/*! 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 1 Suppose \(\mathbf{n}\) is a vect... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 1

Suppose \(\mathbf{n}\) is a vector normal to the tangent plane of the surface \(F(x, y, z)=0\) at a point. How is \(\mathbf{n}\) related to the gradient of \(F\) at that point?

Short Answer

Expert verified
Answer: At a specific point on the surface, the normal vector to the tangent plane is equal to the gradient of the function defining the surface. Thus, $\mathbf{n} = \nabla F$.

Step by step solution

01

Recall the definition of the gradient vector

The gradient of a function \(F(x, y, z)\) is a vector that points in the direction of the maximum rate of change of the function at a given point, and its magnitude represents the rate of change in that direction. The gradient of \(F\) is defined as: $$\nabla F = \left(\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z}\right).$$
02

Compute the tangent plane at a point on the surface

To compute the tangent plane, consider a point \((x_0, y_0, z_0)\) that lies on the surface. The tangent plane at this point can be written as a linear equation: $$\Delta F = \frac{\partial F}{\partial x}(x - x_0) + \frac{\partial F}{\partial y}(y - y_0) + \frac{\partial F}{\partial z}(z - z_0) = 0.$$
03

Derive the normal vector

The normal vector \(\mathbf{n}\) is a vector that is perpendicular to the tangent plane, which can be derived from the tangent plane equation: $$\mathbf{n} = \left(\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z}\right).$$
04

Relate the gradient to the normal vector

We observe that \(\mathbf{n}\) is equal to the gradient vector \(\nabla F\). This means that at a given point on the surface \(F(x, y, z) = 0\), the normal vector \(\mathbf{n}\) to the tangent plane is equal to the gradient of \(F\) at that point: $$\mathbf{n} = \nabla F.$$

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

Let $$f(x, y)=\left\\{\begin{array}{ll}\frac{\sin \left(x^{2}+y^{2}-1\right)}{x^{2}+y^{2}-1} & \text { if } x^{2}+y^{2} \neq 1 \\\b & \text { if } x^{2}+y^{2}=1\end{array}\right.$$ Find the value of \(b\) for which \(f\) is continuous at all points in \(\mathbb{R}^{2}\).

Use the formal definition of a limit to prove that $$\lim _{(x, y) \rightarrow(a, b)} c f(x, y)=c \lim _{(x, y) \rightarrow(a, b)} f(x, y)$$

Determine whether the following statements are true and give an explanation or counterexample. a. The plane passing through the point (1,1,1) with a normal vector \(\mathbf{n}=\langle 1,2,-3\rangle\) is the same as the plane passing through the point (3,0,1) with a normal vector \(\mathbf{n}=\langle-2,-4,6\rangle\) b. The equations \(x+y-z=1\) and \(-x-y+z=1\) describe the same plane. c. Given a plane \(Q\), there is exactly one plane orthogonal to \(Q\). d. Given a line \(\ell\) and a point \(P_{0}\) not on \(\ell\), there is exactly one plane that contains \(\ell\) and passes through \(P_{0}\) e. Given a plane \(R\) and a point \(P_{0},\) there is exactly one plane that is orthogonal to \(R\) and passes through \(P_{0}\) f. Any two distinct lines in \(\mathbb{R}^{3}\) determine a unique plane. g. If plane \(Q\) is orthogonal to plane \(R\) and plane \(R\) is orthogonal to plane \(S\), then plane \(Q\) is orthogonal to plane \(S\).

The closed unit ball in \(\mathbb{R}^{3}\) centered at the origin is the set \(\left\\{(x, y, z): x^{2}+y^{2}+z^{2} \leq 1\right\\} .\) Describe the following alternative unit balls. a. \(\\{(x, y, z):|x|+|y|+|z| \leq 1\\}.\) b. \(\\{(x, y, z): \max \\{|x|,|y|,|z|\\} \leq 1\\},\) where \(\max \\{a, b, c\\}\) is the maximum value of \(a, b,\) and \(c.\)

Among all triangles with a perimeter of 9 units, find the dimensions of the triangle with the maximum area. It may be easiest to use Heron's formula, which states that the area of a triangle with side length \(a, b,\) and \(c\) is \(A=\sqrt{s(s-a)(s-b)(s-c)},\) where \(2 s\) is the perimeter of the triangle.
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Calculus

Read Explanation

Discrete Mathematics

Read Explanation

Logic and Functions

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.