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Rotation of Parabolas

  1. Nov 26, 2011 #1
    What is the maximum angle (degrees or radians) that you can rotate the basic parabola (y=x2) so that it can still be graphed as a function (y=...) with only one possible y-value per x-input.
  2. jcsd
  3. Nov 26, 2011 #2
    I think it's 0, because when you include the xy factor, it doesn't become a function anymore.
  4. Nov 26, 2011 #3
    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.
  5. Nov 27, 2011 #4


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    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 non-zero [itex]\theta[/itex], a quadratic equation which would have two values of y for each x.
  6. Nov 27, 2011 #5


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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.
  7. Nov 27, 2011 #6
    Both what HallsofIvy and phinds said make perfect sense to me. I had a feeling the answer might be zero, but I couldn't convince myself either way, thanks to both of you.
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