Problem 20 - Calculus an Applied Approach

Question

Examine the function for relative extrema and saddle points. $$ f(x, y)=-\frac{3}{x^{2}+y^{2}+1} $$

Step-by-Step Solution

Answer

The function \(f(x, y)=-\frac{3}{x^{2}+y^{2}+1}\) has a saddle point at (0,0). The function does not have a relative extrema.

Step 1: First and Second Partial Derivatives

Begin by finding the first and second partial derivatives of the function. This can be done by setting \(x\) or \(y\) to be constant while differentiating with respect to the other variable. The formula is:\[f_{x}=\frac{\partial f}{\partial x}=-\frac{6x}{(x^{2}+y^{2}+1)^{2}}\]\[f_{y}=\frac{\partial f}{\partial y}=-\frac{6y}{(x^{2}+y^{2}+1)^{2}}\]For the second partial derivatives, we would use the first partial derivatives:\[f_{xx}=\frac{\partial^{2} f}{\partial x^{2}}=\frac{12x^{2}}{(x^{2}+y^{2}+1)^{3}}-\frac{6}{(x^{2}+y^{2}+1)^{2}}\]\[f_{xy}=\frac{\partial^{2} f}{\partial x \partial y}=\frac{12xy}{(x^{2}+y^{2}+1)^{3}}\]\[f_{yy}=\frac{\partial^{2} f}{\partial y^{2}}=\frac{12y^{2}}{(x^{2}+y^{2}+1)^{3}}-\frac{6}{(x^{2}+y^{2}+1)^{2}}\]

Step 2: Finding the Critical Points

The critical points occur where the gradient is zero. In terms of maths, critical points are solutions that satisfy:\[f_{x} = 0\] \[f_{y} = 0\] This gives us \(x = 0\) and \(y = 0\) - so the critical point is at the origin \( (0,0)\).

Step 3: Determination of Extrema and Saddle Points

To confirm whether the critical point found is a maximum, minimum or a saddle point, we check the second derivative using the formula:\[D=f_{xx}(x_{0}, y_{0})f_{yy}(x_{0}, y_{0})-(f_{xy}(x_{0}, y_{0}))^{2}\] If \(D > 0\) and \(f_{xx}(x_{0}, y_{0}) > 0\), then we would have a local minimum. If \(D > 0\) and \(f_{xx}(x_{0}, y_{0}) < 0\), then we would have a local maximum. However, if \(D < 0\), it implies a saddle point. If \(D = 0\), the test is inconclusive.Substitute the critical point coordinate found (0,0) into \(D\), and evaluate \(D\). Since the value of \(D\) will be less than 0, we can confirm that the function has a saddle point at (0,0).

Step 4: Function Behavior Analysis

Consider the nature of the function in the vicinity of the saddle point. Here, you'll notice that the function is positive when absolute values of \(x\) and \(y\) are small but their squares sum to more than 1. Otherwise, the function is negative. This shows the function reaches a minimum at large \(x\) and \(y\), meaning there is no relative extrema but a saddle point at (0,0).

Key Concepts

Partial Derivatives

To understand the behavior of multivariable functions like \(f(x, y)=-\frac{3}{x^{2}+y^{2}+1}\), partial derivatives play a crucial role. They show how the function changes as one variable shifts, holding the other fixed. Think of it as slicing through a 3D surface, capturing change along a single direction.

• To compute the partial derivative with respect to \(x\), regard \(y\) as a constant and differentiate only in terms of \(x\).
• Similarly, the partial derivative with respect to \(y\) treats \(x\) as constant, focusing on changes due exclusively to \(y\).

This method allows us to find the slopes of tangent lines at any point, enabling deeper insights into how the surface bends and curves.
In our case:

• First partial derivatives: \(f_{x} = -\frac{6x}{(x^{2}+y^{2}+1)^{2}}\) and \(f_{y} = -\frac{6y}{(x^{2}+y^{2}+1)^{2}}\).
• Second partial derivatives provide more detailed information about the surface's curvature, leading to terms like \(f_{xx}\), \(f_{yy}\), and \(f_{xy}\).

Understanding partial derivatives is foundational for analyzing multivariable calculus problems, just like how they help us here evaluate the behavior around critical points.

Critical Points

Critical points are vital in finding where functions might reach their peaks, valleys, or behave unusually. For functions of two variables, these points occur where both first partial derivatives are zero, indicating a flat tangent plane.
To find these points:

• Solve \(f_{x} = 0\) and \(f_{y} = 0\).
• These equations set the stage for locating any potential extrema or saddle points.

In the exercise provided, setting \(f_{x} = 0\) and \(f_{y} = 0\) leads us to the solution \(x = 0\) and \(y = 0\), marking the origin \((0,0)\) as a critical point.
This point becomes the focal point for further analysis, helping us decide whether this location is a high point, low point, or something entirely different, like a saddle point.

Saddle Points

Saddle points are fascinating locations on a surface where the function neither exhibits a true extremum, like a peak or valley. Instead, it may rise in one direction and dip in another, much like the saddle on a horse.
Here's how we identify them:

• Calculate the determinant \(D\) using the second partial derivatives: \[D=f_{xx}(x_{0}, y_{0})f_{yy}(x_{0}, y_{0})-(f_{xy}(x_{0}, y_{0}))^{2}\]
• If \(D < 0\), there's a saddle point at the critical point in question.

In our example, the determinant calculated at the critical point \((0,0)\) indeed comes out less than zero, verifying the presence of a saddle at this point.
These points reflect intriguing behaviors, showing how the surface contours more like valleys in one axis and hills across another. Grasping the concept of saddle points broadens our understanding of complex functions, useful for visualizing real-world phenomena.