Proof of the tangent plane

Question

Let $U$ be a proper subset of $\mathbb{R}^2$, and open, $f : U \to \mathbb{R}$ is smooth, $G$ its graph, and $q= (x,y)$ be in $U.$ Prove that the tangent plane to $G$ at $(x, y, f(x,y))$ is the graph of $df_q.$

I'm looking at F(x,y,z) = f(x,y) - z = 0. Found the gradient of F to be $<$df/dx, df/dy, -1$>$. Let (x,y,z) be any point on the tangent plane through (x,y,z). Then get the vector $<$x-x, y-y, z-z $>$ orthogonal to the gradient of F. Taking the partial derivatives of this vector I solved to get z = f(x,y) + (x-x)df/dx + (y-y)df/dy. Is this on the right path?

Answer 1

We need to use different symbols to distinguish points moving on plane form intersection point, that is

• intersection point: $P_0=(x_0,y_0,z_0)$
• normal to the plane: $\vec n=\left(f_x(P_0),f_y(P_0),-1\right)$

then the plane equation is

$$(x-x_0)\cdot f_x(P_0)+(y-y_0)\cdot f_y(P_0)-z_0=0$$