Given the definition of the tangent plane at $(x_0,y_0,z_0)$ to a surface $f(x,y,z) = k$ : $\nabla f(x_0,y_0,z_0) \cdot (x-x_0,y-y_0,z-z_0)$, I wish to show that this is a special case of the formula for the plane tangent to the graph $f(x,y)$ by regarding the graph as a level surface of $F(x,y,z) = f(x,y)-z $.

My attempt: First considering the level surface $F(x,y,z)=0$, the tangent plane is given by $\frac {\partial f} {\partial x}(x_0,y_0)(x-x_0) + \frac {\partial f} {\partial y}(x_0,y_0)(y-y_0) - (z-z_0)=0$, whereas the tangent plane to $f(x,y)$ at $(x_0,y_0,z_0)$ is equal to $z=f(x_0,y_0) + \frac {\partial f} {\partial x}(x_0,y_0)(x-x_0) + \frac {\partial f}{\partial y}(x_0,y_0)(y-y_0)$. I am not sure how to proceed. Any hints much appreciated.


Hint: The definition of $z_0$ is $f(x_0, y_0)$. So the two expressions for the tangent plane are equal.

Also, in the first paragraph, you've got it backwards: I think you mean to say that the tangent plane to the graph of a function is a special case of the more general tangent plane to a surface (defined as a level set).

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