Rotation of cartesian coordinate system

  • #1
xzibition8612
142
0

Homework Statement


Please see the rotation formula in the attachment.

Homework Equations





The Attempt at a Solution


I understand this formula rotates x,y into x',y' by some angle theta. Problem is, how is this formula derived? I cannot for the life of me visualize the cosine and sine transformation physically. Can someone explain to me how you get this formula. Thank you very much.
 

Attachments

  • cartesian counterclockwise rotation.doc
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Answers and Replies

  • #2
gneill
Mentor
20,947
2,892
Consider the vector that extends from the origin to the point (x,y) in the base coordinate system. It has some magnitude R and angle β with respect to the x-axis of the coordinate system. In fact, x = Rcos(β) and y = Rsin(β).

Rotating that point around the origin by some angle θ is equivalent to rotating the vector by θ, so what would the coordinates of its endpoint be?
 
  • #3
xzibition8612
142
0
so the end points would be x=Rcos(β+θ), y=Rsin(β+θ). Then what? I still can't see how this relates to the formula, espcially how in the formula for x' and y' individually there are x and y terms together.
 
  • #4
Muphrid
834
2
What you just found are the [itex]x'[/itex] and [itex]y'[/itex] coordinates. Expand the sines and cosines using angle sum formulas and put any sines or cosines of [itex]\beta[/itex] in terms of the origina [itex]x,y[/itex].
 
  • #5
xzibition8612
142
0
x' = R(cosβcosθ-sinβsinθ)
y' = R(sinβcosθ+sinβcosθ)

x' = R[(x/R)cosθ-(y/R)sinθ]
y' = R[(y/R)cosθ+(y/R)cosθ]

arrrrgh almost there. First term in y' is wrong. I get y' = ycosθ ... instead of y' = xsinθ ...


Can someone point out myt mistake? Thanks a lot for your help!
 
  • #6
Muphrid
834
2
In your second line, you forgot to switch beta and theta in the second term.
 

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