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    Say the surface, M, is defined implicitly by x^2+y^2-z^2=1. I need to show that this is a regular surface....

    An implicitly defined surface is the zero set of a (smooth) function
    f<img src="images/smilies/colone.gif" border="0" alt="" title=":E" class="inlineimg" />^3 \rightarrow R. It is said to be regular if \nabla f does not vanish anywhere on the surface.

    Does anybody have any idea what I should let f be for this case?

    I considered rearranging to get z=\pm \sqrt{x^2+y^2-1}

    and then \nabla z = ( \frac{\partial z}{\partial x},\frac{\partial z}{\partial y},\frac{\partial z}{\partial z})

    But this has two problems:
    (i) i have to choose between the positive or negative root for z and don't know which

    (ii) the third component of grad z is taking the partial deriative of the function with respect to the function and is therefore 1. this seems wrong!
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    \displaystyle \frac{\partial z}{\partial z} = 1, just like \displaystyle \frac{\mathrm{d}x}{\mathrm{d}x} = 1. You would need to consider both the positive and negative square roots - otherwise you would only get half the surface.

    In any case, the function you should be considering is simply f(x,y,z) = x^2 + y^2 - z^2 - 1. Clearly, the zeroes of this function are exactly the points on your surface.
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    well in that case \nabla f = (2x,2y,2z)=0 at the origin only and since (0,0,0) doesn't satisfy f=0 it is not a point on the surface and hence the surface is regular at all points.

    Next part is to sketch it. Now it's clearly a hyperboloid of one sheet and there are plenty examples of what that looks like online but when i use maple to draw it - maple only draws one of the "domes" - would there be any particular reason for this?

    also how did u know to set f=x^2+y^2+z^2-1
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    Just take your original implicit equation and rearrange it so that one side is zero. Then the other side is the function you want.
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    kl. cheers. any ideas on the sketch?
 
 
 
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