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Calc III, what point could this be?

  1. Sep 4, 2005 #1
    Hello everyone, I'm trying to figure out this point on the sphere. The orginal equation is this: x^2 + y^2 + z^2 ≤ 16; 2 ≤ y ≤ 4;

    Here is the picture:
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    If you look where it shows a picture of the sphere and i have an arrow pointing to a point on top of the circle. Has coordinates (0,2, ? ). I"m trying to figure out the point where that ? is. I know the x and y coordinates havn't changed at all, only the z coordinates. Sorry its alittle messy. :bugeye: I know the point below it is (0,2,0). Thanks.
     
  2. jcsd
  3. Sep 4, 2005 #2

    arildno

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    First off, you're dealing with inequalities, not equations.
    Secondly, the solution set to the two given inequalities is a REGION, a segment of the sphere, not just a single point on the sphere.
     
  4. Sep 4, 2005 #3
    I realise the soution set is a region not a single point on my sphere, my question was, what the point was, not what the whole solution set was.
     
  5. Sep 4, 2005 #4

    arildno

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    It is rather unclear what you're after, but:
    Well, if x=0 and y=2, then the solution set is the line segment
    [tex]x=0, y=2,-\sqr{12}\leq{z}\leq\sqrt{12}[/tex]
    If you are interested in the "top most" point, then [tex](0,2,\sqrt{12})[/tex] should be it.
     
  6. Sep 4, 2005 #5

    HallsofIvy

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    [tex]x^2+y^2+z^2\le 16[/tex]
    describes the points inside and on a sphere of radius 4, centered at the origin.

    [tex]1\le y\le 4[/tex]
    describes points between and on the two planes y= 2 and y= 4

    Of course, the plane y= 4 is tangent to the sphere at (0, 4, 0).

    The plane y= 2 crosses the sphere in the circle
    [tex]x^2+ 2^2+ z^2= 16[/tex]
    or [tex]x^2+ z^2= 12[/tex].

    That is a circle in the plane y= 2, centered at (0,0,0) with radius [tex]2\sqrt(3)[/tex].
    In particular, the point you are apparently looking for is (0, 2, 2&radic;(3)).

    The set of all points described by the inequalities are the points inside or on the sphere "beyond" (whatever direction the y-axis is in!) y= 2.
     
  7. Sep 4, 2005 #6
    Thanks, how did you figure out it was [tex](0,2,\sqrt{12})[/tex] for z? My professor requested we figure out that point on the circle thats why its such an odd question. He said, to figure it out all you would use is basic aglebra.
     
  8. Sep 4, 2005 #7
    Thanks Halls, I think your answer is right! It makes sense anyways!
     
  9. Sep 4, 2005 #8

    arildno

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    Okay, I didn't read your post closely enough; you did mention a circle somewhere...
    Anyhow, my answer is the same as HallsofIvy's
     
  10. Sep 4, 2005 #9
    yep thank you both for the help
     
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