/*! 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 32 For the given vector \(\vec{v}\)... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 32

For the given vector \(\vec{v}\), find the magnitude \(\|\vec{v}\|\) and an angle \(\theta\) with \(0 \leq \theta<360^{\circ}\) so that \(\vec{v}=\|\vec{v}\|\langle\cos (\theta), \sin (\theta)\rangle\) (See Definition 11.8.) Round approximations to two decimal places. $$ \vec{v}=\langle 1, \sqrt{3}\rangle $$

Short Answer

Expert verified
The magnitude is 2 and the angle \(\theta\) is \(60^{\circ}\).

Step by step solution

01

Find the Magnitude of the Vector

The magnitude of a vector \( \vec{v} = \langle x, y \rangle \) is given by the formula: \[ \|\vec{v}\| = \sqrt{x^2 + y^2} \]. Substitute \( x = 1 \) and \( y = \sqrt{3} \) into this formula: \[ \|\vec{v}\| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \]. Therefore, the magnitude \( \|\vec{v}\| = 2 \).
02

Express Vector in Terms of Magnitude and Unit Vector

Given that \( \vec{v} = \|\vec{v}\| \langle \cos(\theta), \sin(\theta) \rangle \), we have \( \vec{v} = 2 \langle \cos(\theta), \sin(\theta) \rangle \). This implies \( 2\cos(\theta) = 1 \) and \( 2\sin(\theta) = \sqrt{3} \). Convert this into equations for \( \cos(\theta) \) and \( \sin(\theta) \): \( \cos(\theta) = \frac{1}{2} \) and \( \sin(\theta) = \frac{\sqrt{3}}{2} \).
03

Determine the Angle \(\theta\)

The values \( \cos(\theta) = \frac{1}{2} \) and \( \sin(\theta) = \frac{\sqrt{3}}{2} \) are standard trigonometric ratios for the angle \( 60^{\circ} \) (or \( \frac{\pi}{3} \) radians). We check this angle measure is within the desired range \( 0 \leq \theta < 360^{\circ} \). Thus, \( \theta = 60^{\circ} \).

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trigonometric Functions
Trigonometric functions are essential tools in mathematics, especially when dealing with vectors and angles. They help us relate the angles of a triangle to its side lengths.
In the context of vectors, trigonometric functions such as sine and cosine are used to express the vector components in terms of an angle and a magnitude. For a vector \( \vec{v} = \langle x, y \rangle \), the components can be represented using the angle \( \theta \) with formulas:
  • \( x = \|\vec{v}\| \cos(\theta) \)
  • \( y = \|\vec{v}\| \sin(\theta) \)
These relationships are based on the unit circle and the right triangle definitions of sine and cosine.
Utilizing trigonometric functions simplifies the process of finding unknowns such as angles, by comparing known values with standard trigonometric ratios.
Unit Vector
A unit vector is a vector with a magnitude of 1, and it points in the direction of the vector it represents. Unit vectors are essential because they allow us to express vectors in terms of direction alone, separated from magnitude.
For any vector \( \vec{v} \), the unit vector \( \hat{v} \) pointing in the same direction can be found by dividing each component of \( \vec{v} \) by its magnitude \( \|\vec{v}\| \). Mathematically, this is expressed as:
  • \( \hat{v} = \frac{1}{\|\vec{v}\|} \cdot \vec{v} \)
Using unit vectors, we can express any vector \( \vec{v} \) in terms of its magnitude and direction: \( \vec{v} = \|\vec{v}\| \cdot \hat{v} \).
This concept is key when solving problems like the one from the exercise, where we express a vector using its direction determined by trigonometric functions.
Angle Determination
Determining the angle \( \theta \) of a vector involves using trigonometric ratios to identify the vector's direction with respect to the origin. Given a vector \( \vec{v} = \langle x, y \rangle \), we use the formulas:
  • \( \cos(\theta) = \frac{x}{\|\vec{v}\|} \)
  • \( \sin(\theta) = \frac{y}{\|\vec{v}\|} \)
To find \( \theta \), recognize or calculate these ratios and match them to known trigonometric angles.
In our original problem, \( \cos(\theta) = \frac{1}{2} \) and \( \sin(\theta) = \frac{\sqrt{3}}{2} \), correspond to the angle \( 60^{\circ} \), a standard angle in trigonometry. Thus, determining angles often involves recognizing these common ratios for key angles, making memorization of standard trigonometric values very useful.

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

Every polar curve \(r=f(\theta)\) can be translated to a system of parametric equations with parameter \(\theta\) by \(\\{x=r \cos (\theta)=f(\theta) \cos (\theta), y=r \sin (\theta)=f(\theta) \sin (\theta) .\) Convert \(r=6 \cos (2 \theta)\) to a system of parametric equations. Check your answer by graphing \(r=6 \cos (2 \theta)\) by hand using the techniques presented in Section \(11.5\) and then graphing the parametric equations you found using a graphing utility.

In Exercises \(1-20\), plot the set of parametric equations by hand. Be sure to indicate the orientation imparted on the curve by the parametrization. $$ \left\\{\begin{array}{l} x=2 \tan (t) \\ y=\cot (t) \end{array} \text { for } 0

In Exercises 47 - 52 , we explore the hyperbolic cosine function, denoted \(\cosh (t)\), and the hyperbolic sine function, denoted \(\sinh (t)\), defined below: $$ \cosh (t)=\frac{e^{t}+e^{-t}}{2} \text { and } \sinh (t)=\frac{e^{t}-e^{-t}}{2} $$ Using a graphing utility as needed, verify that the domain of \(\cosh (t)\) is \((-\infty, \infty)\) and the range of \(\cosh (t)\) is \([1, \infty)\)

Convert the equation from polar coordinates into rectangular coordinates. $$ r=1-2 \cos (\theta) $$

Two drunken college students have filled an empty beer keg with rocks and tied ropes to it in order to drag it down the street in the middle of the night. The stronger of the two students pulls with a force of 100 pounds at a heading of \(\mathrm{N} 77^{\circ} \mathrm{E}\) and the other pulls at a heading of \(\mathrm{S} 68^{\circ} \mathrm{E}\). What force should the weaker student apply to his rope so that the keg of rocks heads due east? What resultant force is applied to the keg? Round your answer to the nearest pound.
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation

Pure Maths

Read Explanation

Decision Maths

Read Explanation

Discrete Mathematics

Read Explanation

Geometry

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.