How to interpret parametric equations

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The discussion focuses on interpreting the parametric equations x = cos(t) and y = 3sin(t) for 0 ≤ t ≤ π/2. The equations describe an arc of an ellipse rather than a circle, with the center at the origin (0,0) and a semi-major axis of 3 along the y-axis. The radius is derived from the maximum values of x and y, confirming the semi-major axis length. The positive orientation is determined by evaluating the curve's direction as t increases from 0 to π/2, indicating a counterclockwise movement in the first quadrant. The confusion arises from misinterpreting the shape and orientation of the graph, which is clarified through parameter elimination and analysis of t's effect on the coordinates.
TheKracken
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Homework Statement


Eliminate the parameter to find a description of the following circles or circular arc's in terms of x and y
and find the center and radius and indicate the positive orientation

x=cos(t) , y = 3sin(t) ; 0< t < pi/2 (should be less than or equal to)

Homework Equations


Not sure?

The Attempt at a Solution


When I find t = arccos(x) I then plug it in y=3 sin(t) and I result in
y= sqrt(9-x^2)

The book doesn't even give an answer for the equation part but it does tell me the origin, and radius and the
orientation but I figured you could rewrite my equation to y^2 + x^2 = 9 and that tell me the radius of 3 (sqrt (9) right?) then the orgin is 0,0 because nothing it being done to the x and y

but here is what trips me up, how do they know it is the lower half of a circle going counter clockwise? I mean it has to do something with the t's but I just don't see it.
 
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TheKracken said:

Homework Statement


Eliminate the parameter to find a description of the following circles or circular arc's in terms of x and y
and find the center and radius and indicate the positive orientation

x=cos(t) , y = 3sin(t) ; 0< t < pi/2 (should be less than or equal to)

Homework Equations


Not sure?

The Attempt at a Solution


When I find t = arccos(x) I then plug it in y=3 sin(t) and I result in
y= sqrt(9-x^2)

The book doesn't even give an answer for the equation part but it does tell me the origin, and radius and the
orientation but I figured you could rewrite my equation to y^2 + x^2 = 9
No. y2 + x2 = (3 sin(t))2 + (cos(t))2 = 9sin2(t) + cos2(t). Are you sure you wrote the problem correctly?
TheKracken said:
and that tell me the radius of 3 (sqrt (9) right?) then the orgin is 0,0 because nothing it being done to the x and y

but here is what trips me up, how do they know it is the lower half of a circle going counter clockwise? I mean it has to do something with the t's but I just don't see it.
Based on what you wrote in the problem description, your curve is not a circle.
 
I solved for t and pluged it in for x. Yes I am positive I wrote it down correctly.
 
I should probably mention this is a parametric equation? I am sure you are correct, I may have gotten the right answer by doing something incorrectly by coincidence. Not the first time I've done that :P
 
TheKracken said:
I solved for t and pluged it in for x. Yes I am positive I wrote it down correctly.
Please show what you did to get ##y = \sqrt{9 - x^2}##. Your mistake is in that work.

The parametric equations x = cos(t), y = 3sin(t) do NOT represent a circle.
 
To find the orientation, put t = 0 and t = pi/2 (or put more points if you'd like), and see which way you're going.
 
I want to apologize, I did type the equation incorrectly, it is 3cos(t). I was looking at the next problem it looks like when I typed it out, but my work was for this problem.

In that case is my work correct?
 
Instead of solving for t in one of the parametric equations, I would eliminate the parameter directly. Since (now) x = 3cos(t) and y = 3sin(t), x2 + y2 = ?

Keep in mind that ##0 \leq t \leq \pi/2##. As t increases from 0, what do the points on the curve do? What is the starting point (i.e., when t = 0)? What is the ending point (when t = ##\pi/2##)?
 

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