/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 104 Are the statements true or false... [FREE SOLUTION] | 91Ó°ÊÓ

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Are the statements true or false? Give reasons for your answer. The parametric curve \(x=(3 t+2)^{2}, y=(3 t+2)^{2}-1\) for \(0 \leq t \leq 3\) is a line.

Short Answer

Expert verified
True, the curve is a line given by \(y = x - 1\).

Step by step solution

01

Identify the Parametric Equations

The parametric equations given for the curve are \(x=(3t+2)^2\) and \(y=(3t+2)^2 - 1\). We will use these equations to analyze the nature of the curve.
02

Express y in Terms of x

Since both \(x\) and \(y\) are expressed in terms of \((3t+2)\), we can eliminate \(t\) and express \(y\) directly in terms of \(x\). Note that \(y = x - 1\). This equation \(y = x - 1\) is a linear equation, indicating that the curve is a line.
03

Verify the Range of t

For the range \(0 \leq t \leq 3\), substitute the values of \(t\) in \(x = (3t+2)^2\):When \(t = 0\), \(x = (2)^2 = 4\).When \(t = 3\), \(x = (11)^2 = 121\).Thus, \(x\) ranges from 4 to 121, confirming a continuous linear range.

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Key Concepts

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

Linear Equations
Linear equations represent straight lines in the coordinate plane. They take the general form of \(y = mx + c\) where \(m\) is the slope, and \(c\) is the y-intercept. In linear equations, the relationship between the variables is constant, meaning the graph of the equation is a straight line.

For the given parametric equations, we expressed \(y\) directly in terms of \(x\) and found \(y = x - 1\). This form is similar to \(y = mx + c\) with \(m = 1\) and \(c = -1\). Hence, the equation corresponds to a line.
  • The slope \(m = 1\) indicates the line rises one unit for every unit it moves right.
  • The y-intercept \(c = -1\) tells us the line crosses the y-axis at \(-1\).
This confirms that the parametric curve for these equations indeed forms a straight line.
Parametric Curve
A parametric curve is described by two or more equations where each component is expressed in terms of a common parameter, usually \(t\). The idea is to capture the change in the curve's position over a continuous interval of \(t\). In other words, parametric equations define how \(x\) and \(y\) coordinates position along a curve as \(t\) varies.

For the exercise at hand, the parametric equations:
  • \(x = (3t + 2)^2\)
  • \(y = (3t + 2)^2 - 1\)
are dependent on the parameter \(t\). By manipulating \(t\), the values of \(x\) and \(y\) change, moving the point on a coordinate plane accordingly.

Interestingly, in this case, when substituting \(y\) in terms of \(x\), we see it boils down to a simple linear relationship, defining a straight path or line. This indicates that while parametric equations can represent more complex curves, they can also define simple linear paths in some instances.
Range of Parameter
The range of the parameter \(t\) is vital in determining which part of the parametric curve is being considered. It dictates the start and end points of the curve. Here, our parameter \(t\) is defined to range from 0 to 3.

When \(t = 0\), substituting into \(x = (3t + 2)^2\) gives \(x = 4\). When \(t = 3\), \(x = (3t + 2)^2 = 121\). Hence, \(x\) ranges from 4 to 121 as \(t\) moves from 0 to 3. This continuous interval ensures the line is fully sketched between these positions without any gaps.

Defining the parameter range ensures that we can systematically trace out the segment of the curve we are interested in. In mathematical problems, especially involving parametric equations, knowing the range assists in correctly identifying or excluding parts of curves needed for analysis.

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Most popular questions from this chapter

Explain what is wrong with the statement. A parameterized curve \(\vec{r}(t), A \leq t \leq B,\) has length \(B-A\)

Are the statements in Problems true or false? Give reasons for your answer. If the flow lines for the vector field \(\vec{F}(x, y)\) are all straight lines parallel to the constant vector \(\vec{v}=3 \vec{i}+\) \(5 \vec{j},\) then \(\vec{F}(x, y)=\vec{v}\)

In the middle of a wide, steadily flowing river there is a fountain that spouts water horizontally in all directions. The river flows in the \(\vec{i}\) -direction in the \(x y\) -plane and the fountain is at the origin. (a) If \(A>0, K>0,\) explain why the following expression could represent the velocity field for the combined flow of the river and the fountain: $$\vec{v}=A \vec{i}+K\left(x^{2}+y^{2}\right)^{-1}(x \vec{i}+y \vec{j})$$ (b) What is the significance of the constants \(A\) and \(K ?\) (c) Using a computer, sketch the vector field \(\vec{v}\) for \(K=\) 1 and \(A=1\) and \(A=2,\) and for \(A=0.2, K=2\)

A line has equation \(\vec{r}=\vec{a}+t \vec{b}\) where \(\vec{r}=x \vec{i}+y \vec{j}+z \vec{k}\) and \(\vec{a}\) and \(\vec{b}\) are constant vectors such that \(\vec{a} \neq \overrightarrow{0}, \vec{b} \neq\) \(\overrightarrow{0}, \vec{b}\) not parallel or perpendicular to \(\vec{a} .\) For each of the planes (a)-(c), pick the equation (i)-(ix) which represents it. Explain your choice. (a) A plane perpendicular to the line and through the origin. (b) A plane perpendicular to the line and not through the origin. (c) A plane containing the line. (i) \(\vec{a} \cdot \vec{r}=|| \vec{b}||\) (ii) \(\vec{b} \cdot \vec{r}=|| \vec{a}||\) (iii) \(\vec{a} \cdot \vec{r}=\vec{b} \cdot \vec{r}\) (iv) \((\vec{a} \times \vec{b}) \cdot(\vec{r}-\vec{a})=0\) (v) \(\vec{r}-\vec{a}=\vec{b}\) (vi) \(\vec{a} \cdot \vec{r}=0\) (vii) \(\vec{b} \cdot \vec{r}=0\) (viii) \(\vec{a}+\vec{r}=\vec{b}\) (ix) \((\vec{a} \times \vec{b}) \cdot(\vec{r}-\vec{b})=\|\vec{a}\|\)

Two parameterized lines are given. Are they the same line? $$\begin{array}{l} \vec{r}_{1}(t)=(5-3 t) \vec{i}+(1+t) \vec{j}+2 t \vec{k} \\ \vec{r}_{2}(t)=(2+6 t) \vec{i}+(2-2 t) \vec{j}+(3-4 t) \vec{k} \end{array}$$

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