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Topic: Cartesian  (Read 5068 times)

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missybangsalot

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Cartesian
« on: October 13, 2005, 08:57:06 PM »
Y=r²cos²q + r sinq cosf tanf- (r sinq sinf/ tanf) is the problem given.  I needed to change from sperical to Cartesian when I worked I thought it was to easy to be right I ended up getting


Y=z² + x (y/x) - (y/(y/x))
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Offline Donaldson Tan

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Re:Cartesian
« Reply #1 on: November 10, 2005, 09:47:24 AM »
do you realise that y/x , y/z or x/z are actually trigonometric functions?

they correspond to either sine, cosine, tangent FYI.
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