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Quck question about graphing a parametric equation

  1. Oct 8, 2009 #1
    1. The problem statement, all variables and given/known data


    Graph the parametric equation.

    t is greater than or equal to zero and t is less than or equal to 2 pi.

    x=sint
    y=cost

    2. Relevant equations

    None

    3. The attempt at a solution

    Assuming that I don't have a calculator, how would I graph this?

    I know that there's a point which x=0, y=1, and t=0.

    x=sint
    0=sint
    t=0

    And also, the equation of the line without using trig functions is y=1-2x^2 so putting in x as 0 means y=1.

    Is there any way I can find out the other points?
     
  2. jcsd
  3. Oct 8, 2009 #2

    tiny-tim

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    Hi DMOC! :smile:

    (have a π and a ≤ and a ≥ and try using the X2 tag just above the Reply box :wink:)
    erm :redface: … noooo
    What is the relationship between sint and cost? :smile:
     
  4. Oct 8, 2009 #3

    Mark44

    Staff: Mentor

    Pick a few, well-chosen values for t (like 0, pi/6, pi/4, pi/3, pi/2, etc.), calculate the values for x and y, and graph the x-y pairs.
     
  5. Oct 8, 2009 #4

    jambaugh

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    You don't have a calculator but you can still pick certain standard values for t.

    Try t = 0, π/6 (30°), π/4 (45°), π/3 (60°), π/2 (90°) and so on...
    Figuring out your trig values should also lead you to .... "Oh but of course!" with regard to your original problem.

    If you haven't learned standard trig values at least go back and look up the definition of sine and cosine. (And you ought to go back and learn the standard trig values too.)

    In fact here is a quick way to memorize them:
    [tex][\sqrt{0}/2, \sqrt{1}/2, \sqrt{2}/2, \sqrt{3}/2, \sqrt{4}/2 ][/tex]
    or
    [tex][0,1/2, \sqrt{2}/2, \sqrt{3}/2,1][/tex]
    are the values of sin(θ) for θ = 0°, 30°, 45°, 60°, and 90°. The cosines are the same but in reverse order. To get the rest you still need to review the basic definition of the trig functions for arbitrary angles.
     
  6. Oct 8, 2009 #5
    Thanks for the help everyone.

    All right, let me try to write the equation of the line again.

    By the way, if this matters. .. I made an error in my original post.

    The correct equations are:

    x = sin t same as before

    y = cos (2t) Needed a 2 to add with the cos.

    Here's how I did this:

    Cos 2 theta trig identity.

    [tex]Cos(2\theta) = 1- 2sin^{2}\theta[/tex]

    If x = sint, then x2 = sin2t

    Now just let t = theta.

    [tex]y = 1 - 2x^{2}[/tex]

    Sorry about that error in the original post.

    Is this line equation correct?

    I'm going to start testing out trig values now.
     
  7. Oct 8, 2009 #6

    Mark44

    Staff: Mentor

    Looks good, but it's not a line. Your parametric curve is a parabola.
     
  8. Oct 8, 2009 #7
    So if t = pi/6, then x must be 1/2.

    I think I'll make that a point on my graph. The question is, where will I put it, since it's an x-y coordinate system?
     
  9. Oct 8, 2009 #8

    jambaugh

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    You got x from x=sin(t) = sin(pi/6)
    You just said y = cos(2t) so ....

    Just plot (sin(t), cos(2t) ) for various values. Especially include the end points. Then you can look at the algebraic relationship between x and y via trig identities and see how to fill in the gaps. Remember that the trig functions are bounded between +/- 1 so your shape must be within the square with corners at (+/- 1, +/- 1).

    BTW My comments about "Oh! but of course" reflected your mispost. Obviously x=sin(t) y = cos(t) gives the unit circle centered at the origin. This (with x and y swapped) is the definition of circle trig.
     
  10. Oct 8, 2009 #9
    Yeah, sorry about my screw - up. :(

    Ok, thanks for your advice. I think I've got this problem in the bag.

    Last question: When I plot coordinates on the x-y graph with a parametric equation, do I write out the points like this: (x , y) or (x, y, t)?
     
  11. Oct 8, 2009 #10

    Mark44

    Staff: Mentor

    (x, y) but you could put a side note indicating the value of t for some of the points. On a two-dimension graph, you really have only two coordinates.
     
  12. Oct 8, 2009 #11
    Ok thanks.
     
  13. Oct 8, 2009 #12

    jambaugh

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    (x,y) or if you want to express the t dependence ( x(t) , y(t) )
    x and y functions of t.

    Remember functions can be defined mapping any type of object to any type of object, not just numbers to numbers. In this case the parametrization is a function mapping real numbers (the parameter) to points. We express this single function P(t) by giving the functions for the coordinates in a given coordinate system: P(t) = ( x(t) , y(t) ).
     
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