Answer :
Certainly! Let's convert the given Cartesian equation [tex]\( x^2 + y^2 = 2y \)[/tex] into a polar coordinate equation step by step.
1. Recall the polar coordinates relationships:
[tex]\[ x = r \cos(\theta) \][/tex]
[tex]\[ y = r \sin(\theta) \][/tex]
2. Substitute these polar coordinates into the given Cartesian equation [tex]\( x^2 + y^2 = 2y \)[/tex]:
[tex]\[ (r \cos(\theta))^2 + (r \sin(\theta))^2 = 2 \cdot (r \sin(\theta)) \][/tex]
3. Simplify the left side of the equation:
[tex]\[ r^2 \cos^2(\theta) + r^2 \sin^2(\theta) = 2r \sin(\theta) \][/tex]
4. Factor out [tex]\( r^2 \)[/tex] on the left side:
[tex]\[ r^2 (\cos^2(\theta) + \sin^2(\theta)) = 2r \sin(\theta) \][/tex]
5. Recall that [tex]\( \cos^2(\theta) + \sin^2(\theta) = 1 \)[/tex]:
[tex]\[ r^2 \cdot 1 = 2r \sin(\theta) \][/tex]
[tex]\[ r^2 = 2r \sin(\theta) \][/tex]
6. To solve for [tex]\( r \)[/tex], subtract [tex]\( 2r \sin(\theta) \)[/tex] from both sides:
[tex]\[ r^2 - 2r \sin(\theta) = 0 \][/tex]
7. Factor the equation:
[tex]\[ r (r - 2 \sin(\theta)) = 0 \][/tex]
So, the polar equation equivalent to the given Cartesian equation [tex]\( x^2 + y^2 = 2y \)[/tex] is:
[tex]\[ r (r - 2 \sin(\theta)) = 0 \][/tex]
This is the step-by-step process to convert the equation from Cartesian to polar form.
1. Recall the polar coordinates relationships:
[tex]\[ x = r \cos(\theta) \][/tex]
[tex]\[ y = r \sin(\theta) \][/tex]
2. Substitute these polar coordinates into the given Cartesian equation [tex]\( x^2 + y^2 = 2y \)[/tex]:
[tex]\[ (r \cos(\theta))^2 + (r \sin(\theta))^2 = 2 \cdot (r \sin(\theta)) \][/tex]
3. Simplify the left side of the equation:
[tex]\[ r^2 \cos^2(\theta) + r^2 \sin^2(\theta) = 2r \sin(\theta) \][/tex]
4. Factor out [tex]\( r^2 \)[/tex] on the left side:
[tex]\[ r^2 (\cos^2(\theta) + \sin^2(\theta)) = 2r \sin(\theta) \][/tex]
5. Recall that [tex]\( \cos^2(\theta) + \sin^2(\theta) = 1 \)[/tex]:
[tex]\[ r^2 \cdot 1 = 2r \sin(\theta) \][/tex]
[tex]\[ r^2 = 2r \sin(\theta) \][/tex]
6. To solve for [tex]\( r \)[/tex], subtract [tex]\( 2r \sin(\theta) \)[/tex] from both sides:
[tex]\[ r^2 - 2r \sin(\theta) = 0 \][/tex]
7. Factor the equation:
[tex]\[ r (r - 2 \sin(\theta)) = 0 \][/tex]
So, the polar equation equivalent to the given Cartesian equation [tex]\( x^2 + y^2 = 2y \)[/tex] is:
[tex]\[ r (r - 2 \sin(\theta)) = 0 \][/tex]
This is the step-by-step process to convert the equation from Cartesian to polar form.
