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Math Help - controversy from how to rotate curve thread

  1. #1
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    controversy from how to rotate curve thread

    i have a problem with the equations i was given for curve rotation from/about origin.In the following w1 is the angle between curve and x axis,w2 is the angle the curve is rotated by counterclockwise
    y1=rsinw1
    x1=rcosw1
    sin(w1+w2)=y2/r
    y2=rsin(w1+w2)
    cos(w1+w2)=x2/r
    x2=rcos(w1+w2)
    but
    sin(w1+w2)=sinw1cosw2+cosw1sinw2
    cos(w1+w2)=cosw1cosw2+sinw1sinw2
    therefore
    y2=rsinw1cosw2+rcosw1sinw2
    y2=y1cosw1+x1sinw2
    but i was told
    <br /> <br />
y2=y1cosw1-x1sinw2<br />
    this is quite confusing!
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  2. #2
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    Quote Originally Posted by dynamo View Post
    i have a problem with the equations i was given for curve rotation from/about origin.In the following w1 is the angle between curve and x axis,w2 is the angle the curve is rotated by counterclockwise
    y1=rsinw1
    x1=rcosw1
    sin(w1+w2)=y2/r
    y2=rsin(w1+w2)
    cos(w1+w2)=x2/r
    x2=rcos(w1+w2)
    but
    sin(w1+w2)=sinw1cosw2+cosw1sinw2
    cos(w1+w2)=cosw1cosw2sinw1sinw2 Should be a negative sign in this formula!
    therefore
    y2=rsinw1cosw2+rcosw1sinw2
    y2=y1cosw1+x1sinw2
    but i was told
    <br /> <br />
y2=y1cosw1-x1sinw2<br />
    this is quite confusing!
    .
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  3. #3
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    Quote Originally Posted by Opalg View Post
    .
    but there IS a negative sign in the formula,cos(w1+w2)=cosw1cosw2(<-here)sinw1sinw2
    should there be another?if you agree with my steps up to that point you will agree y2=rsin(w1+w2),then the negative sign in the expansion of cos(w1+w2) does not matter
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  4. #4
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    Quote Originally Posted by dynamo View Post
    i have a problem with the equations i was given for curve rotation from/about origin.In the following w1 is the angle between curve and x axis,w2 is the angle the curve is rotated by counterclockwise
    y1=rsinw1
    x1=rcosw1
    sin(w1+w2)=y2/r
    y2=rsin(w1+w2)
    cos(w1+w2)=x2/r
    x2=rcos(w1+w2)
    but
    sin(w1+w2)=sinw1cosw2+cosw1sinw2
    cos(w1+w2)=cosw1cosw2+sinw1sinw2
    This formula is wrong. It should be
    cos(w1+ w2)= cos(w1)cos(w2)- sin(w1)sin(w2)

    For example, if w1= w2= \pi/4 then cos(w1)= cos(w2)= \frac{\sqrt{2}}{2}. Using your positive sign, you would get cos(w1+ w2)= cos(\pi/2)= \sqrt{2} but, in fact, cos(\pi/2)= 0.

    therefore
    y2=rsinw1cosw2+rcosw1sinw2
    y2=y1cosw1+x1sinw2
    but i was told
    <br /> <br />
y2=y1cosw1-x1sinw2<br />
    this is quite confusing!
    Last edited by HallsofIvy; January 28th 2010 at 02:34 AM.
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    This formula is wrong. It should be
    cos(w1+ w2)= cos(w1)cos(w2)- sin(w1)sin(w2)

    For example, if w1= w2= [itex]\pi/4[/itex], then cos(w1)= cos(w2)= \frac{\sqrt{2}}{2}. Using your positive sign, you would get cos(w1+ w2)= cos(\pi/2)= \sqrt{2} but, in fact, cos(\pi/2)= 0.
    thanks for helping once again.the plus in the position of the minus was a typo
    my point is y2=rsin(w1+w2);<br />
isnt it?
    if so then there is no minus in the expression of y2 since there is no minus in the expression of sin(w1+w2).thats my problem in compressed format.
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  6. #6
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    Quote Originally Posted by dynamo View Post
    the plus in the position of the minus was a typo
    my point is y2=rsin(w1+w2);<br />
isnt it?
    if so then there is no minus in the expression of y2 since there is no minus in the expression of sin(w1+w2). thats my problem in compressed format.
    When I first read the original post in this thread, I saw the typo in the formula for \cos(\omega_1+\omega_2) and I assumed that was the cause of the error. But clearly that was not the case, because it's only the sine formula that is used in the expression of y2.
    Quote Originally Posted by dynamo View Post
    i have a problem with the equations i was given for curve rotation from/about origin.In the following w1 is the angle between curve and x axis,w2 is the angle the curve is rotated by counterclockwise
    y1=rsinw1
    x1=rcosw1
    sin(w1+w2)=y2/r
    y2=rsin(w1+w2)
    cos(w1+w2)=x2/r
    x2=rcos(w1+w2)
    but
    sin(w1+w2)=sinw1cosw2+cosw1sinw2
    cos(w1+w2)=cosw1cosw2+sinw1sinw2
    therefore
    y2=rsinw1cosw2+rcosw1sinw2
    y2=y1cosw1+x1sinw2 Should be cosw2, not cosw1.
    but i was told
    <br /> <br />
y2=y1cosw1-x1sinw2<br />
    this is quite confusing!
    It certainly is confusing, and that is because the question is confusingly worded. In fact, it asks about what happens when a curve is rotated (about the origin), but the rest of the post looks at what happens when a point is rotated.

    If the point (x_1,y_1) is rotated counterclockwise through an angle \omega then it becomes the point (x_2,y_2) = (x_1\cos\omega-y_1\sin\omega,x_1\sin\omega+y_1\cos\omega), which is essentially what you have calculated above.

    But if a curve is rotated through an angle \omega then its equation becomes transformed by replacing x by x\cos\omega + y\sin\omega, and y by -x\sin\omega+y\cos\omega, in the original equation of the curve. I guess that this is effectively because rotating the curve counterclockwise is equivalent to rotating the coordinate axes clockwise.
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  7. #7
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    Quote Originally Posted by Opalg View Post
    When I first read the original post in this thread, I saw the typo in the formula for \cos(\omega_1+\omega_2) and I assumed that was the cause of the error. But clearly that was not the case, because it's only the sine formula that is used in the expression of y2.

    It certainly is confusing, and that is because the question is confusingly worded. In fact, it asks about what happens when a curve is rotated (about the origin), but the rest of the post looks at what happens when a point is rotated.

    If the point (x_1,y_1) is rotated counterclockwise through an angle \omega then it becomes the point (x_2,y_2) = (x_1\cos\omega-y_1\sin\omega,x_1\sin\omega+y_1\cos\omega), which is essentially what you have calculated above.

    But if a curve is rotated through an angle \omega then its equation becomes transformed by replacing x by x\cos\omega + y\sin\omega, and y by -x\sin\omega+y\cos\omega, in the original equation of the curve. I guess that this is effectively because rotating the curve counterclockwise is equivalent to rotating the coordinate axes clockwise.
    thanks for replying.I now see the light.
    Last edited by dynamo; January 28th 2010 at 11:54 PM. Reason: mistake
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