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MATLab Problem

  1. Jul 9, 2007 #1
    a = 60*pi/180;
    a1 = (pi - a)/2;
    a2 = (pi + a)/2;
    theta = a1: a/60: a2;
    rho = ones(size(theta));
    rho1 = rho*sin(a1)./sin(theta);
    polar(theta, rho);
    hold on;
    polar(theta, rho1)

    [​IMG]

    The above commands will draw a segment of a unit circle which starts from 60[tex]^{o}[/tex] to 120[tex]^{o}[/tex].

    I know everything except the line of "rho1 = rho*sin(a1)./sin(theta);" that I don't know why this equation works from 1 --> 0.866 --> 1 and becomes a straight line. Can anyone tell me the reason? Is it related to a similar triangle or inverse proportion?
     
  2. jcsd
  3. Jul 9, 2007 #2
    try converting from polar to euclidean/cartesian...to see if you can figure out how they got that straight line. Remember what does rho,rho1 stand for in polar.
     
  4. Jul 9, 2007 #3
    Cartesian to Polar:
    r = sqrt(x^2 + y^2)
    theta = arctan(y/x)

    Polar to Cartesian:
    x = r cos theta
    y = r sin theta

    rho = 1 (for unit circle)

    rho1 = rho*sin(a1)/sin(theta) = 1*sin(60)/sin(60) to 1*sin(60)/sin(120) = 1 --> 0.866 --> 1
     
    Last edited: Jul 9, 2007
  5. Jul 23, 2007 #4
    Write the equation of a rational function that has vertical asymptotes at x=2 and x=3 and a horizontal asymptote at y=2

    yeah, i really need help with this problem. Thanks so much
     
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