Graph r=6sin(θ)

Question

I'm stuck on this one. I've tried converting it to Cartesian coordinates but I couldn't. I know I could figure it out by testing a bunch of values for θ, but I'd like to know how to do it a better way. Thanks!

The problem:

Sketch the curve in polar coordinates.

r=6sinθ

Answer 1

Note, as fjardon mentioned, that we get $$ \begin {eqnarray*} x &=& 6 \sin \theta \cos \theta, \\ y &=& 6 \sin \theta \sin \theta. \end {eqnarray*} $$Now, rewrite $(x,y)$ as $ \left( 3 \sin 2\theta, 3 \cos 2\theta + 3 \right) $. Now, it is clearly that $$ \left( 3 \sin 2\theta \right)^2 + \left( 3 \cos 2\theta \right)^2 = 3^2, $$ from the Pythagorean identity. This reduces to $x^2+(y-3)^2=9$, which is a circle centered at $(x,y)=(3,0)$ and of radius $3$. $\Box$

Answer 2

When converting to rectangular coordinates you get the following equations:

$$ \left\{ \begin{align} x = r \cdot \cos(\theta) \\ y = r \cdot \sin(\theta) \\ \end{align}\right. $$

You can then replace $r$ by its value and obtain: $$ \left\{ \begin{align} x &= 3 \cdot 2 \cdot \sin(\theta) \cdot \cos(\theta) = 3\cdot \sin(2\cdot\theta) \\ y &= 3 \cdot 2 \cdot \sin(\theta) \cdot \sin(\theta) = 3\cdot(1-\cos(2\cdot\theta)) = 3 - 3\cdot cos(2\cdot\theta)\\ \end{align}\right. $$

It's a circle centered in $(0,3)$ of radius $3$

I know it's a circle from the very definition of $sin(\theta)$ and $cos(\theta)$, see: http://en.wikipedia.org/wiki/Unit_circle#Trigonometric_functions_on_the_unit_circle

Answer 3

Let $x$ and $y$ be Cartesian coordinates of a point and $r$ and $\theta$ be polar coordinate of that point,hence $r=(x^2+y^2)^{\frac{1}{2}}$ and $\sin\theta=\frac{y}{r}$ Hence above equation is $r^2=6y$ where $r=(x^2+y^2)^{\frac{1}{2}}$, hence applying we get answer

Answer 4

If $x$ and $y$ are cartesian coordinates and $r$ and $\theta$ be polar coordinate of one point,

the above equation becomes $(x^2+y^2)=6\,y ,$ or,

$x^2+y^2-6\,y +9 =9 $ or,

$ x^2 + (y-3)^2 = 3^2 $

It a circle on y-axis radius 3, displaced by 3 , so it is tangential to x-axis.