<s>[INST] why is the gradient vector to rotated ellipse wrong? A formula for the rotated ellipse can be found by taking the standard ellipse
$$
\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \qquad\qquad\qquad\qquad {\bf{x}}(t) = (a\cos t, b\sin t) \qquad \text{ for } t\in [0, 2\pi]
$$
and rotating it counter-clockwise using a rotation matrix
$$
R{\bf{x}}(t) = 
\begin{pmatrix}
    \cos\theta & -\sin\theta \\
    \sin\theta & \cos\theta
\end{pmatrix}
\begin{pmatrix}
    a\cos t \\
    b\sin t
\end{pmatrix}
=
\begin{pmatrix}
    a\cos t\cos\theta - b\sin t\sin \theta\\
    a\cos t\sin \theta + b \sin t \cos \theta
\end{pmatrix} 
=
{\bf{x}}_r(t)
$$
I can easily find the gradient of the standard ellipse (non-rotated) by taking the derivative of the first formula.
$$
g = 
\begin{pmatrix}
    \frac{2x}{a^2},  &
    \frac{2y}{b^2}
\end{pmatrix}^\top
$$
Now I tried to methods to compute the gradient at a point on the rotated ellipse.

*

*Method 1: I take the gradient of
$$
\frac{(x\cos\theta - y\sin\theta)^2}{a^2} + \frac{(x\sin\theta + y\cos\theta)^2}{b^2} = 1
$$
obtaining
$$
g_{r1} = 
\begin{pmatrix}
    \frac{2(x\cos\theta - y\sin\theta)\cos\theta}{a^2} + \frac{2(x\sin\theta + y\cos\theta)\sin\theta}{b^2} \\
    \frac{-2(x\cos\theta - y\sin\theta)\sin\theta}{a^2} + \frac{2(x\sin\theta + y\cos\theta)\cos\theta}{b^2}
\end{pmatrix}
$$


*Method 2: Rotating the gradient of the non-rotated ellipse counterclockwise
$$
g_{r2} = \begin{pmatrix}
    \cos\theta & -\sin\theta \\
    \sin\theta & \cos\theta
\end{pmatrix}
\begin{pmatrix}
    \frac{2x}{a^2} \\
    \frac{2y}{b^2}
\end{pmatrix}
 = 
\begin{pmatrix}
    \frac{2x\cos\theta}{a^2} - \frac{2y\sin\theta}{b^2}\\
    \frac{2x\sin\theta}{a^2} + \frac{2y\cos\theta}{b^2}
\end{pmatrix}
$$
To me, method 1 should work while method 2 should be wrong. However, as you can see in the plot below, $g_{r1}$ seems to actually be totally wrong while $g_{r2}$ is right. How come $g_{r2}$ seems to be right but not $g_{r1}$?

 [/INST] Method 1 is wrong because the equation
$$
\frac{(x\cos\theta - y\sin\theta)^2}{a^2} + \frac{(x\sin\theta + y\cos\theta)^2}{b^2} = 1
$$
expresses the fact that the result of rotating the point $(x,y)$ belongs to the ellipse $x^2/a^2+y^2/b^2=1$, but this is not what you need.
</s>