Chapter 12: Problem 13
In what direction \(\mathbf{u}\) does \(f(x, y)=1-x^{2}-y^{2}\) decrease most rapidly at \(\mathbf{p}=(-1,2) ?\)
Short Answer
Expert verified
The direction \( \mathbf{u} \) is \( \left(-\frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}}\right) \).
Step by step solution
01
Find the Gradient of the Function
The gradient of the function \( f(x, y) = 1 - x^2 - y^2 \) is calculated using partial derivatives. Compute the partial derivatives: \( \frac{\partial f}{\partial x} = -2x \) and \( \frac{\partial f}{\partial y} = -2y \). So, the gradient vector is \( abla f = (-2x, -2y) \).
02
Evaluate the Gradient at the Given Point
Substitute the point \( \mathbf{p} = (-1, 2) \) into the gradient vector: \( abla f(-1, 2) = (-2(-1), -2(2)) \). Simplifying this, we get \( abla f(-1, 2) = (2, -4) \).
03
Determine the Direction of Steepest Descent
The direction of steepest descent is given by the negation of the gradient vector. Thus, the direction \( \mathbf{u} \) is \( -abla f(-1, 2) = (-2, 4) \).
04
Normalize the Direction Vector
Normalize the vector \( (-2, 4) \) to make \( \mathbf{u} \) a unit vector. Compute the magnitude: \( \|(-2, 4)\| = \sqrt{(-2)^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \). Divide each component by the magnitude: \( \mathbf{u} = \left(-\frac{2}{2\sqrt{5}}, \frac{4}{2\sqrt{5}}\right) = \left(-\frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}}\right) \).
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.
Directional Derivative
In mathematics, the directional derivative represents the rate at which a function changes as you move in a specific direction. Imagine walking on a hilly terrain; the directional derivative tells you how steep the hill is in the direction you choose to walk. It helps in understanding how a function behaves generally and is particularly useful in optimization problems.
To calculate the directional derivative of a function, you first need to know the gradient of the function. The gradient is a vector that represents the direction of the steepest ascent of the function, and its magnitude describes how steep that ascent is.
To calculate the directional derivative of a function, you first need to know the gradient of the function. The gradient is a vector that represents the direction of the steepest ascent of the function, and its magnitude describes how steep that ascent is.
- The directional derivative of a function at a given point can be computed by taking the dot product of the gradient vector and a unit vector in the direction you're interested in.
- This reveals how quickly the function increases or decreases in any specified direction from a particular point.
Partial Derivatives
Partial derivatives are a fundamental tool in calculus used to analyze functions of multiple variables. They focus specifically on how a function changes as one of the variables is altered while keeping the others constant.
For instance, if you have a function like our example: \[ f(x, y) = 1 - x^2 - y^2 \]
For instance, if you have a function like our example: \[ f(x, y) = 1 - x^2 - y^2 \]
- The partial derivative with respect to \( x \) (denoted as \( \frac{\partial f}{\partial x} \)) measures the rate of change of the function as \( x \) changes, holding \( y \) constant.
- The partial derivative with respect to \( y \) works similarly by analyzing changes in \( f \) as \( y \) varies, while \( x \) is fixed.
Unit Vector
A unit vector is a vector that has a magnitude of one. It is used to indicate direction without any concern for magnitude.
When you deal with functions and their derivatives, particularly in optimization, normalizing a direction vector to become a unit vector helps in simplifying calculations. For instance:
When you deal with functions and their derivatives, particularly in optimization, normalizing a direction vector to become a unit vector helps in simplifying calculations. For instance:
- If you have a direction vector such as \((-2, 4)\), converting it into a unit vector involves dividing each component by the vector's magnitude.
- This normalized vector simply points out the direction, allowing computations like the directional derivative to reflect purely directional trends without magnitude bias.