anonymous
  • anonymous
how do you parametrize 4x^2+y^2=4
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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allank
  • allank
Oh nice, parametrization. First, do you know what we mean by parametrizing a figure?
anonymous
  • anonymous
well its an ellipse, i need to parametrize it so i can use it with stokes theorem
KenLJW
  • KenLJW
To normalize set all coefficients to one 4x^2+y^2=4 x^2 + (y/2)^2 = 1

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Jhannybean
  • Jhannybean
solve for one, substitute it for the other?
anonymous
  • anonymous
KenLJW ive already done that, i need to get it into the form x=rcost y=rsint, but that 4 messes everything up.
allank
  • allank
Alright, the main idea behind parametrization is obtaining a function whose image is the figure...in this case an ellipse (you can read about that later) The parametrization I think you want is for R1 -> R2 Where you have a parametrization that takes in one variable and outputs values on the ellipse. For now, I can only think of how to get the parametrization of the top half of the ellipse. We have 4x^2+y^2=4 Solving for y in terms of x y=sqrt(4-4x^2) Let's create a parametrization x, with variable t. x(t)=(t,sqrt(4-4t^2)) Plugging in different t values yields the upper half of the ellipse.
KenLJW
  • KenLJW
for unit circle x^2 + y'^2 =1 y' = y/2 sin = y' = y/2 cos = x
KenLJW
  • KenLJW
not sure
anonymous
  • anonymous
ok the answer given, is "4x^2 + y 2 = 4 or x^2 + (y^2)/4 = 1, so x = r cos q and y = 2r sin q" can someone show me how they got to this answer ?
KenLJW
  • KenLJW
That's what I had r = 1, angle q
anonymous
  • anonymous
hmm, im still quite confused. thanks for trying though
KenLJW
  • KenLJW
x^2 + y'^2 = r y' =y/2 soh cah toa sin(q) = y'/r =y/2r therefore y =2r sin(q)

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