anonymous
  • anonymous
I have this function: f(x,y)=ln(x^2+y^2) Find the boundary of the function's domain
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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AccessDenied
  • AccessDenied
i can only think of the restriction that ln cannot be 0 x^2 + y^2 = 0 would only occur at (0,0)
anonymous
  • anonymous
why? what makes (0,0) the boundary point?
anonymous
  • anonymous
can you explain?

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AccessDenied
  • AccessDenied
I'm not particularly familiar with domain 'boundaries', but i do know that logarithms are not defined for zero, so the domain would not include (0,0) since ln(0^2 + 0^2) = ln(0)
AccessDenied
  • AccessDenied
just what i'm thinking though, I'm hoping somebody can confirm. :P
anonymous
  • anonymous
yeah cause i still dont get it, cause apparently in my book it sats that a point(x,y) is a boundaary point of R if every disk centered at (x,y) contains the points tha lie outside of T as well as points that lie in R
anonymous
  • anonymous
I mean R not T
anonymous
  • anonymous
and i dont even understand that definition
AccessDenied
  • AccessDenied
yeah, that definition is outside of what I know of math...
anonymous
  • anonymous
i mean i am sure i would get it, if someone could help me understand this question.
TuringTest
  • TuringTest
I'm pretty sure I can answer it if I can figure out exactly what it is asking as well
anonymous
  • anonymous
that is exactly how the question is worded in my book
TuringTest
  • TuringTest
"a point(x,y) is a boundary point of R if every disk centered at (x,y) contains the points tha lie outside of R as well as points that lie in R" is what you say your book says that seems to fit the description of the point (0,0) center a disk at (0,0) every point (x,y) in that disk is in R but within that disk is also the point (0,0), which is not in R so the point is (0,0) I believe
anonymous
  • anonymous
yes (0,0) is the answer, so your explination seems to make sense
TuringTest
  • TuringTest
for instance make a disk of radius r\[x^2+y^2=r^2\]|dw:1332627163982:dw|r can be any size, and any point in it except (0,0) is in R hence "every disk centered at (x,y)=(0,0) contains the points that lied outside of R as well as points that lie in R" as is commanded by your theorem
TuringTest
  • TuringTest
|dw:1332627362029:dw|all that colored region is R (0,0) is not included
anonymous
  • anonymous
But in our case the raidus continues to expand based on the x and y values we choose correct
anonymous
  • anonymous
Thus we do not have a fixed radius
TuringTest
  • TuringTest
yeah we can make r anything, it's arbitrary so it can approach infinity, which means the set off all reals in \(\mathbb R^2\)
TuringTest
  • TuringTest
except (0,0) of course
anonymous
  • anonymous
thus we have an open unbounded region
TuringTest
  • TuringTest
right, it's a boundary \(point\) not a boundary line, or curve, or whatever
anonymous
  • anonymous
Thanks
TuringTest
  • TuringTest
welcome :D
AccessDenied
  • AccessDenied
I give my thanks as well. Makes more sense to me now. :P
TuringTest
  • TuringTest
....by the way, the region R is defined in set notation as\[\left\{(x,y)\in\mathbb R^2:x^2+y^2>0\right\}\]
anonymous
  • anonymous
sure, thats true.Thank you.

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