Question

How do you find the equations for the tangent plane to the surface `g(x,y)=x^2-y^2` through (5,4,9)?

Answer 1

An equation of the tangent plane of the graph of `z=g(x,y)=x^2-y^2` at the point `(x,y,z)=(5,4,9)` is `10x-8y-z=9`.

Explanation:

Write the equation `z=g(x,y)=x^2-y^2` in the equivalent form `G(x,y,z)=x^2-y^2-z=0`.

The gradient vector field of `G` is `nabla G(x,y,z)=(2x,-2y,-1)`. At the given point, this is `nabla G(5,4,9)=(10,-8,-1)`.

Since the gradient vector is perpendicular (normal) to the graph at this point, its components can be used as coefficients for the variables in the equation of the tangent plane (the reason for this is based on the fact that two nonzero vectors are perpendicular if and only if their dot product is zero).

The equation can be written `10(x-5)-8(y-4)-(z-9)=0`. This is equivalent to `10x-8y-z=9`.