
#1
Dec413, 10:01 AM

P: 154

Context: https://www.youtube.com/watch?v=sQM8Oj4Ecg
In the video, he divides out a variable as below: $$r^2=2rsin\theta\\r=2sin\theta$$ When you do a division like this, aren't you dividing out a variable? What happens if you take the square root of both sides like below? $$r^2=2rsin\theta\\r=\sqrt{2rsin\theta}$$ Would that be more correct? 



#2
Dec413, 10:06 AM

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Yes, he is. I particular, he has "lost" the r= 0 for all [itex]\theta[/itex] solution.
I'm not sure what you mean by "more correct" for [itex]r= \sqrt{2r sin(\theta)}[/itex]. It is not at all a "solution" because you have not "solved for r". 



#3
Dec413, 10:07 AM

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#4
Dec413, 10:10 AM

P: 154

Dividing cancels out a solution?
How did you know he cancels out the ##r=0## solution?




#5
Dec413, 02:08 PM

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#6
Dec413, 02:54 PM

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(x1)(x2) = (x1)(2x1) The first thing we do is divide both sides by x1 (x2) = (2x1) Now we solve for x x = 1 OK, but we divided both sides by x1, and we have to be worried about when that is equal to zero. x1= 0 when x=1, so I should go back to the original equation and check that x=1 is a solution, which it is easily seen to be. 



#7
Dec413, 05:28 PM

P: 263

r^2=2rsin\theta\ <> r(r2sin\theta\)=0 <> r = 0 or r = 2sin\theta\ problem solved ! (sorry I dont know how to do the maths letters here ) 


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