Problem 51 - Calculus Early Transcendentals

Question

Use a graphing utility to graph the following curves. Be sure to choose an interval for the parameter that generates all features of interest. $$\text { Folium of Descartes } x=\frac{3 t}{1+t^{3}}, y=\frac{3 t^{2}}{1+t^{3}}$$

Step-by-Step Solution

Answer

Answer: The Folium of Descartes has a single loop with a cusp at the point (3, 3) and has asymptotes along the positive x and positive y-axes when graphed using the parametric equations and interval given.

Step 1: Analyze the Parametric Equations

First, take a look at the parametric equations given for the Folium of Descartes. Notice that as t approaches infinity, both x and y approach 0. This means the curve will approach the x and y axes but never actually touch them. The curve has also a point with both coordinates having the same value, which happens when the denominator equals 1 (giving \(x = y = 3\)). This point happens exactly when \(t = 1\).

Step 2: Determine an Appropriate Interval for the Parameter t

Based on our observation in step 1, the curve clearly has some interesting features near the point \(t=1\). Since we want to capture the full Folium of Descartes, it is reasonable to consider a range of t-values around (but not equal to) 1. We need an interval that goes from a negative value to a positive value, covering a sufficient range of t-values. One such interval could be \(t \in [-2, 2]\) excluding 0.

Step 3: Graph the Parametric Equations

Now that we have determined the interval \(t \in [-2, 2]\) excluding 0, use a graphing utility to input the parametric equations and plot the curve: $$x = \frac{3t}{1 + t^3}$$ $$y = \frac{3t^2}{1 + t^3}$$ Make sure to set the t values to be within the interval [-2, 2] excluding 0. You will see that the Folium of Descartes consists of a single loop with a cusp at (3, 3) and has asymptotes along the positive x and positive y-axes.