/*! 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 22 Use the definition of continuity... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Use the definition of continuity to show that $$f(x, y)=\sqrt{9+x^{2}+y^{2}}$$ is continuous at \((0,0)\).

Short Answer

Expert verified
The function is continuous at \((0, 0)\) because its limit equals its value at that point, which is 3.

Step by step solution

01

Understand the Definition of Continuity

A function \( f(x,y) \) is continuous at a point \((a,b)\) if \[ \lim_{(x,y) \to (a,b)} f(x,y) = f(a,b). \]In this problem, we need to show that \( \lim_{(x,y) \to (0,0)} \sqrt{9 + x^2 + y^2} = \sqrt{9+0^2+0^2} = 3. \)
02

Express the Limit Statement

To show continuity, consider the limit \[ \lim_{(x,y) \to (0,0)} \sqrt{9 + x^2 + y^2}. \]To simplify, we want to show that as \((x,y)\) approaches \((0,0)\), the expression under the square root approaches 9.
03

Rewrite Using \(\epsilon\)-\(\delta\) Definition

We use the \(\epsilon\)-\(\delta\) definition for limits. For a given \(\epsilon > 0\), we want to find a \(\delta > 0\) so that if \[ \sqrt{x^2 + y^2} < \delta, \]then\[ \left| \sqrt{9+x^2+y^2} - 3 \right| < \epsilon. \]
04

Simplify the Difference

Consider \[ \left| \sqrt{9 + x^2 + y^2} - 3 \right|. \]This can be rewritten by multiplying and dividing by the conjugate:\[ \frac{|9 + x^2 + y^2 - 9|}{\sqrt{9 + x^2 + y^2} + 3} = \frac{x^2 + y^2}{\sqrt{9 + x^2 + y^2} + 3}. \]
05

Bounded Denominator

Notice that for small \(x^2 + y^2\), the denominator \(\sqrt{9 + x^2 + y^2} + 3\) is bounded below by 6, because \(\sqrt{9 + x^2 + y^2} \geq 3\).Thus \[ \frac{x^2 + y^2}{\sqrt{9 + x^2 + y^2} + 3} < \frac{x^2 + y^2}{6}. \]
06

Choose \(\delta\) Appropriately

Given \( \epsilon > 0 \), choose \( \delta = \sqrt{6\epsilon}\). Then, if \(\sqrt{x^2 + y^2} < \delta,\) we have \[ x^2 + y^2 < 6\epsilon, \]leading to \[ \frac{x^2 + y^2}{6} < \epsilon. \]
07

Conclude Continuity

Since \( \left| \sqrt{9 + x^2 + y^2} - 3 \right| < \epsilon \), it follows that \[ \lim_{(x,y) \to (0,0)} \sqrt{9 + x^2 + y^2} = 3 \].Thus, the function \( f(x,y) = \sqrt{9 + x^2 + y^2} \) is continuous at \((0,0)\).

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 91Ó°ÊÓ!

Key Concepts

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

Limit of a Function
Understanding the concept of the limit of a function is crucial in calculus, especially when dealing with the continuity of functions. For a given function \( f(x, y) \), the limit describes the behavior of \( f(x, y) \) as \( (x, y) \) approaches a particular point \( (a, b) \). In simpler terms, it answers the question: "As we get infinitesimally close to the point \( (a, b) \), what value does \( f(x, y) \) approach?"

In the context of our function \( \sqrt{9 + x^2 + y^2} \), we are examining what happens as \( (x, y) \to (0,0) \). This function provides a smooth value that smoothly transitions as \( x \) and \( y \) change. Finding this limit involves understanding how the values inside the square root change as \( x \) and \( y \) get closer to zero.

If as \( (x, y) \) approaches \( (0, 0) \), the function's value approaches 3, we say the limit is 3. Hence, our goal becomes demonstrating that \( \lim_{(x, y) \to (0,0)} \sqrt{9 + x^2 + y^2} = 3 \).
Epsilon-Delta Definition
The epsilon-delta definition is a formal way of defining a limit, ensuring precision in mathematical analysis. It gives a precise criterion to show or prove that a function approaches a particular limit as we approach a certain point.

For a function \( f(x, y) \), if we wish to show that \( \lim_{(x, y) \to (a, b)} f(x, y) = L \), we need to demonstrate that for every \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that whenever \( \sqrt{(x-a)^2 + (y-b)^2} < \delta \), we have \( |f(x, y) - L| < \epsilon \).

In our example, the goal is to find this \( \delta \) for each \( \epsilon \) given. After calculating, we strategically chose \( \delta = \sqrt{6\epsilon} \). This choice ensures that \( |\sqrt{9 + x^2 + y^2} - 3| < \epsilon \), thereby proving the function's limit of 3 at the point \( (0, 0) \).
Two-Variable Functions
Two-variable functions are a type of multivariable function where the input consists of two variables, typically denoted as \( x \) and \( y \). These functions often correspond to surfaces in three-dimensional space. The function \( f(x, y) = \sqrt{9 + x^2 + y^2} \) is an example of such a function. Here, for any given \(x\) and \(y\), \(f(x, y)\) outputs a single value, describing a surface over the \(xy\)-plane.

Understanding two-variable functions involves visualizing how the function behaves across the \(xy\) plane. This visualization helps in analyzing the function's continuity and limits.

When we explore continuity with these functions, we must consider how they behave as both \(x\) and \(y\) approach specific values. For the function in discussion, as \( (x, y) \) gets close to \( (0, 0) \), the output—after considering the changes in both directions—must consistently approach a single, definitive value to be continuous. Thus, understanding how to work with two-variable functions and their ensuing limits is essential in higher-level calculus for modeling physical scenarios and for analytic problem-solving.

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

Find \(\partial f / \partial x\) and \(\partial f / \partial y\) for the given functions. \(f(x, y)=\tan (x-2 y)\)

Morphogenesis Most embryos start out as lumps of cells. Cells in these lumps are initially undifferentiated-that is, they start out all in the same state. Over time cells then commit to different functions, e.g., to becoming legs, eyes, and so on. To do this chemicals called morphogens are distributed unequally through the embryo, allowing each cell to tell where in the embryo it is located. How are unequal distributions of morphogens achieved? One model for how morphogens can be distributed through the embryo is that morphogens are continuously produced at one end (also called pole) of the embryo. From there they diffuse through the embryo. As the morphogens diffuse, they are constantly broken down by the cells in the embryo. First let's ignore the process of morphogen degradation, and focus only on diffusion. We will assume that the pole at which the morphogen is produced is located at \(x=0 ;\) and for simplicity's sake the cell occupies the interval \(x \geq 0\). Then our partial differential equation model for the distribution of morphogen becomes: $$ \frac{\partial c}{\partial t}=D \frac{\partial^{2} c}{\partial x^{2}} \quad \text { for } \quad x>0 $$ with $$ -D \frac{\partial c}{\partial x}(0, t)=Q \quad \text { and } \quad c(x, t) \rightarrow 0 \text { as } x \rightarrow \infty $$ where \(Q\) is the rate of morphogen production. (a) Let's try to find a steady state distribution of morphogen. That is, we will assume that over time the morphogen concentration reaches some state that does not change with time, i.e., the concentration is given by a function \(C(x) .\) Then \(C(x)\) will satisfy the partial differential equation if and only if: $$ \begin{aligned} 0 &=D \frac{d^{2} C}{d x^{2}} \quad \text { for } \quad x>0 \\ -D C^{\prime}(0) &=Q \quad \text { and } \quad C(x) \rightarrow 0 \text { as } x \rightarrow \infty \end{aligned} $$ Show that there is no function \(C(x)\) that satisfies this differential equation. [Hint: Start by integrating once \((10.48)\) to find \(d C / d x\) and then again to find \(\mathcal{C}(x)\), then try to impose the constraints at \(x=0\), and as \(x \rightarrow \infty\) on your solution.] (b) Now let's incorporate morphogen degradation into our model. We will assume that the breakdown of morphogen has first order kinetics (see Section \(5.9\) for a discussion of the different kinds of kinetics that chemical reactions may have). This means that in one unit of time a fraction \(r\) of the morphogen contained in each region of the embryo is degraded. Then our partial differential equation must be altered to: $$ \frac{\partial c}{\partial t}=D \frac{\partial^{2} c}{\partial x^{2}}-r c \quad \text { for } \quad x>0 $$ with $$ -D \frac{\partial c}{\partial x}(0, t)=Q \quad \text { and } \quad c(x, t) \rightarrow 0 \text { as } x \rightarrow \infty $$ Show that this partial differential equation does have a steady state solution of the form: $$ C(x)=Q \sqrt{\frac{1}{D r}} \exp \left(-\sqrt{\frac{r}{D}} x\right) $$ That is, check that this function \(C(x)\) satisfies both the steady state form of \((10.49)\) as well as the constraints at \(x=0\) and as \(x \rightarrow \infty\).

Use the properties of limits to calculate the following limits: \(\lim _{(x, y) \rightarrow(1,-2)} \frac{2 x^{2}+y}{2 x y+3}\)

Find \(\partial f / \partial x\) and \(\partial f / \partial y\) for the given functions. \(f(x, y)=\sin ^{2}\left(y^{2} x-x^{3}\right)\)

Show that $$f(x, y)=\left\\{\begin{array}{cc} \frac{3 r^{2} y}{\left(2 x^{4}+y^{2}\right)} & \text { for }(x, y) \neq(0,0) \\\ 0 & \text { for }(x, y)=(0,0)\end{array}\right.$$ is discontinuous at \((0,0) .\)

See all solutions

Recommended explanations on Biology Textbooks

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.