# controversy from how to rotate curve thread

• Jan 27th 2010, 11:51 AM
dynamo
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
$

y2=y1cosw1-x1sinw2
$

this is quite confusing!
• Jan 27th 2010, 12:22 PM
Opalg
Quote:

Originally Posted by dynamo
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
therefore
y2=rsinw1cosw2+rcosw1sinw2
y2=y1cosw1+x1sinw2
but i was told
$

y2=y1cosw1-x1sinw2
$

this is quite confusing!

.
• Jan 27th 2010, 08:49 PM
dynamo
Quote:

Originally Posted by Opalg
.

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
• Jan 27th 2010, 10:51 PM
HallsofIvy
Quote:

Originally Posted by dynamo
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, $cos(\pi/2)= 0$.

Quote:

therefore
y2=rsinw1cosw2+rcosw1sinw2
y2=y1cosw1+x1sinw2
but i was told
$

y2=y1cosw1-x1sinw2
$

this is quite confusing!
• Jan 28th 2010, 12:24 AM
dynamo
Quote:

Originally Posted by HallsofIvy
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, $cos(\pi/2)= 0$.

thanks for helping once again.the plus in the position of the minus was a typo
my point is $y2=rsin(w1+w2);
$
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.
• Jan 28th 2010, 07:42 AM
Opalg
Quote:

Originally Posted by dynamo
the plus in the position of the minus was a typo
my point is $y2=rsin(w1+w2);
$
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
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
$

y2=y1cosw1-x1sinw2
$

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.
• Jan 28th 2010, 11:50 PM
dynamo
Quote:

Originally Posted by Opalg
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.