Problem 56
Question
In Exercises 49-56, use a graphing utility to graph the curve represented by the parametric equations. Folium of Descartes: \(\quad x= \dfrac{3t}{1+t^3}, \quad y= \dfrac{3t^2}{1+t^3}\)
Step-by-Step Solution
Verified Answer
The graph of Folium of Descartes can be obtained by substituting various values of parameter \( t \) into the given parametric equations to generate corresponding (x, y) coordinate points, and plotting these points on a graph using a graphing utility. Connect the points with a smooth curve to complete the graph.
1Step 1: Understand the Parametric Equations
The given parametric equations are \( x = \dfrac{3t}{1+t^3} \) and \( y = \dfrac{3t^2}{1+t^3} \). These equations represent the coordinate points (x, y) on the curve for various values of the parameter \( t \).
2Step 2: Select Values for Parameter \( t \)
Before graphing, choose several values for the parameter \( t \). For this particular curve, it will be most informative to choose both positive and negative values, including \( t = 0 \), and a few values close to 1 and -1 where the curve has interesting features.
3Step 3: Calculate Corresponding Coordinate Points
Substitute each selected value of \( t \) into both parametric equations to get the corresponding (x, y) coordinates for each point. Work carefully to avoid calculation errors, especially in the denominator \( 1+t^3 \) where small changes in \( t \) can lead to large changes in the coordinate points.
4Step 4: Plot the Points and Draw the Curve
After having prepared the series of points consisting of various (x, y) pairs, plot these points on a graphing utility. Join these points with a smooth curve to fully represent the Folium of Descartes.
Key Concepts
Folium of DescartesGraphing UtilityCoordinate Points
Folium of Descartes
The Folium of Descartes is a classic example of a curve described by parametric equations. Its unique form captures the intersection between algebraic curves and parametric equations in a fascinating way. Named after the French philosopher and mathematician René Descartes, the folium is represented by specific parametric equations, where both x and y depend on a third variable, the parameter \( t \).
For the Folium of Descartes, the equations are given by:
For the Folium of Descartes, the equations are given by:
- \( x = \frac{3t}{1+t^3} \)
- \( y = \frac{3t^2}{1+t^3} \)
Graphing Utility
Using a graphing utility is an excellent way to visualize the intricate shapes formed by parametric equations like the Folium of Descartes. These tools allow us to go beyond manual plotting, offering an efficient method to see complex curves instantly.
To graph parametric equations:
To graph parametric equations:
- Input the equations for \( x \) and \( y \) into the utility.
- Set a range for the parameter \( t \). For the folium, including values from negative to positive is crucial, as this range captures the full extent of the curve's features.
- Adjustments in the range of \( t \) can highlight different portions of the curve, offering deeper insight into its shape and behavior.
Coordinate Points
In the study of parametric equations, coordinate points serve as the building blocks for visualizing curves like the Folium of Descartes. Each point on the curve corresponds to a specific set of parametric values, giving us a clear picture of the curve's progression.
To find these points:
To find these points:
- Choose a value for the parameter \( t \).
- Substitute this \( t \) value into the parametric equations for \( x \) and \( y \).
- Calculate the resulting \((x, y)\) coordinate pair.
Other exercises in this chapter
Problem 56
In Exercises 49-58, use a graphing utility to graph the polar equation. Describe your viewing window. \(r=2\ \cos(3\theta-2)\)
View solution Problem 56
In Exercises 55-64, use a graphing utility to find one set of polar coordinates for the point given in rectangular coordinates. \(\left(-4, -2\right)\)
View solution Problem 56
In Exercises 51-56, sketch (if possible) the graph of the degenerate conic. \(x^2-10xy+y^2=0\)
View solution Problem 56
In Exercises 51-60, find the standard form of the equation of the parabola with the given characteristics. Vertex: \((-1, 2) \quad\) focus: \((-1, 0)\)
View solution