MHB Changing variables for defining a region

[Granger]

I was trying to do this exercise but my answer doesn't match with the solution and I'm wondering why:

> Consider the variables transformation defined by $x=2u+v$ and $y=u^2-v$. Being $T$ the triangle with vertices $(0,0)$, $(1,0)$, $(0,2)$ on plan $uv$ determine the image of $T$ on plan $xy$ by the variables transformation.

So what I did was take the vertices and apply the the transformation obtaining $(0,0)$ $(2,1)$ and $(2,-2)$.

Then I represented then in the plane $xy$ and I formed the triangle.

Then I defined my conditions:

- $0<x<2$

- $-x<y<\frac{x}{2}$

However the solutions say the proper answer for $y$ would be:

- $-x<y<\frac{x^2}{4}$

Ok I understand that this might be because of the square value in the change of variables... But I don't know how to get there, can someone help me?

 

[I like Serena]

I was trying to do this exercise but my answer doesn't match with the solution and I'm wondering why:

> Consider the variables transformation defined by $x=2u+v$ and $y=u^2-v$. Being $T$ the triangle with vertices $(0,0)$, $(1,0)$, $(0,2)$ on plan $uv$ determine the image of $T$ on plan $xy$ by the variables transformation.

So what I did was take the vertices and apply the the transformation obtaining $(0,0)$ $(2,1)$ and $(2,-2)$.

Then I represented then in the plane $xy$ and I formed the triangle.

Then I defined my conditions:

- $0<x<2$

- $-x<y<\frac{x}{2}$

However the solutions say the proper answer for $y$ would be:

- $-x<y<\frac{x^2}{4}$

Ok I understand that this might be because of the square value in the change of variables... But I don't know how to get there, can someone help me?

Hi GrangerObliviat! Welcome to MHB! ;)

Did you make a drawing?

The triangle has 3 edges, one of which is given by:
\begin{cases}
0<u<1 \\
v = 0
\end{cases}
Transforming to (x,y), we get:
$$\begin{cases}
x=2u+v \\
y=u^2-v \\
0<u<1 \\
v = 0
\end{cases} \Rightarrow
\begin{cases}
x=2u \\
y=u^2 \\
0<u<1 \\
\end{cases}\Rightarrow
\begin{cases}
u=\frac 12 x \\
y=(\frac 12 x)^2 \\
0<\frac 12 x<1 \\
\end{cases}\Rightarrow
\begin{cases}
y=\frac {x^2}4 \\
0< x<2 \\
\end{cases}$$
From a drawing we can tell that means that $y$ has an upper boundary of $\frac {x^2}4$.

 

[Granger]

Hi GrangerObliviat! Welcome to MHB! ;)

Did you make a drawing?

The triangle has 3 edges, one of which is given by:
\begin{cases}
0<u<1 \\
v = 0
\end{cases}
Transforming to (x,y), we get:
$$\begin{cases}
x=2u+v \\
y=u^2-v \\
0<u<1 \\
v = 0
\end{cases} \Rightarrow
\begin{cases}
x=2u \\
y=u^2 \\
0<u<1 \\
\end{cases}\Rightarrow
\begin{cases}
u=\frac 12 x \\
y=(\frac 12 x)^2 \\
0<\frac 12 x<1 \\
\end{cases}\Rightarrow
\begin{cases}
y=\frac {x^2}4 \\
0< x<2 \\
\end{cases}$$
From a drawing we can tell that means that $y$ has an upper boundary of $\frac {x^2}4$.

Thank you so much!
I didn't try to find the boundaries for u and v.