0
$\begingroup$

Maybe I do not understand what is going on here but I cannot get the right answer.

$y = 2 $

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

$4 + 0^2 = r^2$

$ r = 2$

$y = r \sin \theta$

$1 = \sin \theta$

$\theta = \pi/2$

This is wrong but I do not see anything wrong with my logic.

3 Answers 3

5

Just because $y=2$, it does not mean that $x =0$. In particular, you want that, for any $x$, $y=2$, but $x$ is not $0$. Since you already know that

$y=2$

and that to change to polar coordinates, you can use

$y= \rho \sin \theta$,

we plug in $y=2$, which gives

$2= \rho \sin \theta$

or

$2 \csc \theta= \rho $

Note that we don't consider the variable $x$ and $x= \rho \cos \theta$, since it suffices to use only the $y$-equation above, because it fully describes the dependence between $\theta$ and $\rho$.

In particular note the radius can't be always $2$, else you would get a circle! Everytime you state something such as $x=0$ or $r=2$, think the implications it carries, and it might help you to spot the flaw.

  • 0
    In polar coordinates, one wirtes $\rho$ as a function of $\theta$, that is $\rho = \rho(\theta)$ just as we write $y$ as a function of $x$ in cartesian coordinates, $y =y(x)$. That's why I wrote the equation explicitly.2012-07-01
4

Remember that $y = r \sin(\theta)$. Hence, $r \sin(\theta) = 2$ is the equation in polar coordinates of the straight line $y=2$.

0

Polar equation is

$ y = r \sin \theta = 2 = constant$

Using this equation you computed correctly only one point.

$ x= 0, y=2 ;\, \theta = \pi/2, r = 2 $

$y$ is same for all $\theta$ and $x$ is a variable.