/*! 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 38 Consider the following trajector... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 38

Consider the following trajectories of moving objects. Find the tangential and normal components of the acceleration. $$\mathbf{r}(t)=\left\langle t, t^{2}+1\right\rangle$$

Short Answer

Expert verified
Answer: The tangential component of the acceleration is $$a_\text{T}(t) = \frac{4t}{\sqrt{1+4t^2}}$$, and the normal component of the acceleration is $$a_\text{N}(t) =\sqrt{4-\frac{16t^2}{1+4t^2}}$$.

Step by step solution

01

Find the position vector\(\mathbf{r}(t)\)

The position vector is already given: $$\mathbf{r}(t)=\langle t, t^{2}+1\rangle$$
02

Calculate the velocity vector \(\mathbf{v}(t)\)

Taking the first derivative of \(\mathbf{r}(t)\) with respect to \(t\), we get the velocity vector: $$\mathbf{v}(t)=\frac{d\mathbf{r}}{dt}=\left\langle 1, 2t\right\rangle$$
03

Calculate the acceleration vector \(\mathbf{a}(t)\)

Taking the first derivative of \(\mathbf{v}(t)\) with respect to \(t\), we get the acceleration vector: $$\mathbf{a}(t)=\frac{d\mathbf{v}}{dt}=\left\langle 0, 2\right\rangle$$
04

Calculate the speed \(v(t)\)

The magnitude of the velocity vector \(\mathbf{v}(t)\) is the speed \(v(t)\). We calculate it as follows: $$v(t)=|\mathbf{v}(t)|=\sqrt{1^2+(2t)^2}=\sqrt{1+4t^2}$$
05

Calculate the unit tangent vector \(\mathbf{T}(t)\)

The unit tangent vector \(\mathbf{T}(t)\) is given by dividing the velocity vector by its magnitude: $$\mathbf{T}(t)=\frac{\mathbf{v}(t)}{v(t)}=\left\langle\frac{1}{\sqrt{1+4t^2}},\frac{2t}{\sqrt{1+4t^2}}\right\rangle$$
06

Find the tangential component of the acceleration

The tangential component of the acceleration, \(a_\text{T}(t)\), is given by the dot product of \(\mathbf{a}(t)\) and \(\mathbf{T}(t)\): $$a_\text{T}(t) = \mathbf{a}(t) \cdot \mathbf{T}(t)=\left\langle 0, 2\right\rangle \cdot \left\langle\frac{1}{\sqrt{1+4t^2}},\frac{2t}{\sqrt{1+4t^2}}\right\rangle = (0)(\frac{1}{\sqrt{1+4t^2}})+(2)(\frac{2t}{\sqrt{1+4t^2}})=\frac{4t}{\sqrt{1+4t^2}}$$
07

Find the normal component of the acceleration

Finally, the normal component of the acceleration, \(a_\text{N}(t)\), is given by the square root of the difference between the square of the magnitude of the acceleration vector and the square of the tangential component: $$a_\text{N}(t)=\sqrt{\left|\mathbf{a}(t)\right|^2-a_\text{T}^2(t)}=\sqrt{(0^2+2^2)-\left(\frac{4t}{\sqrt{1+4t^2}}\right)^2}=\sqrt{4-\frac{16t^2}{1+4t^2}}$$ Thus, the tangential component of the acceleration is \(a_\text{T}(t) = \frac{4t}{\sqrt{1+4t^2}}\) and the normal component of the acceleration is \(a_\text{N}(t) =\sqrt{4-\frac{16t^2}{1+4t^2}}\).

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

Jack pulls east on a rope attached to a camel with a force of 40 ib. Jill pulls north on a rope attached to the same camel with a force of 30 Ib. What is the magnitude and direction of the force on the camel? Assume the vectors lie in a horizontal plane.

Let \(\mathbf{u}(t)=\left\langle 1, t, t^{2}\right\rangle, \mathbf{v}(t)=\left\langle t^{2},-2 t, 1\right\rangle\) and \(g(t)=2 \sqrt{t}\). Compute the derivatives of the following functions. $$\mathbf{u}(t) \times \mathbf{v}(t)$$

Consider the curve \(\mathbf{r}(t)=(a \cos t+b \sin t) \mathbf{i}+(c \cos t+d \sin t) \mathbf{j}+(e \cos t+f \sin t) \mathbf{k}\) where \(a, b, c, d, e,\) and \(f\) are real numbers. It can be shown that this curve lies in a plane. Assuming the curve lies in a plane, show that it is a circle centered at the origin with radius \(R\) provided \(a^{2}+c^{2}+e^{2}=b^{2}+d^{2}+f^{2}=R^{2}\) and \(a b+c d+e f=0\).

Compute the following derivatives. $$\frac{d}{d t}\left(\left(t^{3} \mathbf{i}-2 t \mathbf{j}-2 \mathbf{k}\right) \times\left(t \mathbf{i}-t^{2} \mathbf{j}-t^{3} \mathbf{k}\right)\right)$$

Proof of Sum Rule By expressing \(\mathbf{u}\) and \(\mathbf{v}\) in terms of their components, prove that $$\frac{d}{d t}(\mathbf{u}(t)+\mathbf{v}(t))=\mathbf{u}^{\prime}(t)+\mathbf{v}^{\prime}(t)$$
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Discrete Mathematics

Read Explanation

Calculus

Read Explanation

Statistics

Read Explanation

Geometry

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.