Interesting Problem I Found (a set obtained by asymmetric scaling)

[MidgetDwarf]

Homework Statement:

Let $$ A = { (x,y) : x^2 -y^3 +4y \leq 1}$$ . Now let B be the set obtained asymmetrically scaling A by a factor of 3 in the x-direction, and by a factor of 4 in the y-direction, then translating the scale set by the vector <-2,5>. Use a graphing tool to show A and B.

Relevant Equations:

$$ h(x,y) = x^2 -y^3 +4y \leq 1$$

What does asymmetrically scaling mean?

I am unsure of how to do this problem. I am thinking $$ h(3x,4y) = 9x^2 \ - \ 64y^3 \ + \ 16y \leq 1$$, which gives us the scaling we want in the x-direction and y-direction.

Now, for the translation of points of A by the vector <-2, 5>. I am unsure.

I am a bit embarrassed posting this. It appears that this problem is just a linear transformation problem.

This problem is for fun, and not a hw exercise.

I have attached a graph of the set A.

Here we go!

 

[Mark44]

Homework Statement:: Let $$ A = { (x,y) : x^2 -y^3 +4y \leq 1}$$ . Now let B be the set obtained asymmetrically scaling A by a factor of 3 in the x-direction, and by a factor of 4 in the y-direction, then translating the scale set by the vector <-2,5>. Use a graphing tool to show A and B.
Relevant Equations:: $$ h(x,y) = x^2 -y^3 +4y \leq 1$$

What does asymmetrically scaling mean?

Just as a guess, I think it refers to the different scaling in the x and y directions.

 

[haruspex]

which gives us the scaling we want in the x-direction and y-direction.

Amplifying the coefficients reduces the x and y values satisfying the equation. That will shrink the area.
For the translation, you want the bit of the graph that was at x=0 to be at x=-2 now. What do you need to substitute for x in the equation ?

 

[MidgetDwarf]

Thanks for quick response.

So I am thinking of replacing x with x+2 and y with y - 5 into the function $h(3x, 4y) \ = \ 9x^2 \ - \ 64y^3 \ + \ 16y \leq 1$.

Let $k(x,y) = h(3x, 4y) \ = \ 9x^2 \ - \ 64y^3 \ + \ 16y \leq 1$.

So $k(x+2,y-5) = \ 9(x+2)^2 \ - \ 64(y-5)^3 \ + \ 16(y-5) \leq 1$.

I uploaded an image of graph using Desmos. The red graph is the initial set, green graph is the scaled set, and the purple graph is the translated set. Is this correct?

 

[MidgetDwarf]

 

[valenumr]

Isn't this basic linear algebra? B is just Ax + c, with x being a matrix transform and c is a vector offset.

 

[haruspex]

Thanks for quick response.

So I am thinking of replacing x with x+2 and y with y - 5 into the function $h(3x, 4y) \ = \ 9x^2 \ - \ 64y^3 \ + \ 16y \leq 1$.

Let $k(x,y) = h(3x, 4y) \ = \ 9x^2 \ - \ 64y^3 \ + \ 16y \leq 1$.

So $k(x+2,y-5) = \ 9(x+2)^2 \ - \ 64(y-5)^3 \ + \ 16(y-5) \leq 1$.

I uploaded an image of graph using Desmos. The red graph is the initial set, green graph is the scaled set, and the purple graph is the translated set. Is this correct?

You do not seem to have understood the first paragraph in my post #3. Looks to me like you have scaled it by 1/4 in the y direction.

 

[MidgetDwarf]

You do not seem to have understood the first paragraph in my post #3. Looks to me like you have scaled it by 1/4 in the y direction.

Thanks for the reply. Yes, I reread your post #3, and it still is not clear to me.

Since I scaled the set by 1/4 in the y-direction. I am thinking that instead of plugging in 4y into $h(x,y) = x^2 \ - \ y^3 \ + 4y \ \leq 1$ , we plug $\frac {y} {4}$ . Otherwise, was the scaling in the x-direction, and translation ok?

 

[haruspex]

was the scaling in the x-direction, and translation ok?

Obviously the same principle applies in the x direction. To broaden the set by a factor of 3 you must replace x by x/3.
You got the principle right in the translation, though: to move it left by 2 you replace x with x+2, not x-2.

 

[valenumr]

If you consider the set of 3-tuples (x,y,z) as S = { x -> x, y->y, z->f(x, y) } and then apply a map such that S' = { x' -> 3x - 2, y' -> 4x + 5, z->f(x,y) == g(x', y') }, it should be obvious that you need to find a function, g(x', y') that is equal to f(x, y) for all values of z. In other words, if Z = f(1, 1), Z1' = g(1, 9) should have the same value. I'm pretty sure if you just solve x for x' and y for y' it works out.

 

[valenumr]

If you consider the set of 3-tuples (x,y,z) as S = { x -> x, y->y, z->f(x, y) } and then apply a map such that S' = { x' -> 3x - 2, y' -> 4x + 5, z->f(x,y) == g(x', y') }, it should be obvious that you need to find a function, g(x', y') that is equal to f(x, y) for all values of z. In other words, if Z = f(1, 1), Z1' = g(1, 9) should have the same value. I'm pretty sure if you just solve x for x' and y for y' it works out.

That's a little roundabout way of saying that this is a coordinate transform. If you look at the example of z = g(1,9), you can see that g(1,9) is the same as f(h(1), h(9)), where h is a 2d map. Plug any two numbers into h and you will get back to the original coordinates back.