Answer
${\bf{n}} = \left( {\frac{4}{{\sqrt {21} }},\frac{2}{{\sqrt {21} }}, - \frac{1}{{\sqrt {21} }}} \right)$
Work Step by Step
We are given the surface ${z^2} - 2{x^4} - {y^4} = 16$ and $P = \left( {2,2,8} \right)$.
Let $F\left( {x,y,z} \right) \equiv {z^2} - 2{x^4} - {y^4}$.
So, we can write the equation of the surface as
$F\left( {x,y,z} \right) \equiv {z^2} - 2{x^4} - {y^4} = 16$
The gradient of $F\left( {x,y,z} \right)$ is
$\nabla F = \left( {{F_x},{F_y},{F_z}} \right) = \left( { - 8{x^3}, - 4{y^3},2z} \right)$
By Theorem 5, $\nabla {F_P}$ is a vector normal to the surface at $P = \left( {2,2,8} \right)$. So,
$\nabla {F_P} = \left( { - 8\cdot{2^3}, - 4\cdot{2^3},2\cdot8} \right) = \left( { - 64, - 32,16} \right)$
Notice that this vector points in the positive $z$-axis, hence away from the $xy$-plane. To find its unit normal vector ${\bf{n}}$ that points in the direction of the $xy$-plane, we need to reverse the direction of $\nabla {F_P}$. Thus,
${\bf{n}} = - \frac{{\left( { - 64, - 32,16} \right)}}{{\sqrt {\left( { - 64, - 32,16} \right)\cdot\left( { - 64, - 32,16} \right)} }} = - \frac{1}{{16\sqrt {21} }}\left( { - 64, - 32,16} \right)$
So, ${\bf{n}} = \left( {\frac{4}{{\sqrt {21} }},\frac{2}{{\sqrt {21} }}, - \frac{1}{{\sqrt {21} }}} \right)$.
