calc stuff, finding the Cartesian equation for a curve...?

Question

Find the Cartesian equation for the curve with the polar equation r = 2 cos x

can someone explain how to find it?

Answer 1

I guess you were trying to say r= 2 cos t.
x= r cos t so cos t = x/r
r^2=2x so
the equation x^2+y^2-2x=0. Ask again in Y!A if you think that I am unclear

Answer 2

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If x = 9 cos t, then cos t = x/9. If y = 9 sin t, then sin t = y/9. cos^2 t + sin^2 t = 1 (x/9)^2 + (y/9)^2 = 1 x^2/81 + y^2/81 = 1 x^2 + y^2 = 81 is the Cartesian equation. It is a circle, with center (0, 0) and radius 9.

Answer 3

Usually polar equations are expressed as
r = f(Θ)
where:
x = r cos(Θ) and y = r sin(Θ)

r = √(x^2 + y^2)
tan(Θ) = y/x

I will consider your equation as:
r = 2 cos(Θ)
√(x^2 + y^2) = 2 cos(arctan(y/x))
√(x^2 + y^2) = 2(x/√(x^2 +y^2)
cross multiplying:
x^2 + y^2 = 2x
x^2 -2x + __+ y^2 = 0
x^2 -2x + 4+ y^2 =4
( x -2)^2 + y^2 = 4

which is a circle with center at (2,0) and radius 2

Answer 4

you meant r= 2 cos t
x= r cos t so cost = x/r
r^2=2x so
x^2+y^2-2x=0 (End)