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Drawing with trigonometric function

  1. Nov 23, 2011 #1
    1. The problem statement, all variables and given/known data
    Come up with a set of equations to draw the following picture:

    http://img197.imageshack.us/img197/7428/equation.jpg [Broken]

    2. Relevant equations

    3. The attempt at a solution

    The circle is easy.

    1000 = sqrt(x^2+y^2)

    I can just model the equation with sines. So

    y_1 = 300 * Sin(pi * x / 1000) (Assuming 300 is the highest point in the curve)

    The sine graph on the y axis can be easily tilted by switching x and y. Therefore:

    x = 300 * Sin(pi*y_2 / 1000)

    sin^-1 (x/300) = pi * y_2 / 1000

    y_2 = (1000/pi) * sin^-1(x/300)

    Is my y_2 correct?

    And I'm kinda stumbling on how to translate the axis in 45 degree angle to draw the sine graph like that. Any help would be appreciated.
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Nov 23, 2011 #2


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    You should probably add a restriction for x, like -1000 ≤ x ≤ 1000.

    Technically, no, because x has the domain restriction -π/2 ≤ x ≤ π/2. Must all equations be y as functions of x? If not, I would just leave it in terms of x. Also, I think you need a negative in front of the R.H.S.:
    x = -300 * Sin(pi*y_2 / 1000)
  4. Nov 23, 2011 #3
    No it can be in terms of X too.

    How should I modelize the lines which goes at 45 degree angle?
  5. Nov 23, 2011 #4


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    Oops! The radius for that equation is about 31.6227766.

    Are you familiar with rotation of axes? You'll need to make use of it, I think.
  6. Nov 23, 2011 #5


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    No it's not,

  7. Nov 23, 2011 #6
    Yeah, translating the axes by defining a new set of axes such as x' and y', then rotating it via trigonometric function but I was struggling big time about that. Any suggestions?
  8. Nov 23, 2011 #7


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    Wow... that's was embarrassing. :redface: Let's try that again.
    This only gives you the upper semicircle. Define this implicitly instead:

    Do you know the equations that define x' and y' in terms of x, y, and θ? Take the equation
    y' = 300 * sin(pi*x'/1000)
    and replace with what x' and y' equals. Then find a new set of equations for x' and y', in terms of a different angle θ, and replace into the equation above.
    Last edited: Nov 23, 2011
  9. Nov 23, 2011 #8


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    Call your original function y = f(x). The easiest way to rotate is to use the rotation matrix:

    [tex]R(x) = \left[ \begin{array}{cc}
    \cos\theta & -\sin\theta\\
    \sin\theta & \cos\theta

    Pick any θ you wish and plot it parametrically for x from 0 to 1000.
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