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

1. ## 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.

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

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

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

Originally Posted by HallsofIvy
For any surface defined by f(x,y,z)= constant, $\nabla f$ 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, "$\nabla f$...

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

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, "$\nabla f$...
See HallsofIvy's post again. I have corrected the code to $$\nabla f$$.

Thanks.

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

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 $z=f(x,y)$ in the point $(x_{0}, y_{0}, z_{0})$ is given by...

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

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

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

Kind regards

$\chi$ $\sigma$

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

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