/*! 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 23 Find the curvature \(K\) of the ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 23

Find the curvature \(K\) of the curve, where \(s\) is the arc length parameter. Helix in Exercise 19: \(\mathbf{r}(t)=\langle 2 \cos t, 2 \sin t, t\rangle\)

Short Answer

Expert verified
The curvature \(K\) of the curve \(\mathbf{r}(t)=\langle 2 \cos t, 2 \sin t, t\rangle\) is \(\frac{\sqrt{20}}{27}\).

Step by step solution

01

Compute the Derivatives

The first step will be to compute the first derivative, i.e., the velocity, \(\mathbf{r}'(t)\), and the second derivative, i.e., the acceleration, \(\mathbf{r}''(t)\), of the given parametric equation. \(\mathbf{r}'(t)=\langle-2\sin{t}, 2\cos{t}, 1\rangle\) is the first derivative, while \(\mathbf{r}''(t)=\langle-2\cos{t}, -2\sin{t}, 0\rangle\) is the second derivative.
02

Compute the Cross Product of the Derivatives

Next, calculate the cross product of the velocity and the acceleration, \(\mathbf{r}'(t) \times \mathbf{r}''(t)\). Using the formula for the cross product, this will be \(\langle 2\sin{t}, 2\cos{t}, -4\rangle\).
03

Calculate the Magnitude

Now, calculate the magnitude of both vectors \(\mathbf{r}'(t) \times \mathbf{r}''(t)\) and \(\mathbf{r}'(t)\). The magnitude, \(\|\mathbf{r}'(t) \times \mathbf{r}''(t)\| = |\langle 2\sin{t}, 2\cos{t}, -4\rangle| = \sqrt{(2\sin{t})^2+(2\cos{t})^2 + (-4)^2} = \sqrt{20}\). Similarly, \(\|\mathbf{r}'(t)\| = |\langle[-2\sin{t}, 2\cos{t}, 1]\rangle| = sqrt{(-2sin{t})^2 + (2cos{t})^2 + 1^2} = sqrt{9}\).
04

Find the Curvature

Finally, using the formula for curvature: \(K=\frac{\|\mathbf{r}'(t) \times \mathbf{r}''(t)\|}{\|\mathbf{r}'(t)\|^3}\), substitute the calculated magnitudes to get \(K=\frac{\sqrt{20}}{(sqrt{9})^3}= \frac{\sqrt{20}}{27}\)

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

Prove the property. In each case, assume that \(\mathbf{r}, \mathbf{u},\) and \(\mathbf{v}\) are differentiable vector-valued functions of \(t,\) \(f\) is a differentiable real-valued function of \(t,\) and \(c\) is a scalar. $$ D_{t}\left[\mathbf{r}(t) \times \mathbf{r}^{\prime}(t)\right]=\mathbf{r}(t) \times \mathbf{r}^{\prime \prime}(t) $$

The position vector \(r\) describes the path of an object moving in the \(x y\) -plane. Sketch a graph of the path and sketch the velocity and acceleration vectors at the given point. $$ \mathbf{r}(t)=t^{2} \mathbf{i}+t^{3} \mathbf{j},(1,1) $$

Find the tangential and normal components of acceleration for a projectile fired at an angle \(\theta\) with the horizontal at an initial speed of \(v_{0}\). What are the components when the projectile is at its maximum height?

In Exercises \(43-48,\) find the indefinite integral. $$ \int(2 t \mathbf{i}+\mathbf{j}+\mathbf{k}) d t $$

Find the open interval(s) on which the curve given by the vector-valued function is smooth. $$ \mathbf{r}(\theta)=2 \cos ^{3} \theta \mathbf{i}+3 \sin ^{3} \theta \mathbf{j} $$
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Logic and Functions

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

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.