MHB Graph of f(x,y): Contour Curves & Hyperbolas

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The function f(x,y)=x^2-2*y^2 produces contour curves represented by the equation x^2-y^2=k, which are confirmed to be hyperbolas. The surface defined by this function is identified as a hyperbolic paraboloid. To find the tangent plane at the point (sqrt(2),1,0), the correct approach involves using the function g(x,y,z)=f(x,y)-z, leading to dg/dz=-1. The discussion clarifies that df/dz is not applicable since f(x,y) does not include z. Overall, the key points revolve around understanding the nature of the curves and the correct method for determining the tangent plane.
evinda
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Hi!
Let the function be f(x,y)=x^2-2*y^2,which graph is S:z=f(x,y).Which are the contour curves?Are these hyperbolas? :confused:

Thanks in advance!:)
 
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put $$x^2-y^2= k $$

Now , choose some values for $k$ and draw them in the xy-plane . What are these curves ?
 
I did this,and I think that the curves are hyperbolas...Is this correct?

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Or are these hyperbolic paraboloids?
 
evinda said:
I did this,and I think that the curves are hyperbolas...Is this correct?

- - - Updated - - -

Or are these hyperbolic paraboloids?

Heh. The surface is a hyperbolic paraboloid.
The contour curves are indeed hyperbolas.
 
Nice...Thank you very much! :p
 
I have also an other question... :o To find the tangent plane at the point (sqrt(2),1,0),I have to find df/dx,df/dy,df/dz...Is df/dz=d(x^2-2*y62-z)/dz=-1?Or df/dz=dz/dz=1?
 
evinda said:
I have also an other question... :o To find the tangent plane at the point (sqrt(2),1,0),I have to find df/dx,df/dy,df/dz...Is df/dz=d(x^2-2*y62-z)/dz=-1?Or df/dz=dz/dz=1?
An equation of the tangent plane to the surface z = f(x, y) at the point P (http://www.math.ucla.edu/~ronmiech/Calculus_Problems/32A/chap12/section4/793d1/IMG00002.GIF is:
http://www.math.ucla.edu/~ronmiech/Calculus_Problems/32A/chap12/section4/793d1/IMG00003.GIF

Regards.
 
Great...!Thank you very much...! :rolleyes:
 
evinda said:
I have also an other question... :o To find the tangent plane at the point (sqrt(2),1,0),I have to find df/dx,df/dy,df/dz...Is df/dz=d(x^2-2*y62-z)/dz=-1?Or df/dz=dz/dz=1?

Well, this is a bit ambiguous.
The surface z=f(x,y) is a parametric surface in x and y.
It makes no sense to consider df/dz which would be zero, since f(x,y) does not contain z.

So I suspect you're supposed to consider g(x,y,z)=f(x,y)-z.
The equation g(x,y,z)=0 identifies the same surface.
And then yes, dg/dz=-1.
 
  • #10
Oh good!Thank you very much! ;)
 
  • #11
You're welcome! ;)
 

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