xy - 2x = 0

Convert the rectangular equation into polar form.

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- Jun 23rd 2008, 11:49 AMcityisminepolar / parametrics question
xy - 2x = 0

Convert the rectangular equation into polar form. - Jun 23rd 2008, 11:50 AMgalactus
Plug in $\displaystyle x=rcos{\theta}, \;\ y=rsin{\theta}$

- Jun 23rd 2008, 12:37 PMcityismine
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 PMMathstud28
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 PMcityismine
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 PMMathstud28
- Jun 23rd 2008, 06:55 PMcityismine
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 AMMathstud28
- Jun 24th 2008, 09:49 AMwkrepelinBook'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 AMMathstud28
- Jun 24th 2008, 05:52 PMbadgerigarAm 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 PMMathstud28
- Jun 25th 2008, 04:03 PMbadgerigar
Sorry, in my previous post

should have been $\displaystyle \theta = \frac{\pi}{2}+n\pi$, which I think mathstud was trying to point out.