Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![**Exercise 6**:
Given the function \( f(x, y) = e^y x^2 \), find the directional derivative at the point \((-2, 0)\) in the direction of the vector \((2, -3)\).
To find the directional derivative of a function at a given point in a specified direction, follow these steps:
1. **Find the Gradient of the Function**:
The gradient of \( f(x, y) \), denoted \( \nabla f \), is a vector of partial derivatives. For \( f(x, y) = e^y x^2 \),
\[
\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)
\]
2. **Compute the Partial Derivatives**:
- With respect to \( x \):
\[
\frac{\partial f}{\partial x} = 2x e^y
\]
- With respect to \( y \):
\[
\frac{\partial f}{\partial y} = x^2 e^y
\]
3. **Evaluate the Gradient at the Given Point \((-2, 0)\)**:
- At \((-2, 0)\),
\[
\nabla f(-2, 0) = \left( 2(-2)e^0, (-2)^2 e^0 \right) = (-4, 4)
\]
4. **Normalize the Direction Vector \((2, -3)\)**:
The unit vector in the direction of \((2, -3)\) is found by dividing by its magnitude:
\[
\|\mathbf{v}\| = \sqrt{2^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13}
\]
Hence, the unit vector in the direction of \((2, -3)\) is:
\[
\mathbf{u} = \left( \frac{2}{\sqrt{13}}, \frac{-3}{\sqrt{13}} \right)
\]
5. **Compute the Directional Derivative**:
The directional derivative \( D_{\mathbf{u}}f \) at \((-2, 0)\) in](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc303f822-5e20-4d30-8920-178170554c96%2F5d42f49f-52b2-4a49-9f51-16fdc62aea43%2Ft355ukc_processed.png&w=3840&q=75)
Transcribed Image Text:**Exercise 6**:
Given the function \( f(x, y) = e^y x^2 \), find the directional derivative at the point \((-2, 0)\) in the direction of the vector \((2, -3)\).
To find the directional derivative of a function at a given point in a specified direction, follow these steps:
1. **Find the Gradient of the Function**:
The gradient of \( f(x, y) \), denoted \( \nabla f \), is a vector of partial derivatives. For \( f(x, y) = e^y x^2 \),
\[
\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)
\]
2. **Compute the Partial Derivatives**:
- With respect to \( x \):
\[
\frac{\partial f}{\partial x} = 2x e^y
\]
- With respect to \( y \):
\[
\frac{\partial f}{\partial y} = x^2 e^y
\]
3. **Evaluate the Gradient at the Given Point \((-2, 0)\)**:
- At \((-2, 0)\),
\[
\nabla f(-2, 0) = \left( 2(-2)e^0, (-2)^2 e^0 \right) = (-4, 4)
\]
4. **Normalize the Direction Vector \((2, -3)\)**:
The unit vector in the direction of \((2, -3)\) is found by dividing by its magnitude:
\[
\|\mathbf{v}\| = \sqrt{2^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13}
\]
Hence, the unit vector in the direction of \((2, -3)\) is:
\[
\mathbf{u} = \left( \frac{2}{\sqrt{13}}, \frac{-3}{\sqrt{13}} \right)
\]
5. **Compute the Directional Derivative**:
The directional derivative \( D_{\mathbf{u}}f \) at \((-2, 0)\) in
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