# Chaitanya's Random Pages

## January 26, 2015

### Affine Transformations of Cartesian Coordinates

Filed under: mathematics — ckrao @ 11:43 am

A very common exercise in high school mathematics is to plot transformations of some standard functions. For example, to plot $y = 2\sin (5x + \frac{2\pi}{3}) - 1$ we may start with a standard sine curve and apply the following transformations in turn:

• squeeze it by a factor of 5 in the $x$-direction
• shift it left by $\frac{2\pi}{3}$
• stretch it by a factor of 2 in the $y$-direction
• shift it down by $1$

This leads to the plot shown.

For sine and cosine graphs an alternative is to plot successive peaks/troughs of the curve and interpolate accordingly. For example, to plot $y = 2\sin (5x + \frac{2\pi}{3}) - 1$ we may proceed as follows.

• Since $\sin(x)$ has a peak at $\frac{\pi}{2}$, solve $5x + \frac{2\pi}{3} = \frac{\pi}{2}$ to find $x = -\frac{\pi}{30}$ as a point where there is a peak at $y = 2\times 1 - 1 = 1$. Hence plot the point $(-\frac{\pi}{30},1)$.
• Since the angular frequency is 5, the period is $\frac{2\pi}{5}$ and we may plot successive peaks spaced $\frac{2\pi}{5}$ apart from the point $(-\frac{\pi}{30},1)$.
• Troughs will be equally spaced halfway between the peaks at $y = 2\times (-1) - 1 = -3$ (at $x = -\frac{\pi}{30} + \frac{\pi}{5} + k\frac{2\pi}{5}$). Then join the dots with a sinusoidal curve.
• Additionally $x-$ and $y-$ intercepts may be found by setting $y = 0$ and $x = 0$ respectively. We find that the $x-$intercepts are at $x = \frac{2\pi k}{5} - \frac{\pi}{10}, \frac{2\pi k}{5} + \frac{\pi}{30}\ (k \in \mathbb{Z})$ and $y-$intercept is $y = 2\sin (\frac{2\pi}{3}) - 1 = \sqrt{3}-1$.

The first approach is more generalisable to plotting other functions. Instead of thinking of the graph transforming, we also may consider it as a change of coordinates. For example, if we translate the parabola $y = x^2$ so that its turning point is at $(2,1)$, this is equivalent to keeping the parabola fixed and shifting axes so that the new origin is at $(-2,-1)$ with respect to the old coordinates. This is illustrated below where the black coordinates are modified to the red ones. The parabola has equation $y = x^2$ under the black coordinates and $y - 1 = (x-2)^2$ or $y = (x-2)^2 + 1$ under the red coordinates.

As another example suppose we take a unit circle and stretch it by a factor of $2$ in the $x$ direction and a factor of $3$ in the $y$ direction. This is the equivalent of changing scale so that the $x$-axis is squeezed by $2$ and the $y$-axis is squeezed by $3$.

Under this stretching of the circle or squeezing of axes, the unit circle equation $x^2 + y^2 = 1$ transforms to that of the ellipse $\left(\frac{x}{2}\right)^2 + \left(\frac{y}{3}\right)^2 = 1$.

More generally, by stretching a Cartesian graph $f(x,y) = 0$ by $a$ in the $x$-direction and $b$ in the $y$-direction, then shifting it along the vector $(h,k)$, we obtain the equation

$\displaystyle f\left(\frac{x-h}{a}, \frac{y-k}{b}\right) = 0.$

This uses the fact that $\frac{x-h}{a}$ and $\frac{y-k}{b}$ are the inverses of $ax+h$ and $by+k$ respectively. Note that if $|a|$ or $|b|$ are less than 1, the stretch becomes a squeezing of the graph, while $a < 0$ or $b < 0$ correspond to a reflection in the $x$ or $y$ axes.

We can extend this idea to the rotation of a graph. Suppose for example we wish to rotate the hyperbola $y = 1/x$ by 45 degrees anti-clockwise. This is equivalent to a rotation of the axes by 45 degrees clockwise and the matrix corresponding to this linear transform is

$\displaystyle \left[ \begin{array} {c} x' \\ y' \end{array} \right] = \frac{1}{\sqrt{2}}\left[ \begin{array}{rr} 1 & 1\\ -1 & 1 \end{array} \right] \left[ \begin{array} {c} x \\ y \end{array} \right] = \left[ \begin{array} {c} \frac{x+y}{\sqrt{2}} \\ \frac{-x+y}{\sqrt{2}}\end{array} \right].$

(Here the columns of the change of basis matrix correspond to where the basis vectors (1,0) and (0,1) map to under a 45 degree clockwise rotation.)

In other words we replace $x'$ with $(x+y)/\sqrt{2}$ and $y'$ with $(-x+y)/\sqrt{2}$ in the equation $y' = 1/x'$ and obtain $(-x+y)/\sqrt{2} = 1/( (x+y)/\sqrt{2})$ or $y^2 - x^2 = 2$.

Here is the same transformation applied to the parabola $y = x^2$ to obtain $\frac{-x+y}{\sqrt{2}} = \frac{(x+y)^2}{2}$ or $x^2 + y^2 + 2xy + \sqrt{2}(x-y) = 0$:

If a graph is affinely transformed (by an invertible map) so that $(1,0)$ maps to $(a,b)$ and $(0,1)$ maps to $(c,d)$ followed by a shift along the vector $(h,k)$, then this is equivalent to the coordinates shifting by $(-h,-k)$ and then transforming under the inverse mapping $\displaystyle \left[ \begin{array}{cc} a & c\\ b & d\end{array} \right]^{-1} = \frac{1}{ad-bc} \left[ \begin{array}{cc} d & -c\\ -b & a\end{array} \right]$:

\displaystyle \boxed{\begin{aligned} x' &= (x-h) \frac{d}{ad-bc} - (y-k) \frac{c}{ad-bd}\\ y' &= -(x-h) \frac{b}{ad-bc} + (y-k) \frac{a}{ad-bd}\end{aligned}}

Here are some special cases of this formula:

• rotation of the graph by $\theta$ anti-clockwise: $\displaystyle x' = x\cos \theta + y \sin \theta, y' = - x \sin \theta + y\cos \theta$ (the example $\theta = \pi/4$ was done above)
• reflection of the graph in $y = x$: $x' = y, y' = x$
• reflection of the graph in $y = mx$ where $m = \tan \theta$: verify that $a = \cos 2\theta, b = \sin 2\theta, c = \sin 2\theta, d = -\cos 2\theta$ so $\displaystyle x' = x \cos 2\theta + y \sin 2\theta, y' = x \sin 2\theta - y \cos 2\theta$
• reflection of the graph in $y = mx + c$ where $m = \tan \theta$: this is equivalent to a reflection in the line $y = mx$ followed by a shift along the vector $(h,k) = (-c \sin 2\theta, c (1+\cos 2 \theta) )$ so
\displaystyle \begin{aligned} x' &= (x+c \sin 2\theta)\cos 2\theta + (y-c(1+\cos 2\theta))\sin 2\theta\\ &= x \cos 2\theta + (y-c)\sin 2\theta,\\ y' &= (x+c \sin 2\theta)\sin 2\theta - (y-c(1+\cos 2\theta))\cos 2\theta\\ &= x \sin 2\theta - (y-c) \cos 2\theta + c \end{aligned}
• reflection of the graph in $y = x + c$ (special instance of the previous case with $\sin 2\theta = 1, \cos 2\theta = 0$): $x'= y - c, y' = x+c$