How did they get these 3 boundaries?

[unscientific]

Homework Statement

The question is attached in the picture.

The Attempt at a Solution

From the question I deduced that there are 3 boundaries: red, green and blue.

Blue Boundary

This one is the easiest:

That part of the curve of y² = 4a(a-x), which corresponds to u = a.


Red Boundary

This corresponds to x = 0, y > 0.
When x = 0,

y² = 4u² = 4v², which corresponds to u = v.


Blue Boundary

This happens when y = 0, x > 0.

When y = 0,

u = 0 or u = x. ------ (1)

v = 0 or v = -x.

Not sure what is going on here...


To try and vizualize the boundaries in the u-v (u,v) plane, I tried to find the intersection between each boundary.

Blue-red Intersection

u = a AND u = v

Therefore, this corresponds to the point (a, a)

I've yet to find the red-green intersection and the blue-green intersection..

 

[tt2348]

I have a better idea..
Instead of going through and trying to formulate each of these piece by piece, try expressing x=x(u,v) and y=y(u,v)
For instance
$y^{2}=4u(u-x)=4v(v+x)\Rightarrow 4(u^2-v^2)-4x(u+v)=0 \Rightarrow (u^2-v^2)=x(u+v)$ so for x>0, and v≠-u≠0, we get $x=u-v$
Using this, we find
$y^2=4v(v+x)=4v(v+(u-v))=4uv\Rightarrow y=2\sqrt{uv}$ (we can omit the plus or minus, because we are considering y>0.

 

[vela]

Homework Statement

The question is attached in the picture.

The Attempt at a Solution

From the question I deduced that there are 3 boundaries: red, green and blue.

Blue Boundary

This one is the easiest:

That part of the curve of y² = 4a(a-x), which corresponds to u = a.Red Boundary

This corresponds to x = 0, y > 0.
When x = 0,

y² = 4u² = 4v², which corresponds to u = v.Blue Boundary

This happens when y = 0, x > 0.

When y = 0,

u = 0 or u = x. ------ (1)

v = 0 or v = -x.

Not sure what is going on here...

You need to recognize that u and v are both non-negative. Then the only way to satisfy $y^2=4v(v+x)=0$ when x>0 is to have v=0.

 

[unscientific]

You need to recognize that u and v are both non-negative. Then the only way to satisfy $y^2=4v(v+x)=0$ when x>0 is to have v=0.

That is true, but they mentioned nothing about value of u=?

 

[unscientific]

I have a better idea..
Instead of going through and trying to formulate each of these piece by piece, try expressing x=x(u,v) and y=y(u,v)
For instance
$y^{2}=4u(u-x)=4v(v+x)\Rightarrow 4(u^2-v^2)-4x(u+v)=0 \Rightarrow (u^2-v^2)=x(u+v)$ so for x>0, and v≠-u≠0, we get $x=u-v$
Using this, we find
$y^2=4v(v+x)=4v(v+(u-v))=4uv\Rightarrow y=2\sqrt{uv}$ (we can omit the plus or minus, because we are considering y>0.

I understand that this is also a way to solve it. But sometimes the functions aren't as straightforward as these, so I would like to know the analytical method suggested here..

 

[vela]

That is true, but they mentioned nothing about value of u=?

It's easiest to see what's happening in the figure (Figure 6.1) you posted. If you're on the x-axis and x runs from 0 to 2, what values does u have to go through?

You can throw out the u=0 solution because with it, you won't have any restrictions on the value of x. That is, x can be any value and you'd satisfy both equations. But you need to ensure that x is between 0 and 2 on the bottom boundary.