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Changing polar equations to rectangular equations?

  • Thread starter steener
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changing polar equations to rectangular equations????

Can somebody please explain to me, how I would convert:

?=?/2 into a rectangular equation?

Along with: r=sin?, r=6cos+sin?, r(squared)sin2?=2


Your help would be greatly appreciated!!!!!
 

Answers and Replies

malawi_glenn
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Ask in the home-work section, not here.

This thread will be moved, so don't make a new one.
 
HallsofIvy
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Can somebody please explain to me, how I would convert:

?=?/2 into a rectangular equation?

Along with: r=sin?, r=6cos+sin?, r(squared)sin2?=2


Your help would be greatly appreciated!!!!!
Unfortunately your "special characters" just show up as "?" to me. I would guess that the ? in the last three are "theta": [itex]\theta[/itex] in LaTex, but I have no idea what the "?" in ?= ?/2 are- I presume they are different or the equation is trivial.

I presume that you know (or else you wouldn't be attempting these problems) that [itex]x= r cos(\theta)[/itex] and [itex]y= r sin(\theta)[/itex]. Looking at the first one, my thought would be to multiply both sides by r: [itex]r^2= r sin(\theta)[/itex] which, since [itex]r^2= r^2(cos^2(\theta)+ sin^2(\theta)= x^2+ y^2[/itex], is just [itex]x^2+ y^2= y[/itex], the equation of a circle.

For [itex]r= 6cos(\theta)+ sin(theta)[/itex], same thing: multiply both sides by r to get [itex]r^2= 6r cos(\theta)+ r sin(\theta)= x^2+ y^2= 6x+ y[/itex], again the equation of a circle.

For [itex]r^2 sin(2\theta)= 2[/itex], use the fact that [itex]sin(2\theta)= 2sin(\theta)cos(\theta)[/itex].
 

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