/*! 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 11 Exercises \(1-18\) give parametr... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Exercises \(1-18\) give parametric equations and parameter intervals for the motion of a particle in the \(x y\) -plane. Identify the particle's path by finding a Cartesian equation for it. Graph the Cartesian equation. (The graphs will vary with the equation used.) Indicate the portion of the graph traced by the particle and the direction of motion. $$ x=t^{2}, \quad y=t^{6}-2 t^{4}, \quad-\infty < t < \infty $$

Short Answer

Expert verified
The particle's path is given by the curve \( y = x^3 - 2x^2 \) for \( x \geq 0 \), and the motion is from left to right as \( t \) increases.

Step by step solution

01

Set Up the Relationship

We are given the parametric equations \( x = t^2 \) and \( y = t^6 - 2t^4 \). Our goal is to eliminate the parameter \( t \) to find a Cartesian equation.
02

Express t in terms of x

From the equation \( x = t^2 \), solve for \( t \):\[ t = \pm \sqrt{x} \] This expresses \( t \) in terms of \( x \).
03

Substitute t into y equation

Substitute \( t = \pm \sqrt{x} \) into the equation for \( y \):\[ y = (\pm \sqrt{x})^6 - 2(\pm \sqrt{x})^4 \]
04

Simplify the Expression for y

Simplify the equation for \( y \):\[ y = x^3 - 2x^2 \] This is the Cartesian equation for the path of the particle.
05

Graph the Cartesian Equation

Graph the equation \( y = x^3 - 2x^2 \). The graph is a cubic curve that originates from the Cartesian equation derived.
06

Indicate the Traced Portion

Since \( x = t^2 \), \( x \) is always non-negative (i.e., \( x \geq 0 \)). Thus, only the section of the curve where \( x \geq 0 \) is traced by the particle.
07

Indicate Direction of Motion

As \( t \) increases from negative to positive infinity, \( x = t^2 \) starts from zero, increases through all positive values, and \( y = t^6 - 2t^4 \) traces the path following the direction of increasing \( t \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Cartesian equation
When working with parametric equations, our objective is to describe the path of a particle or object using a single equation in terms of Cartesian coordinates, typically denoted as from the form of \( y = f(x) \). This process involves eliminating the parameter—in this case, \( t \)—to express the relationship between \( x \) and \( y \) directly.
  • Starting with the parametric equations \( x = t^2 \) and \( y = t^6 - 2t^4 \), we aim to remove \( t \).
  • To achieve this, solve for \( t \) in terms of \( x \) from one of the parametric equations: \( t = \pm \sqrt{x} \).
  • Then, substitute \( t \) into the equation for \( y \), resulting in \( y = x^3 - 2x^2 \).
This simplified Cartesian equation \( y = x^3 - 2x^2 \) is a powerful representation, as it directly reflects the relationship of the path in terms of \( x \) and \( y \), thus eliminating the need for the parameter \( t \).
By using Cartesian coordinates, it becomes more straightforward to analyze and graph the path of the particle.
graphing curves
Once we have the Cartesian equation, the next step is to graph it to visualize the particle's path. Graphing the equation \( y = x^3 - 2x^2 \) gives us a cubic curve. This graph represents the trajectory of a particle in the \( xy \)-plane.
  • A cubic curve can exhibit various behaviors such as turning points or inflections, making it interesting to study.
  • For \( x = t^2 \), since \( x \) can only be non-negative, we restrict our graph to \( x \geq 0 \).
Graphing aids in illustrating how the path behaves over the domain of interest. In this case, it shows the portion traced by the particle and provides insight into features like the starting and end points of its motion in the positive \( x \)-direction. Graphs are a great tool for conveying critical curve data, enhancing our understanding of the overall particle motion.
particle motion
Understanding particle motion through parametric equations provides insight into how a particle traverses a particular path. By identifying the Cartesian equation, \( y = x^3 - 2x^2 \), and graphing it, we can depict both the pathway and the nature of the particle's journey.
  • Since \( x = t^2 \), the value of \( x \) is always non-negative, which indicates that the particle starts and continues its movement from the origin \( (0, 0) \) in a positive direction along the \( x \)-axis.
  • The trajectory, as established, forms a cubic curve, and as \( t \) progresses from negative to positive infinity, the particle's path is defined by increasing values of both \( x \) and \( y \), as conditionally traced by the equation.
Direction and speed are often inferred from the graph and equations. As seen here, with increasing \( t \), the particle takes on a specific motion path dictated by the relationship between \( x \) and \( y \). This understanding allows for the multifaceted interpretation of particle dynamics in mathematical and physical contexts.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Give equations for hyperbolas. Put each equation in standard form and find the hyperbola's asymptotes. Then sketch the hyperbola. Include the asymptotes and foci in your sketch. \(y^{2}-x^{2}=8\)

If you have a parametric equation grapher, graph the equations over the given intervals in Exercises \(51-58 .\) $$ \begin{array}{l}{\text { Cycloid } x=t-\sin t, \quad y=1-\cos t, \quad \text { over }} \\ {\text { a. } 0 \leq t \leq 2 \pi} \\ {\text { b. } 0 \leq t \leq 4 \pi} \\ {\text { c. } \pi \leq t \leq 3 \pi}\end{array} $$

Give equations for hyperbolas. Put each equation in standard form and find the hyperbola's asymptotes. Then sketch the hyperbola. Include the asymptotes and foci in your sketch. \(y^{2}-3 x^{2}=3\)

Volume The "triangular" region in the first quadrant bounded by the \(x\)-axis, the line \(x=4,\) and the hyperbola \(9 x^{2}-4 y^{2}=36\) is revolved about the \(x\)-axis to generate a solid. Find the volume of the solid.

The asymptotes of \(\left(x^{2} / a^{2}\right)-\left(y^{2} / b^{2}\right)=1\) Show that the vertical distance between the line \(y=(b / a) x\) and the upper half of the right-hand branch \(y=(b / a) \sqrt{x^{2}-a^{2}}\) of the hyperbola \(\left(x^{2} / a^{2}\right)-\left(y^{2} / b^{2}\right)=1\) approaches 0 by showing that $$\lim _{x \rightarrow \infty}\left(\frac{b}{a} x-\frac{b}{a} \sqrt{x^{2}-a^{2}}\right)=\frac{b}{a} \lim _{x \rightarrow \infty}\left(x-\sqrt{x^{2}-a^{2}}\right)=0.$$ Similar results hold for the remaining portions of the hyperbola and the lines \(y=\pm(b / a) x.\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.