Problem 13 - Calculus an Applied Approach

Question

Examine the function for relative extrema and saddle points. $$ f(x, y)=x^{2}-y^{2}+4 x-4 y-8 $$

Step-by-Step Solution

Answer

The function \( f(x, y)=x^{2}-y^{2}+4 x-4 y-8 \) has a saddle point at (-2,-2).

Step 1: Find the first partial derivatives

We need to start by finding the first partial derivatives. For function \( f(x, y) \), these are given by: \n \( f_x = 2x + 4 \) \n \( f_y = -2y - 4 \)

Step 2: Find the critical points

Critical points occur where both first partial derivatives are zero. Set \( f_x = 0 \) and \( f_y = 0 \) and solve for x and y: \n For \( f_x = 2x + 4 = 0 \) gives \( x = -2 \) \n For \( f_y = -2y - 4 = 0 \) gives \( y = -2 \) \n So, the critical point is (-2,-2).

Step 3: Calculate the second partial derivatives

Second partial derivatives will help us to determine whether the critical points are relative minima, maxima or saddle points. They are given by: \n \( f_{xx} = 2 \) \( f_{yy} = -2 \) \( f_{xy} = f_{yx} = 0 \)

Step 4: Apply the Second partial derivative test

The second partial derivative test uses the value of \( D = f_{xx}*f_{yy} - (f_{xy})^2 \). If D > 0 and \( f_{xx} > 0 \) then the point is a relative minimum, if D > 0 and \( f_{xx} < 0 \) then the point is a relative maximum, if D < 0 then the point is a saddle point. So, evaluating D at critical point, we get: \n \( D = 2*(-2) - (0)^2 = -4 \) \n Since D < 0, the point (-2,-2) is a saddle point.

Key Concepts

Partial Derivatives

Partial derivatives allow us to explore how a multivariable function changes with respect to one variable while keeping the others constant. For a function like \( f(x, y) = x^{2} - y^{2} + 4x - 4y - 8 \), the first step is to find these derivatives.
- **For \( x \):** Derivate only with respect to \( x \). We treat \( y \) as a constant.- **For \( y \):** Derivate only with respect to \( y \). We treat \( x \) as a constant.
This results in:

• \( f_x = 2x + 4 \)
• \( f_y = -2y - 4 \)

These equations are essential in identifying points where the surface behavior of the function might shift dramatically.

Critical Points

Critical points are locations where the first partial derivatives of a function are zero. These points can indicate potential maxima, minima, or saddle points in the function.
For our function, set \( f_x = 0 \) and \( f_y = 0 \) and solve for \( x \) and \( y \):

• \( 2x + 4 = 0 \) gives \( x = -2 \)
• \( -2y - 4 = 0 \) gives \( y = -2 \)

Thus, our critical point is \((-2, -2)\). It's crucial to assess this point further; we need to know if it's a peak, a valley, or something else.

Second Partial Derivative Test

The Second Partial Derivative Test helps to classify critical points. It involves calculating the second derivatives and combining them into a determinant:\[ D = f_{xx}f_{yy} - (f_{xy})^2 \]
For our function, calculate:

• \( f_{xx} = 2 \)
• \( f_{yy} = -2 \)
• \( f_{xy} = f_{yx} = 0 \)

Insert these into the formula:
\( D = 2(-2) - (0)^2 = -4 \)
The sign of \( D \) provides crucial information:- If \( D > 0 \) and \( f_{xx} > 0 \): a local minimum.- If \( D > 0 \) and \( f_{xx} < 0 \): a local maximum.- If \( D < 0 \): a saddle point.
In this case, with \( D = -4 \), we anticipate a saddle point.

Saddle Points

A saddle point on a surface is where the curvature changes direction. Specifically, it is a point on the graph which is neither a peak nor a valley but rather a bit of both.
In mathematical terms, it's a point where the Second Partial Derivative Test results in \( D < 0 \), indicating a change in the nature of curvature around the point.
In our example with \( f(x, y) = x^{2} - y^{2} + 4x - 4y - 8 \), the critical point found was \((-2, -2)\). With \( D = -4 \), it confirms the point is a saddle point.
Saddle points can be intuitive in physical models, like a mountain pass, showing a directional change neither entirely upward nor downward. This unique property highlights critical points in analyzing the behavior of complex functions.