# Rotate a linear function

1. Jun 26, 2013

### Niles

1. The problem statement, all variables and given/known data
I have two coordinate system $(x, y)$, $(x', y')$ that differ by a rotation around the $z$-axis by an angle $\alpha$. In the coordinate system $(x', y')$ I have a function $f(x', y') = C$, where $C$ is a constant.

I would like to express $f$ in the coordinate system $(x,y)$, where it is a linear function $x\nabla +y_0$. The gradient $\nabla$ of this function is $1/\tan(\alpha)$.

3. The attempt at a solution
I need to find the shift in $x$ now. I get that this is $C/\cos(\alpha)$. Is there a way for me to test that this function indeed is constant in the coordinate system $(x', y')$?

2. Jun 26, 2013

### SteamKing

Staff Emeritus
Draw a picture of f(x',y') = C in your x'-y' coordinate system. Now imagine that this coordinate system is rotated about the origin by an angle alpha. What happens to the line f(x',y') = C? Don't you need more than one parameter to express this line in the (x,y) system?

3. Jun 26, 2013

### HallsofIvy

Staff Emeritus
Rotation about the z-axis through an angle $\alpha$ is given by the matrix
$$\begin{bmatrix}cos(\alpha) & -sin(\alpha) & 0 \\ sin(\alpha) & cos(\alpha) & 0 \\ 0 & 0 & 1\end{bmatrix}$$