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How did they get these 3 boundaries?

  • #1
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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..
 

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  • #2
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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
[itex]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)[/itex] so for x>0, and v≠-u≠0, we get [itex]x=u-v[/itex]
Using this, we find
[itex] y^2=4v(v+x)=4v(v+(u-v))=4uv\Rightarrow y=2\sqrt{uv}[/itex] (we can omit the plus or minus, because we are considering y>0.
 
  • #3
vela
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Science Advisor
Homework Helper
Education Advisor
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1,149

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.
 
Last edited:
  • #4
1,728
13
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=?
 
  • #5
1,728
13
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
[itex]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)[/itex] so for x>0, and v≠-u≠0, we get [itex]x=u-v[/itex]
Using this, we find
[itex] y^2=4v(v+x)=4v(v+(u-v))=4uv\Rightarrow y=2\sqrt{uv}[/itex] (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..
 
  • #6
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
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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.
 

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