MHB Convert to a equation in RECTANGULAR coordinates

[shamieh]

What am I doing wrong?

Convert $$r = 2sin\theta$$ to an equation in rectangular coordinates..

$$x^2 +y^2 = r^2$$
$$x^2 + y^2 = 2y$$
$$x^2 + y^2 - 2y = 0$$
$$x^2 + y^2 - 2y - 1 = 1$$
$$x^2 + (y-1)^2 = 1$$

Coordinates are $$(0,1)$$ yes?

 

[topsquark]

What am I doing wrong?

Convert $$r = 2sin\theta$$ to an equation in rectangular coordinates..

$$x^2 +y^2 = r^2$$
$$x^2 + y^2 = 2y$$
$$x^2 + y^2 - 2y = 0$$
$$x^2 + y^2 - 2y - 1 = 1$$
$$x^2 + (y-1)^2 = 1$$

Coordinates are $$(0,1)$$ yes?

Typo on line 4...should be [math]x^2 + y^2 - 2y + 1 = 1[/math], otherwise it's just fine.

Coordinates for what? You're picking out the center of the circle, which is just fine, but why are you doing that? Was there more to the question?

-Dan

 

[shamieh]

Typo on line 4...should be [math]x^2 + y^2 - 2y + 1 = 1[/math], otherwise it's just fine.

Coordinates for what? You're picking out the center of the circle, which is just fine, but why are you doing that? Was there more to the question?

-Dan

Exactly as I typed is what it asked, So I'm very confused what the answer is? Should it just be (0,1) the question said to convert it into rectangular coordinates?

 

[topsquark]

Exactly as I typed is what it asked, So I'm very confused what the answer is? Should it just be (0,1) the question said to convert it into rectangular coordinates?

Your equation in rectangular coordinates is [math]x^2 + (y - 1)^2 = 1[/math]. That is the solution to the question you wrote in your first post.

Is there an answer key or something? How do you know that you got it wrong?

-Dan

 

[Deveno]

$r = 2\sin\theta$

Convert using:

$r = \sqrt{x^2 + y^2}$

$\sin\theta = \dfrac{y}{r}$

$\sqrt{x^2 + y^2} = 2\dfrac{y}{\sqrt{x^2 + y^2}}$

$x^2 + y^2 = 2y$

Complete the square:

$x^2 + y^2 - 2y + 1 = 1$

$x^2 + (y - 1)^2 = 1$

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"Rectangular coordinates" means in terms of $x$ and $y$ (the rectangular axes, that are perpendicular to each other), it doesn't mean "some specific coordinates" (a point $(a,b)$).