/*! 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 2 In this problem we show that the... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 2

In this problem we show that the function $$f(x, y)=\frac{6 x-y}{x+y}$$ does not have a limit as \((x, y) \rightarrow(0,0)\). (a) Suppose that we consider \((x, y) \rightarrow(0,0)\) along the curve \(y=4 x\). Find the limit in this case: \(\lim _{(x, 4 x) \rightarrow(0,0)} \frac{6 x-y}{x+y}=\) __________. (b) Now consider \((x, y) \rightarrow(0,0)\) along the curve \(y=5 x\). Find the limit in this case: \(\lim _{(x, 5 x) \rightarrow(0,0)} \frac{6 x-y}{x+y}=\) _________. (c) Note that the results from \((a)\) and \((b)\) indicate that \(f\) has no limit as \((x, y) \rightarrow(0,0)\) (be sure you can explain why!). To show this more generally, consider \((x, y) \rightarrow(0,0)\) along the curve \(y=m x,\) for arbitrary \(m .\) Find the limit in this case: \(\lim _{m x) \rightarrow(0,0)} \frac{6 x-y}{x+y}=\) ____________. (Be sure that you can explain how this result also indicates that \(f\) has no limit as \((x, y) \rightarrow(0,0)\).

Short Answer

Expert verified
(a) \(\lim_{(x, 4x) \rightarrow(0,0)} \frac{6x-4x}{x+4x} = \frac{2}{5}\) (b) \(\lim_{(x, 5x) \rightarrow(0,0)} \frac{6x-5x}{x+5x} = \frac{1}{6}\) (c) \(\lim_{(x, mx) \rightarrow(0,0)} \frac{(6-m)x}{(1+m)x} = \frac{6-m}{1+m}\) The results from (a) and (b) are different, and the limit in (c) depends on the value of \(m\), which indicates that the function \(f(x, y)\) does not have a limit as \((x, y) \rightarrow(0,0)\).

Step by step solution

01

Find the limit along the curve y = 4x

Substitute \(y = 4x\) into the function: $$ f(x, 4x) = \frac{6x - 4x}{x + 4x} = \frac{2x}{5x} $$ Now, find the limit as \((x, 4x) \rightarrow (0,0)\): $$ \lim_{(x, 4x) \rightarrow (0,0)} \frac{2x}{5x} = \lim_{x \rightarrow 0} \frac{2}{5} = \frac{2}{5} $$
02

Find the limit along the curve y = 5x

Substitute \(y = 5x\) into the function: $$ f(x, 5x) = \frac{6x - 5x}{x + 5x} = \frac{x}{6x} $$ Now, find the limit as \((x, 5x) \rightarrow (0,0)\): $$ \lim_{(x, 5x) \rightarrow (0,0)} \frac{x}{6x} = \lim_{x \rightarrow 0} \frac{1}{6} = \frac{1}{6} $$
03

Find the limit along the curve y = mx for an arbitrary constant m

Substitute \(y = mx\) into the function: $$ f(x, mx) = \frac{6x - mx}{x + mx} = \frac{(6 - m)x}{(1 + m)x} $$ Now, find the limit as \((x, mx) \rightarrow (0,0)\): $$ \lim_{(x, mx) \rightarrow (0,0)} \frac{(6 - m)x}{(1 + m)x} = \lim_{x \rightarrow 0} \frac{6 - m}{1 + m} $$ This limit depends on the value of \(m\).
04

Analyze the results and draw conclusions

We found that the limit of the function along the curve \(y = 4x\) was \(\frac{2}{5}\) and the limit along the curve \(y = 5x\) was \(\frac{1}{6}\). These two limits are different, which means that the function does not have a unique limit as \((x, y) \rightarrow (0,0)\). Furthermore, we found that the limit along the curve \(y = mx\) for an arbitrary constant \(m\) depends on the value of \(m\). This result confirms that the function does not have a unique limit as \((x, y) \rightarrow (0,0)\) along different paths, and therefore, does not have a limit at this point.

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Key Concepts

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

Multivariable Calculus
Multivariable calculus is an extension of single-variable calculus to functions of several variables. It involves concepts like limits, derivatives, and integrals but with a complexity added due to the additional dimensions. Unlike in single-variable calculus, where functions take in a single real number and spit out another, multivariable functions have inputs and outputs that are vectors. This multi-dimensional nature leads to richer geometrical interpretations and a wider range of applications from physics to economics.

Understanding limits within this context is crucial as they deal with the behavior of a function as it approaches a certain point from various paths in its domain. The concept of limits in multiple dimensions is similar to that in single-variable calculus, but determining them often requires new techniques because there are infinitely many ways to approach a point in space. Simplification of functions, like factoring or rationalizing, can still be valuable tools when working with limits in multivariable calculus.
Limit of a Function
A limit of a function in multivariable calculus is the value that a function approaches as the input approaches a particular point. For multivariable functions, we consider the limit as the input vector gets infinitely close to a certain point within its domain. As with single-variable functions, if a limit exists, it means we can make the value of the function as close as we want to the limit value by taking inputs sufficiently close to the point of interest.

However, proving that limits exist in multivariable calculus can be trickier. In single-variable calculus, we check the left and right limits to see if they match. In contrast, in multivariable calculus, we must consider all possible paths leading to the point. If the limits along all possible paths exist and are equal, then the limit of the function at that point exists and is equal to that common value. This idea leads us into the notion of path dependence, which is an essential concept when determining limits of multivariable functions.
Path Dependence
Path dependence in the context of multivariable limits refers to the dependency of the limit of a function on the path taken to approach a point. In simpler terms, it means that the value of the limit may differ depending on the way we approach the target point in the domain of the function. If a limit varies with the path, the function does not have a unique limit at that point and, by definition, does not have a limit there at all.

To illustrate with our textbook solution, when approaching the origin \( (0,0) \) along two different lines, \( y = 4x \) and \( y = 5x \) , we obtained two different limits, \( \frac{2}{5} \) and \( \frac{1}{6} \) respectively. This difference in limits for different paths shows the function's limit at the origin depends on the path taken, implying there is no general limit at that point. This demonstrates path dependence and highlights its significance in the study of multivariable limits. Without considering path dependence, we might incorrectly assume the existence of a limit where there isn't one.

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Most popular questions from this chapter

The temperature on a heated metal plate positioned in the first quadrant of the \(x y\) -plane is given by $$C(x, y)=25 e^{-(x-1)^{2}-(y-1)^{3}}$$ Assume that temperature is measured in degrees Celsius and that \(x\) and \(y\) are each measured in inches. a. Determine \(C_{x x}(x, y)\) and \(C_{y y}(x, y) .\) Do not do any additional work to algebraically simplify your results. b. Calculate \(C_{x x}(1.1,1.2) .\) Suppose that an ant is walking past the point (1.1,1.2) along the line \(y=1.2 .\) Write a sentence to explain the meaning of the value of \(C_{x x}(1.1,1.2)\), including units. c. Calculate \(C_{y y}(1.1,1.2) .\) Suppose instead that an ant is walking past the point (1.1,1.2) along the line \(x=1.1 .\) Write a sentence to explain the meaning of the value of \(C_{y y}(1.1,1.2),\) including units. d. Determine \(C_{x y}(x, y)\) and hence compute \(C_{x y}(1.1,1.2) .\) What is the meaning of this value? Explain, in terms of an ant walking on the heated metal plate.

In the following questions, we investigate two different applied settings using the differential. a. Let \(f\) represent the vertical displacement in centimeters from the rest position of a string (like a guitar string) as a function of the distance \(x\) in centimeters from the fixed left end of the string and \(y\) the time in seconds after the string has been plucked. A simple model for \(f\) could be $$f(x, y)=\cos (x) \sin (2 y)$$ Use the differential to approximate how much more this vibrating string is vertically displaced from its position at \((a, b)=\left(\frac{\pi}{4}, \frac{\pi}{3}\right)\) if we decrease \(a\) by \(0.01 \mathrm{~cm}\) and increase the time by 0.1 seconds. Compare to the value of \(f\) at the point \(\left(\frac{\pi}{4}-0.01, \frac{\pi}{3}+0.1\right)\). b. Resistors used in electrical circuits have colored bands painted on them to indicate the amount of resistance and the possible error in the resistance. When three resistors, whose resistances are \(R_{1}, R_{2},\) and \(R_{3},\) are connected in parallel, the total resistance \(R\) is given by $$\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}$$ Suppose that the resistances are \(R_{1}=25 \Omega, R_{2}=40 \Omega,\) and \(R_{3}=\) \(50 \Omega\). Find the total resistance \(R\). If you know each of \(R_{1}, R_{2}\), and \(R_{3}\) with a possible error of \(0.5 \%\), estimate the maximum error in your calculation of \(R\).

The airlines place restrictions on luggage that can be carried onto planes. \- A carry-on bag can weigh no more than 40 lbs. \- The length plus width plus height of a bag cannot exceed 45 inches. \- The bag must fit in an overhead bin. Let \(x, y,\) and \(z\) be the length, width, and height (in inches) of a carry on bag. In this problem we find the dimensions of the bag of largest volume, \(V=x y z,\) that satisfies the second restriction. Assume that we use all 45 inches to get a maximum volume. (Note that this bag of maximum volume might not satisfy the third restriction.) a. Write the volume \(V=V(x, y)\) as a function of just the two variables \(x\) and \(y\) b. Explain why the domain over which \(V\) is defined is the triangular region \(R\) with vertices \((0,0),(45,0),\) and (0,45) . c. Find the critical points, if any, of \(V\) in the interior of the region \(R\). d. Find the maximum value of \(V\) on the boundary of the region \(R\), and then determine the dimensions of a bag with maximum volume on the entire region \(R\). (Note that most carry-on bags sold today measure 22 by 14 by 9 inches with a volume of 2772 cubic inches, so that the bags will fit into the overhead bins.)

Find all the first and second order partial derivatives of \(f(x, y)=3 \sin (2 x+\) \(y)-4 \cos (x-y)\) A. \(\frac{\partial f}{\partial x}=f_{x}=\) ________. B. \(\frac{\partial f}{\partial y}=f_{y}=\) ________. C. \(\frac{\partial^{2} f}{\partial x^{2}}=f_{x x}=\) ________. D. \(\frac{\partial^{2} f}{\partial y^{2}}=f_{y y}=\) ________. E. \(\frac{\partial^{2} f}{\partial x \partial y}=f_{y x}=\) ________. F. \(\frac{\partial^{2} f}{\partial y \partial x}=f_{x y}=\) ________.

Find the maximum and minimum values of the function \(f(x, y, z)=\) \(x^{2} y^{2} z^{2}\) subject to the constraint \(x^{2}+y^{2}+z^{2}=64 .\) Maximum value is ____________, occuring at points (positive integer or "infinitely many"). Minimum value is ____________, occuring at points (positive integer or "infinitely many").
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