How do you find the equations for the tangent plane to the surface g(x,y)=x^2-y^2g(x,y)=x2y2 through (5,4,9)?

1 Answer
Jan 10, 2017

An equation of the tangent plane of the graph of z=g(x,y)=x^2-y^2z=g(x,y)=x2y2 at the point (x,y,z)=(5,4,9)(x,y,z)=(5,4,9) is 10x-8y-z=910x8yz=9.

Explanation:

Write the equation z=g(x,y)=x^2-y^2z=g(x,y)=x2y2 in the equivalent form G(x,y,z)=x^2-y^2-z=0G(x,y,z)=x2y2z=0.

The gradient vector field of GG is nabla G(x,y,z)=(2x,-2y,-1)G(x,y,z)=(2x,2y,1). At the given point, this is nabla G(5,4,9)=(10,-8,-1)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)=010(x5)8(y4)(z9)=0. This is equivalent to 10x-8y-z=910x8yz=9.