Problem 15 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 7-16, sketch the graph of \(f\). $$ f(x, y)=e^{-\left(x^{2}+y^{2}\right)} $$

Step-by-Step Solution

Answer

The graph of \(f(x, y) = e^{-(x^2 + y^2)}\) is a symmetric bell-shaped surface centered at the origin.

Step 1: Analyze the Function

The given function is \(f(x, y) = e^{-(x^2 + y^2)}\). This function represents a surface in three-dimensional space. The term \(x^2 + y^2\) in the exponent suggests that the function depends on the distance from the origin in the \(xy\)-plane.

Step 2: Recognize Symmetry

The function is symmetric around the origin, meaning it has rotational symmetry about the \(z\)-axis. This is because \(x^2 + y^2\) remains the same for any rotation about the z-axis at a fixed distance from the origin.

Step 3: Evaluate Key Points

To understand the shape of the graph, evaluate \(f(x, y)\) at key points. At the origin \((0, 0)\), \(f(0, 0) = e^{0} = 1\). As \(x^2 + y^2\) becomes larger (moving away from the origin), \(f(x, y)\) approaches 0 because the exponent \(- (x^2 + y^2)\) becomes more negative.

Step 4: Identify Contour Levels

Set \(e^{-(x^2 + y^2)} = k\) for some constant \(k\), where \(0 < k < 1\). This equation \(x^2 + y^2 = - ext{ln}(k)\) represents a circle in the \(xy\)-plane with center at the origin. The graph consists of concentric circles of different radii based on different \(k\)-values.

Step 5: Sketch the 3D Surface

Sketch the graph in 3D, using the contours from Step 4 as a guide. The surface resembles a bell shape—a Gaussian surface or a multivariable normal distribution—centered at the origin (\(0,0,1\)) and tapering off symmetrically as it moves away from the origin along any axis.

Key Concepts

3D Surface Graphs

3D surface graphs are a fascinating way to represent functions of two variables, like our function \(f(x, y) = e^{-(x^2 + y^2)}\). This type of graph plots the input values \((x, y)\) on a flat plane and the resulting output \(f(x, y)\) as elevation or height, creating a surface in three-dimensional space.
In our example, the 3D surface graph has a peak at the origin \((0, 0)\) where the function reaches its maximum value of 1. As you move away from the origin in any direction, along the \(x\)- or \(y\)-axes, the elevation decreases, creating a surface that smoothly slopes downwards, eventually approaching zero.
This shape is similar to a gently-sloping hill or mountain in a 3D view, which is characteristic of Gaussian distributions. The surface graph helps visualize how the function distributes its values across the plane, with the peak indicating where it's most concentrated.

Gaussian Distribution

The function \(f(x, y) = e^{-(x^2 + y^2)}\) is an example of a Gaussian distribution in two dimensions. Gaussian distributions, also known as "bell curves," are central in statistics due to their symmetric and predictable shape.
For this function, the highest point or the mode of the distribution is at the origin, given by \(f(0, 0) = 1\). As \(x^2 + y^2\) increases – meaning as you move away from the origin in any direction – the value of the function decreases exponentially. This rate of decrease leads to the bell-shaped curve when visualized as a 3D surface.

• The symmetry and peak at the origin represent a central point of greatest density or probability.
• The smooth tapering away from the peak reflects the likelihood of finding lower values as you move outwards.

This distribution is often used in fields like physics and statistics to model varieties of data and processes.

Symmetry in Functions

Symmetry in functions is a key aspect that can simplify understanding and graphing. For our function \(f(x, y) = e^{-(x^2 + y^2)}\), the symmetry is quite special. It is symmetric about the origin in the \(xy\)-plane and revolves around the \(z\)-axis.
What this means is that the value of \(f(x, y)\) only depends on the distance from the origin, not the specific direction. If you rotate the point around the \(z\)-axis at the same radial distance, the function’s value remains unchanged. This makes the graph rotationally symmetric and results in a surface that looks identical from any perspective around the axis.
Recognizing symmetry helps in sketching the function, predicting behaviors, or simplifying calculations. Knowing that the circle is a contour level, you can predict repeated patterns that help in visualizing the entire graph.

Contour Levels

Contour levels are crucial for understanding functions of two variables without plotting a 3D surface. By setting \(f(x, y) = e^{-(x^2 + y^2)} = k\), where \(0 < k < 1\), we can find contour levels of the function.
These contour levels are described by the equation \(x^2 + y^2 = -\ln(k)\). For any chosen \(k\), this represents a circle centered at the origin with a radius dependent on \(-\ln(k)\).

• For example, if \(k = 0.5\), the contour level is a circle around the origin.
• As \(k\) increases, the circles shrink, illustrating higher function values closer to the peak.

Contour lines enable us to perceive depth and slopes on a flat image of the function, crucial for insights into the behavior and distribution across the plane.