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Chapter 13: Problem 5

In Exercises \(5-8, \mathbf{r}(t)\) is the position of a particle in the \(x y\) -plane at time \(t .\) Find an equation in \(x\) and \(y\) whose graph is the path of the par- ticle. Then find the particle's velocity and acceleration vectors at the given value of \(t\) . $$ \mathbf{r}(t)=(t+1) \mathbf{i}+\left(t^{2}-1\right) \mathbf{j}, \quad t=1 $$

Short Answer

Expert verified
The path equation is \( y = x^2 - 2x \). At \( t=1 \), \( \mathbf{v}(1) = \mathbf{i} + 2\mathbf{j} \) and \( \mathbf{a}(1) = 2 \mathbf{j} \).

Step by step solution

01

Write Position Functions for x and y

The given vector function is \( \mathbf{r}(t)\) which can be expressed as \( \mathbf{r}(t)=(t+1) \mathbf{i}+(t^2-1) \mathbf{j} \). This tells us that the position functions are: - \( x(t) = t+1 \) - \( y(t) = t^2 - 1 \).
02

Eliminate Parameter t to find Path Equation

We need to eliminate \( t \) to find the equation of the path in terms of \( x \) and \( y \). From \( x = t + 1 \), solve for \( t \) to get:\[ t = x - 1 \]. Substitute \( t = x - 1 \) into \( y(t) = t^2 - 1 \):\[ y = (x-1)^2 - 1 \]Expand this to get the equation of the path:\[ y = x^2 - 2x + 1 - 1 \]\[ y = x^2 - 2x \].
03

Find Velocity Vector

The velocity vector \( \mathbf{v}(t) \) is the derivative of the position vector \( \mathbf{r}(t) \). \[ \mathbf{v}(t) = \frac{d}{dt}(t+1) \mathbf{i} + \frac{d}{dt}(t^2-1) \mathbf{j} \]\[ \mathbf{v}(t) = \mathbf{i} + 2t \mathbf{j} \]\ At \( t = 1 \):\[ \mathbf{v}(1) = \mathbf{i} + 2(1) \mathbf{j} = \mathbf{i} + 2\mathbf{j} \].
04

Find Acceleration Vector

The acceleration vector \( \mathbf{a}(t) \) is the derivative of the velocity vector \( \mathbf{v}(t) \). \[ \mathbf{a}(t) = \frac{d}{dt}(1) \mathbf{i} + \frac{d}{dt}(2t) \mathbf{j} \]\[ \mathbf{a}(t) = 0 \mathbf{i} + 2 \mathbf{j} \]Thus, at \( t = 1 \):\[ \mathbf{a}(1) = 0 \mathbf{i} + 2 \mathbf{j} = 2 \mathbf{j} \].

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Key Concepts

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

Parametric Equations
Parametric equations are a way to describe a set of related quantities using one or more parameters. In our exercise, the position of the particle is defined using parameterized equations for both the x and y coordinates:
  • For the x-coordinate, the equation is: \( x(t) = t+1 \)
  • For the y-coordinate, the equation is: \( y(t) = t^2 - 1 \)
With these equations, the path of a particle over time can be visualized without explicitly solving for one variable in terms of another.
To identify the path of the particle in the xy-plane, eliminate the parameter \( t \) from these equations. This transformation helps in analyzing the trajectory or path the particle follows, which is crucial for understanding its motion in two-dimensional space.
Velocity Vector
The velocity vector in vector calculus represents the rate of change of position with respect to time. It describes the direction and speed of a moving particle. For a given position function \( \mathbf{r}(t) \), the velocity vector \( \mathbf{v}(t) \) is its derivative:
  • The derivative of the x-component: \( \frac{d}{dt}(t+1) = 1 \)
  • The derivative of the y-component: \( \frac{d}{dt}(t^2-1) = 2t \)
Thus, the velocity vector can be expressed as:\[ \mathbf{v}(t) = \mathbf{i} + 2t \mathbf{j} \]At any specific time \( t \), such as \( t = 1 \), substitute to find:\[ \mathbf{v}(1) = \mathbf{i} + 2 \mathbf{j} \]The resulting vector indicates that at \( t = 1 \), the particle's velocity is moving one unit in the x-direction and two units in the y-direction.
Acceleration Vector
In the context of vector calculus, the acceleration vector represents the rate of change of the velocity vector over time. It tells us how the velocity of the particle is changing. The acceleration vector \( \mathbf{a}(t) \) is the derivative of the velocity vector \( \mathbf{v}(t) \):
  • Take the derivative of \( \mathbf{i} \): \( \frac{d}{dt}(1) = 0 \)
  • Take the derivative of \( 2t \mathbf{j} \): \( \frac{d}{dt}(2t) = 2 \)
Thus, the acceleration vector becomes:\[ \mathbf{a}(t) = 0 \mathbf{i} + 2 \mathbf{j} \]Substituting \( t = 1 \) gives:\[ \mathbf{a}(1) = 2 \mathbf{j} \]This indicates that the particle experiences constant acceleration solely in the y-direction, causing continuous increase in the y-component of its velocity.
Position Function
The position function, encapsulated by \( \mathbf{r}(t) \), defines the exact position of a particle at any given time \( t \). It's vital for tracking the particle's movement in the plane. For the given problem, the position function is:\[ \mathbf{r}(t) = (t+1) \mathbf{i} + (t^2-1) \mathbf{j} \]Each component of this vector function refers to a coordinate axis:
  • The x-component: \( x(t) = t+1 \)
  • The y-component: \( y(t) = t^2-1 \)
By understanding the position function, one can determine not just the particle's current position, but also compute other critical attributes like velocity and acceleration. The position function assists in deducing paths or trajectories as it includes all position values over continuous time.

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Most popular questions from this chapter

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