Rotate a 2D function I have a 2D function, which traces out an ellipse, 
$f(x, y) = \exp(-x^2)\exp(-(y-1)^2/0.5)$
I would like to rotate it around the point $(x, y)=(0,1)$. For this I guess I need the rotation matrix, but I am unsure of how to use this matrix with my specific example. Can I get a hint to how I should do this?
 A: I will translate my comments into an answer.
The function $f(x,y)=c$ can be rotated about the point $(0,1)$ by considering the change of coordinates
$$
\begin{bmatrix}x'\\y'\end{bmatrix}=\begin{bmatrix}0\\1\end{bmatrix}+\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}\left(\begin{bmatrix}x\\y\end{bmatrix}-\begin{bmatrix}0\\1\end{bmatrix}\right)=\begin{bmatrix}x\cos\theta-(y-1)\sin\theta\\1+x\sin\theta+(y-1)\cos\theta\end{bmatrix}.
$$
The relation $f(x',y')=c$ is a rotation of the original function about the point $(0,1)$ anticlockwise by an angle of $\theta$.
You can also analyse the shape of the ellipse. Taking logs of both sides, we obtain
$$-x^2-\frac{(y-1)^2}{0.5}=\ln c.$$ The LHE is clearly nonpositive, which requires $c\leq1$. Furthermore, we require the restriction $c>0$ for $\ln c$ to be fined on $\mathbb R$. Multiplying both sides by $-1$ and dividing both sides by $\ln (1/c)$, we obtain the standard form for an ellipse
$$\frac{x^2}{\ln(1/c)}+\frac{(y-1)^2}{0.5\ln(1/c)}=1.$$
Thus, the ellipse has a major axis in the $x$ (or $x'$) direction of length $\ln(1/c)$ and a minor axis in the $y$ (or $y'$) direction of length $\dfrac{\ln(1/c)}{2}$.
