anonymous
  • anonymous
parameterized Equations
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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anonymous
  • anonymous
I'm curious. How do you do it?
anonymous
  • anonymous
whoops i need help with question #1

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anonymous
  • anonymous
@Mr.Math maybe he knows how to do it
anonymous
  • anonymous
I think the first graph is \(y=x^2\), but I don't know the parameterization. Maybe \(y=t^2\),\(x=t\). I'm assuming parameterization has something to do with parametric equation.
anonymous
  • anonymous
Thats incorrect i think
anonymous
  • anonymous
Still you gave me a medal?
anonymous
  • anonymous
Joemath is here, he probably knows this stuff.
anonymous
  • anonymous
Like i thought y=t^2 and x= is smth with cost
anonymous
  • anonymous
But if it were like that then, \(y = x^2 + C\). For x = -1 or 1, \(y = 1 + C\). But in the graph y is just 1, so C has to be Zero.
anonymous
  • anonymous
i cant think of a good way to explain it >.< the first one is:\[x=\cos(t+\pi),y=\sin(t+\pi)+1, 0\le t\le \pi\]
anonymous
  • anonymous
joe is kinda on the button
anonymous
  • anonymous
Like this is the books answer
anonymous
  • anonymous
its the bottom half of a circle shifted up one unit. A circle is normally x = cos t, y = sin t. if you move it up one unit, that adds one to the y coordinate. and you want the bottom half, so you add pi to the angle to start on the other side.
anonymous
  • anonymous
Joe is always right! He is the Professor.
anonymous
  • anonymous
x=cost and y=1+sint pi till 2pi
anonymous
  • anonymous
my answer is the same as that. instead of putting (t + pi) and going from 0 to pi, they just put (t), and went from pi to 2 pi. its the same thing.
anonymous
  • anonymous
|dw:1333591281609:dw|
anonymous
  • anonymous
ok i think i need to think abt this stuff

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