/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 44 Sketch the level curve \(z=k\) f... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 44

Sketch the level curve \(z=k\) for the specified values $$ z=y / x ; k=-2,-1,0,1,2 $$

Short Answer

Expert verified
Each level curve is a straight line through the origin, with slope equal to \( k \).

Step by step solution

01

Understand the Function

The function given is \( z = \frac{y}{x} \), where \( z \) represents the level curve value. Level curves are obtained by setting \( z \) equal to constant values \( k \). We will consider \( k = -2, -1, 0, 1, 2 \). This means we need to sketch the curve represented by \( \frac{y}{x} = k \) for each of these values.
02

Rearrange the Equation

For each \( k \), rearrange the equation \( \frac{y}{x} = k \) to express \( y \) in terms of \( x \): \[ y = kx \]This linear relationship simplifies the sketching of the curve for each \( k \).
03

Sketch Each Level Curve for Specific k Values

Draw the curve for each value of \( k \) by considering the equation \( y = kx \):- For \( k = -2 \), the equation becomes \( y = -2x \), a line with slope -2.- For \( k = -1 \), the equation becomes \( y = -x \), a line with slope -1.- For \( k = 0 \), the equation becomes \( y = 0 \), which is the x-axis.- For \( k = 1 \), the equation becomes \( y = x \), a line with slope 1.- For \( k = 2 \), the equation becomes \( y = 2x \), a line with slope 2.Each line passes through the origin (0,0) since \( b = 0 \) in the line equation \( y = mx + b \).
04

Label the Level Curves

Label each line with the corresponding \( k \) value to visually distinguish between them. This helps indicate which line represents which level curve. The steeper the slope, the larger the absolute value of \( k \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Linear Equations
Linear equations form the backbone of many mathematical principles, and they are integral to understanding level curves. The general form of a linear equation is\[ y = mx + b \]where \( m \) represents the slope and \( b \) is the y-intercept. In the exercise, each level curve is represented by a similar linear equation, such as \( y = kx \), where \( b = 0 \), causing the lines to pass through the origin.

Nationally important is how simple transformations like multiplying the slope by \( k \), affect the orientation and steepness of these lines. By observing the constant \( k \), we see there are different slopes (like \(-2, -1, 0, 1, 2\)), leading to diverse line orientations around the origin. Each value of \( k \) gives us crucial insight as it dictates the direction and angle of the line in the coordinate plane.

Understanding linear equations in this manner enables us to quickly assess and sketch changes in complex functions through simpler linear alterations.
Slope
The concept of slope is a core element in interpreting how lines behave within a graph. The slope \( m \) of a line given by the equation \( y = mx + b \) measures the line's steepness. It is calculated as:\[ m = \frac{\text{rise}}{\text{run}} \],indicating the change in \( y \) values over the change in \( x \).

In the provided exercise, slope directly corresponds to the constant \( k \) in the equation \( y = kx \). This means:
  • For \( k = -2 \), the slope is -2, indicating a steep downward direction.
  • For \( k = -1 \), the slope is -1, still a downward, but less steep, line.
  • For \( k = 0 \), the line is flat, running along the x-axis.
  • For \( k = 1 \), the slope is 1, indicating a gentle upward line.
  • For \( k = 2 \), the slope is 2, reflecting a steep upward trend.
The slope \( m \) is not just a number; it physically represents how a line moves across a graph and determines how quickly \( y \) changes with \( x \). This makes understanding slope essential for sketching accurate and meaningful graphs.
Function Sketching
Function sketching is about translating algebraic equations into visual representations. By sketching the curve of a function like \( y = kx \), you convert mathematical expressions into something tangible on graph paper or a screen.

In creating a sketch, it is essential to identify key features:
  • Where the line intersects the axes (primarily through the origin for these lines).
  • The angle and direction of the line, which is determined by the slope \( k \).
For the exercise, sketching each line according to the values of \( k \) reveals different linear behaviors. For example:
  • Lines for \( k > 0 \) incline upwards as they extend from the origin.
  • Lines for \( k < 0 \) descend downwards.

The act of sketching these functions allows more than just visualizing the math. It helps in understanding how such lines can combine or impact one another in space. Simplifying complex mathematical relationships through sketching offers a clearer insight into their nature and purpose.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(f(x, y)=\left(x^{2}+y^{2}\right)^{2 / 3} .\) Show that $$ f_{x}(x, y)=\left\\{\begin{array}{cc}{\frac{4 x}{3\left(x^{2}+y^{2}\right)^{1 / 3}},} & {(x, y) \neq(0,0)} \\ {0,} & {(x, y)=(0,0)}\end{array}\right. $$

When two resistors having resistances \(R_{1}\) ohms and \(R_{2}\) ohms are connected in parallel, their combined resistance \(R\) in ohms is \(R=R_{1} R_{2} /\left(R_{1}+R_{2}\right) .\) Show that $$\frac{\partial^{2} R}{\partial R_{1}^{2}} \frac{\partial^{2} R}{\partial R_{2}^{2}}=\frac{4 R^{2}}{\left(R_{1}+R_{2}\right)^{4}}$$

A manufacturer makes two models of an item, standard and deluxe. It costs \(\$ 40\) to manufacture the standard model and \(\$ 60\) for the deluxe. A market research firm estimates that if the standard model is priced at \(x\) dollars and the deluxe at \(y\) dollars, then the manufacturer will sell \(500(y-x)\) of the standard items and \(45,000+500(x-2 y)\) of the deluxe each year. How should the items be priced to maximize the profit?

A common problem in experimental work is to obtain a mathematical relationship \(y=f(x)\) between two variables \(x\) and \(y\) by "fitting" a curve to points in the plane that correspond to experimentally determined values of \(x\) and \(y,\) say $$ \left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right) $$ The curve \(y=f(x)\) is called a mathematical model of the data. The general form of the function \(f\) is commonly determined by some underlying physical principle, but sometimes it is just determined by the pattern of the data. We are concerned with fitting a straight line \(y=m x+b\) to data. Usually, the data will not lie on a line (possibly due to experimental error or variations in experimental conditions), so the problem is to find a line that fits the data "best" according to some criterion. One criterion for selecting the line of best fit is to choose \(m\) and \(b\) to minimize the function $$ g(m, b)=\sum_{i=1}^{n}\left(m x_{i}+b-y_{i}\right)^{2} $$ This is called the method of least squares, and the resulting line is called the regression line or the least squares line of best fit. Geometrically, \(\left|m x_{i}+b-y_{i}\right|\) is the vertical distance between the data point \(\left(x_{i}, y_{i}\right)\) and the line \(y=m x+b\) These vertical distances are called the residuals of the data points, so the effect of minimizing \(g(m, b)\) is to minimize the sum of the squares of the residuals. In these exercises, we will derive a formula for the regression line. The purpose of this exercise is to find the values of \(m\) and \(b\) that produce the regression line. (a) To minimize \(g(m, b),\) we start by finding values of \(m\) and \(b\) such that \(\partial g / \partial m=0\) and \(\partial g / \partial b=0 .\) Show that these equations are satisfied if \(m\) and \(b\) satisfy the conditions $$ \left(\sum_{i=1}^{n} x_{i}^{2}\right) m+\left(\sum_{i=1}^{n} x_{i}\right) b=\sum_{i=1}^{n} x_{i} y_{i} $$ \(\left(\sum_{i=1}^{n} x_{i}\right) m+n b=\sum_{i=1}^{n} y_{i}\) (b) Let \(\bar{x}=\left(x_{1}+x_{2}+\cdots+x_{n}\right) / n\) denote the arithmetic average of \(x_{1}, x_{2}, \ldots, x_{n} .\) Use the fact that $$ \sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2} \geq 0 $$ to show that $$ n\left(\sum_{i=1}^{n} x_{i}^{2}\right)-\left(\sum_{i=1}^{n} x_{i}\right)^{2} \geq 0 $$ with equality if and only if all the \(x_{i}\) 's are the same. (c) Assuming that not all the \(x_{i}\) 's are the same, prove that the equations in part (a) have the unique solution $$ \begin{aligned} m=& \frac{n \sum_{i=1}^{n} x_{i} y_{i}-\sum_{i=1}^{n} x_{i} \sum_{i=1}^{n} y_{i}}{n \sum_{i=1}^{n} x_{i}^{2}-\left(\sum_{i=1}^{n} x_{i}\right)^{2}} \\ b=& \frac{1}{n}\left(\sum_{i=1}^{n} y_{i}-m \sum_{i=1}^{n} x_{i}\right) \end{aligned} $$ [Note: We have shown that \(g\) has a critical point at these values of \(m\) and \(b\). In the next exercise we will show that \(g\) has an absolute minimum at this critical point. Accepting this to be so, we have shown that the line \(y=m x+b\) is the regression line for these values of \(m \text { and } b .]\)

Let \(\alpha, \beta,\) and \(\gamma\) be the angles of a triangle. (a) Use Lagrange multipliers to find the maximum value of \(f(\alpha, \beta, \gamma)=\cos \alpha \cos \beta \cos \gamma,\) and determine the angles for which the maximum occurs. (b) Express \(f(\alpha, \beta, \gamma)\) as a function of \(\alpha\) and \(\beta\) alone, and use a CAS to graph this function of two variables. Confirm that the result obtained in part (a) is consistent with the graph.
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Calculus

Read Explanation

Statistics

Read Explanation

Decision Maths

Read Explanation

Geometry

Read Explanation

Pure Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.