Rotate a curve by an angle I’m trying to find an equation of a rotated curve about the origin and I’m confused about an issue, hoping for some help.
So to rotate a point $(x,y)$ about the origin counter clockwise by theta, we use the following mapping
$X\to \cos(\theta)-y\sin(\theta)$
$Y\to x\sin(\theta)+y\cos(\theta)$
For example if I want to rotate by 90 degrees counterclockwise then the mapping becomes:
$(x,y)\to (-y,x)$
Now, if I have the curve $y= x^2+3x$ and If I want to rotate it by 90 degrees counter clockwise with respect to the origin, I replace x with -y, and y with x. The equation of the rotated curve becomes:
$X=y^2-3y$
Now if I graph the rotated curve, I see that its rotated by 90 degrees clockwise, and not counterclockwise!!where is my mistake?
 A: You have to plug in the inverse of your coordinate transformation. The reason is that if $(x_0,y_0)$ lies on the original curve, then you want $(-y_0,x_0)$ to lie on the new curve. $(y,-x)$ is the inverse transformation, so we get the equation $-x=y^2+3y$. Now if you plug in $(-y_0,x_0)$ you get $y_0=x_0^2+3x_0$ which is true because $(x_0,y_0)$ belongs to the first curve.
The idea of plugging in the inverse transformation is that when you then plug in your point, it now cancels out with the inverse and you get your old curve back.
A: Suppose the original curve is given by $f(x,y) = 0$.  The image a point $(x, y)$ on this curve is $(x', y') = (x \cos \theta - y \sin \theta, x \sin \theta + y \cos \theta)$.  Since we want the relation between $x'$, and $y'$ and we only have the relation between $x$ and $y$, we have write $x$ and $y$ in terms of $x'$ and $y'$, then plug these expressions into $f(x,y)=0$.  From the above expressions, it follows that,
$ x = \cos \theta  x' + \sin \theta y' $
and
$ y = -\sin \theta x' + \cos \theta y' $
And then just plug in these expressions into $f(x,y) = 0$ to get the relation between $x'$ and $y'$.  So with your example, $y = x^2 + 3 x$, and $\theta = 90^\circ$, hence
$x = 0 \cdot x' + 1 \cdot y' = y'$ and $ y = -1 \cdot x' + 0 \cdot y' = -x' $
Plug these in, you get
$ - x' = y'^2 + 3 y' $
so that
$ x' = - y'^2 - 3y' $
Finally, replacing $x'$ and $y'$ with the standard names of the variables
we get $ x = - y^2 - 3 y $  as the equation of the rotated curve.
