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# Regular Surfaces watch

1. Say the surface, M, is defined implicitly by . I need to show that this is a regular surface....

An implicitly defined surface is the zero set of a (smooth) function
. It is said to be regular if 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

and then

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!
2. , just like . 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 . Clearly, the zeroes of this function are exactly the points on your surface.
3. well in that case at the origin only and since doesn't satisfy 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
4. Just take your original implicit equation and rearrange it so that one side is zero. Then the other side is the function you want.
5. kl. cheers. any ideas on the sketch?

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Updated: January 28, 2009
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