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

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?

• Yes, sorry that was just a typo – joseph Apr 19 at 3:03

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.

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$$\int_{4c}^{4d}\int_{2a}^{2b}2\,\text{d}u\,\text{d}v=16(d-c)(b-a).$$

• How is the absolute value of the determinant of the Jacobian Matrix $2$? Wouldn't it be $(1)(-2)-(2)(1)=-4$ so the absolute value is $4$? – joseph Apr 19 at 20:45
• Oh I see you had the equations wrong but I understand what you did... thank you for the help – joseph Apr 19 at 20:46