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Chapter 14: Problem 13

Describe the graph of the equation. $$r=(2-3 t) i-4 t j$$

Short Answer

Expert verified
The graph represents a straight line with slope \( \frac{4}{3} \) and y-intercept \( -\frac{8}{3} \).

Step by step solution

01

Understanding the Equation

The equation given is a vector equation described by a parameter, \( t \). The vector \( \mathbf{r} \) is a function of the parameter, where \( i \) and \( j \) are unit vectors in the Cartesian coordinate system.
02

Identify Components

Identify the vector components. The vector \( \mathbf{r}(t) \) is given by \( (2-3t) \mathbf{i} - 4t \mathbf{j} \). This means the \( x \)-component is \( 2-3t \) and the \( y \)-component is \( -4t \).
03

Restate as Parametric Equations

Express the vector components as parametric equations: \( x = 2-3t \) and \( y = -4t \). These equations describe a line in the plane parameterized by \( t \).
04

Eliminate the Parameter

To find a single equation without the parameter \( t \), solve for \( t \) in one of the parametric equations. From \( y = -4t \), solve for \( t \): \( t = -\frac{y}{4} \). Substitute into the \( x \) equation: \( x = 2 - 3(-\frac{y}{4}) = 2 + \frac{3y}{4} \).
05

Simplify to Cartesian Form

Simplify the expression from the previous step to find the equation of the line: rearrange it to \( x = 2 + \frac{3y}{4} \) or alternatively, \( 4x = 8 + 3y \), leading to the line equation \( 3y = 4x - 8 \), or \( y = \frac{4}{3}x - \frac{8}{3} \).
06

Interpret the Result

This result represents a straight line with a slope of \( \frac{4}{3} \) and a y-intercept of \( -\frac{8}{3} \). The graph is a line in the Cartesian plane.

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Key Concepts

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

Vector Equation
In the realm of mathematics, a vector equation is an equation where each component of the vector depends upon one or more parameters. The given problem presents a vector equation for curve description: \( \mathbf{r}(t) = (2 - 3t) \mathbf{i} - 4t \mathbf{j} \). Here, \( \mathbf{r} \) denotes the vector dependent on the parameter \( t \).

  • Unit vectors \( \mathbf{i} \) and \( \mathbf{j} \) represent the x and y axes, respectively.
  • The component equations \( (2 - 3t) \mathbf{i} \) and \( -4t \mathbf{j} \) describe motion along each axis.
Understanding vector equations is the foundation for describing the behavior of objects in spaces like fields of physics and computer graphics. They allow us to start from a parameter-based description and move towards a more intuitive graphical representation.
Cartesian Coordinates
Cartesian coordinates provide a way of expressing each point in an n-dimensional space as an ordered sequence of numbers, commonly using two-dimensional (2D) for simplicity. The vector given translates into Cartesian coordinates as parametric equations relating to \( t \), namely \( x = 2 - 3t \) and \( y = -4t \).

  • These equations simplify the process of graphing since each value of \( t \) corresponds to a point \( (x, y) \) in the plane.
  • Cartesian coordinates make it easier to visualize complex mathematical descriptions.
By using Cartesian coordinates, the abstract nature of vectors is grounded into more visual and relatable plane geometry, aiding in understanding how mathematical structures relate to real-world scenarios.
Line Equation
A line equation in its simplest form, \( y = mx + b \), shows the relationship between each point along a straight line. In our vector problem, we derive a single line equation that defines the trajectory described by the vector equation. Through parameter elimination and transformations, we get:

  • \( y = \frac{4}{3}x - \frac{8}{3} \) in slope-intercept form.
  • The slope \( m = \frac{4}{3} \) conveys the steepness and direction of the line.
  • The y-intercept \( b = -\frac{8}{3} \) indicates where the line crosses the y-axis.
Line equations form the backbone of linear algebra and geometry, describing lines succinctly and allowing for quick comparisons and calculations in diagnostics and design applications.
Parameter Elimination
Parameter elimination involves removing the parameter from parametric equations to connect them as a single, non-parametric equation. From our solution, we took the equations \( x = 2 - 3t \) and \( y = -4t \) and made \( t \) the subject in terms of \( y \): \( t = -\frac{y}{4} \), substituting into the equation for \( x \).

Steps to Eliminate:
  • Solve one equation for the parameter \( t \).
  • Substitute this expression into the other equation.
  • Simplify to obtain a familiar form, such as \( y = mx + b \).
This step simplifies understanding and interpretation by providing a conventional Cartesian equation. Parameter elimination allows the translation of curves and paths into concise mathematical expressions, facilitating further analysis and problem solving.

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Most popular questions from this chapter

(a) Suppose that at time \(t=t_{0}\) an electron has a position vector of \(\mathbf{r}=3.5 \mathbf{i}-1.7 \mathbf{j}+\mathbf{k},\) and at a later time \(t=t_{1}\) it has a position vector of \(\mathbf{r}=4.2 \mathbf{i}+\mathbf{j}-2.4 \mathbf{k} .\) What is the displacement of the electron during the time interval from \(t_{0}\) to \(t_{1} ?\) (b) Suppose that during a certain time interval a proton has a displacement of \(\Delta \mathbf{r}=0.7 \mathbf{i}+2.9 \mathbf{j}-1.2 \mathbf{k}\) and its final position vector is known to be \(\mathbf{r}=3.6 \mathbf{k} .\) What was the initial position vector of the proton?

Calculate $$ \left.\frac{d}{d t} | \mathbf{r}_{1}(t) \cdot \mathbf{r}_{2}(t)\right] \quad \text { and } \quad \frac{d}{d t}\left[\mathbf{r}_{1}(t) \times \mathbf{r}_{2 (t)\right] $$ first by differentiating the product directly and then by applying Formulas (8) and (9) $$\mathbf{r}_{1}(t)=2 t \mathbf{i}+3 t^{2} \mathbf{j}+t^{3} \mathbf{k}, \quad \mathbf{r}_{2}(t)=t^{4} \mathbf{k}$$

Find two elevation angles that will enable a shell, fired from ground level with a muzzle speed of \(800 \mathrm{ft} / \mathrm{s}\), to hit a groundlevel target \(10,000 \mathrm{ft}\) away.

Find the displacement and the distance traveled over the indicated time interval. $$\mathbf{r}=e^{t} \mathbf{i}+e^{-t} \mathbf{j}+\sqrt{2} t \mathbf{k} ; 0 \leq t \leq \ln 3$$

Determine whether \(\mathbf{r}(t)\) is continuous at \(t=0 .\) Explain your reasoning. A. $$\mathbf{r}(t)=e^{t} \mathbf{i}+\mathbf{j}+\csc t \mathbf{k}$$ B. $$\mathbf{r}(t)=5 \mathbf{i}-\sqrt{3 t+1} \mathbf{j}+e^{2 t} \mathbf{k}$$
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