Finding a tangent plane of a surface z = f(x,y)

Assume the tangent plane of the surface z = f(x,y) in the point (0,0,2) is given by z = Ax + By + 2. Find the equation for the tangent plane of the surface z = 1 / f(x,y) in the point (0,0,1/2).

No clue how to go about it. A hint or two, not a complete solution, would be appreciated.

Re: Finding a tangent plane of a surface z = f(x,y)

For any surface defined by f(x,y,z)= constant, $\displaystyle \nabla f$ is perpendicular to the surface at each point. You know how to find a plane given a point and normal vector?

Re: Finding a tangent plane of a surface z = f(x,y)

Quote:

Originally Posted by

**HallsofIvy** For any surface defined by f(x,y,z)= constant, [itex]\nabla f[/itex] is perpendicular to the surface at each point. You know how to find a plane given a point and normal vector?

I don't know if off-hand since I have done any linear algebra in a while, but I've done it before, so Google will help me out in that regards. Now, if I only could figure out what you meant to write there, "[itex]\nabla f[/itex]...

Re: Finding a tangent plane of a surface z = f(x,y)

Quote:

Originally Posted by

**Sabo** I don't know if off-hand since I have done any linear algebra in a while, but I've done it before, so Google will help me out in that regards. Now, if I only could figure out what you meant to write there, "[itex]\nabla f[/itex]...

See HallsofIvy's post again. I have corrected the code to [tex]\nabla f[/tex].

Re: Finding a tangent plane of a surface z = f(x,y)

Re: Finding a tangent plane of a surface z = f(x,y)

Quote:

Originally Posted by

**Sabo** Assume the tangent plane of the surface z = f(x,y) in the point (0,0,2) is given by z = Ax + By + 2. Find the equation for the tangent plane of the surface z = 1 / f(x,y) in the point (0,0,1/2).

No clue how to go about it. A hint or two, not a complete solution, would be appreciated.

A simple approach that requires only the basic concepts of partial derivative: if the tangent plane of the surface $\displaystyle z=f(x,y)$ in the point $\displaystyle (x_{0}, y_{0}, z_{0})$ is given by...

$\displaystyle z=z_{0} + x\ \frac{\partial{f}}{\partial{x}} (x_{0}, y_{0}) + y\ \frac{\partial{f}}{\partial{y}} (x_{0}, y_{0})$ (1)

... then the tangent plane of the surface $\displaystyle z=\frac{1}{f(x,y)}$ in the point $\displaystyle (x_{0}, y_{0}, \frac{1}{z_{0}})$ is given by...

$\displaystyle z=\frac{1}{z_{0}} + \frac{1}{f^{2}(x_{0}, y_{0})}\ \{ x\ \frac{\partial{f}}{\partial{x}} (x_{0}, y_{0})+ y\ \frac{\partial{f}}{\partial{y}} (x_{0}, y_{0})\}$ (2)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

Re: Finding a tangent plane of a surface z = f(x,y)

Thanks. The partial derivative approach was what I was looking for.