xy Let f: R² →R where f(x, y) = 0 if (x, y) = (0,0) and f(x, y) = 2² + y² if (x, y) ‡ (0,0). (a) Prove that Duf(0,0) = 0 if u = (1,0). (b) Prove that Duf(0,0) = 0 if u = (0,1). (c) Prove that Duf(0, 0) does not exist if u = (a, b) where ab 0. (d) Prove that Df(0,0) does not exist (this can be done by proving that f is not continuous at (0, 0)).
xy Let f: R² →R where f(x, y) = 0 if (x, y) = (0,0) and f(x, y) = 2² + y² if (x, y) ‡ (0,0). (a) Prove that Duf(0,0) = 0 if u = (1,0). (b) Prove that Duf(0,0) = 0 if u = (0,1). (c) Prove that Duf(0, 0) does not exist if u = (a, b) where ab 0. (d) Prove that Df(0,0) does not exist (this can be done by proving that f is not continuous at (0, 0)).
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Transcribed Image Text:**Problem 1: Analysis of a Function**
Consider the function \( f: \mathbb{R}^2 \to \mathbb{R} \) defined by \( f(x, y) = 0 \) if \((x, y) = (0, 0)\) and \( f(x, y) = \frac{xy}{x^2+y^2} \) if \((x, y) \neq (0, 0)\).
**Tasks:**
(a) Prove that \( D_u f(0, 0) = 0 \) if \( u = (1, 0) \).
(b) Prove that \( D_u f(0, 0) = 0 \) if \( u = (0, 1) \).
(c) Prove that \( D_u f(0, 0) \) does not exist if \( u = (a, b) \) where \( ab \neq 0 \).
(d) Prove that \( Df(0, 0) \) does not exist (this can be done by proving that \( f \) is not continuous at \( (0, 0) \)).
![**Mathematical Function and Linear Transformation Example**
**Example:**
Let \( f: \mathbb{R}^2 \rightarrow \mathbb{R}^2 \) be defined by \( f(x, y) = (x^2 y, 3x + 2y) \). Let \( u = (1, 0) \) and \( c = (1, 1) \).
**Claim:**
\( Lu = (2, 3) \)
**Proof:**
1. Fix \( \varepsilon > 0 \). Choose \( \delta = \frac{\varepsilon}{3} \). Assume \( 0 < |t| < \delta \). Then, we have:
\[
\left| \frac{f(c + tu) - f(c) - Lu}{t} \right| = \left| \frac{f(1 + t, 1) - (1, 5) - (2, 3)}{t} \right|
\]
2. Expand and simplify:
\[
= \left| \frac{(1 + t)^2 + 3t + 5 - (1, 5) - (2, 3)}{t} \right|
\]
3. Further simplification gives:
\[
= \left| \frac{(2t + t^2, 3t) - (2, 3)}{t} \right|
\]
4. Simplifying yields:
\[
= \left| (t, 0) \right| \quad \text{when} \quad |t| < \delta = \varepsilon
\]
**Query:**
Why is \( Lu = (2, 3) \)?
\[
L = \begin{bmatrix}
\frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} \\
\frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y}
\end{bmatrix}
=
\begin{bmatrix}
2xy & x^2 \\
3 & 2
\end{bmatrix}
,\ Lu = \begin{bmatrix}
\frac{2xy}{3} \\
\frac{x^2}{3}
\end{bmatrix}
\]
\[
L](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F79599c56-a340-49a0-b0ff-829b3947a798%2Faf236013-5e75-436b-b706-9342051890dd%2Fs3qem7i_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Mathematical Function and Linear Transformation Example**
**Example:**
Let \( f: \mathbb{R}^2 \rightarrow \mathbb{R}^2 \) be defined by \( f(x, y) = (x^2 y, 3x + 2y) \). Let \( u = (1, 0) \) and \( c = (1, 1) \).
**Claim:**
\( Lu = (2, 3) \)
**Proof:**
1. Fix \( \varepsilon > 0 \). Choose \( \delta = \frac{\varepsilon}{3} \). Assume \( 0 < |t| < \delta \). Then, we have:
\[
\left| \frac{f(c + tu) - f(c) - Lu}{t} \right| = \left| \frac{f(1 + t, 1) - (1, 5) - (2, 3)}{t} \right|
\]
2. Expand and simplify:
\[
= \left| \frac{(1 + t)^2 + 3t + 5 - (1, 5) - (2, 3)}{t} \right|
\]
3. Further simplification gives:
\[
= \left| \frac{(2t + t^2, 3t) - (2, 3)}{t} \right|
\]
4. Simplifying yields:
\[
= \left| (t, 0) \right| \quad \text{when} \quad |t| < \delta = \varepsilon
\]
**Query:**
Why is \( Lu = (2, 3) \)?
\[
L = \begin{bmatrix}
\frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} \\
\frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y}
\end{bmatrix}
=
\begin{bmatrix}
2xy & x^2 \\
3 & 2
\end{bmatrix}
,\ Lu = \begin{bmatrix}
\frac{2xy}{3} \\
\frac{x^2}{3}
\end{bmatrix}
\]
\[
L
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