# Thread: First-order equations by introducing new coordinates.

1. ## First-order equations by introducing new coordinates.

$3u_x+4u_y-2u=1\Rightarrow\omega_{\xi}+k\omega=\varphi(\xi,\e ta)$

The book states, "If the $(\xi,\eta)$-axes are obtained from $(x,y)$-axes by rotating through an angle $\alpha$, then $(\xi,\eta)$ and $(x,y)$ are related by either of the pair of equations:

$\xi=x\cos{\alpha}+y\sin{\alpha}, \ \ x=\xi\cos{\alpha}-\eta\sin{\alpha},$

$\eta=-x\sin{\alpha}+y\cos{\alpha}, \ \ y=\xi\sin{\alpha}+\eta\sin{\alpha}.\mbox{"}$

How did the book obtain those equations?

2. Draw the orthogonal axes for $x,y$, and the rotated axes of $\xi, \eta$ by $\alpha$.
The equations are obtained by simple trig and vector addition.

3. Originally Posted by snowtea
Draw the orthogonal axes for $x,y$, and the rotated axes of $\xi, \eta$ by $\alpha$.
The equations are obtained by simple trig and vector addition.
I did that, but unfortunately, I still don't see how it works.

4. A drawing normally explains this very easily, but since I suck at drawing (and cannot seem to find a picture online for this common transformation), I will try to explain it analytically.

Any axis can be defined by a unit vector parallel to it in the postive direction.

So let $e_x,e_y,e_\xi,e_\eta$ be the unit vectors parallel to the corresponding axes.

A point in the $x,y$ coordinate system can be represented by the vector $xe_x + ye_y$.

For any vector $\vec{v}$, its representation in the $\xi, \eta$ (orthogonal) coordinate system is:
$(\vec{v}\cdot e_\xi)e_\xi + (\vec{v}\cdot e_\eta)e_\eta$.

So the $\xi$ coordinate is given by $\vec{v}\cdot e_\xi$.

So for $xe_x + ye_y$, this is $x(e_x\cdot e_\xi) + y(e_y \cdot e_\xi) = x\cos{\alpha}+y\sin{\alpha}$.

Similarly for $\eta$...

[Edit: Found an image here: http://www.tutornext.com/system/file...Fig.1.36_0.GIF,
perhaps this will help with intuition using simple trigonometry]

5. Originally Posted by snowtea

$(e_x\cdot e_\xi)= \cos{\alpha}$.
Why is that equal to cosine alpha?

6. $e_x, e_\xi$ are unit vectors parallel to the axes. Their dot product is cosine of the angle between them (they have unit length).

7. Ok that makes since since $\displaystyle\cos{\alpha}=\frac{u\cdot v}{||u|| \ ||v||}$ but I have never seen sine defined in this manner so how does sine come into play?

8. $\cos(\frac{\pi}{2} - \alpha) = \sin(\alpha)$