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How to interpret parametric equations

  1. Oct 24, 2014 #1
    1. The problem statement, all variables and given/known data
    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)



    2. Relevant equations
    Not sure?

    3. 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 dont see it.
     
    Last edited by a moderator: Oct 24, 2014
  2. jcsd
  3. Oct 24, 2014 #2

    Mark44

    Staff: Mentor

    No. y2 + x2 = (3 sin(t))2 + (cos(t))2 = 9sin2(t) + cos2(t). Are you sure you wrote the problem correctly?
    Based on what you wrote in the problem description, your curve is not a circle.
     
  4. Oct 24, 2014 #3
    I solved for t and pluged it in for x. Yes I am positive I wrote it down correctly.
     
  5. Oct 24, 2014 #4
    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 ive done that :P
     
  6. Oct 24, 2014 #5

    Mark44

    Staff: Mentor

    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.
     
  7. Oct 24, 2014 #6
    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.
     
  8. Oct 24, 2014 #7
    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?
     
  9. Oct 24, 2014 #8

    Mark44

    Staff: Mentor

    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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