Chapter 3: Chapter 3

Problem 59

Let $\mathrm{f}(\mathrm{x}, \mathrm{y})=\mathrm{x}^{2}+3 \mathrm{y}$ and let $\varepsilon>0$ be given. Prove that $\mathrm{F}$ is continuous in the whole plane by finding $\delta>0$ such that for $\left|(x, y)-\left(x_{0}, y_{0}\right)\right|<\delta$ $\left|F(x, y)-F\left(x_{0}, y_{0}\right)\right|<\varepsilon$ where $x_{0}, y_{0}$ is an arbitrary point in the plane.

Short Answer

Expert verified

For the given function $F(x, y)=x^2+3y$, we can find a bound $\delta>0$ such that if the distance between $(x, y)$ and $(x_0, y_0)$ is less than $\delta$, then the distance between $F(x, y)$ and $F(x_0, y_0)$ is less than $\varepsilon$. We have found that $\delta = \frac{\varepsilon}{2 (|x_0| + 1) + 7}$ satisfies this condition. Therefore, the function $F(x, y)=x^2+3y$ is continuous in the whole plane.

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Step by step solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Analyze the difference between function values

For the given function, $\mathrm{F}(x, y)=x^2+3y$, we will first analyze the difference between the function values at $(x, y)$ and $(x_0, y_0)$: \[|F(x, y) - F(x_0, y_0)| = |(x^2 - x_0^2) + 3(y - y_0)|\]

Step 2: Apply triangle inequality

By using triangle inequality, we can say that $(x^2 - x_0^2) + 3(y - y_0)| \leq |x^2 - x_0^2| + 3|y - y_0|$. Now, using the identity, $||a| - |b|| \leq |a - b| \leq |a| + |b|$, we get, \[|x^2 - x_0^2| \leq |x + x_0||x - x_0|\]

Step 3: Bound the difference in function values

Combining the bounds from the previous steps, we get: \[|F(x, y) - F(x_0, y_0)| \leq |x + x_0||x - x_0| + 3|y - y_0|\] Notice that we have the terms $|x - x_0|$ and $|y - y_0|$, which can be related to the distance between $(x, y)$ and $(x_0, y_0)$. So, let's calculate this distance and bound it by some $\delta$.

Step 4: Calculate the distance between (x, y) and (x_0, y_0)

The distance between $(x, y)$ and $(x_0, y_0)$ is given by: \[|(x, y) - (x_0, y_0)| = \sqrt{(x - x_0)^2 + (y - y_0)^2}\] Now, we need to find a bound for this distance so that it influences the inequality involving the function values.

Step 5: Find the bound to relate the distance and function values

Let's find $\delta$ such that \[\delta = \sqrt{(x - x_0)^2 + (y - y_0)^2}\] thus \[|(x, y) - (x_0, y_0)| < \delta \Rightarrow |x - x_0|^2 + |y - y_0|^2 < \delta^2\] and also, \[|x - x_0| < \delta, |y - y_0| < \delta\] Now, choose $\delta = \frac{\varepsilon}{2 (|x_0| + 1) + 7}$. Observe that: \[|x + x_0||x - x_0| + 3|y - y_0|\] \[\leq (|x_0| + |x - x_0|)|x - x_0| + 3|y - y_0|\] \[\leq (|x_0| + \delta)|\delta| + 3\delta\] \[\leq (|x_0| + 1)\delta + 3\delta\] \[\leq |x_0| \delta + \delta + 3\delta\] \[\leq (2 |x_0| + 4)\delta\] \[\leq \varepsilon\] Therefore, we have shown that there exists a $\delta > 0$ such that for any $(x, y)$ with $|(x, y)-(x_0, y_0)| < \delta$, $|F(x, y)-F(x_0, y_0)| < \varepsilon$. This means the function $\mathrm{f}(\mathrm{x}, \mathrm{y})=\mathrm{x}^{2}+3 \mathrm{y}$ is continuous in the whole plane.

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Chapter 3: Chapter 3

Short Answer

Step by step solution

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Step 1: Analyze the difference between function values

Step 2: Apply triangle inequality

Step 3: Bound the difference in function values

Step 4: Calculate the distance between (x, y) and (x_0, y_0)

Step 5: Find the bound to relate the distance and function values

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Chapter 1

Chapter 2

Chapter 4

Chapter 5

Chapter 6

Chapter 7

Chapter 8

Chapter 9

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 14

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