How to apply coordinate transformations

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Lets say I want to rotate a parabola by $\pi/4$ degrees counterclockwise.
Wikipedia tells me a counterclockwise transformation would mean: $$ x'=x\cos t-y\sin t \\ y'=x\sin t+y\cos t $$ however when I substitute these new expressions into $y=x^2$, the graph appears rotated clockwise
$$ (x\sin(\pi/4)+y\cos(\pi/4)) = (x\cos(\pi/4)-y\sin(\pi/4))^2 $$ (try it in wolframalpha)

why is it working in reverse, how should these transformations work?


to be more specific regarding my problem I need to do a coordinate transformation to model a distortion in an image but I do not understand what the expression means.

if I wanted to apply the transformation $x' =$ sign$(x)x^2$ where $x'$ corresponds to the "image plane", what would I get?

my intuition tells me that if I had a point at $(0.5,0.5)$, it would be pulled into $(0.25,0.25)$ but this website tells me the opposite

http://paulbourke.net/miscellaneous/imagewarp/

I am aware of $r_u=r_d+kr_d$ 3 transformations, I want to use $x^2$

but if I substitute $x^2$ and $y^2$ into the equation of a circle, I see that the circle looks inflated $$ [\mbox{sign}(x)x^2]^2+[\mbox{sign}(y)y^2]^2=1 $$

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Your substitution is not rotating $y=x^2$ -- on the contrary you're starting with $y'=x'^2$ and rewriting that to express the same points in the $xy$ coordinate system.

With the transformation you're quoting, the $xy$ coordinate axes are rotated an angle of $t$ clockwise from the $x'y'$ coordinate axes.

Thus when you take a parabola with its axis along the $y'$ axis and "turn the paper" such that the $y$ axis points due north, the $y'$ axis will now point to the northeast.

If you want to rotate an image counterclockwise, then you need its original coordinates in the $xy$ coordinate system, and find a new description for it in the $x'y'$ system. That means that you need to invert the transformation to express $x$ and $y$ as functions of $x'$ and $y'$ -- or just negate the angle you rotate through.


Expressed in other words, there's a difference between rotating a point and rotating an equation. If your transformation is $T$, and $\phi(v)$ is an equation whose solution set you want to rotate, then you are after the set $$ \{ T(v) \mid \phi(v) \} $$ In order to get an equation for the rotated set, you can change the variable to $u=T(v)$, and then you get $$ \{ u \mid \phi(T^{-1}(u)) \} $$ So you construct an equation for the transformed set by composing the existing equation with the inverse of the transformation.