/*! 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 21 What is vector field \(\mathbf{F... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

What is vector field \(\mathbf{F}(x, y)\) with a value at \((x, y)\) that is of unit length and points toward (1,0) ?

Short Answer

Expert verified
\( \mathbf{F}(x, y) = \left( \frac{1-x}{\sqrt{(1-x)^2+y^2}}, \frac{-y}{\sqrt{(1-x)^2+y^2}} \right) \)

Step by step solution

01

Understand the Problem

We need to find a vector field \( \mathbf{F}(x, y) \) that has two properties: it must have a unit length and it must always point towards the point \( (1, 0) \) from any given point \( (x, y) \).
02

Direction of Vector Field

To point towards \( (1, 0) \) from a point \( (x, y) \), the vector should be \( (1-x, -y) \). This vector would point from \( (x, y) \) to \( (1, 0) \).
03

Calculate Magnitude

Find the magnitude, or length, of the vector \( (1-x, -y) \): \[ \text{Magnitude} = \sqrt{(1-x)^2 + (-y)^2} = \sqrt{(1-x)^2 + y^2} \]
04

Normalize the Vector

To ensure the vector has a unit length, divide it by its magnitude: \[ \mathbf{F}(x, y) = \left( \frac{1-x}{\sqrt{(1-x)^2+y^2}}, \frac{-y}{\sqrt{(1-x)^2+y^2}} \right) \]
05

Verify Solution

Check that the vector \( \mathbf{F}(x, y) \) points in the correct direction and has a unit length. The direction is determined correctly as we derived from the vector pointing towards \( (1, 0) \), and since each component of \( \mathbf{F}(x, y) \) is divided by its magnitude, the vector has a unit length.

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.

Understanding Unit Vector
A unit vector is simply a vector that has a magnitude or length of 1. It's like a pointer indicating direction without concern for its size. In mathematical terms, if we have a vector \( \mathbf{v} \), a unit vector \( \mathbf{u} \) in the direction of \( \mathbf{v} \) is expressed as \( \mathbf{u} = \frac{\mathbf{v}}{||\mathbf{v}||} \), where \( ||\mathbf{v}|| \) is the magnitude of \( \mathbf{v} \).
This idea makes unit vectors a handy tool for describing directions, especially when you’re interested in what direction a vector points, but not in how far it extends. They often appear in physics to represent directions of forces or velocities.
Importantly, in our solution, the vector field needs to have unit vectors at every point to consistently have a length of one, ensuring it only dictates direction.
Exploring Magnitude
The magnitude of a vector, often visualized as its length, is calculated through the Pythagorean Theorem in a coordinate system. For example, a vector \( \mathbf{v} = (a, b) \) has a magnitude given by \( ||\mathbf{v}|| = \sqrt{a^2 + b^2} \). This formula forms the basis of determining the length of the vector.
Magnitude comes into play critically when we are tasked to scale vectors to become unit vectors. By calculating the original size, we can adjust the vector appropriately, ensuring it fits the desired unit-length condition.
  • Helps determine if a vector is a unit vector when checking if length is 1.
  • Used in normalizing vectors, as seen in deriving the solution for our problem.
Understanding this concept allows pinpointing the magnitude’s significance in practically reshaping vectors without altering their direction.
Directional Vector Overview
A directional vector exclusively focuses on the direction in which it points, leaving out consideration of length or magnitude initially. When tasked with finding a directional vector, the essential action is ensuring it points exactly where intended.
From the step-by-step solution, to direct towards the point \((1, 0)\) from any \((x, y)\), the natural choice of vector is \((1-x, -y)\). This selection stemmed from vector subtraction, capturing the position shift needed from \((x, y)\) to reach \((1, 0)\).
This directional vector sets the stage for later modifications, like normalizing, ensuring it achieves the ultimate goal of a unit vector with correct directional alignment. When working with directional vectors, the steps taken should prioritize accuracy in direction before considering magnitude.
The Process of Normalization
Normalization is the process of converting a vector into a unit vector—a vector with a magnitude of 1—while maintaining its direction. The aim is to standardize its magnitude, so only direction remains.
The typical process involves dividing each component of a vector by its magnitude. For a vector \( \mathbf{v} = (a, b) \), it becomes a unit vector \( \mathbf{u} = \left(\frac{a}{||\mathbf{v}||}, \frac{b}{||\mathbf{v}||}\right) \).
In the example provided, the vector \((1-x, -y)\) is normalized to obtain \( \mathbf{F}(x, y) = \left( \frac{1-x}{\sqrt{(1-x)^2+y^2}}, \frac{-y}{\sqrt{(1-x)^2+y^2}} \right) \), guaranteeing it maintains its direction but scales to the required unit length.
  • Essential for ensuring vectors only convey direction without size influence.
  • Frequently used in computer graphics for constructing objects correctly oriented irrespective of original scale.
Placing the final touches on vectors, normalization completes them, aligning with given constraints smoothly.

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

In the following exercises, find the work done by force field \(\mathbf{F}\) on an object moving along the indicated path. Let \(\mathbf{F}\) be vector field \(\mathbf{F}(x, y)=\left(y^{2}+2 x e^{y}+1\right) \mathbf{i}+\left(2 x y+x^{2} e^{y}+2 y\right) \mathbf{j}\) Compute the work of integral \(\int_{C} \mathbf{F} \cdot d \mathbf{r}, \quad\) where \(C\) is the path \(\mathbf{r}(t)=\sin t \mathbf{i}+\cos t \mathbf{j}, 0 \leq t \leq \frac{\pi}{2}\).

. & \text { Consider } & \text { radial } & \text { vector }\end{array}\( field \)\mathbf{F}=\frac{\mathbf{r}}{|\mathbf{r}|}=\frac{\langle x, y, z\rangle}{\left(x^{2}+y^{2}+z^{2}\right)^{1 / 2}}, \quad\( Compute \)\quad\( the \)\quad\( surface integral, where \)S\( is the surface of a sphere of radius \)a$ centered at the origin.

Use the divergence theorem to compute the value of flux integral \(\iint_{S} \mathbf{F} \cdot d \mathbf{S}\), where \(\mathbf{F}(x, y, z)=\left(y^{3}+3 x\right) \mathbf{i}+(x z+y) \mathbf{j}+\left[z+x^{4} \cos \left(x^{2} y\right) \mathbf{k}\right.\) and \(S\) is the area of the region bounded by \(x^{2}+y^{2}=1, x \geq 0, y \geq 0,\) and \(0 \leq z \leq 1\)

Take paraboloid \(z=x^{2}+y^{2},\) for \(0 \leq z \leq 4,\) and slice it with plane \(y=0\). Let \(S\) be the surface that remains for \(y \geq 0\), including the planar surface in the \(x z\) -plane. Let \(C\) be the semicircle and line segment that bounded the cap of \(S\) in plane \(z=4\) with counterclockwise orientation. Let \(\quad \mathbf{F}=\langle 2 z+y, 2 x+z, 2 y+x\rangle\). Evaluate \(\iint_{S}(\nabla \times \mathbf{F}) \cdot \mathbf{n} d S\)

For the following exercises, use a CAS and the divergence theorem to compute the net outward flux for the vector fields across the boundary of the given regions \(D\). [T] \(\mathbf{F}=\left\langle x^{2},-y^{2}, z^{2}\right\rangle ; D\) is the region in the first octant between planes \(z=4-x-y\) and \(z=2-x-y\)

See all solutions

Recommended explanations on Math Textbooks

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.