/*! 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 29 If \(\mathbf{r}(t)=t \mathbf{i}+... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

If \(\mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j}+t^{3} \mathbf{k}\) is the position vector for a moving particle at time \(t\), find the tangential and normal components, \(a_{T}\) and \(a_{N},\) of the acceleration vector at \(t=1\)

Short Answer

Expert verified
\(a_T = \frac{22}{\sqrt{14}}\), \(a_N \approx 0.93\).

Step by step solution

01

Compute the velocity vector

The velocity vector \( \mathbf{v}(t) \) is the derivative of the position vector \( \mathbf{r}(t) \) with respect to time \( t \). This is given by:\[ \mathbf{v}(t) = \frac{d}{dt}(t \mathbf{i} + t^2 \mathbf{j} + t^3 \mathbf{k}) = \mathbf{i} + 2t \mathbf{j} + 3t^2 \mathbf{k} \]At \( t = 1 \), the velocity vector is:\[ \mathbf{v}(1) = \mathbf{i} + 2 \mathbf{j} + 3 \mathbf{k} \]
02

Compute the acceleration vector

The acceleration vector \( \mathbf{a}(t) \) is the derivative of the velocity vector \( \mathbf{v}(t) \) with respect to \( t \). We find:\[ \mathbf{a}(t) = \frac{d}{dt}(\mathbf{i} + 2t \mathbf{j} + 3t^2 \mathbf{k}) = 2 \mathbf{j} + 6t \mathbf{k} \]At \( t = 1 \), the acceleration vector is:\[ \mathbf{a}(1) = 2 \mathbf{j} + 6 \mathbf{k} \]
03

Compute the magnitude of the velocity vector

To find the tangential component of the acceleration, we first compute the magnitude of the velocity vector at \( t = 1 \). This is given by:\[ \|\mathbf{v}(1)\| = \sqrt{(1)^2 + (2)^2 + (3)^2} = \sqrt{14} \]
04

Find the tangential component \(a_T\)

The tangential component of the acceleration \( a_T \) is given by projecting the acceleration vector onto the velocity vector and dividing by the magnitude of the velocity vector:\[ a_T = \frac{\mathbf{a}(1) \cdot \mathbf{v}(1)}{\|\mathbf{v}(1)\|} \]First, compute the dot product:\[ \mathbf{a}(1) \cdot \mathbf{v}(1) = (0)(1) + (2)(2) + (6)(3) = 22 \]Thus, \( a_T \) is:\[ a_T = \frac{22}{\sqrt{14}} \]
05

Find the normal component \(a_N\)

The normal component of the acceleration \( a_N \) can be found using:\[ a_N = \sqrt{\|\mathbf{a}(1)\|^2 - a_T^2} \]First, calculate \( \|\mathbf{a}(1)\| \):\[ \|\mathbf{a}(1)\| = \sqrt{0^2 + 2^2 + 6^2} = \sqrt{40} \]Now calculate \( a_N \):\[ a_N = \sqrt{40 - \left(\frac{22}{\sqrt{14}}\right)^2} = \sqrt{40 - \frac{484}{14}} \approx 0.93 \]

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 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Position Vector
In vector calculus, the position vector is a fundamental concept that allows us to describe the position of a point in space with respect to a fixed origin. Think of it as a directed line segment from the origin of your coordinate system to the point in question. This is mathematically expressed as: \[ \mathbf{r}(t) = x(t) \mathbf{i} + y(t) \mathbf{j} + z(t) \mathbf{k} \]where \(x(t)\), \(y(t)\), and \(z(t)\) indicate the coordinates of the point in space as functions of time. In the original exercise, the position vector is given as \(\mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} + t^3 \mathbf{k}\). This shows that as time progresses, the point's location changes in three-dimensional space, following a certain path. The position vector helps us track how this point moves over time, allowing for calculations regarding its speed and direction.
Velocity Vector
The velocity vector is derived by differentiating the position vector with respect to time. It provides vital information about the rate of change of the position vector. Essentially, the velocity vector describes both how fast and in which direction the particle is moving. Mathematically, it is represented as:\[ \mathbf{v}(t) = \frac{d\mathbf{r}(t)}{dt} = \frac{d}{dt}(x(t) \mathbf{i} + y(t) \mathbf{j} + z(t) \mathbf{k}) \]In our example, by differentiating \(\mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} + t^3 \mathbf{k}\), we get the velocity vector:\[ \mathbf{v}(t) = \mathbf{i} + 2t \mathbf{j} + 3t^2 \mathbf{k} \]At \(t = 1\), the velocity vector is \(\mathbf{v}(1) = \mathbf{i} + 2 \mathbf{j} + 3 \mathbf{k}\), describing the velocity of the particle at this specific moment. This vector is crucial for understanding how the particle moves along its trajectory.
Acceleration Vector
The acceleration vector is a second derivative level calculation following the velocity vector. It indicates how the velocity of the object changes over time. This not only involves changes in the speed but also how the direction of the velocity changes. Formally, the acceleration vector is represented by:\[ \mathbf{a}(t) = \frac{d\mathbf{v}(t)}{dt} \]Applying this to our example, the velocity vector \(\mathbf{v}(t) = \mathbf{i} + 2t \mathbf{j} + 3t^2 \mathbf{k}\) gives an acceleration vector of:\[ \mathbf{a}(t) = 2 \mathbf{j} + 6t \mathbf{k} \]When \(t = 1\), the acceleration vector becomes \(\mathbf{a}(1) = 2 \mathbf{j} + 6 \mathbf{k}\). This reveals the vector's magnitude and direction of acceleration at that instant—a crucial step in finding both tangential and normal components of the acceleration.
Magnitude of a Vector
The magnitude of a vector represents its length or size, regardless of its direction. It is fundamentally important when studying vectors to understand how strong or significant their impact is. The magnitude for a vector \( \mathbf{v} = a \mathbf{i} + b \mathbf{j} + c \mathbf{k} \) is found using:\[ \|\mathbf{v}\| = \sqrt{a^2 + b^2 + c^2} \]In the exercise, calculating the magnitude of the velocity vector \(\mathbf{v}(1) = \mathbf{i} + 2 \mathbf{j} + 3 \mathbf{k}\) gives:\[ \|\mathbf{v}(1)\| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14} \]This numerical value is essential for further calculations, such as determining the tangential and normal components of the acceleration. It helps in understanding how pronounced the velocity of the particle is at the given instance.
Dot Product
The dot product is an operation that takes two vectors and returns a scalar quantity. This mathematical operation is essential when projecting one vector onto another or when finding angles between vectors. Mathematically, it is expressed as:\[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \]In our task, the dot product is used to find the tangential component of acceleration. Following the example, for vectors \(\mathbf{a}(1) = 2 \mathbf{j} + 6 \mathbf{k}\) and \(\mathbf{v}(1) = \mathbf{i} + 2 \mathbf{j} + 3 \mathbf{k}\), the dot product is:\[ \mathbf{a}(1) \cdot \mathbf{v}(1) = 0 \times 1 + 2 \times 2 + 6 \times 3 = 22 \]This value contributes to finding the tangential component of the acceleration by measuring how much of the acceleration vector aligns with the velocity vector.

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 the curvature \(\kappa,\) the unit tangent vector \(\mathbf{T},\) the unit normal vector \(\mathbf{N},\) and the binormal vector \(\mathbf{B}\) at \(t=t_{1}\). $$ \mathbf{r}(t)=3 \cosh (t / 3) \mathbf{i}+t \mathbf{j} ; t_{1}=1 $$

Find the unit tangent vector \(\mathbf{T}(t)\) and the curvature \(\kappa(t)\) at the point where \(t=t_{1} .\) For calculating \(\kappa,\) we suggest using Theorem \(A\), as in Example \(5 .\) $$ \mathbf{r}(t)=\frac{1}{3} t^{3} \mathbf{i}+\frac{1}{2} t^{2} \mathbf{j} ; t_{1}=1 $$

Draw the graph of \(x=4 \cos t, y=3 \sin (t+0.5)\), \(0 \leq t \leq 2 \pi .\) Estimate its maximum and minimum curvature by looking at the graph (curvature is the reciprocal of the radius of curvature). Then use a graphing calculator or a CAS to approximate these two numbers to four decimal places.

As you may have guessed, there is a simple formula for expressing great-circle distance directly in terms of longitude and latitude. Let \(\left(\alpha_{1}, \beta_{1}\right)\) and \(\left(\alpha_{2}, \beta_{2}\right)\) be the longitude- latitude coordinates of two points on the surface of the earth, where we interpret \(\mathrm{N}\) and \(\mathrm{E}\) as positive and \(\mathrm{S}\) and \(\mathrm{W}\) as negative. Show that the great-circle distance between these points is \(3960 \gamma\) miles, where \(0 \leq \gamma \leq \pi\) and $$ \cos \gamma=\cos \left(\alpha_{1}-\alpha_{2}\right) \cos \beta_{1} \cos \beta_{2}+\sin \beta_{1} \sin \beta_{2} $$

Sketch the curve over the indicated domain for \(t\). Find \(\mathbf{v}, \mathbf{a}, \mathbf{T},\) and \(\kappa\) at the point where \(t=t_{1}\). $$ \mathbf{r}(t)=\frac{t^{2}}{4} \mathbf{i}+2 \cos t \mathbf{j}+2 \sin t \mathbf{k} ; \quad 0 \leq t \leq 4 \pi ; t_{1}=\pi $$

See all solutions

Recommended explanations on Math Textbooks

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.