/*! 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 12 Let \(v\) and a denote the veloc... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 12

Let \(v\) and a denote the velocity and acceleration vectors of a particle moving on a path \(c .\) Suppose the initial position of the particle is \(\mathfrak{c}(0)=(3,4,0),\) the initial velocity is \(\mathbf{v}(0)=(1,1,-2),\) and the acceleration function is \(a(t)=(0,0,6) .\) Find \(v(t)\) and \(c(t)\).

Short Answer

Expert verified
The velocity function is \(v(t) = (1, 1, 6t - 2)\), and the position function is \(c(t) = (t + 3, t + 4, 3t^2 - 2t)\).

Step by step solution

01

Determine the velocity function

The velocity function is found by integrating the acceleration function. The given acceleration is \(a(t) = (0, 0, 6)\). Integrating each component:\[\int 0 \, dt = C_1,\]\[\int 0 \, dt = C_2,\]\[\int 6 \, dt = 6t + C_3.\]Thus, \(v(t) = (C_1, C_2, 6t + C_3)\).
02

Use initial condition to find constants for velocity

Use the initial condition for velocity \(v(0) = (1, 1, -2)\) to solve for the constants \(C_1\), \(C_2\), and \(C_3\):- Since \(v(0) = C_1 = 1\), we have \(C_1 = 1\).- Since \(v(0) = C_2 = 1\), we have \(C_2 = 1\).- For the third component, \(v(0) = 6 imes 0 + C_3 = -2\), so \(C_3 = -2\).Thus, the velocity function is \(v(t) = (1, 1, 6t - 2)\).
03

Determine the position function

To find the position function \(c(t)\), integrate the velocity function \(v(t) = (1, 1, 6t - 2)\):\[\int 1 \, dt = t + D_1,\]\[\int 1 \, dt = t + D_2,\]\[\int (6t - 2) \, dt = 3t^2 - 2t + D_3.\]Thus, \(c(t) = (t + D_1, t + D_2, 3t^2 - 2t + D_3)\).
04

Use initial condition to find constants for position

Use the initial position \(c(0) = (3, 4, 0)\) to solve for \(D_1\), \(D_2\), and \(D_3\):- From the first component, \(0 + D_1 = 3\) gives \(D_1 = 3\).- From the second component, \(0 + D_2 = 4\) gives \(D_2 = 4\).- From the third component, \(3 \times 0^2 - 2 \times 0 + D_3 = 0\) gives \(D_3 = 0\).Thus, the position function is \(c(t) = (t + 3, t + 4, 3t^2 - 2t)\).

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.

Velocity Vector
Velocity vectors are crucial in understanding the motion of an object along a pathway. A vector describes both the speed and the direction of the object at any given time. In our given problem, the velocity vector, denoted as \( v(t) \), shows us how fast and in which direction the particle is traveling along the path \( c \). Having both magnitude and direction makes the vector an essential tool in vector calculus.

In the exercise, the velocity vector is found by integrating the acceleration vector. The result yields a formula \( v(t) = (C_1, C_2, 6t + C_3) \). To determine specific values for these constants \( C_1, C_2, \) and \( C_3 \), we use initial conditions. In this case, when time \( t = 0 \), the velocity is given as \( (1,1,-2) \). This allows us to calculate the constants, finalizing the velocity function as \( v(t) = (1,1,6t-2) \).

This function indicates the particle has a steady velocity of 1 unit in the x and y directions, and an increasing velocity (6t-2) in the z direction as time progresses.
Acceleration Vector
The acceleration vector tells us how the velocity of an object changes over time. It is a derivative of the velocity vector with respect to time. Utilizing vectors to describe acceleration helps in understanding changes in motion comprehensively, including the direction in which speed changes.

In vector calculus, the acceleration vector is crucial, as it helps explain how, why, and in what direction a particle's velocity alters. In the problem presented, the acceleration vector \( a(t) = (0, 0, 6) \) suggests that acceleration occurs only in the z direction at a constant rate of 6 units per second squared. This implies that the object's velocity in the z direction is increasing constantly over time, resulting in a change of motion in that respective direction.

Being aware that acceleration might vary across different directions is essential, especially if we are tasked with predicting future positions and velocities of objects in physics.
Integration in Calculus
Integration is a fundamental concept in calculus, often perceived as the reverse operation to differentiation. In the context of vector calculus, integration is used to derive position and velocity vectors from acceleration vectors.
  • Integral of acceleration: By integrating the acceleration vector, we determine the velocity vector. This integration process is straightforward, particularly when the acceleration is a constant vector, as seen in our problem with \(a(t) = (0, 0, 6)\).
  • Initial conditions: To solve definitively, we integrate the acceleration, which often introduces constants. These are resolved using known conditions or initial values, such as initial velocity or position. In our case, the initial velocity \(v(0) = (1, 1, -2)\) assists in finding these constants.
  • Integral of velocity: Similarly, integrating the velocity vector as the final step gives us the position vector. Again, constants appear and are solved with initial conditions, like the initial position \(c(0) = (3, 4, 0)\).
Understanding how integration is used to transition from acceleration to velocity and then to position is key in solving motion problems in vector calculus.

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

In meteorology, the negative pressure gradient \(\mathbf{G}\) is a vector quantity that points from regions of high pressure to regions of low pressure, normal to the lines of constant pressure (isobars). (a) In an \(x y\) coordinate system, $$\mathbf{G}=-\frac{\partial P}{\partial x} \mathbf{i}-\frac{\partial P}{\partial y} \mathbf{j}$$ Write a formula for the magnitude of the negative pressure gradient. (b) If the horizontal pressure gradient provided the only horizontal force acting on the air, the wind would blow directly across the isobars in the direction of G, and for a given air mass, with acceleration proportional to the magnitude of \(\mathbf{G}\). Explain, using Newton's second law. (c) Because of the rotation of the earth, the wind does not blow in the direction that part (b) would suggest. Instead, it obeys Buys- Ballot's law, which states: "If in the Northern Hemisphere, you stand with your back to the wind, the high pressure is on your right and the low pressure is on your left." Draw a figure and introduce \(x y\) coordinates so that \(G\) points in the proper direction. (d) State and graphically illustrate Buys-Ballot's law for the Southern Hemisphere, in which the orientation of high and low pressure is reversed.

Find the are length of \(c(t)=(t,|t|)\) for \(-1 \leq t \leq 1\)

(a) Let \(\mathbf{F}=2 x y e^{z} \mathbf{i}+e^{z} x^{2} \mathbf{j}+\left(x^{2} y e^{z}+z^{2}\right) \mathbf{k} .\) Compute \(\nabla \cdot \mathbf{F}\) and \(\nabla \times \mathbf{F}\) (b) Find a function \(f(x, y, z)\) such that \(\mathbf{F}=\nabla f\)

A particle of mass \(m\) moves under the influence of a force \(\mathbf{F}=-k \mathbf{r},\) where \(k\) is a constant and \(\mathbf{r}(t)\) is the position of the particle at time \(t\) (a) Write down differential equations for the components of \(\mathbf{r}(t)\) (b) Solve the equations in part (a) subject to the initial conditions \(\mathbf{r}(0)=\mathbf{0}, \mathbf{r}^{\prime}(0)=2 \mathbf{j}+\mathbf{k}\)

Find the are length of the given curve on the specified interval. $$\left(t+1, \frac{2 \sqrt{2}}{3} t^{3 / 2}+7, \frac{1}{2} t^{2}\right), \text { for } 1 \leq t \leq 2$$
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Calculus

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Statistics

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.