/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 50 In your own words, give a geomet... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 50

In your own words, give a geometric description of the directional derivative of \(z=f(x, y)\).

Short Answer

Expert verified
Geometrically, the directional derivative of a function \(z=f(x, y)\) at a point \((x, y)\) in the direction of a unit vector \(u\) provides the rate of change (slope) of the function in the direction of the vector \(u\).

Step by step solution

01

Define the Directional Derivative

The directional derivative of a function \(z=f(x, y)\) in the direction of a unit vector \(u = \) is a derivative that measures the rate at which the function changes at a point in the direction of that vector. Mathematically, it is defined as \(D_{u}f(x, y) = f_{x}(x, y)a + f_{y}(x, y)b\).
02

Describe the Geometric Interpretation

Geometrically, the directional derivative of \(z=f(x, y)\) at a point \((x, y)\), in the direction of a unit vector \(u\), provides the slope of the function at that point in that specific direction. More intuitively, it measures 'how fast' the function is increasing or decreasing at point \((x, y)\) if you start moving in the direction of the given vector \(u\). This is essentially the projection of the gradient vector \(\nabla f\) onto \(u\).

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 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Differentiate implicitly to find \(d y / d x\). \(\frac{x}{x^{2}+y^{2}}-y^{2}=6\)

Consider the function \(w=f(x, y),\) where \(x=r \cos \theta\) and \(y=r \sin \theta .\) Prove each of the following. (a) \(\frac{\partial w}{\partial x}=\frac{\partial w}{\partial r} \cos \theta-\frac{\partial w}{\partial \theta} \frac{\sin \theta}{r}\) \(\frac{\partial w}{\partial y}=\frac{\partial w}{\partial r} \sin \theta+\frac{\partial w}{\partial \theta} \frac{\cos \theta}{r}\) (b) \(\left(\frac{\partial w}{\partial x}\right)^{2}+\left(\frac{\partial w}{\partial y}\right)^{2}=\left(\frac{\partial w}{\partial r}\right)^{2}+\left(\frac{1}{r^{2}}\right)\left(\frac{\partial w}{\partial \theta}\right)^{2}\)

Differentiate implicitly to find the first partial derivatives of \(z\) \(e^{x z}+x y=0\)

Find \(d w / d t\) (a) using the appropriate Chain Rule and (b) by converting \(w\) to a function of \(t\) before differentiating. \(w=x^{2}+y^{2}+z^{2}, \quad x=e^{t} \cos t, \quad y=e^{t} \sin t, \quad z=e^{t}\)

In Exercises \(39-42,\) find \(\partial w / \partial r\) and \(\partial w / \partial \theta\) (a) using the appropriate Chain Rule and (b) by converting \(w\) to a function of \(r\) and \(\boldsymbol{\theta}\) before differentiating. \(w=x^{2}-2 x y+y^{2}, x=r+\theta, \quad y=r-\theta\)
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Applied Mathematics

Read Explanation

Logic and Functions

Read Explanation

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Calculus

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.