Question

Special curves The following classical curves have been studied by generations of mathematicians. Use analytical methods (including implicit differentiation) and a graphing utility to graph the curves. Include as much detail as possible. $$x^{4}-x^{2}+y^{2}=0 \text { (Figure- } 8 \text { curve) }$$

Short answer

Question: Describe the symmetry, critical points, and slope at any given point of the curve defined by the equation \(x^4 - x^2 + y^2 = 0\). Answer: The curve is symmetric about both the x-axis and the y-axis. The critical points are \((0, 0)\), \(\left(\sqrt{\frac{1}{2}}, 0\right)\), and \(\left(-\sqrt{\frac{1}{2}}, 0\right)\). The slope at any given point on the curve can be found using the implicit derivative \(y' = \frac{x(2 - 4x^2)}{y}\).

Step-by-step solution

Symmetry Analysis

The given equation is: $$x^4 - x^2 + y^2 = 0$$ Observe that if we replace \(x\) with \(-x\), the equation remains the same. Similarly, when we replace \(y\) with \(-y\), the equation also remains the same. Therefore, the given curve is symmetric about both the x-axis and the y-axis.

Finding Critical Points

To find the critical points, we need to determine where the gradient of the curve is zero or undefined: (i) Set the equation as a function \(f(x, y)\): $$f(x, y) = x^4 - x^2 + y^2$$ (ii) Compute the partial derivatives, \(f_x\) and \(f_y\): $$f_x = \frac{\partial f}{\partial x} = 4x^3 - 2x$$ $$f_y = \frac{\partial f}{\partial y} = 2y$$ (iii) Identify the critical points by finding when \(f_x = 0\) and \(f_y = 0\): $$4x^3 - 2x = 0 \Rightarrow x(4x^2 - 2) = 0$$ Here, we have x = 0 and \(4x^2 - 2 = 0\) which gives us \(x=\pm\sqrt{\frac{1}{2}}\). $$2y = 0 \Rightarrow y = 0$$ The critical points are \((0, 0)\), \(\left(\sqrt{\frac{1}{2}}, 0\right)\), and \(\left(-\sqrt{\frac{1}{2}}, 0\right)\).

Implicit Differentiation

To find the slope at a given point on the curve, we'll use implicit differentiation: Given, $$x^4 - x^2 + y^2 = 0$$ Differentiating both sides with respect to x: $$4x^3 - 2x + 2yy' = 0$$ Now, we can find the derivative \(y'\): $$ y' = \frac{2x - 4x^3}{2y} = \frac{x(2 - 4x^2)}{y} $$

Graphing the Curve

We can now use the information from the previous steps to graph the curve: 1. The curve is symmetric about the x-axis and y-axis. 2. The critical points are \((0, 0)\), \(\left(\sqrt{\frac{1}{2}}, 0\right)\), and \(\left(-\sqrt{\frac{1}{2}}, 0\right)\). 3. The implicit derivative \(y' = \frac{x(2 - 4x^2)}{y}\) gives us the slope at any given point on the curve. Use a graphing utility (e.g., Desmos, GeoGebra) to plot the given equation, and you will see the Figure-8 curve with the above properties. It will have two loops on either side of the y-axis with critical points \((0, 0)\), \(\left(\sqrt{\frac{1}{2}}, 0\right)\), and \(\left(-\sqrt{\frac{1}{2}}, 0\right)\).

Key concepts

Symmetry Analysis

Exploring the symmetry of a curve is a fundamental step in understanding its overall shape and behavior. When dealing with the equation \(x^4 - x^2 + y^2 = 0\), known visually as a Figure-8 curve, we begin by examining its symmetry.

By replacing \(x\) with \(-x\) and observing that the equation remains unchanged, we establish that the curve is symmetric about the y-axis. Similarly, substituting \(y\) with \(-y\) and seeing no change in the equation indicates symmetry about the x-axis. This tells us that we need only to consider one quarter of the curve when plotting it, as the rest can be mirrored across these axes.

Tip for Ease of Understanding: Imagine folding a paper along the y-axis or the x-axis; if the curve matches perfectly, it possesses that line of symmetry. Symmetry analysis greatly simplifies the graphing process and provides insight into the inherent structure of the curve.

Critical Points

Identifying critical points on a curve is essential for understanding where the curve changes direction or where it has horizontal or vertical tangents. For the Figure-8 curve represented by \(x^4 - x^2 + y^2 = 0\), we analyze the gradient to find these points.

By setting the equation as a function \(f(x, y)\) and calculating the partial derivatives \(f_x\) and \(f_y\), we are preparing to find where the slope of the curve, or the first derivative, is equal to zero. Solving the equations \(4x^3 - 2x = 0\) and \(2y = 0\), we determine that the critical points are at \((0, 0)\), \(\left(\sqrt{\frac{1}{2}}, 0\right)\), and \(\left(-\sqrt{\frac{1}{2}}, 0\right)\).

Explanation Enhancement: When examining critical points, envision them as 'rest stops' on a hill—the points where you'd pause during an ascent or descent. These are significant for plotting the curve accurately, as they are the high points, low points, or cusps in the graph's landscape.

Graphing Curves

The process of graphing a curve transforms the abstract equations we've derived into a visual representation that captures the essence of the mathematical relationship. With the Figure-8 curve, graphing is facilitated by implicit differentiation and the symmetry and critical points already determined.

Using the derivative \(y' = \frac{x(2 - 4x^2)}{y}\), we calculate the slope at any point of the curve, which is essential for understanding the steepness and direction of the curve at that location. Then, we put everything together to sketch the curve: the symmetry around the axes reduces the number of points we need to plot, the critical points indicate where we have horizontal tangents (or turns), and the derivative provides the slope information needed to draw the curve accurately.

Graphing Tip: Utilize a graphing utility to input the expression and view the Figure-8 curve. Observe how the critical points mark the transitions in the curve and how the curves emanate symmetrically from these points. Analyzing the graph you plotted can be uniquely rewarding, as it helps to solidify the abstract concepts you've been studying into a tangible and visual form.