Question 9.6.4: Find an equation of the tangent plane to the graph of z = 1/......

Define F(x, y, z)=\frac{1}{2} x^{2}+\frac{1}{2} y^{2}-z+4 so that the level surface of F passing through the given point is F(x, y, z)=F(1,-1,5) or F(x, y, z)=0. Now, F_{x}=x, F_{y}=y, and F_{z}=-1, so that

\nabla F(x, y, z)=x \mathrm{i}+y \mathrm{j}-\mathrm{k} \quad \text { and } \quad \nabla F(1,-1,5)=\mathrm{i}-\mathrm{j}-\mathrm{k}.

Thus, from (5) the desired equation is

F_x\left(x_0, y_0, z_0\right)\left(x-x_0\right)+F_y\left(x_0, y_0, z_0\right)\left(y-y_0\right)+F_z\left(x_0, y_0, z_0\right)\left(z-z_0\right)=0 .     (5)

(x+1)-(y-1)-(z-5)=0 \quad \text { or } \quad-x+y+z=7 .

See FIGURE 9.6.6.