# polar / parametrics question

• Jun 23rd 2008, 11:49 AM
cityismine
polar / parametrics question
xy - 2x = 0

Convert the rectangular equation into polar form.
• Jun 23rd 2008, 11:50 AM
galactus
Plug in $\displaystyle x=rcos{\theta}, \;\ y=rsin{\theta}$
• Jun 23rd 2008, 12:37 PM
cityismine
I get...

r^2 cosx sinx - 2r cosx = 0

what do i do next, since this answer doesn't match the one in the back fo the book.
• Jun 23rd 2008, 12:47 PM
Mathstud28
Quote:

Originally Posted by cityismine
xy - 2x = 0

Convert the rectangular equation into polar form.

We first parametrize it

$\displaystyle x=r\cos(\theta)$

and $\displaystyle y=r\sin(\theta)$

Inputting this we get

$\displaystyle (r\cos(\theta))(r\sin(\theta))-2r\cos(theta)=0$

This can be rewritten as
$\displaystyle \frac{r^2}{2}\sin(2\theta)-2r\cos(\theta)=0$

Or we could have factored out a $\displaystyle r\cos(\theta)$

$\displaystyle r\cos(\theta)(r\sin(\theta)-1)=0$

Or finally we can solve this using the quadratic equation

If we let $\displaystyle u=\cos(\theta)\Rightarrow\sqrt{1-u^2}=\sin(\theta)$

so we have

$\displaystyle r^2u\sqrt{1-u^2}-2ru$

Now using the quadratic equation we get

$\displaystyle r=\frac{-u\sqrt{1-u^2}\pm\sqrt{u^2(1-u^2)-4(-2u)}}{2}$

So $\displaystyle r=\frac{-\cos(\theta)\sin(\theta)\pm\sqrt{\cos^2(\theta)\si n^2(\theta)+8\cos(\theta)}}{2}$
• Jun 23rd 2008, 04:35 PM
cityismine
The answer in the book is r^2= 2 sec(theta)csc(theta)

Is there a simpler way of deriving this answer?
• Jun 23rd 2008, 05:23 PM
Mathstud28
Quote:

Originally Posted by cityismine
The answer in the book is r^2= 2 sec(theta)csc(theta)

Is there a simpler way of deriving this answer?

This cannot be right

$\displaystyle r^2=2\sec(\theta)\csc(\theta)\Rightarrow{r\cos(\th eta)r\sin(\theta)=2}$

Or $\displaystyle xy=2$
• Jun 23rd 2008, 06:55 PM
cityismine
I guess the answer in the book is wrong then.

r^2= 4 / sin(theta)^2 is this the right answer?
• Jun 24th 2008, 09:05 AM
Mathstud28
Quote:

Originally Posted by cityismine
I guess the answer in the book is wrong then.

r^2= 4 / sin(theta)^2 is this the right answer?

$\displaystyle r^2=\frac{4}{\sin^2(\theta)}\Rightarrow{(r\sin(\th eta))(r\sin(\theta))=4}$

So $\displaystyle y^2=4$

So no.
• Jun 24th 2008, 09:49 AM
wkrepelin
Book's wrong
xy-2x=0

xy=2x

y=2, x does not equal zero

OR

xy-2x=0

t=theta

x=r*cos(t), -pi < t <pi

y=r*sin(t)

xy-2x=r*sin(t)*r*cos(t)-2*r*cos(t)

r*sin(t)*r*cos(t)-2*r*cos(t)=0

r*sin(t)*r*cos(t)=2*r*cos(t)

r*sin(t)*cos(t)=2*cos(t), r does not equal zero

r=sec(t)*csc(t)*(2*cos(t))

r=2*sec(t)*csc(t)*[cos(t)]

maybe this is where the computation in the book went wrong, you can see it's the same except for a factor of cos(t). Simplifying we get,

r=2*csc(t)

r=2/sin(t), r does not equal zero, -pi < t <pi

Even editors make mistakes(Smirk)
• Jun 24th 2008, 10:12 AM
Mathstud28
Quote:

Originally Posted by wkrepelin
xy-2x=0

xy=2x

y=2, x does not equal zero

OR

xy-2x=0

t=theta

x=r*cos(t), -pi < t <pi

y=r*sin(t)

xy-2x=r*sin(t)*r*cos(t)-2*r*cos(t)

r*sin(t)*r*cos(t)-2*r*cos(t)=0

r*sin(t)*r*cos(t)=2*r*cos(t)

r*sin(t)*cos(t)=2*cos(t), r does not equal zero

r=sec(t)*csc(t)*(2*cos(t))

r=2*sec(t)*csc(t)*[cos(t)]

maybe this is where the computation in the book went wrong, you can see it's the same except for a factor of cos(t). Simplifying we get,

r=2*csc(t)

r=2/sin(t), r does not equal zero, -pi < t <pi

Even editors make mistakes(Smirk)

*Ahem* I think you have made some undue assumptions here friend.
• Jun 24th 2008, 05:52 PM
Am I missing something here?
xy-2x=0
x(y-2) = 0
By null factor theorem, x=0 or y =2

in polar

$\displaystyle \theta = n\pi, n\in\mathbb{Z}$
or
r=0
or
$\displaystyle r = 2cosec(\theta)$
• Jun 24th 2008, 07:12 PM
Mathstud28
Quote:

xy-2x=0
x(y-2) = 0
By null factor theorem, x=0 or y =2

in polar

$\displaystyle \theta = n\pi, n\in\mathbb{Z}$
or
r=0
or
$\displaystyle r = 2cosec(\theta)$

Convert back

$\displaystyle r=2\csc(\theta)\Rigtharrow{r\sin(\theta)=2}$

So $\displaystyle y=2$

Which is not at all what the original problem was.
• Jun 25th 2008, 04:03 PM
should have been $\displaystyle \theta = \frac{\pi}{2}+n\pi$, which I think mathstud was trying to point out.