Problem 333 - Calculus 3

Question

For the following exercises, use the second derivative test to identify any critical points and determine whether each critical point is a maximum, minimum, saddle point, or none of these. $$ f(x, y)=e^{-\left(x^{2}+y^{2}+2 x\right)} $$

Step-by-Step Solution

Answer

The critical point \((-1, 0)\) is a saddle point.

Step 1: Find the First Partial Derivatives

The first step is to calculate the first partial derivatives of \( f(x, y) = e^{-(x^2 + y^2 + 2x)} \) with respect to \( x \) and \( y \). - \( f_x = \frac{\partial}{\partial x} (e^{-(x^2 + y^2 + 2x)}) = e^{-(x^2 + y^2 + 2x)}(-2x - 2) \)- \( f_y = \frac{\partial}{\partial y} (e^{-(x^2 + y^2 + 2x)}) = e^{-(x^2 + y^2 + 2x)}(-2y) \)

Step 2: Solve for Critical Points

Set the first partial derivatives equal to zero to find the critical points:- \( f_x = 0 \Rightarrow e^{-(x^2 + y^2 + 2x)}(-2x - 2) = 0 \) - \( e^{-(x^2 + y^2 + 2x)} eq 0 \) for any real \( x \) and \( y \), so \( -2x - 2 = 0 \Rightarrow x = -1 \). - \( f_y = 0 \Rightarrow e^{-(x^2 + y^2 + 2x)}(-2y) = 0 \) implies \( -2y = 0 \Rightarrow y = 0 \).Thus, the critical point is \((-1, 0)\).

Step 3: Find the Second Partial Derivatives

Calculate the second partial derivatives which are needed for the second derivative test:- \( f_{xx} = \frac{\partial}{\partial x}(f_x) = \frac{\partial}{\partial x}(e^{-(x^2 + y^2 + 2x)}(-2x - 2)) \). Using the product and chain rule, this gives: \( f_{xx} = (4x + 2)e^{-(x^2 + y^2 + 2x)} - 2e^{-(x^2 + y^2 + 2x)} \).- \( f_{yy} = \frac{\partial}{\partial y}(f_y) = \frac{\partial}{\partial y}(e^{-(x^2 + y^2 + 2x)}(-2y)) = (4y^2 - 2)e^{-(x^2 + y^2 + 2x)} \).- \( f_{xy} = \frac{\partial}{\partial y}(f_x) = \frac{\partial}{\partial y}(e^{-(x^2 + y^2 + 2x)}(-2x - 2)) \). Simplifying, we find:\( f_{xy} = 4xe^{-(x^2 + y^2 + 2x)} \).

Step 4: Evaluate the Second Derivative Test Determinant

Find the determinant of the Hessian matrix at the critical point \((-1, 0)\).The Hessian matrix \( H \) is:\[H = \begin{bmatrix} f_{xx} & f_{xy} \ f_{xy} & f_{yy} \end{bmatrix}\]Calculate the determinant of \( H \) at \((-1, 0)\):\[D = f_{xx}(-1, 0) f_{yy}(-1, 0) - (f_{xy}(-1, 0))^2\]

Step 5: Calculate and Interpret the Determinant

Substitute the second derivatives at the critical point \((-1, 0)\):- \( f_{xx}(-1, 0) = -2e^{-(1)} \) implies it is positive.- \( f_{yy}(-1, 0) = -2e^{-(1)} \) implies it is positive.- \( f_{xy}(-1, 0) = 4\times(-1)\times e^{-(1)} = -4e^{-(1)} \).Now calculate \( D \):\[ D = (-2e^{-1})(-2e^{-1}) - (-4e^{-1})^2 = 4e^{-2} - 16e^{-2} = -12e^{-2} \]Since \( D < 0 \), the point \((-1, 0)\) is a saddle point.

Key Concepts

Partial Derivatives

Partial derivatives are like the magnifying glasses of calculus. They help us understand how a multi-variable function changes when we tweak just one variable, treating all other variables like constants.
For a function like \( f(x, y) \), the first partial derivative with respect to \( x \) , denoted as \( f_x \), shows the rate of change of the function as \( x \) changes, but it's like pressing a pause button on \( y \). Similarly, \( f_y \) tells how \( f \) changes as \( y \) varies. By differentiating \( f(x, y) = e^{-(x^2 + y^2 + 2x)} \), we get two expressions:

• \( f_x = e^{-(x^2 + y^2 + 2x)}(-2x - 2) \)
• \( f_y = e^{-(x^2 + y^2 + 2x)}(-2y) \)

These derivatives give us insights into the slope in different directions on the surface defined by \( f(x, y) \). They are essential first steps in finding critical points of the function.

Critical Points

Imagine you're hiking on a landscape, and you come across flat spots where you can't decide whether you're at a peak, a valley, or just a flat ledge. These are like critical points in calculus.
Critical points occur where the first partial derivatives \( f_x \) and \( f_y \) are both zero. At these points, the surface is flat with no immediate slope.
For the function \( f(x, y) = e^{-(x^2 + y^2 + 2x)} \), setting \( f_x = 0 \) and \( f_y = 0 \) leads us to solve for \( x \) and \( y \), yielding a critical point at \( (-1, 0) \).
This critical point is where we delve deeper using the second derivative test to determine the nature of this flat spot—whether it's a peak (maximum), a valley (minimum), a saddle point, or none of these.

Hessian Matrix

The Hessian matrix is a tool that gathers all second partial derivatives into a neat package. Think of it as a careful analysis of the terrain at a critical point, helping us decide whether it's a maximum, a minimum, or something in between.
This matrix for a function \( f(x, y) \) includes entries for \( f_{xx} \), \( f_{yy} \), and \( f_{xy} \), where each piece examines the curvature in a particular direction.

• \( f_{xx} \) and \( f_{yy} \) assess the curvature along the x-axis and y-axis.
• \( f_{xy} \) evaluates how the two directions influence each other.

Calculating these derivatives for \( f(x, y) = e^{-(x^2 + y^2 + 2x)} \), we form the Hessian matrix \( H \): \[H = \begin{bmatrix} f_{xx} & f_{xy} \ f_{xy} & f_{yy} \end{bmatrix} \] This matrix helps analyze the critical point further, seeing if the surface curves upwards, downwards, or saddles.

Saddle Point

A saddle point is a peculiar spot on a surface that doesn't sit squarely in any camp of being a high point or low point. It's kind of like a mountain pass—it's partway up and partway down.
The distinguishing feature of a saddle point is the change in concavity around it. In particular, it means you'll have a situation where moving along one direction might seem like you're ascending a hill, while moving in another direction feels like descending into a valley.
To identify such points using the Hessian, we look at the determinant of this matrix, calculated at the critical point.
For our function, the determinant \( D \) was calculated as:\[D = (-2e^{-1})(-2e^{-1}) - (-4e^{-1})^2 = -12e^{-2}\]Since \( D < 0 \), this tells us there's a saddle point at \((-1, 0)\). Understanding saddle points is key as they indicate neither maxima nor minima in their immediate vicinity but a more intricate picture of the function surface.