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

In summary: 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.
  • #1
Fatima Hasan
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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##

Homework Equations


-

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.
3333.png

Is my answer correct ?
 

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  • #2
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##

Homework Equations


-

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
Is my answer correct ?
Looks good!
 
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Likes Fatima Hasan
  • #3
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.
 

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

What is the equation for the graph given by x=4t^2 and y=2t?

The equation for the graph is a parametric equation, where the x-coordinate is defined by x=4t^2 and the y-coordinate is defined by y=2t.

How do I plot the points for the graph given by x=4t^2 and y=2t?

To plot the points, choose a range of values for t and plug them into the equations to find corresponding values for x and y. Then, plot the points (x,y) on a Cartesian plane.

What is the shape of the graph for x=4t^2 and y=2t?

The graph is a parabola, as the equation for x is in the form of y=ax^2, which is the standard form for a parabola.

How can I find the vertex of the parabola for x=4t^2 and y=2t?

To find the vertex, we can use the formula (-b/2a, -b^2/4a), where a and b are the coefficients of x and t, respectively. In this case, the vertex is (0,0).

Can I change the values of a and b in the equations x=4t^2 and y=2t to alter the shape of the graph?

Yes, changing the values of a and b will change the shape of the graph. The value of a determines how narrow or wide the parabola is, while the value of b determines the direction the parabola opens (up or down).

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