BloomLocke367
  • BloomLocke367
Is it possible to have a piecewise function with only part of a circle?.. I know the equation isn't a function so I'm not sure if it's an actual thing...
Mathematics
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
I believe it's possible if you specify the domain.
anonymous
  • anonymous
The only issue is the co-variant nature of that circle.
BloomLocke367
  • BloomLocke367
I tried and it didn't do it :O that's why I was wondering.. I have a project where I have to make a picture using piecewise functions and I only need part of a circle but I can't figure out how to do it.

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BloomLocke367
  • BloomLocke367
@Nnesha
BloomLocke367
  • BloomLocke367
@Shalante
anonymous
  • anonymous
I am too trying to figure out how piecewise function works for my brother
anonymous
  • anonymous
It appears pretty intuitive until you figure out each specified domain must also account for different equations.
BloomLocke367
  • BloomLocke367
okay.. hmm
BloomLocke367
  • BloomLocke367
@King.Void.
King.Void.
  • King.Void.
Not good at math
BloomLocke367
  • BloomLocke367
dang ok
beginnersmind
  • beginnersmind
Think about the function \[f(x)= \sqrt{1-x^2}\] compare it with the equation of a circle of radius 1 centered on the origin: \[x^2+y^2 = 1\]
BloomLocke367
  • BloomLocke367
@Shalante
jdoe0001
  • jdoe0001
you can have a function alone, and be only a semi-circle a circle is compounded of two semicircles really each semi-circle occurs when you use the square root the square root has a + and a - part, thus \(\large \pm \sqrt{\quad}\)
BloomLocke367
  • BloomLocke367
that only shows half of the circle..
BloomLocke367
  • BloomLocke367
yea okay XD I was about to say..
BloomLocke367
  • BloomLocke367
but I'm working on a piecewise project and I have to create a picture on a graph using piecewise functions and that's why I was wondering XD

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