Given f(x, y) = then V 4e4* sin(2y) ▼ ƒ(0, 7) = 2
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
2.3
![### Problem Statement
**Given:**
\( f(x, y) = 4e^{4x} \sin(2y) \)
**Question:**
Calculate the gradient \(\nabla f(0, \frac{\pi}{2})\).
### Explanation
This problem requires finding the gradient of the function \( f(x, y) = 4e^{4x} \sin(2y) \) at the point \( (0, \frac{\pi}{2}) \). The gradient, \(\nabla f(x, y)\), is a vector consisting of partial derivatives of \( f \) with respect to \( x \) and \( y \).
**Steps to Solve:**
1. **Find Partial Derivative with respect to \(x\):**
\[
\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \left( 4e^{4x} \sin(2y) \right) = 16e^{4x} \sin(2y)
\]
2. **Find Partial Derivative with respect to \(y\):**
\[
\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \left( 4e^{4x} \sin(2y) \right) = 8e^{4x} \cos(2y)
\]
3. **Evaluate at \( (0, \frac{\pi}{2}) \):**
- For \(\frac{\partial f}{\partial x}\),
\[
\frac{\partial f}{\partial x}(0, \frac{\pi}{2}) = 16e^{0} \sin(\pi) = 0
\]
- For \(\frac{\partial f}{\partial y}\),
\[
\frac{\partial f}{\partial y}(0, \frac{\pi}{2}) = 8e^{0} \cos(\pi) = -8
\]
**Gradient Vector:**
\[
\nabla f(0, \frac{\pi}{2}) = \left( 0, -8 \right)
\]
The answer to the problem is the gradient vector \( (0, -8) \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F876c9bbd-0074-4123-babe-119feafc9f73%2Fd4ced0aa-881e-4748-a0c9-8aa839199669%2F9ja6sib_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
**Given:**
\( f(x, y) = 4e^{4x} \sin(2y) \)
**Question:**
Calculate the gradient \(\nabla f(0, \frac{\pi}{2})\).
### Explanation
This problem requires finding the gradient of the function \( f(x, y) = 4e^{4x} \sin(2y) \) at the point \( (0, \frac{\pi}{2}) \). The gradient, \(\nabla f(x, y)\), is a vector consisting of partial derivatives of \( f \) with respect to \( x \) and \( y \).
**Steps to Solve:**
1. **Find Partial Derivative with respect to \(x\):**
\[
\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \left( 4e^{4x} \sin(2y) \right) = 16e^{4x} \sin(2y)
\]
2. **Find Partial Derivative with respect to \(y\):**
\[
\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \left( 4e^{4x} \sin(2y) \right) = 8e^{4x} \cos(2y)
\]
3. **Evaluate at \( (0, \frac{\pi}{2}) \):**
- For \(\frac{\partial f}{\partial x}\),
\[
\frac{\partial f}{\partial x}(0, \frac{\pi}{2}) = 16e^{0} \sin(\pi) = 0
\]
- For \(\frac{\partial f}{\partial y}\),
\[
\frac{\partial f}{\partial y}(0, \frac{\pi}{2}) = 8e^{0} \cos(\pi) = -8
\]
**Gradient Vector:**
\[
\nabla f(0, \frac{\pi}{2}) = \left( 0, -8 \right)
\]
The answer to the problem is the gradient vector \( (0, -8) \).
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