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

[evinda]

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?

Thanks in advance!:)

 

[alyafey22]

put \(\displaystyle x^2-y^2= k \)

Now , choose some values for $k$ and draw them in the xy-plane . What are these curves ?

 

[evinda]

I did this,and I think that the curves are hyperbolas...Is this correct?

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Or are these hyperbolic paraboloids?

 

[I like Serena]

I did this,and I think that the curves are hyperbolas...Is this correct?

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Or are these hyperbolic paraboloids?

Heh. The surface is a hyperbolic paraboloid.
The contour curves are indeed hyperbolas.

 

[evinda]

Nice...Thank you very much! :p

 

[evinda]

I have also an other question... 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?

 

[Petrus]

I have also an other question... 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.

 

[evinda]

Great...!Thank you very much...!

 

[I like Serena]

I have also an other question... 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.

 

[evinda]

Oh good!Thank you very much! ;)

 

[I like Serena]

You're welcome! ;)