Question

How do you find an equation for the horizontal tangent plane to the surface $z=4{{\left( x-1 \right)}^{2}}+3{{\left( y+1 \right)}^{2}}$ ?

Answer 1

Hint: In order to find a solution for this question, we have to remember that a horizontal plane is tangent to a curve in the space in its points of maximum, minimum or saddle. Tangent planes to a surface are planes that just touch the surface at the point and are “parallel” to the surface at the point.

Complete step by step solution:
As we have our equation as:
$z=4{{\left( x-1 \right)}^{2}}+3{{\left( y+1 \right)}^{2}}$
We will first have to calculate the two partial derivative,
For $x$ , since we have: $4{{\left( x-1 \right)}^{2}}$ so we get:
$\Rightarrow \dfrac{\partial z}{\partial x}=4\times 2\left( x-1 \right)$
For $y$ , since we have: $3{{\left( y+1 \right)}^{2}}$ so we get:
$\Rightarrow \dfrac{\partial z}{\partial y}=3\times 2\left( y+1 \right)$
$\Rightarrow \dfrac{\partial z}{\partial y}=6\left( y+1 \right)$
Now put $ =0$ in the system of two previous equations (that is in place of derivative) and thus it will give us the stationary points:
$8\left( x-1 \right)=0$
$6\left( y+1 \right)=0$
And therefore we get:
$\begin{align}
  & \Rightarrow x=1 \\
 & \Rightarrow y=-1 \\
\end{align}$
From this we can conclude that $z=0$
Now we have to find what type of stationary point \[P\left( 1,-1,0 \right)~\] is.
So, to do that we have to calculate the four partial derivative of the second order:
$\Rightarrow \dfrac{\partial z}{\partial x}=8\left( x-1 \right)$
By applying second order of above derivative, we get:
$\Rightarrow \dfrac{{{\partial }^{2}}z}{\partial {{x}^{2}}}=8$
Also, for:
$\Rightarrow \dfrac{\partial z}{\partial y}=6\left( y+1 \right)$
By applying second order of above derivative, we get:
$\Rightarrow \dfrac{{{\partial }^{2}}z}{\partial {{y}^{2}}}=6$
With this we can conclude as:
$\dfrac{{{\partial }^{2}}z}{\partial x\partial y}=\dfrac{{{\partial }^{2}}z}{\partial y\partial x}=0$
Also, the hessian matrix will be:
$\left[ \begin{matrix}
   8 & 0 \\
   0 & 6 \\
\end{matrix} \right ]$
Since its determinant in P is positive $48$ and $\dfrac{{{\partial }^{2}}z}{\partial {{x}^{2}}}$ is positive $8$
Therefore, the point P is a local minimum, so there exists a horizontal tangent plane.
Also, since it has to pass from P which \[z=0\], its equation is \[z=0\]

Note: If the tangent plane is horizontal, the gradient must pointed in the $z$-direction, therefore the $x$ and $y$ components are $0$. So it follows $x=1,y=-1$. Therefore, this gives us a point that is on the plane. Since the tangent plane and the surface touch at $\left( {{x}_{0}},{{y}_{0}} \right)$the following point will be on both the surface and the plane \[({{x}_{0}},{{y}_{0}},{{z}_{0}})=({{x}_{0}},{{y}_{0}},f({{x}_{0}},{{y}_{0}}))\].
Alternate method:
If we can recognize the curve of an equation:
As we can see this function is the equation of an elliptic paraboloid with concavity upwards.
Also, Since $z$ is surely positive or zero because the sum of two quantities will be positive or zero.
Hence the horizontal plane is tangent to a curve in the space in its points is minimum, then the vertex, is zero, and this happens where:
\[x=1\] and \[y=-1\]
So the vertex is \[V\left( 1,-1,0 \right)~\]and therefore the plane requested is the floor of the $3$-dimensional space:
Therefore,
$z=0$ is the answer.