Problem 314 - Calculus 3

Question

For the following exercises, find the critical points of the function by using algebraic techniques (completing the square) or by examining the form of the equation. Verify your results using the partial derivatives test. $$ f(x, y)=\sqrt{x^{2}+y^{2}+1} $$

Step-by-Step Solution

Answer

The critical point is at (0, 0), and it is a local minimum.

Step 1: Understand the Function

The function given is \( f(x, y) = \sqrt{x^2 + y^2 + 1} \). It's a multivariable function that involves a square root of a sum of squares.

Step 2: Find Partial Derivatives

To find the critical points, first compute the partial derivatives with respect to \(x\) and \(y\). The partial derivative with respect to \(x\) is \( f_x = \frac{x}{\sqrt{x^2 + y^2 + 1}} \), and with respect to \(y\) is \( f_y = \frac{y}{\sqrt{x^2 + y^2 + 1}} \).

Step 3: Solve for Critical Points

Set the partial derivatives equal to zero to find critical points: 1. \( \frac{x}{\sqrt{x^2 + y^2 + 1}} = 0 \), which implies \( x = 0 \).2. \( \frac{y}{\sqrt{x^2 + y^2 + 1}} = 0 \), which implies \( y = 0 \).Thus, the critical point is \((0, 0)\).

Step 4: Verify with Second Derivative Test

For the second derivative test, compute the second order partial derivatives: 1. \( f_{xx} = \frac{y^2 + 1}{(x^2 + y^2 + 1)^{3/2}} \)2. \( f_{yy} = \frac{x^2 + 1}{(x^2 + y^2 + 1)^{3/2}} \)3. \( f_{xy} = -\frac{xy}{(x^2 + y^2 + 1)^{3/2}} \). Evaluate at the critical point \((0,0)\): \( f_{xx}(0, 0) = f_{yy}(0, 0) = 1 \), and \( f_{xy}(0, 0) = 0 \). The determinant of the Hessian is \( D = f_{xx}f_{yy} - (f_{xy})^2 = 1 \times 1 - 0^2 = 1 \). Since \(D > 0\) and \( f_{xx} > 0 \), it is a local minimum.

Key Concepts

Partial Derivatives

In the world of multivariable calculus, partial derivatives act like a powerful magnifying glass, allowing us to examine how a function changes with respect to one variable, while keeping other variables constant. Imagine them as directional insights into the function's behavior.

When working with functions of multiple variables, such as the one in this problem, you have partial derivatives for each variable. For our function, given as \( f(x, y) = \sqrt{x^2 + y^2 + 1} \), we calculate partial derivatives \( f_x \) and \( f_y \):

• \( f_x = \frac{x}{\sqrt{x^2 + y^2 + 1}} \) focuses on the rate of change in the direction of \(x\).
• \( f_y = \frac{y}{\sqrt{x^2 + y^2 + 1}} \) captures how the function changes as \(y\) increases or decreases.

To find critical points, we set these partial derivatives equal to zero. These conditions help highlight specific points where the function isn't changing in any of the studied directions, known as critical points.

Multivariable Function

A multivariable function like \( f(x, y) \) involves more than one input. It assigns a single output based on the combined effect of both \(x\) and \(y\).
The task is to discover the critical points, key locations where the function's rate of change is zero in all directions. Here, understanding the shape and behavior of the function across both variables is essential.

In the case of our example, the function is a square root of a sum involving \(x^2\) and \(y^2\). Such expressions often define a surface in three-dimensional space, characterized by its radius from a central point, revealing rich insights about their geometrical properties.For solving this problem, we aim to determine how and where the terrain of this surface reaches a peak, valley, or flattens out, all thanks to the interplay of \(x\) and \(y\). Finding where these variables interact critically allows us to fully appreciate the landscape formed by the function.

Second Derivative Test

The second derivative test is a pivotal tool to classify the nature of critical points. Once the critical points are identified using partial derivatives, we turn to the second derivative test to determine if these points are maxima, minima, or saddle points.

For the function \( f(x, y) = \sqrt{x^2 + y^2 + 1} \), we compute second-order partial derivatives. These derivatives provide the necessary measures to perform the test:

• \( f_{xx} = \frac{y^2 + 1}{(x^2 + y^2 + 1)^{3/2}} \)
• \( f_{yy} = \frac{x^2 + 1}{(x^2 + y^2 + 1)^{3/2}} \)
• \( f_{xy} = -\frac{xy}{(x^2 + y^2 + 1)^{3/2}} \)

Evaluating these at the critical point \((0, 0)\), we check the determinant of the Hessian matrix: \[ D = f_{xx}f_{yy} - (f_{xy})^2 = 1 \times 1 - 0^2 = 1 \]With \(D > 0\) and \(f_{xx} > 0\), the critical point is confirmed as a local minimum.
This process provides a clear classification, helping us understand the function's behavior at the critical point. Thus, with critical examination and thoughtful calculations, we navigate the landscape of multivariable functions successfully.