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Find the points at which the surface $$ x^2 +2y^2+z^2 -2x -2z -2 = 0 $$ has horizontal tangent planes. Find the equation of these tangent planes.

I found that $$ \nabla f = (2x-2,4y) $$ I'm thinking that the gradient vector must be equal to $(0,0)$ so $x = 1, y=0$ which implies $z = 3,-1$. So the points are $(1,0,3)$ and $(1,0,-1)$. How would we write the equation of these tangent planes?

rmzep
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  • if they were horizontal, the plane would have no x,y dependence. thus, the plane is described by the equation z=3, etc... – reluctant mathematician Oct 24 '14 at 03:25
  • Remember that the gradient is normal to such planes! So, you have the normal vectors and its points, what do you waiting to start the computations? I'm kidding! Do you understanding? – DiegoMath Oct 24 '14 at 03:26

1 Answers1

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Your answer is correct, but the logic is wrong. You should consider the above surface as a level surface of the function:

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

Then the gradient should be a vector with 3 components.

$$ \nabla f(x, y, z) = (2x - 2, 4y, 2z - 2) $$

If the tangent plane is horizontal, the gradient must point in the $z$-direction, therefore the $x$ and $y$ components are $0$. So it follows that $x = 1$ and $y = 0$

Finally, the equations are just $z = 3$ and $z = -1$ since horizontal planes have the same $z$ coordinates everywhere.

Another way is to treat $z$ as a function of $x$ and $y$, then set $\partial z / \partial x$ and $\partial z / \partial y$ equal to $0$. However, those expressions are not the same as what you wrote, since they have to be found through implicit differentiation. I'm not sure which method you were trying to use.

Dylan
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  • Thanks, I was confused by whether we have to use implicit differentiation or not. You clarified everything – rmzep Oct 24 '14 at 04:12