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Chapter 17: Problem 16

Draw a sketch of the graph of the given vector equation and find a cartesian equation of the graph. $$ \mathbf{R}(t)=\cos t \mathbf{i}+\cos t \mathbf{j} ; t \text { in }\left[0, \frac{1}{2} \pi\right] $$

Short Answer

Expert verified
The Cartesian equation is \(x = y\). The graph is a line segment from \(0,0\) to \(1,1\).

Step by step solution

01

Understand the Given Vector Equation

The given vector equation is \(\textbf{R}(t) = \text{cos} (t) \textbf{i} + \text{cos} (t) \textbf{j}\). Here, \(\textbf{i}\) and \(\textbf{j}\) are the unit vectors along the x-axis and y-axis, respectively. We need to express this in Cartesian coordinates.
02

Express in Terms of x and y

Observe that \(\text{cos} (t)\) is the x-component and \(\text{cos} (t)\) is also the y-component. Thus, we can write \(\textbf{R}(t)\) as: \(\textbf{R}(x, y) = (x, y)\) where \(x = \text{cos}(t)\) and \(y = \text{cos}(t)\).
03

Find the Cartesian Equation

Since \(x = \text{cos}(t)\) and \(y = \text{cos}(t)\), we can set these equal to each other: \(x = y\). This is the Cartesian equation of the graph.
04

Determine the Domain

Given that \(t\) is in the interval \([0, \frac{1}{2} \pi)\), the range for \( \text{cos}(t) \) is \([0, 1]\). Therefore, \(x\) and \(y\) must both range between 0 and 1.
05

Sketch the Graph

Draw an \(xy\)-plane. Plot the line \(x=y\) in the first quadrant between \(x=0\) to \(x=1\), since the domain is \()0 \leq x \leq 1\). The resulting graph is a 45-degree line segment from (0,0) to (1,1).

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Key Concepts

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

Vector Equations
Vector equations are a way to describe geometric objects using vectors. A vector equation represents a point in a plane or space as a combination of direction vectors and scalar parameters. In our exercise, the vector equation is \(\textbf{R}(t) = \text{cos}(t) \textbf{i} + \text{cos}(t) \textbf{j}\). Here, \(\textbf{i}\) and \(\textbf{j}\) are unit vectors pointing in the x and y directions, respectively. This means the position of any point on our graph depends on the parameter \(\text{t}\), with \(\text{cos}(t)\) as both the x and y components. Vectors are a robust tool for analyzing movement and direction, making vector equations helpful in physics and engineering, among other fields.
Cartesian Coordinates
Cartesian coordinates allow us to describe the position of points on a plane using ordered pairs (x, y). In our exercise, we transformed the given vector equation \(\textbf{R}(t) = \text{cos}(t) \textbf{i} + \text{cos}(t) \textbf{j}\) into Cartesian coordinates. To do this, we observed that both x and y components are equal to \(\text{cos}(t)\). Thus, we can describe this as \(\textbf{R}(x, y) = (x, y)\) where \(\text{x = \text{cos}(t)}\) and \(\text{y = \text{cos}(t)}\). This simplifies to the Cartesian equation \(\text{x = y}\). Cartesian coordinates are extremely useful in graphing equations and understanding geometric shapes on the coordinate plane. They form the basis of many fields including geometry, physics, and engineering.
Parametric Equations
Parametric equations describe curves by expressing the coordinates of the points as functions of a variable, typically denoted as \(\text{t}\). In our problem, the vector equation \(\textbf{R}(t) = \text{cos}(t) \textbf{i} + \text{cos}(t) \textbf{j}\) is a parametric representation of the curve in terms of \(\text{t}\). Here, \(\text{t}\) varies between 0 and \(\frac{1}{2} \text{蟺}\). This gives us a range for \(\text{cos}(t)\) between 0 and 1. By using parametric equations, we can neatly describe more complicated paths and motions, making these equations essential in mathematics and physics.
Graphing in 2D
Graphing in 2D involves plotting points or curves on an x-y plane. To create the graph of our transformed Cartesian equation \(\text{x = y}\), we need to plot points where the x and y values are equal. The given domain for \(\text{x = y}\) is within \([0, 1]\). This results in a line segment that starts at (0,0) and ends at (1,1). For visualization:
  • Draw x and y axes.
  • Plot points (0,0) and (1,1).
  • Draw a straight line connecting these points.
This line segment represents all points where \(\text{x}\) equals \(\text{y}\) within the specified domain. Graphing helps us see the solutions and understand the relationships between variables clearly and intuitively.

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Most popular questions from this chapter

If a polar equation of a curve is \(r=F(\theta)\), prove that the curvature \(K\) is given by the formula $$ K=\frac{\left|r^{2}+2(d r / d \theta)^{2}-r\left(d^{2} r / d \theta^{2}\right)\right|}{\left[r^{2}+(d r / d \theta)^{2}\right]^{3 / 2}} $$

A particle is moving along the curve having the given vector equation. In each problem, find the vectors \(\mathbf{V}(t), \mathbf{A}(t), \mathbf{T}(t)\), and \(\mathbf{N}(t)\), and the following scalars for an arbitrary value of \(t:|\mathbf{V}(t)|, A_{T}, A_{N}, K(t) .\) Also find the particular values when \(t=t_{1} .\) At \(t=t_{1}\), draw a sketch of a portion of the curve and representations of the vectors \(\mathbf{V}\left(t_{1}\right)\), \(\mathbf{A}\left(t_{1}\right), A_{T} \mathbf{T}\left(t_{1}\right)\), and \(A_{N} \mathbf{N}\left(t_{1}\right)\). $$ \mathbf{R}(t)=e^{t} \mathbf{i}+e^{-t} \mathbf{j} ; t_{1}=0 $$

Two forces represented by the vectors \(F_{1}\) and \(F_{2}\) act on a particle and cause it to move along a straight line from the point \((2,5)\) to the point \((7,3) .\) If \(F_{1}=3 \mathbf{i}-j\) and \(F_{2}=-4 \mathbf{i}+5 \mathbf{j}\), the magnitudes of the forces are measured in pounds, and distance is measured in feet, find the work done by the two forces acting together.

In Exercises 7 through 12, the position of a moving particle at \(t\) sec is determined from a vector equation. Find: (a) \(\mathbf{V}\left(t_{1}\right)\) (b) \(\mathbf{A}\left(t_{1}\right) ;\) (c) \(\left|\mathbf{V}\left(t_{1}\right)\right| ;\) (d) \(\left|\mathbf{A}\left(t_{1}\right)\right| .\) Draw a sketch of a portion of the path of the particle containing the position of the particle at \(t=t_{1}\), and draw the representations of \(\mathbf{V}\left(t_{1}\right)\) and \(\mathbf{A}\left(t_{1}\right)\) having initial point where \(t=t_{1}\). $$ \mathbf{R}(t)=\cos 2 t \mathbf{i}-3 \sin t \mathbf{j} ; t_{1}=\pi $$

Find the length of arc in each of the following exercises. When \(a\) appears, \(a>0\). \(x=e^{-t} \cos t, y=e^{-t} \sin t ;\) from \(t=0\) to \(t=\pi\).
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