, y) \) points. For example, for \( t = 0 \), \( x = 1 \) and \( y = 0 \); for \( t = 1 \), \( x = 2 \) and \( y = \frac{1}{2} \); for \( t = -1 \), \( x = 0 \) and \( y = 1 \). Plot these points on a graph to create a smooth curve.
3Step 3: Eliminate Parameter \( t \)
To find a rectangular-coordinate equation, solve the equation for \( t \) in terms of \( x \): \( x = t + 1 \), so \( t = x - 1 \). Substitute this value for \( t \) in the equation for \( y \), yielding \( y = \frac{x-1}{x} \).
4Step 4: Simplify Rectangular Equation
Simplify the expression \( y = \frac{x-1}{x} \) by dividing each term by \( x \). This gives \( y = 1 - \frac{1}{x} \). This is the rectangular form of the curve.
5Step 5: Analyze the Rectangular Equation
The equation \( y = 1 - \frac{1}{x} \) describes a hyperbola with an asymptote along the x-axis as \( x \to \infty \) and another asymptote at the line \( y = 1 \). The graph has a discontinuity (undefined point) at \( x = 0 \), which corresponds to \( t = -1 \), as the denominator in the expression \( t+1 \) becomes zero.
Key Concepts
Rectangular-Coordinate EquationEliminating the ParameterCurve Sketching
Rectangular-Coordinate Equation
The concept of converting parametric equations into rectangular-coordinate equations involves finding a relationship between the Cartesian coordinates
This is quite valuable as it often gives us a clearer picture of the curve's characteristics within a standard xy-plane.
In our case, the given parametric equations were \( x = t + 1 \) and \( y = \frac{t}{t+1} \).
To eliminate the parameter \( t \) and find the rectangular-coordinate equation, we manipulate the equation for \( x \):
By eliminating the parameter, we've gained a more direct understanding of how \( y \) varies with \( x \).
- \( x \)
- \( y \)
This is quite valuable as it often gives us a clearer picture of the curve's characteristics within a standard xy-plane.
In our case, the given parametric equations were \( x = t + 1 \) and \( y = \frac{t}{t+1} \).
To eliminate the parameter \( t \) and find the rectangular-coordinate equation, we manipulate the equation for \( x \):
- \( t = x - 1 \)
- \( y = \frac{(x - 1)}{x} \)
- \( y = 1 - \frac{1}{x} \)
By eliminating the parameter, we've gained a more direct understanding of how \( y \) varies with \( x \).
Eliminating the Parameter
Eliminating the parameter is a process used to convert parametric equations into a single equation in rectangular form. This involves solving one of the parametric equations for the parameter \( t \).
Here, we started with the equation \( x = t + 1 \).
By rearranging this equation:
By focusing on a single equation, analyzing and understanding the shape of the curve becomes simpler and more straightforward.
This equation, \( y = 1 - \frac{1}{x} \), is more accessible for sketching and interpreting the details of the curve.
Here, we started with the equation \( x = t + 1 \).
By rearranging this equation:
- \( t = x - 1 \)
- \( y = \frac{x-1}{x} \)
By focusing on a single equation, analyzing and understanding the shape of the curve becomes simpler and more straightforward.
This equation, \( y = 1 - \frac{1}{x} \), is more accessible for sketching and interpreting the details of the curve.
Curve Sketching
Curve sketching with parametric equations involves plotting points on a graph and connecting them to form a smooth curve. When sketching a curve from parametric equations, you compute several \((x, y)\) coordinates by choosing values for the parameter \( t \).
For example, let's consider some specific values:
Curve sketching in this way provides a visual representation of the path traced by the equations as \( t \) changes.
Studying characteristics like where the curve might be undefined or approach certain lines (asymptotes) helps in understanding the complete behavior.
For the curve represented by \( y = 1 - \frac{1}{x} \), we notice it behaves like a hyperbola:
For example, let's consider some specific values:
- When \( t = 0 \): \( x = 1 \), \( y = 0 \)
- When \( t = 1 \): \( x = 2 \), \( y = \frac{1}{2} \)
- When \( t = -1 \): \( x = 0 \), \( y = 1 \)
Curve sketching in this way provides a visual representation of the path traced by the equations as \( t \) changes.
Studying characteristics like where the curve might be undefined or approach certain lines (asymptotes) helps in understanding the complete behavior.
For the curve represented by \( y = 1 - \frac{1}{x} \), we notice it behaves like a hyperbola:
- There is a discontinuity at \( x = 0 \).
- The curve approaches the line \( y = 1 \) as \( x \to \infty \).
Other exercises in this chapter
Problem 7
Find the vertices, foci, and asymptotes of the hyperbola, and sketch its graph. \(y^{2}-\frac{x^{2}}{25}=1\)
View solution Problem 7
Find the focus, directrix, and focal diameter of the parabola, and sketch its graph. $$y^{2}=4 x$$
View solution Problem 8
Find the vertices, foci, and eccentricity of the ellipse. Determine the lengths of the major and minor axes, and sketch the graph. $$ 4 x^{2}+25 y^{2}=100 $$
View solution Problem 8
Determine the equation of the given conic in \(X Y\) -coordinates when the coordinate axes are rotated through the indicated angle. $$y=(x-1)^{2}, \quad \phi=45
View solution