Problem 18 - Calculus an Applied Approach

Question

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

Step-by-Step Solution

Answer

The function \(f(x, y) = 3 e^{-\left(x^{2}+y^{2}\right)}\) has a relative maximum at the point (0,0). There are no saddle points.

Step 1: Compute First Order Partial Derivatives

The first order partial derivatives of the function \(f(x, y)=3 e^{-\left(x^{2}+y^{2}\right)}\) are calculated as follows: \[f_x = -6xe^{-\left(x^{2}+y^{2}\right)}\] \[f_y = -6ye^{-\left(x^{2}+y^{2}\right)}\] These are obtained by differentiating \(f(x, y)\) with respect to \(x\) and \(y\) individually by using chain rule.

Step 2: Find Critical Points

Critical points happen where the first order derivatives equal to zero, i.e., \(-6xe^{-\left(x^{2}+y^{2}\right)} = 0\) and \(-6ye^{-\left(x^{2}+y^{2}\right)} = 0\). This yields the only solution (0,0).

Step 3: Compute Second Order Partial Derivatives

The second order partial derivatives of the function are calculated as follows: \[f_{xx} = -6e^{-\left(x^{2}+y^{2}\right)} + 12x^{2}e^{-\left(x^{2}+y^{2}\right)}\] \[f_{yy} = -6e^{-\left(x^{2}+y^{2}\right)} + 12y^{2}e^{-\left(x^{2}+y^{2}\right)}\] \[f_{xy} = f_{yx} = 12x y e^{-\left(x^{2}+y^{2}\right)}\]

Step 4: Apply the Second Derivative Test

The second derivative test uses type of a quadratic form to determine the nature of the critical point. The quadratic term is denoted with the Hessian Matrix: \[H = \begin{bmatrix} f_{xx} & f_{xy} \ f_{yx} & f_{yy} \end{bmatrix}\] which yields \[H = \begin{bmatrix} f_{xx}(0,0) & f_{xy}(0,0) \ f_{yx}(0,0) & f_{yy}(0,0) \end{bmatrix} = \begin{bmatrix} -6 & 0 \ 0 & -6 \end{bmatrix}\] The determinant, denoted as D, of this matrix is calculated as D = f_{xx} * f_{yy} - (f_{xy})^2. For critical point (0,0), D = (-6*(-6)) - (0^2) = 36. Since D is greater than zero and f_{xx} is less than zero, the point (0,0) is determined to be a relative maximum.

Key Concepts

First Order Partial Derivatives

Understanding first order partial derivatives is essential for analyzing the behavior of multivariable functions. In the case of our function, f(x, y) = 3e^{-(x^2 + y^2)}, these derivatives measure the rate at which the function changes as we move in the x or y direction independently.

The calculation involves taking the derivative of the function with respect to one variable at a time while the other variable is held constant. For our function, we obtained f_x = -6xe^{-(x^2 + y^2)} and f_y = -6ye^{-(x^2 + y^2)}.

These directional derivatives represent the slope in the direction of the x-axis and y-axis, respectively. Having calculated these, we can move on to identifying the critical points of the function.

Critical Points

Critical points are where the first order partial derivatives simultaneously equate to zero. This means there is no immediate tendency for the function to increase or decrease when infinitesimally moving away from these points in any direction of the domain variables.

In our exercise, setting f_x and f_y equal to zero and solving for x and y reveals that (0,0) is a critical point. Critical points can potentially be locations of relative extrema or saddle points, making them a focus of thorough investigation using further mathematical tools, such as the second derivative test.

Second Order Partial Derivatives

To glean more insight about a function at critical points, we calculate the second order partial derivatives. These are the derivatives of the first order partial derivatives and they indicate the curvature of the function in various directions.

In our function, we derive f_{xx}, f_{yy}, and f_{xy} to be -6e^{-(x^2 + y^2)} + 12x^2e^{-(x^2 + y^2)}, -6e^{-(x^2 + y^2)} + 12y^2e^{-(x^2 + y^2)} and 12xye^{-(x^2 + y^2)}, respectively. This information is crucial for the second derivative test, which helps determine the type of critical point we have identified.

Second Derivative Test

The second derivative test is an invaluable tool for categorizing critical points. It uses the values of the second order partial derivatives at the critical point to decide whether it is a relative maximum, minimum, or a saddle point. The nature of the critical point is deduced by evaluating a determinant, which stems from the Hessian matrix.

For the critical point at (0,0) in our function, the determinant is positive and f_{xx} is negative, indicating a relative maximum. If the determinant were negative, we would have a saddle point; if positive with a positive f_{xx}, a relative minimum.

Hessian Matrix

The Hessian matrix plays a pivotal role in the second derivative test. It is a square matrix of all the second order partial derivatives of a function. In a two-variable function like ours, the Hessian matrix is a 2x2 matrix.

For our function, the Hessian at the critical point is H = [[-6, 0], [0, -6]]. We calculate the determinant of this matrix to assess the nature of the critical point. The sign and value of this determinant, alongside the signs of the second order partial derivatives, give us the conclusive characteristics of the critical point.