# Graph the Cartesian equation: x=4t^2, y=2t

• Fatima Hasan
Also, the direction of motion is not specified in the given equations, but we can assume it is in the positive x-direction. Therefore, in summary, the given parametric equations represent a parabola with the equation ##y^2 = x## in the xy-plane. The portion of the graph traced by the particle is the parabola itself and the direction of motion is in the positive x-direction.

## Homework Statement

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.

(1)##x = 4t^2##
(2) ##y = 2t## , ## - \infty ≤ t ≤ \infty##

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## The Attempt at a Solution

Square the second equation :
##y^2 = 4 t^2 ## (3)
Subtract (3) - (1):
##y^2 - x = 0 ##
##y^2 = x##
This equation forms a parabola .
When ##t = - 1 ## , ##x=4## and ##y=-2##
When ## t = 1 ## , ##x = 4## and ##y=2 ##
From bottom to top.

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Fatima Hasan said:

## Homework Statement

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.

(1)##x = 4t^2##
(2) ##y = 2t## , ## - \infty ≤ t ≤ \infty##

-

## The Attempt at a Solution

Square the second equation :
##y^2 = 4 t^2 ## (3)
Subtract (3) - (1):
##y^2 - x = 0 ##
##y^2 = x##
This equation forms a parabola .
When ##t = - 1 ## , ##x=4## and ##y=-2##
When ## t = 1 ## , ##x = 4## and ##y=2 ##
From bottom to top.
View attachment 231783
Looks good!

Fatima Hasan
Yes, it is. I'm not quite sure what is meant by "Indicate the portion of the graph traced by the particle", but I assume it is the parabola itself.