Math Problem Statement

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Part 1

Parametric equations and a parameter interval for the motion of a particle in the​ xy-plane are given. Identify the​ particle's path by finding a Cartesian equation for it. Graph the Cartesian equation. Indicate the portion of the graph traced by the particle and the direction of motion.

xequals=99t squaredt2​,

yequals=33​t,

minus−infinity∞less than<tless than<infinity∞

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Part 1

Solution

To identify the particle's path and find the corresponding Cartesian equation, we start with the given parametric equations:

x=9t2x = 9t^2 y=3ty = 3t

Step 1: Express tt in terms of yy

From the equation y=3ty = 3t, we can solve for tt:

t=y3t = \frac{y}{3}

Step 2: Substitute tt into the equation for xx

Next, substitute t=y3t = \frac{y}{3} into the equation x=9t2x = 9t^2:

x=9(y3)2x = 9\left(\frac{y}{3}\right)^2

Simplify the expression:

x=9(y29)x = 9\left(\frac{y^2}{9}\right)

x=y2x = y^2

Step 3: Identify the Cartesian Equation

The Cartesian equation of the particle's path is:

x=y2x = y^2

Step 4: Determine the Portion of the Graph Traced and the Direction of Motion

The parametric equations describe the path of a particle along the parabola x=y2x = y^2. Since tt ranges from -\infty to \infty, and given that x=9t2x = 9t^2 (which is always non-negative), the particle moves along the parabola in the positive xx-direction.

  • For t>0t > 0: y=3ty = 3t is positive, so the particle moves upward along the right branch of the parabola.
  • For t<0t < 0: y=3ty = 3t is negative, so the particle moves downward along the left branch of the parabola.

Thus, the graph x=y2x = y^2 is a parabola opening to the right, and the particle moves from the lower left (downward) branch to the upper right (upward) branch as tt increases from negative infinity to positive infinity.

Step 5: Graph the Cartesian Equation

To graph x=y2x = y^2:

  • The parabola opens to the right with its vertex at the origin (0,0)(0, 0).
  • The portion traced by the particle includes the entire parabola since tt ranges over all real numbers.
  • The direction of motion starts from the left branch of the parabola (as tt is negative) and moves towards the right branch as tt becomes positive.

Would you like to see a sketch of the graph, more details, or have any other questions? Here are some follow-up questions:

  1. How would the graph change if tt was limited to a specific range?
  2. What happens to the graph if the sign of xx is changed in the parametric equations?
  3. How do you determine the direction of motion for other parametric curves?
  4. Can you describe the motion of a particle for parametric equations with a trigonometric function?
  5. What is the significance of parametric equations in modeling real-world motion?

Tip: When converting parametric equations to a Cartesian equation, always solve for one parameter in terms of the other to eliminate the parameter effectively.

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Math Problem Analysis

Mathematical Concepts

Parametric Equations
Cartesian Equations
Graphing

Formulas

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Theorems

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Suitable Grade Level

Grades 11-12