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A particle traveling in a straight line is located at the point (1,-1,2) and has speed 2 at time \(t=0 .\) The particle moves toward the point (3,0,3) with constant acceleration \(2 \mathbf{i}+\mathbf{j}+\mathbf{k}\) Find its position vector \(\mathbf{r}(t)\) at time \(t\)

Short Answer

Expert verified
The position vector is \( \mathbf{r}(t) = \left( 1 + \frac{4}{\sqrt{6}}t + t^2, -1 + \frac{2}{\sqrt{6}}t + \frac{1}{2}t^2, 2 + \frac{2}{\sqrt{6}}t + \frac{1}{2}t^2 \right) \).

Step by step solution

01

Determine Initial Velocity Direction

The particle initially moves from the point (1, -1, 2) to (3, 0, 3). The direction vector \( \mathbf{d} \) from the initial point to the destination can be calculated as (3-1, 0-(-1), 3-2) = (2, 1, 1). To ensure the initial velocity is in the direction of \( \mathbf{d} \), normalize this vector to get \( \mathbf{d_{unit}} = \left( \frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}} \right) \).
02

Calculate Initial Velocity Vector

Since the speed of the particle at time \( t = 0 \) is 2, multiply the unit direction vector by 2 to get the initial velocity vector: \( \mathbf{v_0} = 2 \cdot \mathbf{d_{unit}} = \left( \frac{4}{\sqrt{6}}, \frac{2}{\sqrt{6}}, \frac{2}{\sqrt{6}} \right) \).
03

Expression for Velocity Vector

Using the initial velocity and constant acceleration, the velocity vector at time \( t \), \( \mathbf{v}(t) \), is given by \( \mathbf{v}(t) = \mathbf{v_0} + \mathbf{a}t \), where \( \mathbf{a} = 2\mathbf{i} + \mathbf{j} + \mathbf{k} \). Thus, \( \mathbf{v}(t) = \left( \frac{4}{\sqrt{6}} + 2t, \frac{2}{\sqrt{6}} + t, \frac{2}{\sqrt{6}} + t \right) \).
04

Integrate Velocity to Find Position

Integrate the velocity vector \( \mathbf{v}(t) \) with respect to \( t \) to obtain the position vector \( \mathbf{r}(t) \). Start with the known initial position \( \mathbf{r}(0) = (1, -1, 2) \). The integration yields: \( \mathbf{r}(t) = \mathbf{r}(0) + \int \mathbf{v}(t) \, dt \).
05

Evaluate the Integral

Perform the integration: \[ \mathbf{r}(t) = \left( 1, -1, 2 \right) + \int \left( \frac{4}{\sqrt{6}} + 2t, \frac{2}{\sqrt{6}} + t, \frac{2}{\sqrt{6}} + t \right) \, dt \] gives us \( \mathbf{r}(t) = \left( 1 + \frac{4}{\sqrt{6}}t + t^2, -1 + \frac{2}{\sqrt{6}}t + \frac{1}{2}t^2, 2 + \frac{2}{\sqrt{6}}t + \frac{1}{2}t^2 \right) \).
06

Final Step: Write Down Resulting Position Vector

The final expression for the position vector \( \mathbf{r}(t) \) is: \( \mathbf{r}(t) = \left( 1 + \frac{4}{\sqrt{6}}t + t^2, -1 + \frac{2}{\sqrt{6}}t + \frac{1}{2}t^2, 2 + \frac{2}{\sqrt{6}}t + \frac{1}{2}t^2 \right) \).

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Key Concepts

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

Position vector
The position vector, denoted as \( \mathbf{r}(t) \), is a fundamental concept in motion in 3D. It provides the location of a particle in space at a given time \( t \). For any moving object in three-dimensional space, the position vector is represented by coordinates \( (x(t), y(t), z(t)) \). This vector indicates the exact point within the 3D space where the particle is located.

The position vector changes over time as the particle moves. At time \( t=0 \), we often have an initial position vector. For the given exercise, the initial position of the particle is \( (1, -1, 2) \). As time progresses and the particle moves due to its initial velocity and acceleration, the position changes. Thus, the position vector \( \mathbf{r}(t) \) captures this continuous motion and is derived from integrating the velocity vector.
Velocity vector
The velocity vector \( \mathbf{v}(t) \) describes the speed and direction in which a particle is moving at any instant in time. It is a key vector in understanding motion because it tells us not only how fast but also in what direction the particle is moving. In 3D space, the velocity vector is expressed as \( (v_x(t), v_y(t), v_z(t)) \).

This problem starts by determining the direction of the initial velocity. Once we know the direction toward the destination, the velocity is set by taking the unit direction vector and multiplying it by the initial speed. For this exercise, the initial speed is 2. Over time, the constant acceleration affects the velocity, hence changing the velocity vector. This relationship is mathematically captured by \( \mathbf{v}(t) = \mathbf{v_0} + \mathbf{a}t \), where \( \mathbf{v_0} \) is the initial velocity vector and \( \mathbf{a} \) is the acceleration.
Acceleration
Acceleration is a crucial vector that measures the rate of change of the velocity vector. For this exercise, the acceleration is constant and is given by the vector \( 2\mathbf{i} + \mathbf{j} + \mathbf{k} \), which signals a steady change in velocity over time.

In practical terms, acceleration affects how fast the velocity is increasing or decreasing in each of the three spatial dimensions. A positive component in the acceleration vector indicates an increase in velocity, while a negative component indicates a deceleration in that dimension. Understanding how acceleration impacts motion allows us to predict future movement.

In calculations, acceleration is typically used to adjust the velocity over time, forming expressions like \( \mathbf{v}(t) = \mathbf{v_0} + \mathbf{a}t \), where the result of this expression is then used to find the next position of the particle.
Integration in calculus
Integration is a fundamental tool in calculus, particularly useful in the study of kinematics for converting velocity vectors to position vectors. Since velocity is the derivative of position, integration is the natural inverse process to recover the position function from a known velocity.

In the given exercise, to find the position vector \( \mathbf{r}(t) \), we integrate the velocity vector \( \mathbf{v}(t) \) with respect to time \( t \). The integration involves handling each component of the velocity vector separately, adding the initial conditions found at \( t=0 \).

The integration results in the final expression for \( \mathbf{r}(t) \) which includes terms like \( t^2 \) that indicate changes due to both the velocity and the constant acceleration. Thus, through integration, we translate the dynamic movement into a tangible expression of position over time.

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Most popular questions from this chapter

Each of the following equations in parts (a)-(e) describes the motion of a particle having the same path, namely the unit circle \(x^{2}+y^{2}=1 .\) Although the path of each particle in parts (a)-(e) is the same, the behavior, or "dynamics," of each particle is different. For each particle, answer the following questions. i) Does the particle have constant speed? If so, what is its constant speed? ii) Is the particle's acceleration vector always orthogonal to its velocity vector? iii) Does the particle move clockwise or counterclockwise around the circle? iv) Does the particle begin at the point (1,0)\(?\) a. \(\mathbf{r}(t)=(\cos t) \mathbf{i}+(\sin t) \mathbf{j}, \quad t \geq 0\) b. \(\mathbf{r}(t)=\cos (2 t) \mathbf{i}+\sin (2 t) \mathbf{j}, \quad t \geq 0\) c. \(\mathbf{r}(t)=\cos (t-\pi / 2) \mathbf{i}+\sin (t-\pi / 2) \mathbf{j}, \quad t \geq 0\) d. \(\mathbf{r}(t)=(\cos t) \mathbf{i}-(\sin t) \mathbf{j}, \quad t \geq 0\) e. \(\mathbf{r}(t)=\cos \left(t^{2}\right) \mathbf{i}+\sin \left(t^{2}\right) \mathbf{j}, \quad t \geq 0\)

Derive the equations $$\begin{array}{l} x=\frac{v_{0}}{k}\left(1-e^{-k t}\right) \cos \alpha \\ y=\frac{v_{0}}{k}\left(1-e^{-k t}\right)(\sin \alpha)+\frac{g}{k^{2}}\left(1-k t-e^{-k t}\right) \end{array}$$ by solving the following initial value problem for a vector \(\mathbf{r}\) in the plane. $$\text { Differential equation: } \frac{d^{2} \mathbf{r}}{d t^{2}}=-g \mathbf{j}-k \mathbf{v}=-g \mathbf{j}-k \frac{d \mathbf{r}}{d t}$$ Initial conditions: \(\quad \mathbf{r}(0)=\mathbf{0}\) $$ \left.\frac{d \mathbf{r}}{d t}\right|_{f-0}=\mathbf{v}_{0}=\left(v_{0} \cos \alpha\right) \mathbf{i}+\left(v_{0} \sin \alpha\right) \mathbf{j} $$ The drag coefficient \(k\) is a positive constant representing resistance due to air density, \(v_{0}\) and \(\alpha\) are the projectile's initial speed and launch angle, and \(g\) is the acceleration of gravity.

Can anything be said about the speed of a particle whose acceleration is always orthogonal to its velocity? Give reasons for your answer.

The length \(2 \pi \sqrt{2}\) of the turn of the helix in Example 1 is also the length of the diagonal of a square \(2 \pi\) units on a side. Show how to obtain this square by cutting away and flattening a portion of the cylinder around which the helix winds.

You will explore graphically the behavior of the helix $$\mathbf{r}(t)=(\cos a t) \mathbf{i}+(\sin a t) \mathbf{j}+b t \mathbf{k}$$ as you change the values of the constants \(a\) and \(b\). Use a CAS to perform the steps in each exercise. Set \(b=1 .\) Plot the helix \(\mathbf{r}(t)\) together with the tangent line to the curve at \(t=3 \pi / 2\) for \(a=1,2,4,\) and 6 over the interval \(0 \leq t \leq 4 \pi .\) Describe in your own words what happens to the graph of the helix and the position of the tangent line as \(a\) increases through these positive values.

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