Finding what a line does under the function $\mathbf f:\mathbf R^2\rightarrow\mathbf R^3,\mathbf f(x,y)=(x+2y,x-2y) $

Question

Let $\mathbf f:\mathbf R^2\rightarrow\mathbf R^2,\mathbf f(x,y)=(x+2y,x-2y) $. I'm trying to estimate the area of a rectangle under $\mathbf f$ with the points $$P_1(1,1),\ P_2(1+\Delta x,1), \ P_3(1+\Delta x,1+\Delta y),\ P_4(1,1+\Delta y)$$ However, I first have to sketch what the rectangle looks like under $\mathbf f$. I found what the image of the points would be with $Q_i=\mathbf f(P_i)$: $$ Q_1(3,-1),\ Q_2(3+\Delta x,\Delta x-1), \ Q_3(3+\Delta x+2\Delta y,\Delta x-2\Delta y-1),\ Q_4(3+2\Delta y,-1-2\Delta y) $$ My question is how do I find what the lines connecting the points look like under this function. Specifically, how do I find the equations of them?

Answer 1

It appears you are asking what is the image of a straight line $Ax+By+C=0$ under the transformation $f(x,y)=(x+2y,x-2y)$.

Simplify $A(x+2y)+B(x-2y)+C=0$ and you will have your answer.

Then you will see that image of a rectangle under the transformation must be a quadrilateral.

Then take a look at the dot-product of pairs of intersecting sides of the transformed rectangle, for example, $(Q_2-Q_1)\cdot(Q_3-Q_2)$

\begin{eqnarray} (Q_2-Q_1)\cdot(Q_3-Q_2)&=&(\Delta x,\Delta x)\cdot(2\Delta y,-2\Delta y)\\ &=&0 \end{eqnarray}

The intersecting sides will intersect at right angles. So the transformed rectangle is a rectangle.

Answer 2

Let $\omega$ be a parametrization of the boundary of the rectangle $[a,b]\times[c,d]$.

It is piecewise linear. Grab one of its pieces, say $\omega_1(t)=(t,c)$ as $t$ ranges from $a$ to $b$.

The image of $\omega_1$ under $f$ is the image of $f(\omega_1(t))=(t+2c,t-2c)=(t,t)+(2c,-2c)$, which has an equation of $y=x-4c$ as $x$ ranges from $a+2c$ to $b+2c$. You can get an idea for what that looks like. Similarly, you can find that$$\begin{align}x-y&=4c,\\x-y&=4d,\\x+y&=2a,\text{ and}\\x+y&=2b\end{align}$$are the equations of the four lines.

Note that the absolute value of the determinant of the Jacobian matrix of $(x,y)\mapsto(x-y,x+y)$ is $2$. That is the constant of proportionality of how an infinitesimal unit of area in the plane changes under this map. Therefore, the area of $[a,b]\times[c,d]$ under $f$ is$$\int_{4c}^{4d}\int_{2a}^{2b}2\,\text{d}u\,\text{d}v=16(d-c)(b-a).$$