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Rotation of Parabolas 
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#1
Nov2611, 08:42 PM

P: 784

What is the maximum angle (degrees or radians) that you can rotate the basic parabola (y=x^{2}) so that it can still be graphed as a function (y=...) with only one possible yvalue per xinput.



#2
Nov2611, 09:18 PM

P: 2,568

I think it's 0, because when you include the xy factor, it doesn't become a function anymore.



#3
Nov2611, 09:46 PM

P: 784

But more abstractly, I think it's possible to do a slight rotation, but there's an obvious cutoff point. I'm curious where that cutoff point is, it could be zero, I can't quite picture it well enough.



#4
Nov2711, 07:17 AM

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Rotation of Parabolas
No, that's not correct. Any rotation at all makes it no longer a function.
Start with [itex]y= x^2[/itex]. With a rotation through an angle [itex]\theta[/itex] we can write [itex]x= x' cos(\theta)+ y' sin(\theta)[/itex], [itex]y= x' sin(\theta) y' cos(\theta)[/itex] where x' and y' are the new, tilted coordinates. In this new coordinate system, the parabola becomes [tex]x'sin(\theta) y'cos(\theta)= (x'cos(\theta)+ y'sin(\theta))^2= x'^2 cos^2(\theta)+ 2x'y'sin(\theta)cos(\theta)+ y'^2 sin^2(\theta)[/tex]. Now, if we were to fix x' and try to solve for y' we would get, for any nonzero [itex]\theta[/itex], a quadratic equation which would have two values of y for each x. 


#5
Nov2711, 08:35 AM

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Vorde, I could not fault the logic presented by Hallsofivy, but it didn't FEEL right, so I played w/ it a bit from what I thought of as a more intuitive way of looking at it thinking it would show that at least a small rotation would work, but it clearly doesn't.
Here's how I got there. Think of a line that goes through the origin but really hugs the y axis. Let's say it has a slope of 1,000, and it has a sister line just on the other side of the y axis with a slope of 1,000. If neither of them hit the parabola, then clearly you could rotate it by that much. It's trivially easy to show though that they both DO hit the parabola (at x = 1,000 and x=1,000 assuming the given example of y = x^2) so Hallofivy obviously had it right and that was all a waste of time mathematically, but it DID help me see more graphically why he is right. 


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