Question

how do you parametrize 4x^2+y^2=4

Answer 1

Oh nice, parametrization. First, do you know what we mean by parametrizing a figure?

Answer 2

well its an ellipse, i need to parametrize it so i can use it with stokes theorem

Answer 3

To normalize set all coefficients to one
4x^2+y^2=4
x^2 + (y/2)^2 = 1

Answer 4

solve for one, substitute it for the other?

Answer 5

KenLJW ive already done that, i need to get it into the form x=rcost y=rsint, but that 4 messes everything up.

Answer 6

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.

Answer 7

for unit circle
x^2 + y'^2 =1 y' = y/2
sin = y' = y/2
cos = x

Answer 8

not sure

Answer 9

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 ?

Answer 10

That's what I had r = 1, angle q

Answer 11

hmm, im still quite confused. thanks for trying though

Answer 12

x^2 + y'^2 = r y' =y/2
soh cah toa
sin(q) = y'/r =y/2r therefore y =2r sin(q)