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Chapter 12: Problem 69

Limits at (0,0) may be easier to evaluate by converting to polar coordinates. Remember that the same limit must be obtained as \(r \rightarrow 0\) along all paths in the domain to (0, 0). Evaluate the following limits or state that they do not exist. $$\lim _{(x, y) \rightarrow(0,0)} \frac{x^{2} y}{x^{2}+y^{2}}$$

Short Answer

Expert verified
$$ Answer: The limit as (x, y) approaches (0, 0) for the given expression is 0.

Step by step solution

01

Convert the expression to polar coordinates

Substitute x = r*cos(胃) and y = r*sin(胃) into the expression: $$\frac{x^2y}{x^2+y^2} = \frac{\left(r\cos\theta\right)^2\left(r\sin\theta\right)}{\left(r\cos\theta\right)^2 + \left(r\sin\theta\right)^2}$$
02

Simplify the expression

Simplify the expression: $$\frac{\left(r^2\cos^2\theta\right)\left(r\sin\theta\right)}{r^2\cos^2\theta + r^2\sin^2\theta}$$ Factor out an r虏 from the denominator: $$\frac{r^3\cos^2\theta\sin\theta}{r^2\left(\cos^2\theta+\sin^2\theta\right)}$$
03

Identify a useful trigonometric identity

The trigonometric identity for sin虏(胃) + cos虏(胃) is equal to 1. So, our expression becomes: $$\frac{r^3\cos^2\theta\sin\theta}{r^2}$$
04

Simplify the expression further

Divide both the numerator and the denominator by r虏: $$\frac{r\cos^2\theta\sin\theta}{1}$$
05

Evaluate the limit as r approaches 0

Find the limit of the expression as r approaches 0: $$\lim_{r \rightarrow 0} r\cos^2\theta\sin\theta$$ As r approaches 0, the whole expression will approach 0 because r will be 0. This limit does not depend on the value of 胃, so the limit is the same along all paths to (0, 0).
06

Write the final answer

The limit exists and is equal to 0: $$\lim_{(x, y) \rightarrow (0, 0)} \frac{x^2 y}{x^2 + y^2} = 0$$

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Key Concepts

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

Limit Evaluation
Evaluating limits is an essential concept in calculus. It's about finding the value that a function approaches as its input gets closer to a specific point. Sometimes, limits can be tricky, especially when they involve two variables, as with functions in \(x\) and \(y\). When dealing with limits at a point like \((0,0)\), it's crucial to ensure the limit is consistent along all paths leading to that point. If the value differs depending on the path, the limit doesn鈥檛 exist.

In our exercise, after converting the function to polar coordinates, we noticed that as \(r\) approaches zero, the whole expression depends on \(r\) itself, apart from the angle \(\theta\). Since \(r\) tends towards zero, and the function remains zero for all angles \(\theta\), the limit becomes zero, and consistent across all paths. This means that the limit exists and is \(0\).
Trigonometric Identities
Trigonometric identities are equations involving trigonometric functions that are true for all values of the variable involved. These identities are invaluable in simplifying expressions, especially when working with polar coordinates, as they connect angles and side lengths in a triangle.

One of the most fundamental identities is \(\sin^2(\theta) + \cos^2(\theta) = 1\). In our problem, this identity directly helps to simplify the expressions when we replaced \(x\) and \(y\) with their polar equivalents \(r\cos(\theta)\) and \(r\sin(\theta)\).

By substituting these values and using the identity, we collapse the denominator \(r^2(\cos^2(\theta) + \sin^2(\theta))\) into just \(r^2\), which significantly simplifies the limit evaluation process. Remembering and identifying when to apply such identities is a key skill when dealing with trigonometry and polar coordinates.
Cartesian to Polar Conversion
Converting between Cartesian and polar coordinates is a handy technique that can simplify solving problems involving curves and limits. In many cases, expressions involving \(x\) and \(y\) become more manageable in polar coordinates, using \(r\) and \(\theta\).

The conversion is straightforward: \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\). Here, \(r\) represents the distance from the origin to the point, while \(\theta\) is the angle between the positive \(x\)-axis and the line connecting the origin with the point.

In the original exercise, switching from Cartesian to polar coordinates helped in simplifying the given expression and made the limit evaluation easier at \((0,0)\). Polar conversion particularly shines in evaluating limits where traditional Cartesian methods become cumbersome. Understanding when to switch between these coordinate systems is an advantageous strategy that enhances problem-solving efficiency.

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Most popular questions from this chapter

The flow of heat along a thin conducting bar is governed by the one- dimensional heat equation (with analogs for thin plates in two dimensions and for solids in three dimensions) $$\frac{\partial u}{\partial t}=k \frac{\partial^{2} u}{\partial x^{2}},$$ where \(u\) is a measure of the temperature at a location \(x\) on the bar at time t and the positive constant \(k\) is related to the conductivity of the material. Show that the following functions satisfy the heat equation with \(k=1\). \(u(x, t)=A e^{-\alpha^{2} t} \cos a x,\) for any real numbers \(a\) and \(A\)

Find the absolute maximum and minimum values of the following functions over the given regions \(R\). Use Lagrange multipliers to check for extreme points on the boundary. $$f(x, y)=(x-1)^{2}+(y+1)^{2} ; R=\left\\{(x, y): x^{2}+y^{2} \leq 4\right\\}$$

An identity Show that if \(f(x, y)=\frac{a x+b y}{c x+d y},\) where \(a, b, c,\) and \(d\) are real numbers with \(a d-b c=0,\) then \(f_{x}=f_{y}=0,\) for all \(x\) and \(y\) in the domain of \(f\). Give an explanation.

Economists model the output of manufacturing systems using production functions that have many of the same properties as utility functions. The family of Cobb-Douglas production functions has the form \(P=f(K, L)=C K^{a} L^{1-a},\) where \(K\) represents capital, \(L\) represents labor, and C and a are positive real numbers with \(0

Consider the following functions \(f\). a. Is \(f\) continuous at (0,0)\(?\) b. Is \(f\) differentiable at (0,0)\(?\) c. If possible, evaluate \(f_{x}(0,0)\) and \(f_{y}(0,0)\). d. Determine whether \(f_{x}\) and \(f_{y}\) are continuous at (0,0). e. Explain why Theorems 12.5 and 12.6 are consistent with the results in parts \((a)-(d)\). $$f(x, y)=\sqrt{|x y|}$$
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