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Chapter 12: Problem 46

Find parametric equations for the following curves. Include an interval for the parameter values. Answers are not unique. The horizontal line segment starting at \(P(8,2)\) and ending at \(Q(-2,2)\)

Short Answer

Expert verified
Answer: The parametric equations for the given horizontal line segment are: $x(t) = 8 - 10t \\ y(t) = 2$ with an interval of \(0 \leq t \leq 1\).

Step by step solution

01

Calculate the direction vector

The direction vector of the line segment \(\overrightarrow{PQ}\) is the difference of the coordinates of point Q and point P: \(\overrightarrow{PQ} = Q - P\). So, the direction vector is: \(\overrightarrow{PQ} = (-2 - 8, 2 - 2) = (-10, 0)\)
02

Set up a parameterized vector form

To set up the parameterized vector form for the line segment, we will use the starting point, P, the direction vector, and a parameter, t. The parameterized vector form is: \(\overrightarrow{r}(t) = P + t\overrightarrow{PQ} = (8,2) + t(-10, 0)\)
03

Extract the parametric equations

To find the parametric equations for the line segment, we will extract the equations for the x and y coordinates from the parameterized vector form: $x(t) = 8 - 10t \\ y(t) = 2$
04

Provide an interval for the parameter values

To find an appropriate interval for the parameter values, we need to consider the starting point, P, and the ending point, Q. When t = 0, the parameterized vector form gives us the starting point: \(\overrightarrow{r}(0) = (8,2) + 0(-10, 0) = (8,2)\) To find the value of t that corresponds to the ending point, Q, we can set the x-coordinate of the parameterized vector form equal to the x-coordinate of point Q and solve for t: \(-2 = 8 - 10t\) \(t = 1\) So, a suitable interval for the parameter values is \( 0 \leq t \leq 1\). The parametric equations for the given horizontal line segment are: $x(t) = 8 - 10t \\ y(t) = 2$ with an interval of \(0 \leq t \leq 1\).

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Key Concepts

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

Direction Vector
Understanding the direction vector is crucial when working with parametric equations. It is defined as a vector that gives the direction of a line segment or ray in space. In essence, it's like an arrow pointing from one point to another, showing the path along which you travel from the start point to the end point.

In our exercise, the direction vector \(\overrightarrow{PQ}\) is found by subtracting the coordinates of the starting point P from the ending point Q. This results in \(\overrightarrow{PQ} = (-10, 0)\), which tells us that the movement from P to Q is entirely horizontal, with no vertical change, as the y-component is zero.
Parameterized Vector Form
The parameterized vector form is a method of expressing a line or curve using a vector equation. It combines a fixed point, usually the starting point of the line segment, with the direction vector scaled by a parameter, commonly denoted as \(t\).

In our step-by-step solution, we use the starting point \(P(8,2)\) and the direction vector \(\overrightarrow{PQ} = (-10, 0)\) to write the parameterized vector form, \(\overrightarrow{r}(t) = (8,2) + t(-10, 0)\). This compact form encapsulates all possible points on the line segment as \(t\) varies.
Parametric Representation of Lines
The parametric representation of lines translates the idea of a path between two points into a set of equations dependent on a parameter. It's a powerful way to describe the position of any point along a line at any given instance.

In our example, the parametric equations extracted from the parameterized vector form are \(x(t) = 8 - 10t\) and \(y(t) = 2\). The first equation gives us the x-coordinate of any point along the line segment as \(t\) varies, while the second equation tells us that the y-coordinate remains constant at 2, consistent with the horizontal line segment.
Intervals for Parameter Values
Determining the correct interval for parameter values is essential to accurately define the segment of the curve or line we are interested in. The parameter \(t\) has different values that correspond to different points on the line. By setting bounds on \(t\), we define the specific portion of the line we're dealing with.

In the provided solution, the interval \(0 \leq t \leq 1\) was found by ensuring that \(t = 0\) corresponds to point P and \(t = 1\) corresponds to point Q. Within this interval, \(t\) smoothly transitions from the start to the end point of the line segment, perfectly encapsulating the entire path of our horizontal line.

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Most popular questions from this chapter

Suppose the function \(y=h(x)\) is nonnegative and continuous on \([\alpha, \beta],\) which implies that the area bounded by the graph of h and the x-axis on \([\alpha, \beta]\) equals \(\int_{\alpha}^{\beta} h(x) d x\) or \(\int_{\alpha}^{\beta} y d x .\) If the graph of \(y=h(x)\) on \([\alpha, \beta]\) is traced exactly once by the parametric equations \(x=f(t), y=g(t),\) for \(a \leq t \leq b,\) then it follows by substitution that the area bounded by h is $$\begin{array}{l}\int_{\alpha}^{\beta} h(x) d x=\int_{\alpha}^{\beta} y d x=\int_{a}^{b} g(t) f^{\prime}(t) d t \text { if } \alpha=f(a) \text { and } \beta=f(b) \\\\\left(\text { or } \int_{\alpha}^{\beta} h(x) d x=\int_{b}^{a} g(t) f^{\prime}(t) d t \text { if } \alpha=f(b) \text { and } \beta=f(a)\right)\end{array}$$. Find the area of the region bounded by the astroid \(x=\cos ^{3} t, y=\sin ^{3} t,\) for \(0 \leq t \leq 2 \pi\) (see Example 8 Figure 12.17 ).

A plane is 150 miles north of a radar station, and 30 minutes later it is \(60^{\circ}\) east of north at a distance of 100 miles from the radar station. Assume the plane flies on a straight line and maintains constant altitude during this 30 -minute period. a. Find the distance traveled during this 30 -minute period. b. Determine the average velocity of the plane (relative to the ground) during this 30 -minute period.

Derivatives Consider the following parametric curves. a. Determine \(dy/dx\) in terms of t and evaluate it at the given value of \(t\). b. Make a sketch of the curve showing the tangent line at the point corresponding to the given value of \(t\). $$x=t+1 / t, y=t-1 / t ; t=1$$

Find the arc length of the following curves on the given interval. $$x=t^{4}, y=\frac{t^{6}}{3} ; 0 \leq t \leq 1$$

Suppose the function \(y=h(x)\) is nonnegative and continuous on \([\alpha, \beta],\) which implies that the area bounded by the graph of h and the x-axis on \([\alpha, \beta]\) equals \(\int_{\alpha}^{\beta} h(x) d x\) or \(\int_{\alpha}^{\beta} y d x .\) If the graph of \(y=h(x)\) on \([\alpha, \beta]\) is traced exactly once by the parametric equations \(x=f(t), y=g(t),\) for \(a \leq t \leq b,\) then it follows by substitution that the area bounded by h is $$\begin{array}{l}\int_{\alpha}^{\beta} h(x) d x=\int_{\alpha}^{\beta} y d x=\int_{a}^{b} g(t) f^{\prime}(t) d t \text { if } \alpha=f(a) \text { and } \beta=f(b) \\\\\left(\text { or } \int_{\alpha}^{\beta} h(x) d x=\int_{b}^{a} g(t) f^{\prime}(t) d t \text { if } \alpha=f(b) \text { and } \beta=f(a)\right)\end{array}$$. Find the area under one arch of the cycloid \(x=3(t-\sin t)\) \(y=3(1-\cos t)(\text { see Example } 5)\)
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