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Equation into Matrix

  1. Jul 17, 2009 #1
    I have the following equations:

    [tex]x* = C\frac{(xcos(t)+ysin(t))/cos(s)}{f-xtan(s)cos(t)-ytan(s)sin(t)}[/tex]

    [tex]y* = C\frac{-xsin(t)+ycos(t)}{f-xtan(s)cos(t)-ytan(s)sin(t)}[/tex]

    I want to transform it into a matrix so that the x and y are separated.

    I can easily separate the numerator. But I'm having trouble separating the denominator. I tried to do it using partial fraction, but I don't know what to do with the f. I got as far as taking tan(s) out.

    [tex]f-xtan(s)cos(t)-ytan(s)sin(t) = f-tan(s)(xcos(t)-ysin(t))[/tex]

    I don't even know where to start.
     
  2. jcsd
  3. Jul 17, 2009 #2

    tiny-tim

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    Hi sourlemon! :wink:

    (I haven't actually tried this, but …)

    does it help if you rotate, and put p = xcos(t) + ysin(t), q = ycos(t) - xsin(t)? :smile:
     
  4. Jul 18, 2009 #3
    thank you for your quick response tim. I don't exactly understand what you mean though. As you said, if I set the variable to p and q, I would have the following.

    [tex]x* = C\frac{p/cos(s)}{f-tan(s)(p)}[/tex]

    [tex]y* = C\frac{q}{f-tan(s)(p)}[/tex]

    But how would that help me extract x and y out of the equation with the denominator being f-tan(s)p. This is the result I want. (I don't know how to write matrix in here, so I'll write it in matlab format, hopefully you'll understand.)

    [x*; y*] = [matrix] [x; y]
     
    Last edited: Jul 18, 2009
  5. Jul 18, 2009 #4

    tiny-tim

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    Hi sourlemon! :wink:

    (again, I haven't tried it, but …)

    the next step would be finding p* and q* (and maybe tidying a bit by multiplying top and bottom by cos(s)) :smile:
     
  6. Jul 18, 2009 #5
    thank you again for your quick response tim. You don't have to worry about it working or not, I'm just happy you're pointing me to a direction. I hope you won't lose your patience with me as I'm slow with this. But can I ask what is p* and q*?

    [tex]x* = \frac{Cp}{fcos(s)-tan(s)(p)cos(s)}[/tex]

    [tex]y* = \frac{Cq}{f-tan(s)(p)}[/tex]
     
  7. Jul 19, 2009 #6

    tiny-tim

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    Sure! We defined p = xcos(t) + ysin(t), q = ycos(t) - xsin(t),

    so p* = x*cos(t) + y*sin(t), q* = y*cos(t) - x*sin(t). :smile:
     
  8. Jul 19, 2009 #7
    Thank you tim. So now do I substitute x* and y* in terms of p* and q*?

    Tim, is this a special type of substitution? Is there an example that I can see?
     
    Last edited: Jul 19, 2009
  9. Jul 19, 2009 #8

    tiny-tim

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    To get p*, just multiply the equation for x* by cos(t), and the equation for y* by sin(t), and add.

    And similarly to get q*.
     
  10. Jul 19, 2009 #9
    Thank you again for your patience, Tim :) Hopefully I followed your instruction correctly. This is what I got for p* and q*.

    [tex]p* = \frac{Cpcos(t) + Cqsin(t)cos(s)}{cos(s)(f-tan(s)(p))}[/tex]

    [tex]q* = \frac{Cqcos(t)cos(s) - CPsin(t)}{cos(s)(f-tan(s)(p))}[/tex]


    And if I multiply the cos(s)

    [tex]p* = \frac{Cpcos(t) + Cqsin(t)cos(s)}{fcos(s)-psin(s)}[/tex]

    [tex]q* = \frac{Cqcos(t)cos(s) - Cpsin(t)}{fcos(s)-psin(s)}[/tex]

    So what do I do next?
     
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