Chapter 10: Problem 26
Compute the directional derivative of \(f(x, y)\) at the given point in the indicated direction. $$ f(x, y)=x^{2} \sin y \text { at }(-1,0) \text { in the direction }\left[\begin{array}{r} 2 \\ -1 \end{array}\right] $$
Short Answer
Expert verified
The directional derivative is \( \frac{-1}{\sqrt{5}} \).
Step by step solution
01
Understand the Problem
We need to calculate the directional derivative of the function \( f(x, y) = x^2 \sin y \) at the point \( (-1, 0) \) in the direction of the vector \( \begin{bmatrix} 2 \ -1 \end{bmatrix} \). Directional derivatives measure how a function changes as we move in a specific direction.
02
Calculate the Gradient of the Function
The gradient of \( f \) is given by \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). Compute these partial derivatives:1. \( \frac{\partial f}{\partial x} = 2x \sin y \)2. \( \frac{\partial f}{\partial y} = x^2 \cos y \)Thus, \( abla f = \left( 2x \sin y, x^2 \cos y \right) \).
03
Evaluate the Gradient at the Given Point
Substitute \( x = -1 \) and \( y = 0 \) into the gradient:\[abla f(-1,0) = \left( 2(-1) \sin 0, (-1)^2 \cos 0 \right) = (0, 1)\]
04
Normalize the Direction Vector
The direction vector is \( \begin{bmatrix} 2 \ -1 \end{bmatrix} \). Normalize it to get a unit direction vector. The magnitude of the vector is:\[\| \begin{bmatrix} 2 \ -1 \end{bmatrix} \| = \sqrt{2^2 + (-1)^2} = \sqrt{5}\]So the unit vector is:\[\begin{bmatrix} \frac{2}{\sqrt{5}} \ \frac{-1}{\sqrt{5}} \end{bmatrix}\]
05
Compute the Directional Derivative
The directional derivative of \( f \) in the direction of a unit vector \( \mathbf{u} \) is given by:\[D_{\mathbf{u}}f = abla f \cdot \mathbf{u} \]Substitute \( abla f(-1,0) = (0, 1) \) and the unit vector:\[D_{\mathbf{u}}f = \begin{bmatrix} 0 \ 1 \end{bmatrix} \cdot \begin{bmatrix} \frac{2}{\sqrt{5}} \ \frac{-1}{\sqrt{5}} \end{bmatrix} = 0 \cdot \frac{2}{\sqrt{5}} + 1 \cdot \frac{-1}{\sqrt{5}} = \frac{-1}{\sqrt{5}}\]
06
Final Result
The directional derivative of \( f(x, y) \) at \( (-1, 0) \) in the specified direction is \( \frac{-1}{\sqrt{5}} \).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Gradient Calculation
The gradient of a function is a vital concept in calculus used to describe the direction and rate of the quickest increase of a function at a point. In our context, for the function \( f(x, y) = x^2 \sin y \), the gradient \( abla f \) is a vector that combines the partial derivatives with respect to \( x \) and \( y \). It's expressed as the vector:
\[ abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \]
Let’s break this down:
\[ abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \]
Let’s break this down:
- \( \frac{\partial f}{\partial x} = 2x \sin y \) captures how the function changes as \( x \) varies, keeping \( y \) constant.
- \( \frac{\partial f}{\partial y} = x^2 \cos y \) describes the function's change with variations in \( y \) while \( x \) remains fixed.
Partial Derivatives
Partial derivatives are a cornerstone technique in calculus, especially when dealing with functions of multiple variables. They show how a function changes as we vary one of the variables, keeping all others constant. For the function \( f(x, y) = x^2 \sin y \), two partial derivatives are crucial:
- The partial derivative with respect to \( x \), \( \frac{\partial f}{\partial x} = 2x \sin y \), tells us how the function changes as we move along the \( x \)-axis, while \( y \) is held constant.
- The partial derivative with respect to \( y \), \( \frac{\partial f}{\partial y} = x^2 \cos y \), informs us about changes in the function along the \( y \)-axis, with \( x \) fixed.
Unit Vector Normalization
Unit vector normalization is an essential method used in vector calculus to factor out the scale of a vector, focusing only on its direction. This process turns any vector into a unit vector, which means it has a length of one.
When dealing with directional derivatives, such as in the problem at hand, the direction vector provided is \( \begin{bmatrix} 2 \ -1 \end{bmatrix} \). To normalize it:
When dealing with directional derivatives, such as in the problem at hand, the direction vector provided is \( \begin{bmatrix} 2 \ -1 \end{bmatrix} \). To normalize it:
- First, calculate its magnitude using the formula \( \| \mathbf{v} \| = \sqrt{x^2 + y^2} \). For our vector:
- \( \| \begin{bmatrix} 2 \ -1 \end{bmatrix} \| = \sqrt{2^2 + (-1)^2} = \sqrt{5} \)
- Now, divide each component by the magnitude: \( \begin{bmatrix} \frac{2}{\sqrt{5}} \ \frac{-1}{\sqrt{5}} \end{bmatrix} \)