Chapter 13: Problem 24
Describe the graph of the function \(f\). $$ f(x, y)=\sqrt{x^{2}+y^{2}} $$
Short Answer
Expert verified
The graph of \(f(x, y) = \sqrt{x^2 + y^2}\) is a cone opening upward, centered on the z-axis.
Step by step solution
01
Analyze the Function
The function given is \(f(x, y) = \sqrt{x^2 + y^2}\). This is a two-variable function, meaning it takes points \((x, y)\) in 2D and maps them to a value in 3D space.
02
Recognize the Type of Surface
Notice that the function \(\sqrt{x^2 + y^2}\) is similar in form to the equation for the distance from the origin in polar coordinates \(r = \sqrt{x^2 + y^2}\). This suggests that the function describes distances from the origin, forming a circular or radial pattern.
03
Consider the Domain and Range
The domain of \(f(x, y)\) is all \((x, y)\) in \(\mathbb{R}^2\) since any real number can be squared and under the square root. The range of \(f(x, y)\) is \([0, \infty)\) because distances (and thus, square roots of squared terms) are non-negative.
04
Describe the Level Curves
Level curves of the function, where \(f(x, y) = k\) for some constant \(k\), give \(\sqrt{x^2 + y^2} = k\), equivalent to \(x^2 + y^2 = k^2\). These are circles centered at the origin, with radius \(k\).
05
Visualize the Graph
The graph of \(f(x, y) = \sqrt{x^2 + y^2}\) creates a surface known as a cone, which extends in the positive z-axis direction, widening as \(x^2+y^2\) increases. It is symmetric around the z-axis.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
3D Graphing
3D graphing is a powerful tool that helps us visualize functions with two variables. When we graph a function like \(f(x, y) = \sqrt{x^2 + y^2}\), it maps every point \((x, y)\) on a 2D plane to a point on a 3D surface. This particular function results in a cone-shaped graph that stretches outward and upward from the origin.
To imagine this surface, start by visualizing a flat circular area centered at the origin in 2D. Each circle's distance from the origin forms a rising line in 3D, creating a cone.
To imagine this surface, start by visualizing a flat circular area centered at the origin in 2D. Each circle's distance from the origin forms a rising line in 3D, creating a cone.
- The base of the cone lies flat in the xy-plane.
- As we move further from the center, the value of the function increases.
- The surface is smooth, with each circle lifting higher than the last.
Level Curves
Level curves offer a unique way to understand the contours of a 3D surface on a 2D plane. For \(f(x, y) = \sqrt{x^2 + y^2}\), level curves are circles centered at the origin. These curves are derived from setting the function equal to a constant value \(k\), or \(\sqrt{x^2 + y^2} = k\). This simplifies to the equation of a circle: \(x^2 + y^2 = k^2\).
Level curves provide several advantages:
Level curves provide several advantages:
- They reduce complex 3D shapes to simpler 2D representations.
- Each curve corresponds to a particular height on the 3D graph.
- When level curves are close, the surface is steep in those areas.
- If they are spaced out, the surface is relatively flat.
Domain and Range
In multivariable calculus, understanding the domain and range is crucial for evaluating how a function behaves across different inputs. For the function \(f(x, y) = \sqrt{x^2 + y^2}\), the domain is all pairs of real numbers \((x, y)\). That's because squaring any real number yields a valid input for the square root.
However, the function's range is confined to non-negative values \([0, \infty)\), as the square root outputs positives or zero. Key points to remember include:
However, the function's range is confined to non-negative values \([0, \infty)\), as the square root outputs positives or zero. Key points to remember include:
- The domain encompasses the entire xy-plane, meaning any real number combination of \(x\) and \(y\) is valid.
- The range ensures that no negative values are produced, as you can’t take a square root of a negative number without involving imaginary numbers.
- Edge cases like \(x = 0\) and \(y = 0\) result in \(f(x, y) = 0\), a valid output within the range.