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Chapter 9: Problem 12

(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. $$\begin{aligned}&x=\frac{t^{3}}{10} ; t \in[-5,5]\\\&y=|t|\end{aligned}$$

Short Answer

Expert verified
Graph resembles two cubic branches; rectangular form is \(x = \pm \frac{y^3}{10}\).

Step by step solution

01

Parametric Equations

The parametric equations provided are \(x = \frac{t^3}{10}\) and \(y = |t|\), with \(t\) ranging from \([-5, 5]\). First, understand the behavior of these equations: as \(t\) varies, it affects both \(x\) and \(y\).
02

Graphing the Parametric Equations

To graph these equations, compute key points by substituting values of \(t\) within the interval \([-5, 5]\). For example, at \(t = -5\), \(x = \frac{(-5)^3}{10} = -12.5\) and \(y = 5\). Repeat this with several points (e.g., \(t = -3, 0, 3, 5\)) to capture the shape of the graph. This graph resembles two branches of a cubic curve where \(y\) is always non-negative.
03

Eliminate the Parameter

To eliminate the parameter \(t\), express \(t\) in terms of \(y\) from the equation \(y = |t|\). This gives \(t = y\) or \(t = -y\). Substitute these relations into the equation for \(x\): \(x = \frac{t^3}{10}\). This provides two equations: \(x = \frac{y^3}{10}\) and \(x = -\frac{y^3}{10}\).
04

Combine to Rectangular Form

Combine the two equations from step 3 to capture the full graph: \(x = \pm \frac{y^3}{10}\). This rectangular equation describes the relationship between \(x\) and \(y\) without the parameter \(t\). It depicts a set of symmetrical branches with respect to the \(y\)-axis.

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Key Concepts

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

Rectangular Form
Transforming parametric equations into a rectangular form is crucial when you want to express the relationship between coordinates without relying on a parameter. For the given parametric equations, \(x = \frac{t^3}{10}\) and \(y = |t|\), we start by expressing the parameter \(t\) in terms of \(y\) so that we can substitute it back into the equation for \(x\). Let's break it down:
  • Given \(y = |t|\), we can express \(t\) as \(t = y\) or \(t = -y\) because absolute value means \(t\) could be positive or negative.
  • Substitute these into \(x = \frac{t^3}{10}\), which results in two possible equations: \(x = \frac{y^3}{10}\) and \(x = -\frac{y^3}{10}\).
Now, we have the rectangular form \(x = \pm \frac{y^3}{10}\), which neatly encapsulates the relationship between \(x\) and \(y\) and represents the entirety of the graph derived from the original parametric equations.
Eliminate the Parameter
Eliminating the parameter in a set of parametric equations means expressing the relationship entirely in terms of \(x\) and \(y\), without using \(t\). This process involves solving one of the equations for \(t\) and substituting it into the other equation.
  • Start with \(y = |t|\). The absolute value ensures \(t\) is always non-negative, giving us two cases: \(t = y\) or \(t = -y\).
  • Insert \(t = y\) and \(t = -y\) into \(x = \frac{t^3}{10}\) to yield: \(x = \frac{y^3}{10}\) and \(x = -\frac{y^3}{10}\).
As a result, the parameter \(t\) has been successfully eliminated, and we're left with a pair of equations that define the graph symmetrically in terms of \(x\) and \(y\).
Graphing
Graphing parametric equations can help visualize the path described by the parameter \(t\). With the provided equations \(x = \frac{t^3}{10}\) and \(y = |t|\), the graph can be understood by finding particular points and observing their pattern. Here's a step-by-step method:
  • Choose several values of \(t\) within the interval \([-5, 5]\). For example, \(t = -5, -3, 0, 3, 5\).
  • For each value of \(t\), calculate \(x\) and \(y\) using the parametric equations.
  • At \(t = -5\): \(x = \frac{(-5)^3}{10} = -12.5\) and \(y = 5\). Similarly, calculate for other values of \(t\).
The resulting points, such as (-12.5, 5) or (0, 0), form a curve that consists of two symmetrical branches when plotted on a coordinate plane. The graph is akin to parts of a cubic function, showing the path defined as \(t\) shifts through negative and positive values.

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Most popular questions from this chapter

Parabolic car headlights: The cross section of a typical car headlight can be modeled by an equation similar to \(25 x=16 y^{2},\) where \(x\) and \(y\) are in inches and \(x \in[0,4] .\) Use this information to graph the relation for the indicated domain.

The midpoint formula in polar coordinates: \(M=\left(\frac{r \cos \alpha+R \cos \beta}{2}, \frac{r \sin \alpha+R \sin \beta}{2}\right)\) The midpoint of a line segment connecting the points \((r, \alpha)\) and \((R, \beta)\) in polar coordinates can be found using the formula shown. Find the midpoint of the line segment between \((r, \alpha)=\left(6,45^{\circ}\right)\) and \((R, \beta)=\left(8,30^{\circ}\right),\) then convert these points to rectangular coordinates and find the midpoint using the "standard" formula. Do the results match?

Complete the square in both \(x\) and \(y\) to write each equation in standard form. Then draw a complete graph of the relation and identify all important features. $$x^{2}+4 y^{2}-8 y+4 x-8=0$$

Planetary motion: The perihelion, aphelion, and orbital period of the planets Jupiter, Saturn, Uranus, and Neptune are shown in the table. Use the information to answer or complete the following exercises. The formula $$L=2 \pi \sqrt{0.5\left(a^{2}+b^{2}\right)}$$ can be used to estimate the length of the orbital path. Recall for an cllipse, \(c^{2}=a^{2}-b^{2}\). $$\begin{array}{|l|c|c|c|} \hline \text { Planet } & \begin{array}{c} \text { Perihelion } \\ \left(10^{6} \mathrm{mi}\right) \end{array} & \begin{array}{c} \text { Aphelion } \\ \left(10^{6} \mathrm{mi}\right) \end{array} & \begin{array}{c} \text { Period } \\ (y r) \end{array} \\ \hline \text { Jupiter } & 460 & 507 & 11.9 \\ \hline \text { Saturn } & 840 & 941 & 29.5 \\ \hline \text { Uranus } & 1703 & 1866 & 84 \\ \hline \text { Neptune } & 2762 & 2824 & 164.8 \\ \hline \end{array}$$ Find the eccentricity of the planets Jupiter and Saturn.

The reflector of an industrial spot light has the shape of a parabolic dish with a diameter of \(120 \mathrm{cm} .\) What is the depth of the dish if the correct placement of the bulb is \(11.25 \mathrm{cm}\) above the vertex (the lowest point of the dish)? What equation will the engineers and technicians use for the manufacture of the dish? (Hint: Analyze the information using a coordinate system.)
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