Find the directional derivative of f(x, y) = 3x² - y² + 5xy at (2,-1) in the direction of -3i - 4j. Find on equation of the plane tongo
Find the directional derivative of f(x, y) = 3x² - y² + 5xy at (2,-1) in the direction of -3i - 4j. Find on equation of the plane tongo
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Finding the Directional Derivative**
In this example, we are given a function \( f(x, y) = 3x^2 - y^2 + 5xy \), and we need to find its directional derivative at the point \((2, -1)\) in the direction of the vector \(\mathbf{v} = -3\mathbf{i} - 4\mathbf{j}\).
To find the directional derivative of \( f \) in the direction of \(\mathbf{v}\), we need to follow these steps:
1. **Find the gradient of the function \( f(x, y) \), denoted as \(\nabla f\).**
The gradient \(\nabla f\) is given by the vector of partial derivatives:
\[
\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)
\]
For our function \( f(x, y) = 3x^2 - y^2 + 5xy \):
\[
\frac{\partial f}{\partial x} = 6x + 5y
\]
\[
\frac{\partial f}{\partial y} = -2y + 5x
\]
2. **Evaluate the gradient at the given point \((2, -1)\).**
Substitute \( x = 2 \) and \( y = -1 \) into the partial derivatives:
\[
\nabla f(2, -1) = \left( 6(2) + 5(-1), -2(-1) + 5(2) \right) = \left( 12 - 5, 2 + 10 \right) = (7, 12)
\]
3. **Normalize the direction vector \(\mathbf{v}\).**
The given direction vector is \(\mathbf{v} = -3\mathbf{i} - 4\mathbf{j}\). The magnitude of \(\mathbf{v}\) is:
\[
\| \mathbf{v} \| = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5
\]
The unit vector in](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9981eaaa-92cc-40db-b668-6adab491a4d6%2F0af8c66b-d1e7-4b71-93e6-7292740fb0f0%2Fxycnz6_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Finding the Directional Derivative**
In this example, we are given a function \( f(x, y) = 3x^2 - y^2 + 5xy \), and we need to find its directional derivative at the point \((2, -1)\) in the direction of the vector \(\mathbf{v} = -3\mathbf{i} - 4\mathbf{j}\).
To find the directional derivative of \( f \) in the direction of \(\mathbf{v}\), we need to follow these steps:
1. **Find the gradient of the function \( f(x, y) \), denoted as \(\nabla f\).**
The gradient \(\nabla f\) is given by the vector of partial derivatives:
\[
\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)
\]
For our function \( f(x, y) = 3x^2 - y^2 + 5xy \):
\[
\frac{\partial f}{\partial x} = 6x + 5y
\]
\[
\frac{\partial f}{\partial y} = -2y + 5x
\]
2. **Evaluate the gradient at the given point \((2, -1)\).**
Substitute \( x = 2 \) and \( y = -1 \) into the partial derivatives:
\[
\nabla f(2, -1) = \left( 6(2) + 5(-1), -2(-1) + 5(2) \right) = \left( 12 - 5, 2 + 10 \right) = (7, 12)
\]
3. **Normalize the direction vector \(\mathbf{v}\).**
The given direction vector is \(\mathbf{v} = -3\mathbf{i} - 4\mathbf{j}\). The magnitude of \(\mathbf{v}\) is:
\[
\| \mathbf{v} \| = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5
\]
The unit vector in
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