horizontal tangent planes: how to find their equations...?

[petrol.veem]

how does one go about finding equations of the horizontal tangent planes to a surface?

right now i'm thinking i need to set the gradient vector to 0, almost like a critical point or something like that.

am i on the right path?

 

[pka]

Re: horizontal tangent planes

how does one go about finding equations of the horizontal tangent planes to a surface?

I suppose the answer depends on what one means by “horizontal” in \(\displaystyle \Re ^3\)?
If it means parallel to the xy-plane then the normal vector would be parallel to \(\displaystyle \left\langle {0,0,1} \right\rangle\).

 

[petrol.veem]

the question in the text book reads:

Find the equation(s) of the horizontal tangent plane(s) to the surface f(x,y) = 4(x-1)^2 + 3(y+1)^2

Obviously the gradient is equal to (8x-8,6y+6)

So pka, are you suggesting that (x,y,z) = (0,0,1), thus making the gradient (-8,6)?

 

[pka]

In order to have a plane we must be in \(\displaystyle \Re^3\), i.e. in three dimensions.
It is usual to write \(\displaystyle z = f(x,y) = 4\left( {x - 1} \right)^2 + 3\left( {y - 1} \right)^2\) which becomes \(\displaystyle f(x,y,z) = 4\left( {x - 1} \right)^2 + 3\left( {y - 1} \right)^2 - z\).
So the gradient is \(\displaystyle \nabla f = \left\langle {8\left( {x - 1} \right),6\left( {y - 1} \right), - 1} \right\rangle\) from which we get the point \(\displaystyle \left( {1, - 1,0} \right)\) where there is a horizontal plane.