/*! 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 14 Velocity and acceleration from p... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 14

Velocity and acceleration from position Consider the following position functions. a. Find the velocity and speed of the object. b. Find the acceleration of the object. $$r(t)=\langle 3 \cos t, 4 \sin t\rangle, \text { for } 0 \leq t \leq 2 \pi$$

Short Answer

Expert verified
Based on the position function $$r(t) = \langle 3 \cos t, 4 \sin t\rangle$$, determine the following: a. The velocity vector and speed of the object b. The acceleration vector of the object. Answer: a. The velocity vector is: $$v(t) = \langle -3\sin t,4\cos t\rangle$$, and the speed is: $$s(t) = \sqrt{9\sin^2 t + 16\cos^2 t}$$ b. The acceleration vector is: $$a(t) = \langle -3\cos t, -4\sin t\rangle$$

Step by step solution

01

Find the velocity vector

To find the velocity vector, we need to calculate the derivative of the position function with respect to t. We do this by taking the derivative of each component of the vector separately, then putting these together to form the velocity vector. The derivative will be: $$\frac{d}{dt}\langle 3 \cos t, 4 \sin t\rangle = \langle -3\sin t,4\cos t\rangle$$ So the velocity vector is given by: $$v(t) = \langle -3\sin t,4\cos t\rangle$$
02

Find the speed

To find the speed, we need to calculate the magnitude of the velocity vector. We do this using the Pythagorean theorem for vector magnitudes: $$|v(t)| = \sqrt{(-3\sin t)^2 + (4\cos t)^2}$$ Simplifying, we get: $$|v(t)| = \sqrt{9\sin^2 t + 16\cos^2 t}$$ So the speed of the object is given by the magnitude of the velocity vector: $$s(t) = \sqrt{9\sin^2 t + 16\cos^2 t}$$
03

Find the acceleration vector

To find the acceleration vector, we need to calculate the derivative of the velocity vector with respect to t. We do this by taking the derivative of each component of the vector separately, then putting these together to form the acceleration vector. The derivative will be: $$\frac{d}{dt}\langle -3\sin t, 4\cos t\rangle = \langle -3\cos t,-4\sin t\rangle$$ So the acceleration vector is given by: $$a(t) = \langle -3\cos t, -4\sin t\rangle$$
04

Summary

In summary, we have found the following results for the object: a. The velocity vector is: $$v(t) = \langle -3\sin t,4\cos t\rangle$$, and the speed is: $$s(t) = \sqrt{9\sin^2 t + 16\cos^2 t}$$ b. The acceleration vector is: $$a(t) = \langle -3\cos t, -4\sin t\rangle$$

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!

Key Concepts

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

Position Function
The position function is a fundamental concept in vector calculus, particularly when dealing with motion in physics. It describes the position of an object at any given time, typically represented as a vector that changes over time. For the exercise at hand, the position function is given as \[ r(t) = \langle 3 \cos t, 4 \sin t \rangle \] where each component of the vector function, \(3 \cos t\) and \(4 \sin t\), describes a parametric system. This means it tells us how the object moves in space, with \(t\) being the parameter or the time variable. Understanding the position function allows us to derive other important aspects like velocity and acceleration. The position function鈥檚 components are expressed in terms of trigonometric functions, which makes it suitable for circular motion. The range of \(t\) from \(0\) to \(2\pi\) indicates a complete cycle of motion, often related to circles or ellipses in geometry.
Derivative
Understanding derivatives is crucial in calculating both velocity and acceleration, as these are derived by differentiating the position function. The derivative essentially measures how a function changes as its input changes.
  • Velocity as a Derivative: It is the first derivative of the position function. By differentiating each component of the position function, you obtain the velocity vector. For our function \( r(t) = \langle 3 \cos t, 4 \sin t \rangle \), the velocity vector is computed as: \[ v(t) = \langle -3 \sin t, 4 \cos t \rangle \]

  • Acceleration as a Derivative: It is the derivative of the velocity vector, providing insight into how velocity changes over time. Differentiating the velocity vector \( v(t) = \langle -3 \sin t, 4 \cos t \rangle \) yields the acceleration vector: \[ a(t) = \langle -3 \cos t, -4 \sin t \rangle \]
Derivatives help us transition from motion description (position) to the dynamics (velocity and acceleration). In vector calculus, taking derivatives of vector functions is basically taking derivatives of each component separately, simplifying calculation.
Vector Calculus
Vector calculus is a branch of mathematics focused on differential and integral calculus dealing with vector fields. It applies heavily in physics for analyzing motion-related problems. In this context, vector calculus helps describe how objects move through space using vector functions. This involves computing derivatives to find velocity and acceleration, just as we did in the problem.
  • Velocity Vector: Through vector calculus, the velocity provides direction and magnitude with which the object moves. With \( v(t) = \langle -3 \sin t, 4 \cos t \rangle \), you see it accounts for how fast and in what direction an object is moving along a path.

  • Magnitude as Speed: Calculating the magnitude of the velocity vector gives the speed, representing the rate of change of position magnitude: \[ s(t) = \sqrt{9 \sin^2 t + 16 \cos^2 t} \]

  • Acceleration Vector: Acceleration tells us how the velocity changes, which directly results from further application of derivatives: \[ a(t) = \langle -3 \cos t, -4 \sin t \rangle \]
In summary, vector calculus provides a comprehensive framework to analyze multi-dimensional motion, making it an essential tool for anyone studying physics and engineering fields.

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

Designing a baseball pitch A baseball leaves the hand of a pitcher 6 vertical feet above and 60 horizontal feet from home plate. Assume the coordinate axes are oriented as shown in the figure. Figure cannot copy a. Suppose a pitch is thrown with an initial velocity of (130,0,-3) ft/s (about \(90 \mathrm{mi} / \mathrm{hr}\) ). In the absence of all forces except gravity, how far above the ground is the ball when it crosses home plate and how long does it take the pitch to arrive? b. What vertical velocity component should the pitcher use so that the pitch crosses home plate exactly 3 ft above the ground? c. A simple model to describe the curve of a baseball assumes the spin of the ball produces a constant sideways acceleration (in the \(y\) -direction) of \(c \mathrm{ft} / \mathrm{s}^{2} .\) Suppose 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 \((130,0,-3) \mathrm{ft} / \mathrm{s} ?\) 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 \langle 0,-3,6\rangle with initial velocity \(\langle 130,0,-3\rangle .\) What value of the spin parameter \(c\) is needed to put the ball over home plate passing through the point (60,0,3)\(?\)

Nonuniform straight-line motion Consider the motion of an object given by the position function $$\mathbf{r}(t)=f(t)\langle a, b, c\rangle+\left\langle x_{0}, y_{0}, z_{0}\right\rangle, \quad \text { for } t \geq 0$$ where \(a, b, c, x_{0}, y_{0},\) and \(z_{0}\) are constants, and \(f\) is a differentiable scalar function, for \(t \geq 0\) a. Explain why \(r\) describes motion along a line. b. Find the velocity function. In general, is the velocity constant in magnitude or direction along the path?

Curves on spheres Graph the curve \(\mathbf{r}(t)=\left\langle\frac{1}{2} \sin 2 t, \frac{1}{2}(1-\cos 2 t), \cos t\right\rangle\) and prove that it lies on the surface of a sphere centered at the origin.

Tilted ellipse Consider the curve \(\mathbf{r}(t)=\langle\cos t, \sin t, c \sin t\rangle,\) for \(0 \leq t \leq 2 \pi,\) where \(c\) is a real number. Assuming the curve lies in a plane, prove that the curve is an ellipse in that plane.

Suppose the vector-valued function \(\mathbf{r}(t)=\langle f(t), g(t), h(t)\rangle\) is smooth on an interval containing the point \(t_{0}\) The line tangent to \(\mathbf{r}(t)\) at \(t=t_{0}\) is the line parallel to the tangent vector \(\mathbf{r}^{\prime}\left(t_{0}\right)\) that passes through \(\left(f\left(t_{0}\right), g\left(t_{0}\right), h\left(t_{0}\right)\right) .\) For each of the following functions, find an equation of the line tangent to the curve at \(t=t_{0} .\) Choose an orientation for the line that is the same as the direction of \(\mathbf{r}^{\prime}.\) $$\mathbf{r}(t)=\langle\sqrt{2 t+1}, \sin \pi t, 4\rangle ; t_{0}=4$$
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Applied Mathematics

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.