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I am given the equation of a surface: $$x^3+y^3+z^3-3xyz =0$$ And I need to find the equation of the plane tangent to this surface at $(1,1,1).$
At first, this task did not look easy for me as we are not given an explicit equation of a surface, but I tried using implicit differentiation assuming that $z$ depends on $x$ and $y$, therefore:

  1. Differentiate with respect to $x$ $$3x^2+3z^2 \frac {\partial z}{\partial x} - 3yz - 3xy \frac{\partial z}{\partial x} = 0$$ and I use this equation to solve for $\frac{\partial z}{\partial x}$

  2. Differentiate with respect to $y$: $$3y^2 + 3z^2 \frac{\partial z}{\partial y} - 3xz - 3xy \frac{\partial z}{\partial y} = 0 $$ And I solve for $\frac{\partial z}{\partial y}$

Now, I have both partial derivatives which I can evaluate at $(1, 1, 1)$ to obtain the equation of the plane.
Is it the correct way to solve this problem?

Aemilius
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    $z-1= \frac {\partial z}{\partial x} (x-1)+ \frac {\partial z}{\partial y} (y-1)$ – imranfat May 03 '18 at 15:49
  • @imranfat, the values of $ \frac {\partial z}{\partial x} $ and $ \frac {\partial z}{\partial y} $ can't be solved for using the OP's equations at the point (1,1,1). – Cedron Dawg May 03 '18 at 16:49
  • @CedronDawg why can't they be solved for? Is it because of the domains of these partial derivatives I solve for using my equations? – Aemilius May 03 '18 at 16:58
  • They can be solved for, as I hope you have, but when you evaluate them at (1,1,1) you get 0/0 which is indeterminate. It is not a pluggable discontinuity either as you get different limits when you approach it from different directions. – Cedron Dawg May 03 '18 at 17:34
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    I find it odd that you accepted the answer that has an error in it. @Cesar Eo miscalculated the gradient. – Cedron Dawg May 03 '18 at 18:08
  • Check out the answer by Will Jagy in this followup question by @Legoman. The points on the "surface" are either in the plane $x+y+z=1$, which (1,1,1) is not, or on the line $x=y=z$ which (1,1,1) is. – Cedron Dawg May 03 '18 at 18:47
  • Oops, I forgot to add in the link: https://math.stackexchange.com/questions/2765351 – Cedron Dawg May 03 '18 at 19:08
  • @CedronDawg. My formula is a general approach to the problem. But...if the functions is not differentiable at that point (hole or whatever) then that changes it. Makes me wonder if this problem would be a book problem? – imranfat May 03 '18 at 19:10
  • @imranfat, I don't disagree. I think the approach that considers it a level curve is a little slicker. It turns out to be a kind of degenerate case, which factors into a plane or a line. See my two comments just prior to this one. It makes me wonder what the correct answer is at the point (0,0,0). – Cedron Dawg May 03 '18 at 19:17

2 Answers2

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The surface can be considered a level surface of:

$$ F(x,y,z) = x^3+y^3+z^3-3xyz $$

The gradient is always orthogonal to the level surface, and thus will serve to define the normal to the tangent plane. When the gradient is zero, this doesn't apply.

The gradient of $F$ is:

$$ \nabla F = (F_x,F_y,F_z) = \vec n = ( 3x^2 - 3yz, 3y^2 -3xz, 3z^2 - 3xy ) $$

Evaluated at the point (1,1,1) the gradient is the zero vector:

$$ \nabla F(1,1,1) = (0,0,0) $$

This is a trick question, there is no tangent plane at that point. Think of the two dimensional analog with a contour plot (level curves instead of a level surface). At any given level curve, I can find the tangent line. But at a peak, which is a point on the contour map, the idea of a tangent line is undefinable.

Ced

Cedron Dawg
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The so called surface $f(x,y,z) = x^3+y^3+z^3-3 x y z= 0$ is the product of a plane and a null radius cylinder or

$$ f(x,y,z) = (x+y+z)(x^2+y^2+z^2-x y- y z - x z) = 0 $$

The plane doesn't contain the point $(1,1,1)$ and the cylinder contains it but it has null radius so it is reduced to a line (generatrix) and a line has not unique normal vector.

Cesareo
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