/*! 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 4 Find the curve's unit tangent ve... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 4

Find the curve's unit tangent vector. Also, find the length of the indicated portion of the curve. $$\mathbf{r}(t)=(2+t) \mathbf{i}-(t+1) \mathbf{j}+t \mathbf{k}, \quad 0 \leq t \leq 3$$

Short Answer

Expert verified
The unit tangent vector is \( \frac{1}{\sqrt{3}} \mathbf{i} - \frac{1}{\sqrt{3}} \mathbf{j} + \frac{1}{\sqrt{3}} \mathbf{k} \) and the curve length is \( 3\sqrt{3} \).

Step by step solution

01

Find the Curve's Derivative

To find the unit tangent vector, we first need to differentiate the vector function \( \mathbf{r}(t) \). The derivative \( \mathbf{r}'(t) \) is calculated as follows:- Differentiate the \( i \)-component: \( \frac{d}{dt}(2+t) = 1 \).- Differentiate the \( j \)-component: \( \frac{d}{dt}(-(t+1)) = -1 \).- Differentiate the \( k \)-component: \( \frac{d}{dt}(t) = 1 \).Thus, \( \mathbf{r}'(t) = 1 \mathbf{i} - 1 \mathbf{j} + 1 \mathbf{k} \).
02

Calculate the Magnitude of the Derivative

The magnitude of \( \mathbf{r}'(t) \) is given by \( ||\mathbf{r}'(t)|| = \sqrt{(1)^2 + (-1)^2 + (1)^2} = \sqrt{3} \).
03

Find the Unit Tangent Vector

The unit tangent vector \( \mathbf{T}(t) \) is computed by dividing the derivative by its magnitude:\[ \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{||\mathbf{r}'(t)||} = \frac{1}{\sqrt{3}} \mathbf{i} - \frac{1}{\sqrt{3}} \mathbf{j} + \frac{1}{\sqrt{3}} \mathbf{k} \].
04

Find the Length of the Curve over the Given Interval

The length \( L \) of the curve over the interval from \( t = 0 \) to \( t = 3 \) is determined by integrating the magnitude of \( \mathbf{r}'(t) \) over this interval:\[ L = \int_{0}^{3} ||\mathbf{r}'(t)|| \, dt = \int_{0}^{3} \sqrt{3} \, dt = [\sqrt{3} t]_{0}^{3} \].Evaluating the definite integral yields:\[ L = \sqrt{3} \times 3 - \sqrt{3} \times 0 = 3\sqrt{3} \].

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.

Understanding the Unit Tangent Vector
In vector calculus, a unit tangent vector is used to describe the direction of a curve at any given point and is always of unit length, meaning its magnitude is 1. The process involves a few straightforward steps:
  • Start by finding the derivative of the vector function that describes the curve. In our example, the curve is given by \( \mathbf{r}(t) = (2+t) \mathbf{i} -(t+1) \mathbf{j}+t \mathbf{k} \).
  • The derivative of this vector function, denoted as \( \mathbf{r}'(t) \), gives us a new vector \( 1 \mathbf{i} - 1 \mathbf{j} + 1 \mathbf{k} \), which approximates the direction of the curve's tangent line at each point \( t \).
  • To convert this tangent direction into a unit tangent vector (\( \mathbf{T}(t) \)), divide the derivative by its magnitude.
  • The magnitude of \( \mathbf{r}'(t) \) is found to be \( \sqrt{3} \). Therefore, the unit tangent vector becomes \( \mathbf{T}(t) = \frac{1}{\sqrt{3}} \mathbf{i} - \frac{1}{\sqrt{3}} \mathbf{j} + \frac{1}{\sqrt{3}} \mathbf{k} \).
This unit tangent vector offers a consistent frame of reference for analyzing how a segment of the curve behaves in space.
Calculating Curve Length Using Integral Calculus
Finding the length of a curve is an application of integral calculus, determining how far you travel along the curve between two points. For the vector function\( \mathbf{r}(t) \) provided:
  • We use its derivative \( \mathbf{r}'(t) \) to find the rate of change along the curve, known as its velocity.
  • The magnitude of this derivative \( \|\mathbf{r}'(t)\| = \sqrt{3} \) represents how much distance is covered as \( t \) changes.
  • To find the total length of the curve from \( t=0 \) to \( t=3 \), integrate this magnitude over the given interval.
This results in the integral \( \int_{0}^{3} \sqrt{3} \, dt \). Evaluating this definite integral gives us \[ L = \left[\sqrt{3} t\right]_{0}^{3} = 3\sqrt{3} \].So, the total length of the curve over the specified range is \( 3\sqrt{3} \). This use of integrals elegantly connects the dots 鈥� or in this case, the segments 鈥� along a curve, calculating the exact path length.
Integral Calculus in Vector Applications
Integral calculus plays a pivotal role in analyzing curves described by vector functions. It involves two primary actions: integrating or summing small quantities to find total measurements.
  • When applied to vector functions, integral calculus helps in determining accumulative properties like curve lengths or area under curves.
  • In our case, where the curve \( \mathbf{r}(t) \) is defined by a vector function, we looked at the integral of the magnitude of the derivative to find how far a traveler would move between points.
  • By computing \( \int_{0}^{3} \|\mathbf{r}'(t)\| \, dt \), we collapsed a potentially infinite number of decimal points between \( t=0 \) and \( t=3 \) into a single result: \( 3\sqrt{3} \).
This power gives us not only numerical results but also fosters a deeper understanding of the 'journey' taken along the curve. Integral calculus transforms the continuous elegance of curves into understandable mathematics.
Exploring Vector Function Derivatives
Derivatives are a core concept in calculus, representing the rate of change. For vector functions, this involves differentiating each component separately to form a new vector.
  • For \( \mathbf{r}(t) = (2+t) \mathbf{i} -(t+1) \mathbf{j}+t \mathbf{k} \), we differentiate component-by-component to find \( \mathbf{r}'(t) \). This results in the vector \( 1 \mathbf{i} - 1 \mathbf{j} + 1 \mathbf{k} \).
  • The derivative vector shows how each component of \( \mathbf{r}(t) \) changes with respect to \( t \).
  • This new vector not only provides insights into the tangent direction at points along the curve but also reveals the character of motion when considered over time.
Understanding vector function derivatives is crucial to translating the continuous, flowing nature of curves into discrete, manageable pieces of data. Each component tells its own story, allowing us to map out complex trajectories in space.

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

Products of scalar and vector functions Suppose that the scalar function \(u(t)\) and the vector function \(\mathbf{r}(t)\) are both defined for \(a \leq t \leq b.\) a. Show that ur is continuous on \([a, b]\) if \(u\) and \(r\) are continuous on \([a, b].\) b. If \(u\) and \(r\) are both differentiable on \([a, b],\) show that \(u \mathbf{r}\) is differentiable on \([a, b]\) and that $$\frac{d}{d t}(u \mathbf{r})=u \frac{d \mathbf{r}}{d t}+\mathbf{r} \frac{d u}{d t}$$

Solve the initial value problems for \(\mathbf{r}\) as a vector function of \(t.\) Differential equation: \(\quad \frac{d^{2} \mathbf{r}}{d t^{2}}=-(\mathbf{i}+\mathbf{j}+\mathbf{k})\) Initial conditions: \(\quad \mathbf{r}(0)=10 \mathbf{i}+10 \mathbf{j}+10 \mathbf{k}\) and \(\left.\frac{d \mathbf{r}}{d t}\right|_{t=0}=0\)

Volleyball A volleyball is hit when it is 4 \(\mathrm{ft}\) above the ground and 12 ft from a \(6-f t\) -high net. It leaves the point of impact with an initial velocity of \(35 \mathrm{ft} / \mathrm{sec}\) at an angle of \(27^{\circ}\) and slips by the opposing team untouched. a. Find a vector equation for the path of the volleyball. b. How high does the volleyball go, and when does it reach maximum height? c. Find its range and flight time. d. When is the volleyball 7 ft above the ground? How far (ground distance) is the volleyball from where it will land? e. Suppose that the net is raised to 8 ft. Does this change things? Explain.

Projectile flights in the following exercises are to be treated as ideal unless stated otherwise. All launch angles are assumed to be measured from the horizontal. All projectiles are assumed to be launched from the origin over a horizontal surface unless stated otherwise. For some exercises, a calculator may be helpful. Firing golf balls \(\quad\) A spring gun at ground level fires a golf ball at an angle of \(45^{\circ} .\) The ball lands \(10 \mathrm{m}\) away. a. What was the ball's initial speed? b. For the same initial speed, find the two firing angles that make the range \(6 \mathrm{m}\)

the point \(\left(f\left(t_{0}\right), g\left(t_{0}\right), h\left(t_{0}\right)\right)\) parallel to \(\mathbf{v}\left(t_{0}\right),\) the curve's velocity vector at \(t_{0} .\) In Exercises \(23-26,\) find parametric equations for the line that is tangent to the given curve at the given parameter value \(t=t_{0}\). $$\mathbf{r}(t)=t^{2} \mathbf{i}+(2 t-1) \mathbf{j}+t^{3} \mathbf{k}, \quad t_{0}=2$$
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Calculus

Read Explanation

Decision Maths

Read Explanation

Discrete Mathematics

Read Explanation

Logic and Functions

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.