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Chapter 14: Problem 29

Sketch the graph of the function. $$f(x, y)=\sqrt{x^{2}+y^{2}}$$

Short Answer

Expert verified
The graph of \( f(x, y) = \sqrt{x^2 + y^2} \) is a cone with its tip at the origin (0,0,0).

Step by step solution

01

Understand the function

The function given is \( f(x, y) = \sqrt{x^2 + y^2} \). This function takes two variables, \( x \) and \( y \), and computes the square root of the sum of their squares. This is a common expression used to describe the Euclidean distance from the origin (0, 0) to the point (x, y) in a coordinate plane.
02

Identify the Nature of the Graph

The equation \( z = \sqrt{x^2 + y^2} \) represents a three-dimensional surface. This is the formula for the radius of a circle centered at the origin in the xy-plane, but for \( z \), it represents the surface of a cone with its tip at the origin.
03

Analyze the Level Curves

In order to better understand the surface, consider fixing \( z \). When \( z \) is constant (say \( z = c \)), the equation becomes \( c = \sqrt{x^2 + y^2} \) or equivalently \( x^2 + y^2 = c^2 \). These are equations of circles of radius \( c \) centered at the origin in the xy-plane. As \( c \) increases, these circles get larger.
04

Consider the Surface in 3D

Imagine the circles \( x^2 + y^2 = c^2 \) being extruded upward along the z-axis as \( c \) increases, forming an upside-down cone. As both \( x \) and \( y \) increase in value, \( z \) also increases, maintaining the shape of a cone.
05

Draw the Sketch

Draw the graph by sketching a 3D cone. The tip of the cone should be at the origin (0, 0, 0). The cones surface represents the points \( (x, y, z) \) that satisfy \( z = \sqrt{x^2 + y^2} \), expanding outward with increasing z.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Function Graphing
Function graphing involves visualizing mathematical expressions, usually as curves or surfaces, on a set of axes. In multivariable calculus, we often extend this to represent functions with more than one variable. For the function \( f(x, y) = \sqrt{x^2 + y^2} \), this involves a 3D graph. This function is a classic example, converting from a formula in two variables \( x \) and \( y \) into a meaningful visual form.
  • When we say 'graph the function,' we're looking to portray the set of points \((x, y, z)\) that satisfy \( z = f(x, y) \).
  • The graph in question here is a 3D surface, which is often more intricate than simple 2D graphs.
Understanding the shape and nature of this surface is key to interpreting multivariable functions.
3D Surfaces
3D surfaces are crucial in multivariable calculus, providing a tangible representation of functions with two inputs. The function \( z = \sqrt{x^2 + y^2} \) forms a cone in three-dimensional space. This means for each point \( (x, y) \), we calculate \( z \) to locate a point on the cone's surface.
  • Visualize the xy-plane as a horizontal flat surface.
  • The height, \( z \), is directly related to the distance from the origin on the xy-plane.
  • This builds a sloping surface which increases continuously as you move further from the center.
3D surfaces bring mathematical functions into the physical world, allowing easier conceptualization of how variables interact.
Level Curves
Level curves are 2D slices of a 3D surface, showing lines along which the function takes constant values. Fixing \( z \) gives curves like \( x^2 + y^2 = c^2 \), which are circles in this context. Each circle represents a constant height on the cone's surface.
  • These circles emerge from the xy-plane horizontally, showing constant values of \( z \).
  • As \( z \) increases, the radius of these circles increases, representing larger distances from the origin.
Level curves are useful for understanding the structure of a surface by studying these constant value lines. They are like contour lines on a topographic map, which indicates height.
Euclidean Distance
The concept of Euclidean distance is central in this exercise. It gives the straight-line distance between two points in the calculation \( \sqrt{x^2 + y^2} \). This is essentially the function's definition鈥攁 measure from the origin \( (0,0) \) to any point \( (x, y) \) on a plane.
  • This distance formula helps in understanding why the graph is a cone.
  • Every point \( z \) at height represents the distance from the origin to the point \( (x, y) \) on the circle \( x^2 + y^2 = c^2 \).
Therefore, the Euclidean distance gives this function its shape, forming the foundational geometry of the cone. It is one of the simplest yet most profound concepts in understanding spatial relationships.

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Most popular questions from this chapter

Find all the second partial derivatives. $$f(x, y)=x^{3} y^{5}+2 x^{4} y$$

The gas law for a fixed mass \(m\) of an ideal gas at absolute temperature \(T,\) pressure \(P,\) and volume \(V\) is \(P V=m R T,\) where \(R\) is the gas constant. Show that $$\frac{\partial P}{\partial V} \frac{\partial V}{\partial T} \frac{\partial T}{\partial P}=-1$$

\(45-48\) Assume that all the given functions are differentiable. If \(z=f(x, y),\) where \(x=s+t\) and \(y=s-t,\) show that $$\left(\frac{\partial z}{\partial x}\right)^{2}-\left(\frac{\partial z}{\partial y}\right)^{2}=\frac{\partial z}{\partial s} \frac{\partial z}{\partial t}$$

Use a computer to investigate the family of surfaces $$z=\left(a x^{2}+b y^{2}\right) e^{-x^{2}-y^{2}}$$ How does the shape of the graph depend on the numbers a and \(b ?\)

Use a graph and/or level curves to estimate the local maximum and minimum values and saddle point(s) of the function. Then use calculus to find these values precisely. $$f(x, y)=x y e^{-x^{2}-y^{2}}$$
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