Problem 38 - Calculus with Concepts in Calculus

Question

Let \(f(x, y)=A x^{2}+2 B x y+C y^{2}\), as in (1). Assume that \(A \neq 0\) and \(A C-B^{2}<0\). Verify that there are points \((x, y)\) as close to \((0,0)\) as one wishes such that \(f(x, y)>0\), and other points \((x, y)\) as close to \((0,0)\) as one wishes such that \(f(x, y)<0 .\) Conclude that \(f\) has a saddle point at \((0,0) .\) (Hint: Consider points of the form \((x, 0)\) and \((-B y / A, y) .)\)

Step-by-Step Solution

Answer

At \((0,0)\), \(f(x, y)\) shows mixed behavior, confirming a saddle point.

Step 1: Examine Points on the x-axis

First, consider points of the form \((x, 0)\). Substitute \(y = 0\) into \(f(x, y)\), resulting in \(f(x, 0) = A x^2\). Since \(A eq 0\), \(f(x, 0)\) is positive for all \(x eq 0\), indicating points as close to \((0,0)\) as one wishes such that \(f(x, 0) > 0\) exist.

Step 2: Examine Points on the Line \(y = -\frac{Bx}{A}\)

Next, consider points of the form \((x, y) = (-\frac{By}{A}, y)\). Substitute into \(f(x, y)\), yielding \(f\left(-\frac{By}{A}, y\right) = A\left(-\frac{By}{A}\right)^2 + 2B\left(-\frac{By}{A}\right)y + C y^2\). Simplifying gives \(f\left(-\frac{By}{A}, y\right) = \left(\frac{B^2}{A} - \frac{2B^2}{A} + C\right)y^2 = \left(\frac{C A - B^2}{A}\right) y^2\). Since \(A C - B^2 < 0\), the expression simplifies to \(\frac{A C - B^2}{A}\cdot y^2 < 0\), meaning \(f(x, y) < 0\) for \(y eq 0\), showing points as close to \((0,0)\) as one wishes such that \(f(x, y) < 0\) exist.

Step 3: Conclude Saddle Point at \((0, 0)\)

Given the points \((x, 0)\) where \(f(x, 0) > 0\) and points \((x, y) = (-\frac{By}{A}, y)\) where \(f(x, y) < 0\), both exist as close to \((0,0)\) as one wishes, it shows the mixed behavior of \(f(x, y)\). This mixed behavior around \((0,0)\) is characteristic of a saddle point, confirming that \(f\) has a saddle point at \((0,0)\).

Key Concepts

Saddle Point

In multivariable calculus, a "saddle point" refers to a specific type of critical point that exhibits a mix of behaviors. In simple terms, if you visualize the graph of a function, a saddle point occurs at a spot that looks like a mountain pass or a saddle for a horse. In this location, the function tilts upwards in one direction and downwards in another direction.

For the function in our exercise, defined as \(f(x, y) = Ax^2 + 2Bxy + Cy^2\), we are asked to show it has a saddle point at \((0, 0)\). Here’s how:

• First, evaluate the function along the x-axis. When substituting \(y = 0\), the function becomes \(Ax^2\). This is always positive for values of \(x\) close to zero because \(A eq 0\).
• Then, consider a line parallel to the x-axis using points of the form \((-\frac{By}{A}, y)\). The function simplifies to an expression that is negative, \( \frac{A C - B^2}{A} \cdot y^2 \), proving it can be negative for small \(y\).
• This mix of positive and negative values around \((0, 0)\) confirms the presence of a saddle point, as it represents various directions of shifts in the slope of the function graph.

Quadratic Forms

Quadratic forms are a way of capturing polynomial functions of degree two that involve multiple variables. They are particularly important in understanding how certain functions behave in multivariable calculus settings, such as determining the nature of critical points, including saddle points.

Looking into the quadratic form \(Ax^2 + 2Bxy + Cy^2\), each term can tell us something:

• \(Ax^2\) represents the behavior of the function along the x-axis.
• The term \(2Bxy\) contains the link between both axes, providing information on the interaction between the variables \(x\) and \(y\).
• \(Cy^2\) indicates how the function behaves along the y-axis.

Analyzing these elements helps identify how a function changes in two dimensions. By examining criteria like those in the exercise, where \(AC - B^2 < 0\), we understand the behavior of quadratic forms regarding their directionality and curvature. The given criteria here indicate that instead of primarily opening upwards or downwards, the function is saddle-shaped.

Critical Points

Critical points in calculus are locations on the graph of a function where certain properties hold, such as zero derivative or undefined slopes. For multivariable calculus, these points help us determine where key behaviors happen, and they often occur when the partial derivatives are zero.

For our example function, a critical point is at \((0, 0)\) since it involves examining the curvature pointed by the derivatives of the quadratic function \(f(x, y) = Ax^2 + 2Bxy + Cy^2\).

• By calculating the partial derivatives and setting them to zero, we find potential critical points.
• However, merely having zero at these points is not enough—we investigate further using quadratic forms to determine the nature of these points (like minima, maxima, or saddle points).

The presence of a saddle point at \((0, 0)\)—confirmed by examining function behavior on the axes and other lines—demonstrates how critical points allow for deeper insights into the function's behavior around those key moments.