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Chapter 14: Problem 18

(A) find the function's domain, (b) find the function's range, (c) describe the function's level curves, (d) find the boundary of the function's domain, (e) determine if the domain is an open region, a closed region, or neither, and (f) decide if the domain is bounded or unbounded. $$f(x, y)=\sqrt{y-x}$$

Short Answer

Expert verified
The domain of the function is closed and unbounded; the range is \([0, \infty)\).

Step by step solution

01

Identify Constraints for Domain

The function is defined as \( f(x, y) = \sqrt{y-x} \). The square root function requires \( y-x \geq 0 \) to avoid complex numbers. Thus, the domain consists of all points \((x, y)\) such that \( y \geq x \).
02

Describe the Function's Domain

Given the inequality from Step 1, the domain of \( f \) is the set of all points in the \(xy\)-plane where \( y \geq x \). In interval notation, it can be expressed as \{(x,y) \mid y \geq x\}.
03

Find the Function's Range

Since the square root function produces values from \(0\) to \(\infty\), and \(y - x \geq 0\) for points in the domain, the range of \(f(x, y)\) is \([0, \infty)\).
04

Describe the Level Curves

Level curves of the function are defined by the equation \(\sqrt{y-x} = k\) where \(k\) is a constant. Solving for \(y\), we get \(y = x + k^2\). Each level curve is a line parallel to \(y = x\) with a vertical shift depending on \(k^2\).
05

Find the Boundary of the Domain

The boundary of the domain \(y \geq x\) is the line \(y = x\). This line is included in the domain as points on it satisfy \(y \geq x\).
06

Determine if the Domain is Open, Closed, or Neither

The domain \(y \geq x\) includes its boundary \(y = x\), hence it is a closed region.
07

Determine if the Domain is Bounded or Unbounded

The domain, extending indefinitely in the direction \(y > x\), is unbounded as there are no limits to how large \(x\) or \(y\) can get within the inequality \(y \geq x\).

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Key Concepts

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

Domain of a Function
Understanding the domain of a function is crucial because it tells us where the function is defined. For functions in multivariable calculus, like \( f(x, y) = \sqrt{y-x} \), we need to consider how the variables interact. The square root function only works for non-negative inputs. This means:
  • We need \( y - x \geq 0 \).
  • Simply put, \( y \) must be greater than or equal to \( x \).
Thus, the domain includes all points \((x, y)\) that satisfy this condition. To visualize this, imagine shading the area in the \(xy\)-plane above the line \( y = x \). This region represents all the points where the function can "live" and be calculated without issues.
Range of a Function
The range of a function tells us all the possible output values. For \( f(x, y) = \sqrt{y-x} \), understanding the range requires looking at the values the square root can yield.
  • Clearly, since we're dealing with a square root, the smallest value is \(0\).
  • The largest value can go to infinity, as \( y \) can be much larger than \( x \).
Combining these insights means the range is \[ [0, \infty) \]. This range tells us the function can produce any value from \(0\) all the way up to as large as needed, akin to unlocking a door with limitless possibilities as long as the domain condition \(y \geq x \) is met.
Level Curves
Level curves are an essential way to understand functions in multivariable calculus, providing a snapshot of the function at constant output values. For \( f(x, y) = \sqrt{y-x} \), we assess these curves by setting the function equal to a constant \( k \).
  • The equation becomes \( \sqrt{y-x} = k \), leading to \( y = x + k^2 \).
  • This represents a family of lines parallel to \( y = x \), shifted vertically by \( k^2 \).
Each level curve gives a different "slice" of the function, painting a picture of how the function behaves at various heights. It's like taking a flat map of a hilly terrain, helping to visualize the landscape of the function.
Bounded and Unbounded Regions
In multivariable calculus, understanding whether a domain is bounded or unbounded helps determine the scope of the function. The domain of \( f(x, y) = \sqrt{y-x} \) was identified as \( y \geq x \).
  • Since both \( x \) and \( y \) can increase indefinitely as long as \( y \) remains greater than or equal to \( x \), the domain is unbounded.
  • There are no walls or borders that eventually stop \( x \) or \( y \) from growing.
Understanding bounded versus unbounded regions is like knowing whether a park has a fence (bounded) or is open to the vast countryside (unbounded). Here, the domain stretches infinitely, just like an endless landscape.

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Most popular questions from this chapter

You will explore functions to identify their local extrema. Use a CAS to perform the following steps: a. Plot the function over the given rectangle. b. Plot some level curves in the rectangle. c. Calculate the function's first partial derivatives and use the CAS equation solver to find the critical points. How do the critical points relate to the level curves plotted in part (b)? Which critical points, if any, appear to give a saddle point? Give reasons for your answer. d. Calculate the function's second partial derivatives and find the discriminant \(f_{x x} f_{y y}-f_{x y}^{2}\). e. Using the max-min tests, classify the critical points found in part (c). Are your findings consistent with your discussion in part (c)? $$\begin{aligned} &f(x, y)=\left\\{\begin{array}{ll} x^{5} \ln \left(x^{2}+y^{2}\right), & (x, y) \neq(0,0) \\ 0, & (x, y)=(0,0) \end{array}\right.,\\\ &-2 \leq x \leq 2,-2 \leq y \leq 2 \end{aligned}$$

Find and sketch the level curves \(f(x, y)=c\) on the same set of coordinate axes for the given values of \(c .\) We refer to these level curves as a contour map. $$f(x, y)=x y, \quad c=-9,-4,-1,0,1,4,9$$

Find and sketch the domain for each function. $$f(x, y)=\frac{(x-1)(y+2)}{(y-x)\left(y-x^{3}\right)}$$

Show that each function satisfies a Laplace equation. \(f(x, y, z)=\left(x^{2}+y^{2}+z^{2}\right)^{-1 / 2}\)

Find and sketch the domain for each function. $$f(x, y)=\frac{\sin (x y)}{x^{2}+y^{2}-25}$$
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