/*! 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 14 Find \(\mathbf{T}, \mathbf{N},\)... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 14

Find \(\mathbf{T}, \mathbf{N},\) and \(\kappa\) for the space curves. $$\mathbf{r}(t)=\left(\cos ^{3} t\right) \mathbf{i}+\left(\sin ^{3} t\right) \mathbf{j}, \quad 0

Short Answer

Expert verified
\( \mathbf{T}(t) = -\cos t \mathbf{i} + \sin t \mathbf{j}, \mathbf{N}(t) = \sin t \mathbf{i} + \cos t \mathbf{j}, \kappa(t) = \frac{2}{3\sin(2t)} \).

Step by step solution

01

Find the Velocity Vector

The velocity vector \( \mathbf{v}(t) \) is the derivative of \( \mathbf{r}(t) \) with respect to \( t \). Calculate \( \mathbf{v}(t) = \frac{d}{dt}\left( \cos^3 t \right)\mathbf{i} + \frac{d}{dt}\left( \sin^3 t \right)\mathbf{j} \). \[ \mathbf{v}(t) = -3\cos^2 t \sin t\mathbf{i} + 3\sin^2 t \cos t\mathbf{j} \]
02

Find the Unit Tangent Vector \( \mathbf{T}(t) \)

The unit tangent vector \( \mathbf{T}(t) \) is given by \( \mathbf{T}(t) = \frac{\mathbf{v}(t)}{\|\mathbf{v}(t)\|} \). First, find \( \|\mathbf{v}(t)\| = \sqrt{(-3\cos^2 t \sin t)^2 + (3\sin^2 t \cos t)^2} = 3\sin t \cos t \). Thus, \[ \mathbf{T}(t) = \frac{-\cos t \sin t\mathbf{i} + \sin^2 t \mathbf{j}}{\sin t \cos t} \] Simplifies to \( \mathbf{T}(t) = -\cos t \mathbf{i} + \sin t \mathbf{j} \).
03

Find the Derivative of the Unit Tangent Vector

Once we have \( \mathbf{T}(t) = -\cos t \mathbf{i} + \sin t \mathbf{j} \), calculate its derivative with respect to \( t \). \[ \frac{d\mathbf{T}(t)}{dt} = \sin t \mathbf{i} + \cos t \mathbf{j} \]
04

Find the Unit Normal Vector \( \mathbf{N}(t) \)

The unit normal vector is \( \mathbf{N}(t) = \frac{\frac{d\mathbf{T}(t)}{dt}}{\left\|\frac{d\mathbf{T}(t)}{dt}\right\|} \). Calculate the magnitude: \[ \left\|\frac{d\mathbf{T}(t)}{dt}\right\| = \sqrt{(\sin t)^2 + (\cos t)^2} = 1 \] Therefore, \( \mathbf{N}(t) = \sin t \mathbf{i} + \cos t \mathbf{j} \).
05

Compute the Curvature \( \kappa(t) \)

Curvature \( \kappa(t) \) is found using the formula \[ \kappa(t) = \frac{\left\|\frac{d\mathbf{T}(t)}{dt}\right\|}{\left\|\mathbf{v}(t)\right\|} \]. Substitute the found values: \[ \kappa(t) = \frac{1}{3\sin t \cos t} = \frac{1}{\frac{3}{2} \sin(2t)} \].

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.

Unit Tangent Vector
The unit tangent vector, denoted as \( \mathbf{T}(t) \), is essential in vector calculus for understanding the direction of a curve in space. It's derived from the velocity vector \( \mathbf{v}(t) \). The velocity vector represents the instantaneous rate of change of the position vector \( \mathbf{r}(t) \). To get the unit tangent vector, we normalize the velocity vector by dividing it by its magnitude.

Here's how the process works:
  • Calculate the derivative of the position vector \( \mathbf{r}(t) \) to find the velocity vector \( \mathbf{v}(t) = -3\cos^2 t \sin t \mathbf{i} + 3\sin^2 t \cos t \mathbf{j} \).
  • Find the magnitude of the velocity vector: \( \|\mathbf{v}(t)\| = 3\sin t \cos t \).
  • Normalize \( \mathbf{v}(t) \) to get \( \mathbf{T}(t) = \frac{\mathbf{v}(t)}{\|\mathbf{v}(t)\|} \), which simplifies to \( \mathbf{T}(t) = -\cos t \mathbf{i} + \sin t \mathbf{j} \).
Essentially, \( \mathbf{T}(t) \) gives the curve's direction, maintaining a unit length that helps us focus solely on direction without magnitude distortion.
Unit Normal Vector
The unit normal vector \( \mathbf{N}(t) \) plays a significant role in identifying how a curve changes direction. It is perpendicular to the unit tangent vector and lies within the plane of curvature. To derive it, we first find the derivative of the unit tangent vector with respect to \( t \), which we denote as \( \frac{d\mathbf{T}(t)}{dt} \).

Here is how you determine the unit normal vector:
  • Differentiate the unit tangent vector \( \mathbf{T}(t) \) to get \( \frac{d\mathbf{T}(t)}{dt} = \sin t \mathbf{i} + \cos t \mathbf{j} \).
  • Calculate the magnitude of this derivative, which is \( \left\| \frac{d\mathbf{T}(t)}{dt} \right\| = 1 \).
  • Normalize \( \frac{d\mathbf{T}(t)}{dt} \) to find \( \mathbf{N}(t) = \sin t \mathbf{i} + \cos t \mathbf{j} \).
The unit normal vector is crucial as it indicates the direction in which the curve is turning or bending. Unlike \( \mathbf{T}(t) \), which hints at directional flow, \( \mathbf{N}(t) \) provides insights into the curve's behavior more deeply.
Curvature
Curvature, denoted as \( \kappa(t) \), measures how sharply a curve bends at a particular point. It is a crucial concept in vector calculus that shows the change in directionrate of the tangent vector along a curve. High curvature implies sharper bending, while low curvature indicates a smoother, less pronounced turn.

To find the curvature, you can follow these steps:
  • Use the unit tangent vector's derivative magnitude \( \left\| \frac{d\mathbf{T}(t)}{dt} \right\| \) which has already been calculated as 1.
  • Find the magnitude of the velocity vector \( \| \mathbf{v}(t) \| = 3\sin t \cos t \).
  • Compute the curvature using the formula \( \kappa(t) = \frac{ \left\| \frac{d\mathbf{T}(t)}{dt} \right\| }{ \| \mathbf{v}(t) \| } = \frac{1}{3\sin t \cos t} \). Ultimately, the formula simplifies to \( \kappa(t) = \frac{1}{\frac{3}{2} \sin(2t)} \).
Curvature helps in understanding the geometrical properties of the curve, such as where it might loop or have higher turns, thus providing a complete picture of its layout.

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

Find an equation for the circle of curvature of the curve \(\mathbf{r}(t)=\) \((2 \ln t) \mathbf{i}-[t+(1 / t)] \mathbf{j}, e^{-2} \leq t \leq e^{2},\) at the point (0,-2) where \(t=1\)

Find the are length parameter along the curve from the point where \(t=0\) by evaluating the integral \(s(t)=\int_{0}^{t}|\mathbf{v}(\tau)| d \tau\) from Equation (3). Then use the formula for \(s(t)\) to find the length of the indicated portion of the curve. $$\mathbf{r}(t)=\left(e^{t} \cos t\right) \mathbf{i}+\left(e^{t} \sin t\right) \mathbf{j}+e^{t} \mathbf{k}, \quad-\ln 4 \leq t \leq 0$$

Use a CAS to perform the following steps. a. Plot the space curve traced out by the position vector \(\mathbf{r}\). b. Find the components of the velocity vector \(d \mathbf{r} / d t\) c. Evaluate \(d \mathbf{r} / d t\) at the given point \(t_{0}\) and determine the equation of the tangent line to the curve at \(\mathbf{r}\left(t_{0}\right)\) d. Plot the tangent line together with the curve over the given interval. $$\begin{array}{l} \mathbf{r}(t)=(\sin t-t \cos t) \mathbf{i}+(\cos t+t \sin t) \mathbf{j}+t^{2} \mathbf{k} \\ 0 \leq t \leq 6 \pi, \quad t_{0}=3 \pi / 2 \end{array}$$

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. Flight time and height \(\quad\) A projectile is fired with an initial speed of \(500 \mathrm{m} / \mathrm{sec}\) at an angle of elevation of \(45^{\circ}\) a. When and how far away will the projectile strike? b. How high overhead will the projectile be when it is \(5 \mathrm{km}\) downrange? c. What is the greatest height reached by the projectile?

Show that the parabola \(y=a x^{2}, a \neq 0,\) has its largest curvature et its vertex and has no minimum curvature. (Note: since the curvature of a curve remains the same if the curve is translated or rotated, this result is true for any parabola.)
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Logic and Functions

Read Explanation

Statistics

Read Explanation

Pure Maths

Read Explanation

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

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.