In the question Orthonormal basis for a tangent plane, one of the commenters notes that for a manifold $M$ described by the graph of an arbitrary smooth function $f:U\subseteq \mathbb{R}^2 \to\mathbb{R}$, one possible basis of the tangent plane is given by $$ \begin{align} v_1 &= (-f_y, f_x , 0) \\ v_2 &= (f_x, f_y, f_{x}^2 + f_{y}^2) \end{align} $$ but I can't understand how those vectors were arrived at, i.e. what is meant by ''one vector along the level curves of $f$ and another vector along the direction of greatest change''. Can anyone please explain?

  • $\begingroup$ One thing that may be confusing is that there was a mistake in the OP that you refer to. $f$ is a function $\mathbb{R}^2\to \mathbb{R}$. Only its graph hence lies in $\mathbb{R}^3$. Just fixed it. $\endgroup$ Feb 14, 2018 at 19:45
  • $\begingroup$ Thanks for pointing that error out @ArnaudMortier. $\endgroup$ Feb 14, 2018 at 19:49
  • $\begingroup$ Unfortunately, that error aside, I still don't understand how the commenter arrived at their solution. $\endgroup$ Feb 14, 2018 at 19:52

2 Answers 2


For any $g(x,y,z)=0$ describing a plane in $\Bbb R^3$, we know that the gradient vector $\left (g_x(x,y,z),g_y(x,y,z),g_z(x,y,z)\right )$ is perpendicular to any tangent plane intersecting with $g$ at any point (e.g. all gradient vectors of a sphere are its diameters which are perpendicular to any tangent plane on the sphere in any point of its perimeter).

Now take $g(x,y,z)=z-f(x,y)=0$ therefore the gradient vector is $$\vec{\nabla g}=(f_x,f_y,-1)$$it suffices to show that both $v_1$ and $v_2$ are independent and perpendicular on the gradient vector. Proving independence is simple and for normality we have$$v_1\cdot\vec{\nabla g}=(f_x,f_y,-1)\cdot(-f_y,f_x,0)=0\\v_2\cdot\vec{\nabla g}=(f_x,f_y,-1)\cdot(f_x,f_y,f_x^2+f_y^2)=0$$which proves what we want.

  • $\begingroup$ Thank you for offering this explanation @Mostafa. $\endgroup$ Feb 14, 2018 at 20:19
  • $\begingroup$ That's very kind of you! $\endgroup$ Feb 14, 2018 at 20:19

The graph of $f$ is also the graph of the map $F$, this time really $\mathbb{R}^2\to \mathbb{R}^3$:


The tangent plane to $M$ is then spanned by $F_x$ and $F_y$, which are respectively $$(1,0,f_x)\qquad \text{and}\qquad (0,1,f_y)$$

Now you can see that the vectors given in the OP are linear combinations of these: $$v_1=-f_yF_x+f_xF_y$$ and $$v_2=f_xF_x+f_yF_y$$

The matrix that takes $F_x$ and $F_y$ to $v_1$ and $v_2$ has determinant $-(f_x^2+f_y^2)\neq0$, so these vectors are still a basis of the tangent space.

Now since $v_1$'s last coordinate is zero and since we know by the above that $v_1$ is tangent to $M$, these two facts imply that $v_1$ is tangent to the level curves of $f$.

Finally, the fact that $v_1\perp v_2$, together with $v_2\in TM$, implies that $v_2$ is indeed the direction of greatest change of $f$.

  • $\begingroup$ My compliments on an excellent answer: very helpful and instructive. Thank you for taking the time to help! $\endgroup$ Feb 14, 2018 at 20:08
  • $\begingroup$ @AnCoileanach You're welcome :) $\endgroup$ Feb 14, 2018 at 20:09

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