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f(x, y) = iy² – x², P(1,4), v = 2i + j

f(x, y) = iy² – x², P(1,4), v = 2i + j

Calculus: Early Transcendentals
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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## Finding a Directional Derivative In Exercises 57 and 58, use Theorem 13.9 to find the directional derivative of the function at \( P \) in the direction of \( \mathbf{v} \). ### Exercise 57 Given: - Function \( f(x, y) = x^2 y \) - Point \( P(-5, 5) \) - Direction vector \( \mathbf{v} = 3i - 4j \) ### Exercise 58 Given: - Function \( f(x, y) = \frac{1}{4}y^2 - x^2 \) - Point \( P(1, 4) \) - Direction vector \( \mathbf{v} = 2i + j \) #### Explanation of Theorem 13.9 Theorem 13.9 typically involves using the gradient of the function \( \nabla f \) and the given direction vector \( \mathbf{v} \) to find the directional derivative. The directional derivative \( D_{\mathbf{v}} f \) at a point \( P \) is given by: \[ D_{\mathbf{v}} f = \nabla f \cdot \mathbf{u} \] where \( \mathbf{u} \) is the unit vector in the direction of \( \mathbf{v} \) (i.e., \( \mathbf{u} = \frac{\mathbf{v}}{|\mathbf{v}|} \)) and \( \cdot \) represents the dot product. #### Steps to Solve: 1. **Compute the Gradient \( \nabla f \)**: Find the partial derivatives of \( f(x, y) \) with respect to \( x \) and \( y \). 2. **Evaluate the Gradient at Point \( P \)**: Substitute the coordinates of \( P \) into \( \nabla f \) to obtain \( \nabla f(P) \). 3. **Normalize the Direction Vector \( \mathbf{v} \)**: Find the unit vector \( \mathbf{u} = \frac{\mathbf{v}}{|\mathbf{v}|} \). 4. **Compute the Dot Product**: Find the dot product of \( \nabla f(P) \) and \( \mathbf{u} \). By following these steps, the directional derivatives for Exercises 57 and 58 can be calculated
Transcribed Image Text:## Finding a Directional Derivative In Exercises 57 and 58, use Theorem 13.9 to find the directional derivative of the function at \( P \) in the direction of \( \mathbf{v} \). ### Exercise 57 Given: - Function \( f(x, y) = x^2 y \) - Point \( P(-5, 5) \) - Direction vector \( \mathbf{v} = 3i - 4j \) ### Exercise 58 Given: - Function \( f(x, y) = \frac{1}{4}y^2 - x^2 \) - Point \( P(1, 4) \) - Direction vector \( \mathbf{v} = 2i + j \) #### Explanation of Theorem 13.9 Theorem 13.9 typically involves using the gradient of the function \( \nabla f \) and the given direction vector \( \mathbf{v} \) to find the directional derivative. The directional derivative \( D_{\mathbf{v}} f \) at a point \( P \) is given by: \[ D_{\mathbf{v}} f = \nabla f \cdot \mathbf{u} \] where \( \mathbf{u} \) is the unit vector in the direction of \( \mathbf{v} \) (i.e., \( \mathbf{u} = \frac{\mathbf{v}}{|\mathbf{v}|} \)) and \( \cdot \) represents the dot product. #### Steps to Solve: 1. **Compute the Gradient \( \nabla f \)**: Find the partial derivatives of \( f(x, y) \) with respect to \( x \) and \( y \). 2. **Evaluate the Gradient at Point \( P \)**: Substitute the coordinates of \( P \) into \( \nabla f \) to obtain \( \nabla f(P) \). 3. **Normalize the Direction Vector \( \mathbf{v} \)**: Find the unit vector \( \mathbf{u} = \frac{\mathbf{v}}{|\mathbf{v}|} \). 4. **Compute the Dot Product**: Find the dot product of \( \nabla f(P) \) and \( \mathbf{u} \). By following these steps, the directional derivatives for Exercises 57 and 58 can be calculated
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