Contour Curves

evinda

Well-known member
MHB Site Helper
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???

ZaidAlyafey

Well-known member
MHB Math Helper
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

Well-known member
MHB Site Helper
I did this,and I think that the curves are hyperbolas...Is this correct???

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Or are these hyperbolic paraboloids???

Klaas van Aarsen

MHB Seeker
Staff member
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

Well-known member
MHB Site Helper
Nice...Thank you very much!!!

evinda

Well-known member
MHB Site Helper
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

Well-known member
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 (
is:

Regards.

evinda

Well-known member
MHB Site Helper
Great...!!!Thank you very much...!!!

Klaas van Aarsen

MHB Seeker
Staff member
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

Well-known member
MHB Site Helper
Oh good!!!!!Thank you very much!!!!

Klaas van Aarsen

MHB Seeker
Staff member
You're welcome!!!!!!!