1

Let $C$ be a curve on the $xy$-plane described by the equation $x^2 + 4y^2 = 16$. If every point $(x, y)$ on $C$ is replaced by the point $\left(\frac{1}{2}x, y\right)$, what is the area enclosed by the resulting curve?

Of course, my initial reaction is to just find the area of the shape described by $\left(\frac{1}{2}x\right)^2 + 4y^2 = 16~(\text{eqn. 1)}$, but what I really should be doing is finding the area of the shape described by $(2x)^2 + 4y^2 = 16$. What is the intuition behind this so that it feels right? What am I doing to the graph by replacing $x$ with $\frac{1}{2}x$, and why is this not the same as "replacing $(x, y)$ by $(\frac{1}{2}x, y)$?

I understand that, if $(x, y) \mapsto (x', y')$, where $x' = \frac{1}{2}x$ and $y' = y$, then the equation becomes $(2x')^2 + 4y'^2 = 16$—and that makes perfect sense to me. But why is doing this the same thing as "replacing $(x, y)$ with $(\frac{1}{2}x, y)$"? There's some weird geometric disconnect that I'm just not following here.

In fact, $(\text{eqn. 1})$ has a larger area than the correct equation—which obviously shouldn't be the case if $x$ were replaced by $\frac{1}{2}x$, so that follows intuition pretty well. Am I dwelling on a semantic issue? The idea of "replacing $(x, y)$ by $\left(\frac{1}{2}x, y\right)$ and not being able to do literally that (in the equation) just doesn't jive well with me, and I'm not sure why.

(I feel as if: "You need twice as much $x$ when $x \mapsto \frac{1}{2}x$" is close. It still doesn't feel right with the equation, though. :l )

AmagicalFishy
  • 1,990

4 Answers4

3

I will do a similar example, being as informal as possible. Take the circle $x^2+y^2=1$. When you do $(x,y) \mapsto (x,y/2)$ you are shrinking the coordinates toward to $x$ axis. And numerically you obtain $x^2+(2y)^2=1$, and not $x^2+(y/2)^2=1$. Why?

Because it is as if you have a new $y$-axis where $0,1,2,3,\ldots$ is replaced by $0,\frac{1}{2},1,\frac{3}{2},\ldots$. This means that if you want to write the new curve with the old coordinates you have to multiply by $2$ the new $y$.

Paolo Leonetti
  • 15,554
  • 3
  • 26
  • 60
2

I think that this figure can help the intuition.

enter image description here

The ellipse is the curve $x^2+4y^2=16$ and the circle is the curve after the contraction $x \to \frac{x}{2}$.

Emilio Novati
  • 64,377
1

The ellipse is stretched to become a circle. The area of course is different. If you want to keep the same area then you have to stretch x and y in a way that makes the ellipse become an equivalent circle.

This can be done, actually, by setting $$x\to \frac{x}{\sqrt b},y\to \frac{y}{\sqrt a}\quad(*)$$ into the equation of the ellipse $$a^2 y^2+b^2 x^2=a^2b^2$$ getting the circle $$x^2+y^2=ab$$ whose area is $\pi ab$, which is the area of the ellipse having semi axis $a$ and $b$.

It must be explained the reason why a transformation like the one I defined in $(*)$ keeps the area unchanged.

Try to apply it to the rectangle having vertices $$(0;\;0),\;(a;\;0),\;(a;\;b),\;(0;\;b)$$

Raffaele
  • 26,731
1

Only through non-dimensional coordinate relations should the scaling be done

$$x^2 + 4y^2 = 16$$

is unsuitable to do that. To do this express it in the form

$$(x/4)^2 +(y/2)^2 =1$$

Halving major axis results in

$$(x/2)^2 +(y/2)^2 =1,$$

a circle of radius $2$ centered at origin.

Narasimham
  • 42,260