I was reading this answer regarding the span of a tangent plane here.

The answer says the graph of $f$ is also the graph of the map $F(x,y) = (x,y,f(x,y))$. The tangent plane is spanned by $(1,0,f_x), (0,1,f_y)$.

However, I don't understand why. If I were to find the tangent plane at a point $c = (x_0,y_0)$, I would do the following

$$\nabla F(c) ( (x,y) -c) = f_x(x - x_0) + f_y(y - y_0) - (f(x,y) - f(c)) = 0$$

The tangent plane would be set $T$ that satisfies the equation above.

Two questions:

(1) Is $f(x,y) = f(c)$?

(2) Suppose that I have a set $T$ that satisfies the equation above. Then each element of $T$ can be written as $[a_1 (1,0,f_x), \ a_2 (0,1,f_y)]$. If I plug that into the equation, I get $$f_x(a_1 - x_0) + f_y (a_2 - y_0) - (f(x,y) - f(c)) = 0$$

But $f(x,y) = a_1 f_x + a_2 f_y$, then I have

$$f_x(a_1 - x_0) + f_y (a_2 - y_0) - (a_1 f_x + a_2 f_y - f(c)) = 0$$ which simplifies to

$$ f(c) - (f_x x_0 + f_y y_0) = 0$$

How do you show that this equality holds?


I’m not sure how you’re calculating your tangent plane above. The general approach is to observe that the tangent plane should contain every tangent line: given any direction $(dx, dy)$, restrict $F$ to the line passing though $(x_0, y_0)$ in that direction:

$$G(t) = F(x_0 + tdx, y_0+ tdy).$$

This is now a one-dimensional function whose derivative gives the slope of a tangent line to $G$; therefore the vector $(dx, dy, G’(0))$ should lie on the tangent plane of $F$.

To find a basis for the tangent plane, pick any linearly independent pair of directions $(dx, dy)$. $(1,0)$ and $(1,0)$ are particularly convenient.

  • $\begingroup$ Why do you have $G'(0)$ in $(dx,dy,G'(0))$? $\endgroup$ – user1691278 Feb 18 '18 at 23:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.