/*! 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 8 (a) graph the curves defined by ... [FREE SOLUTION] | 91Ó°ÊÓ

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(a) graph the curves defined by the parametric equations using the specified interval and identify the graph (if possible) and (b) eliminate the parameter (Exercises \(7 \text { to } 16 \text { only })\) and write the corresponding rectangular form. $$\begin{aligned}&x=t-3 ; t \in[-5,5]\\\&y=2-0.5 t^{2}\end{aligned}$$

Short Answer

Expert verified
The graph is a downward-opening parabola, and the rectangular form is \(y = -0.5x^2 - 3x - 2.5\).

Step by step solution

01

Graph the Parametric Equations

Start by plotting points from the parametric equations \(x = t - 3\) and \(y = 2 - 0.5t^2\) for \(t\) in the interval \([-5, 5]\). Compute values of \((x, y)\) by substituting different \(t\) values. For example: - For \(t = -5\), \(x = -5 - 3 = -8\) and \(y = 2 - 0.5(-5)^2 = 2 - 12.5 = -10.5\).- Repeat this for several values of \(t\) within the range to find other points such as \((-3, 2)\) for \(t = 0\) and \((2, -10.5)\) for \(t = 5\). The graph will be a downward-opening parabola that is centered a bit to the left in the Cartesian plane, starting from left-most at \(t = -5\) to right-most at \(t = 5\).
02

Eliminate the Parameter

To express the relationship between \(x\) and \(y\) without the parameter \(t\), solve for \(t\) from the equation \(x = t - 3\). Thus, \(t = x + 3\). Substitute \(t = x + 3\) into the equation for \(y\):\[y = 2 - 0.5(t)^2 = 2 - 0.5(x + 3)^2.\]Simplify:\[y = 2 - 0.5(x^2 + 6x + 9) = 2 - 0.5x^2 - 3x - 4.5.\]Therefore, the rectangular equation is:\[y = -0.5x^2 - 3x - 2.5.\]
03

Identify the Graph

The rectangular form of the equation \(y = -0.5x^2 - 3x - 2.5\) corresponds to a downward-opening parabola. Compared with the standard parabola form \(y = ax^2 + bx + c\), this parabola is translated and inverted based on the coefficients.

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Key Concepts

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

Rectangular Form
Converting parametric equations into a rectangular form is essentially about removing the parameter, which is often represented by a variable like \(t\). This process simplifies the expression, making it a clearer reflection of the relationship between \(x\) and \(y\). In this case, the parametric equations are \(x = t - 3\) and \(y = 2 - 0.5t^2\). We start by isolating \(t\) in the first equation: \(t = x + 3\). Then, we substitute this expression for \(t\) into the second equation for \(y\), resulting in \(y = 2 - 0.5(x + 3)^2\). After expanding and simplifying, we find the rectangular form: \(y = -0.5x^2 - 3x - 2.5\).
This form directly relates \(x\) and \(y\) without requiring computation of \(t\) values, making it easier to work with and analyze the curve in broader mathematical applications.
Graphing
Graphing parametric equations often involves calculating various \((x, y)\) pairs by plugging in different values of \(t\) within the given interval. It allows us to visualize the path that the parameters describe. In this example, by using values for \(t\) from \([-5, 5]\), we attain a collection of points that trace the path of the curve. When \(t = -5\), we find that \(x = -8\) and \(y = -10.5\). Similarly, as \(t\) progresses to 5, we derive points like \((2, -10.5)\).
This set of calculated points helps sketch the general shape of the curve which, in this situation, is a downward-facing parabola. Ensuring a wide range of \(t\) values allows the entire structure of the graph to be accurately represented.
Parabola
A parabola describes a symmetric curve that can open upwards or downwards depending on the coefficient of its highest degree term. In the context of this exercise, the rectangle equation derived, \(y = -0.5x^2 - 3x - 2.5\), represents a downward-opening parabola.
  • The negative coefficient in front of \(x^2\) indicates the parabola opens downwards.
  • This transformation involves a vertical "stretch" or "shrink" depending on the magnitude of the coefficient and a horizontal "shift" based on the linear term.
Parabolas are significant in various mathematical fields as they graphically solve quadratic functions and model numerous real-world phenomena, from physics trajectories to modeling reflective properties.

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