/*! 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 26 Find the unit tangent vector for... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 26

Find the unit tangent vector for the following parameterized curves. $$\mathbf{r}(t)=\left\langle e^{2 t}, 2 e^{2 t}, 2 e^{-3 t}\right\rangle, \text { for } t \geq 0$$

Short Answer

Expert verified
Answer: The unit tangent vector for the given parameterized curve is \(\mathbf{T}(t) = \left\langle \frac{2e^{2t}}{\sqrt{4e^{4t} + 16e^{4t} + 36e^{-6t}}}, \frac{4e^{2t}}{\sqrt{4e^{4t} + 16e^{4t} + 36e^{-6t}}}, \frac{-6e^{-3t}}{\sqrt{4e^{4t} + 16e^{4t} + 36e^{-6t}}} \right\rangle\).

Step by step solution

01

Calculate the derivative of the curve

To find the derivative, take the derivative of each component with respect to \(t\): $$ \frac{d\mathbf{r}(t)}{dt} = \left\langle \frac{de^{2t}}{dt}, \frac{d(2e^{2t})}{dt}, \frac{d(2e^{-3t})}{dt} \right\rangle $$ Using the chain rule, we get: $$ \frac{d\mathbf{r}(t)}{dt} = \left\langle 2e^{2t}, 4e^{2t}, -6e^{-3t} \right\rangle $$
02

Calculate the magnitude of the curve's derivative

Next, find the magnitude of \(\frac{d\mathbf{r}(t)}{dt}\) by taking the square root of the sum of the squares of its components: $$ ||\frac{d\mathbf{r}(t)}{dt}|| = \sqrt{(2e^{2t})^2 + (4e^{2t})^2 + (-6e^{-3t})^2} $$ Simplify to find the magnitude: $$ ||\frac{d\mathbf{r}(t)}{dt}|| = \sqrt{4e^{4t} + 16e^{4t} + 36e^{-6t}} $$
03

Find the unit tangent vector

To find the unit tangent vector, divide the curve's derivative by its magnitude, component-wise: $$ \mathbf{T}(t) = \frac{\frac{d\mathbf{r}(t)}{dt}}{||\frac{d\mathbf{r}(t)}{dt}||} = \frac{\left\langle 2e^{2t}, 4e^{2t}, -6e^{-3t} \right\rangle}{\sqrt{4e^{4t} + 16e^{4t} + 36e^{-6t}}} $$ Thus, the unit tangent vector for the given parameterized curve is: $$ \mathbf{T}(t) = \left\langle \frac{2e^{2t}}{\sqrt{4e^{4t} + 16e^{4t} + 36e^{-6t}}}, \frac{4e^{2t}}{\sqrt{4e^{4t} + 16e^{4t} + 36e^{-6t}}}, \frac{-6e^{-3t}}{\sqrt{4e^{4t} + 16e^{4t} + 36e^{-6t}}} \right\rangle $$

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

Carry out the following steps to determine the (smallest) distance between the point \(P\) and the line \(\ell\) through the origin. a. Find any vector \(\mathbf{v}\) in the direction of \(\ell\) b. Find the position vector u corresponding to \(P\). c. Find \(\operatorname{proj}_{\mathbf{v}} \mathbf{u}\). d. Show that \(\mathbf{w}=\mathbf{u}-\) projy \(\mathbf{u}\) is a vector orthogonal to \(\mathbf{v}\) whose length is the distance between \(P\) and the line \(\ell\) e. Find \(\mathbf{w}\) and \(|\mathbf{w}| .\) Explain why \(|\mathbf{w}|\) is the distance between \(P\) and \(\ell\). \(P(1,1,-1) ; \ell\) has the direction of $$\langle-6,8,3\rangle$$.

Let \(\mathbf{u}=\left\langle u_{1}, u_{2}, u_{3}\right\rangle\) \(\mathbf{v}=\left\langle v_{1}, v_{2}, v_{3}\right\rangle\), and \(\mathbf{w}=\) \(\left\langle w_{1}, w_{2}, w_{3}\right\rangle\). Let \(c\) be a scalar. Prove the following vector properties. \(c(\mathbf{u} \cdot \mathbf{v})=(c \mathbf{u}) \cdot \mathbf{v}=\mathbf{u} \cdot(c \mathbf{v})\)

Maximum curvature Consider the "superparabolas" \(f_{n}(x)=x^{2 n},\) where \(n\) is a positive integer. a. Find the curvature function of \(f_{n},\) for \(n=1,2,\) and 3 b. Plot \(f_{n}\) and their curvature functions, for \(n=1,2,\) and 3 and check for consistency. c. At what points does the maximum curvature occur, for \(n=1,2,3 ?\) d. Let the maximum curvature for \(f_{n}\) occur at \(x=\pm z_{n} .\) Using either analytical methods or a calculator determine \(\lim _{n \rightarrow \infty} z_{n}\) Interpret your result.

The definition \(\mathbf{u} \cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}| \cos \theta\) implies that \(|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u}||\mathbf{v}|(\text {because}|\cos \theta| \leq 1) .\) This inequality, known as the Cauchy-Schwarz Inequality, holds in any number of dimensions and has many consequences. Consider the vectors \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{u}+\mathbf{v}\) (in any number of dimensions). Use the following steps to prove that \(|\mathbf{u}+\mathbf{v}| \leq|\mathbf{u}|+|\mathbf{v}|\). a. Show that \(|\mathbf{u}+\mathbf{v}|^{2}=(\mathbf{u}+\mathbf{v}) \cdot(\mathbf{u}+\mathbf{v})=|\mathbf{u}|^{2}+\) \(2 \mathbf{u} \cdot \mathbf{v}+|\mathbf{v}|^{2}\). b. Use the Cauchy-Schwarz Inequality to show that \(|\mathbf{u}+\mathbf{v}|^{2} \leq(|\mathbf{u}|+|\mathbf{v}|)^{2}\). c. Conclude that \(|\mathbf{u}+\mathbf{v}| \leq|\mathbf{u}|+|\mathbf{v}|\). d. Interpret the Triangle Inequality geometrically in \(\mathbb{R}^{2}\) or \(\mathbb{R}^{3}\).

An object moves along an ellipse given by the function \(\mathbf{r}(t)=\langle a \cos t, b \sin t\rangle,\) for \(0 \leq t \leq 2 \pi,\) where \(a > 0\) and \(b > 0\) a. Find the velocity and speed of the object in terms of \(a\) and \(b\) for \(0 \leq t \leq 2 \pi\) b. With \(a=1\) and \(b=6,\) graph the speed function, for \(0 \leq t \leq 2 \pi .\) Mark the points on the trajectory at which the speed is a minimum and a maximum. c. Is it true that the object speeds up along the flattest (straightest) parts of the trajectory and slows down where the curves are sharpest? d. For general \(a\) and \(b\), find the ratio of the maximum speed to the minimum speed on the ellipse (in terms of \(a\) and \(b\) ).
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Logic and Functions

Read Explanation

Discrete Mathematics

Read Explanation

Calculus

Read Explanation

Statistics

Read Explanation

Pure Maths

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.