/*! 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 25 Speed and arc length For the fol... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 25

Speed and arc length For the following trajectories, find the speed associated with the trajectory and then find the length of the trajectory on the given interval. $$\mathbf{r}(t)=\langle 13 \sin 2 t, 12 \cos 2 t, 5 \cos 2 t\rangle, \text { for } 0 \leq t \leq \pi$$

Short Answer

Expert verified
Answer: The speed associated with the trajectory is given by the magnitude of the first derivative of 饾憻(饾憽) with respect to time (t), which is $$ |\mathbf{r'}(t)| = \sqrt{676 + 576\cos^2{2t} + 100\cos^2{2t}}$$. The arc length of the trajectory on the interval [0, 饾湅] can be approximated using numerical methods to solve the integral $$length \approx \int_{0}^{\pi} \sqrt{338(1 + \cos{4t})} dt$$, as an analytical solution is too complex to obtain.

Step by step solution

01

Find the first derivative of 饾憻(饾憽)

To find the first derivative of 饾憻(饾憽) with respect to time (t), we will differentiate each component of 饾憻(饾憽) separately: $$\mathbf{r}(t)=\langle 13 \sin 2 t, 12 \cos 2 t, 5 \cos 2 t\rangle$$ Differentiate with respect to t: $$\mathbf{r'}(t)=\langle 26 \cos 2t(-1), -24 \sin 2t, -10 \sin 2t\rangle$$
02

Calculate the magnitude of 饾憻'(饾憽)

Now we'll calculate the speed at any time t by finding the magnitude of 饾憻'(饾憽): $$|\mathbf{r'}(t)| = \sqrt{(26 \cos 2t(-1))^2 + (-24 \sin 2t)^2 + (-10 \sin 2t)^2}$$ After simplification, we get: $$|\mathbf{r'}(t)| = \sqrt{676 + 576\cos^2{2t} + 100\cos^2{2t}}$$
03

Setup the arc length integral

To find the arc length, we will integrate the magnitude of the first derivative over the given interval [0, 饾湅]: $$length = \int_{0}^{\pi} |\mathbf{r'}(t)| dt$$ Now, substitute the expression for the magnitude of 饾憻'(饾憽): $$length = \int_{0}^{\pi} \sqrt{676 + 576\cos^2{2t} + 100\cos^2{2t}} dt$$
04

Solve the integral for arc length

At this point, solving the integral for the arc length may be very difficult analytically, so it might be necessary to use a numerical method (like a calculator or a software package) to evaluate the integral and obtain an approximation for the arc length. However, let's try to simplify the integrand further: Note that the integrand can be written as: $$\sqrt{676 + 676\cos^2{2t}}$$ Now, let's use the trigonometric identity \(\cos^2x = \frac{1 + \cos{2x}}{2}\): $$length = \int_{0}^{\pi} \sqrt{676 + 676\cdot\frac{1 + \cos{4t}}{2}} dt$$ Now we can further simplify as: $$length = \int_{0}^{\pi} \sqrt{338(1 + \cos{4t})} dt$$ Unfortunately, even after this simplification, solving the integral analytically is still challenging. It is recommended to use computational tools such as a calculator, WolframAlpha, or other software tools to find an approximate value for the arc length. To summarize, the speed associated with the trajectory is given by the magnitude of 饾憻'(饾憽) which is $$ |\mathbf{r'}(t)| = \sqrt{676 + 576\cos^2{2t} + 100\cos^2{2t}}$$ The arc length of the trajectory on the interval [0, 饾湅] can be approximated using numerical methods since the integral is too complex to solve analytically: $$length \approx \int_{0}^{\pi} \sqrt{338(1 + \cos{4t})} dt$$

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.

Speed
When we talk about speed in the context of a trajectory, we refer to the rate at which the position changes with respect to time. To determine this for a given vector-valued function like \(\mathbf{r}(t)\), we find its derivative with respect to time, \(\mathbf{r'}(t)\). This derivative essentially gives us the velocity vector. The speed is then the magnitude of this velocity vector.
  • In our example, the vector function \(\mathbf{r}(t)\) defines a trajectory in three dimensions.
  • By differentiating each component of \(\mathbf{r}(t)\) with respect to time, we get \(\mathbf{r'}(t) = \langle 26 \cos 2t(-1), -24 \sin 2t, -10 \sin 2t \rangle\).
  • To find the speed, we calculate the magnitude: \(|\mathbf{r'}(t)| = \sqrt{(26 \cos 2t)^2 + (-24 \sin 2t)^2 + (-10 \sin 2t)^2}\).
Simplifying this expression gives insight into how fast an object travels along the trajectory without considering its direction.
Trigonometric Identities
Trigonometric identities are fundamental tools in simplifying expressions involving trigonometric functions. They can make complex calculations more manageable. In our problem, we encounter \(\cos^2{2t}\), which can be simplified using trigonometric identities.
Using the identity for \(\cos^2 x\): \
  • \(\cos^2 x = \frac{1 + \cos{2x}}{2}\)
we transform the original integrand from \(\sqrt{676 + 676\cos^2{2t}}\) to \(\sqrt{338(1 + \cos{4t})}\).
This simplification helps in setting up the integral for evaluating the arc length, especially when analytical methods of solving are complex or impractical.Trigonometric identities allow us to express functions in alternate forms, which can be key to solving integrals and other mathematical problems efficiently.
Numerical Integration
Numerical integration comes into play when evaluating integrals that are otherwise too complicated for analytical solutions. For instance, in our exercise, the integral for arc length: \(\int_{0}^{\pi} \sqrt{338(1 + \cos{4t})} dt\) is challenging due to its complexity.
In such cases:
  • We employ numerical methods to approximate the integral's value instead of solving it exactly.
  • Popular techniques include Simpson's Rule, the Trapezoidal Rule, and using computational tools like calculators or software (e.g., Matlab, Mathematica).
These methods break down the integral into smaller intervals, compute approximations, and sum them to estimate the total area under the curve.
Though these methods don't provide exact solutions, they offer close approximations that are useful in practical applications. Numerical integration is essential when facing real-world problems where analytical integration is infeasible.
Differentiation
Differentiation is a mathematical process that determines the rate of change of a function with respect to its variables. In the context of vector-valued functions, like in our exercise, differentiation helps find the velocity vector from the position function \(\mathbf{r}(t)\).
Here's how it works:
  • Each component of the vector is differentiated with respect to time \(t\).
  • This results in the derivative \(\mathbf{r'}(t) = \langle 26 \cos 2t(-1), -24 \sin 2t, -10 \sin 2t \rangle\).
Differentiation reveals crucial information about motion, such as speed and acceleration. In physics and engineering, this forms the basis for studying dynamics and kinematics.
By understanding differentiation, one can predict how objects move, how fast they travel, or how they accelerate over time. This powerful tool is not only pivotal in solving mathematical problems but also in understanding and modeling real-world scenarios.

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

The points \(O(0,0,0), P(1,4,6),\) and \(Q(2,4,3)\) lie at three vertices of a parallelogram. Find all possible locations of the fourth vertex.

Evaluate the following limits. $$\lim _{t \rightarrow 0}\left(\frac{\sin t}{t} \mathbf{i}-\frac{e^{t}-t-1}{t} \mathbf{j}+\frac{\cos t+t^{2} / 2-1}{t^{2}} \mathbf{k}\right)$$

A baseball leaves the hand of a pitcher 6 vertical feet above home plate and \(60 \mathrm{ft}\) from home plate. Assume that the coordinate axes are oriented as shown in the figure. a. In the absence of all forces except gravity, assume that a pitch is thrown with an initial velocity of \(\langle 130,0,-3\rangle \mathrm{ft} / \mathrm{s}\) (about \(90 \mathrm{mi} / \mathrm{hr}\) ). How far above the ground is the ball when it crosses home plate and how long does it take for the pitch to arrive? b. What vertical velocity component should the pitcher use so that the pitch crosses home plate exactly \(3 \mathrm{ft}\) above the ground? c. A simple model to describe the curve of a baseball assumes that the spin of the ball produces a constant sideways acceleration (in the \(y\) -direction) of \(c \mathrm{ft} / \mathrm{s}^{2}\). Assume a pitcher throws a curve ball with \(c=8 \mathrm{ft} / \mathrm{s}^{2}\) (one-fourth the acceleration of gravity). How far does the ball move in the \(y\) -direction by the time it reaches home plate, assuming an initial velocity of (130,0,-3) ft/s? d. In part (c), does the ball curve more in the first half of its trip to the plate or in the second half? How does this fact affect the batter? e. Suppose the pitcher releases the ball from an initial position of (0,-3,6) with initial velocity \((130,0,-3) .\) What value of the spin parameter \(c\) is needed to put the ball over home plate passing through the point (60,0,3)\(?\)

Assume that \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\) are vectors in \(\mathrm{R}^{3}\) that form the sides of a triangle (see figure). Use the following steps to prove that the medians intersect at a point that divides each median in a 2: 1 ratio. The proof does not use a coordinate system. a. Show that \(\mathbf{u}+\mathbf{v}+\mathbf{w}=\mathbf{0}\) b. Let \(\mathbf{M}_{1}\) be the median vector from the midpoint of \(\mathbf{u}\) to the opposite vertex. Define \(\mathbf{M}_{2}\) and \(\mathbf{M}_{3}\) similarly. Using the geometry of vector addition show that \(\mathbf{M}_{1}=\mathbf{u} / 2+\mathbf{v} .\) Find analogous expressions for \(\mathbf{M}_{2}\) and \(\mathbf{M}_{3}\) c. Let \(a, b,\) and \(c\) be the vectors from \(O\) to the points one-third of the way along \(\mathbf{M}_{1}, \mathbf{M}_{2},\) and \(\mathbf{M}_{3},\) respectively. Show that \(\mathbf{a}=\mathbf{b}=\mathbf{c}=(\mathbf{u}-\mathbf{w}) / 3\) d. Conclude that the medians intersect at a point that divides each median in a 2: 1 ratio.

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. What conditions on \(\mathbf{u}\) and \(\mathbf{v}\) lead to equality in the Cauchy-Schwarz Inequality?
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Applied Mathematics

Read Explanation

Geometry

Read Explanation

Decision Maths

Read Explanation

Pure Maths

Read Explanation

Calculus

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.