Rewrite the given equation using the substitutions x=rcosΘ and y=rsinΘ?

Question

given equation: x^2+y^2-2y=0

show work please so i understand

Answer 1

(rcosΘ)^2 + (rsinΘ)^2 - 2(rsinΘ) = 0

(r^2)(cos^2 Θ + sin^2 Θ ) -2(rsinΘ) = 0

(r^2)(1) - 2(rsinΘ) = 0

r^2 = 2(rsinΘ)

r = 2sinΘ

   radius always equals 1

1/2 = sinΘ

30ﾟ= Θ



proof

(r^2)(cos^2 Θ + sin^2 Θ ) -2(rsinΘ) = 0

(1)(cos^2 Θ + (1/2)^2) -2(1/2) = 0

cos^2 Θ + 1/4 -1 = 0

cos^2 Θ = 3/4

cosΘ = √(3/4) =√(3)/2

30ﾟ= Θ  

Answer 2

r^2 * cos^2Θ + r ^2 * sin^2Θ - 2*r*sinΘ = 0 ...substitute

r^2 * (cos^2Θ + sin^2Θ) - 2*r*sinΘ = 0    ...factor

r^2 * (1) - 2*r*sinΘ = 0      ...Pythagorean theorem

r^2 = 2*r*sinΘ  ...bring negative to other side

r = 2*sinΘ  ...divide by and solve for r

Matt... this is in polar coordinates... it's a graphical equation.  you can't solve for theta because it is not a specific value, but varries upon radius and radius varries upon theta.

Answer 3

r² cos²Ө + r² sin²Ө - 2 r sin Ө = 0

r² - 2 r sin Ө = 0