First-order equations by introducing new coordinates.

**dwsmith** · January 7th 2011, 12:28 PM

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

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?

**snowtea** · January 7th 2011, 12:46 PM

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.

**dwsmith** · January 7th 2011, 01:46 PM

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.

**snowtea** · January 7th 2011, 02:00 PM

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]

**dwsmith** · January 7th 2011, 02:19 PM

Originally Posted by snowtea

$(e_x\cdot e_\xi)= \cos{\alpha}$ .

Why is that equal to cosine alpha?

**snowtea** · January 7th 2011, 02:21 PM

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

**dwsmith** · January 7th 2011, 02:24 PM

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?

**snowtea** · January 7th 2011, 02:36 PM

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

Thread: First-order equations by introducing new coordinates.

LinkBack

Thread Tools

Display

First-order equations by introducing new coordinates.

Similar Math Help Forum Discussions

Introducing myself.

Decomposition of 2 fourth order differential equations on 4 equations of second order

Writing two Second-Order equations as a System of First-Order Equations

DE Tutorial - Part I: First Order Equations and Homogeneous Second Order Equations

Higher-order differential equations, the Laplace transform and exponential order