Problem on finding a tangent plane
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Give the equation for the tangent plane to the surface $z x^2 + x y^2 + y z^2 = 5$ at the point $(-1,1,2)$.
Solution
Because we seek a plane, recall that
A plane is specified by a point and a normal vector
To specify a plane, we need a point $\bfx_0$ and a normal vector $\bfn$.
The plane going through $\bfx_0$ with normal vector $\bfn$ is given by $$(\mathbf{x} - \mathbf{x_0}) \cdot \mathbf{n} = 0$$
We know that the plane contains the point $\mathbf{x_0} = \langle -1,1,2 \rangle$. Thus, all we need is a normal vector to the plane.
The normal vector to our plane will be given by a normal vector to the surface.
Recall that
Normal vector to an implicitly defined surface
Gradient is perpendicular to level curves
Level curves and surfaces
To find a normal vector to a surface, view that surface as a level set of some function $g(x,y,z)$.
A normal vector to the implicitly defined surface $g(x,y,z) = c$ is $\nabla g(x,y,z)$.
We observe that the surface is given implicitly. That is, it is of the form $g(x,y,z) = c$, where $g(x,y,z) = z x^2 + x y^2 + y z^2 $ and $c=5$.
Definition of the gradient
A normal vector to the surface is given by by the gradient \begin{align}\nabla g(x,y,z) &= \langle 2xz + y^2, 2xy + z^2, 2yz+x^2 \rangle, \\
\nabla g(-1,1,2) &= \langle -3, 2, 5 \rangle.
\end{align}
Letting $\bfx = \langle x, y, z \rangle$, the normal plane is given by \begin{align} \Bigl ( \langle x,y,z \rangle - \langle -1,1, 2 \rangle \Bigr) \cdot \langle -3, 2, 5 \rangle &= 0 \\ -3(x+1) +2 (y-1) + 5 (z-2) &= 0 \\ -3 x + 2 y + 5z &=15. \end{align}
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