Answer
(a) $z = - 4 \mp \sqrt {{x^2} + {y^2} - 12x + 12} $
(b) The formula $z = - 4 + \sqrt {{x^2} + {y^2} - 12x + 12} $ defines the portion of the surface satisfying $z \ge - 4$.
The formula $z = - 4 - \sqrt {{x^2} + {y^2} - 12x + 12} $ defines the portion satisfying $z \le - 4$.
(c)
1. using the formula $z = f\left( {x,y} \right)$
For $z = f\left( {x,y} \right) = - 4 + \sqrt {{x^2} + {y^2} - 12x + 12} $
$\frac{{\partial z}}{{\partial x}} = \frac{{x - 6}}{{\sqrt {{x^2} + {y^2} - 12x + 12} }}$
For $z = f\left( {x,y} \right) = - 4 - \sqrt {{x^2} + {y^2} - 12x + 12} $
$\frac{{\partial z}}{{\partial x}} = - \frac{{x - 6}}{{\sqrt {{x^2} + {y^2} - 12x + 12} }}$
2. via implicit differentiation
$\frac{{\partial z}}{{\partial x}} = - \frac{{2x - 12}}{{ - 2z - 8}} = \frac{{x - 6}}{{z + 4}}$
For $z = f\left( {x,y} \right) = - 4 + \sqrt {{x^2} + {y^2} - 12x + 12} $
$\frac{{\partial z}}{{\partial x}} = \frac{{x - 6}}{{\sqrt {{x^2} + {y^2} - 12x + 12} }}$
For $z = f\left( {x,y} \right) = - 4 - \sqrt {{x^2} + {y^2} - 12x + 12} $
$\frac{{\partial z}}{{\partial x}} = - \frac{{x - 6}}{{\sqrt {{x^2} + {y^2} - 12x + 12} }}$
The two methods give the same results. Hence, the answers agree.
Work Step by Step
(a) We have $F\left( {x,y,z} \right) = {x^2} + {y^2} - {z^2} - 12x - 8z - 4 = 0$.
We arrange the terms such that
$F\left( {x,y,z} \right) = - {z^2} - 8z + \left( {{x^2} + {y^2} - 12x - 4} \right) = 0$
Using the quadratic formula we solve for $z$ as a function of $x$ and $y$:
$z = \frac{{8 \pm \sqrt {{{\left( { - 8} \right)}^2} - 4\left( { - 1} \right)\left( {{x^2} + {y^2} - 12x - 4} \right)} }}{{2\left( { - 1} \right)}}$
$z = - 4 \mp \sqrt {{x^2} + {y^2} - 12x + 12} $
(b) From part (a) we obtain
$z = - 4 \mp \sqrt {{x^2} + {y^2} - 12x + 12} $.
The formula $z = - 4 + \sqrt {{x^2} + {y^2} - 12x + 12} $ defines the portion of the surface satisfying $z \ge - 4$.
The formula $z = - 4 - \sqrt {{x^2} + {y^2} - 12x + 12} $ defines the portion satisfying $z \le - 4$.
(c)
1. Calculate $\frac{{\partial z}}{{\partial x}}$ using the formula $z = f\left( {x,y} \right)$
For $z = f\left( {x,y} \right) = - 4 + \sqrt {{x^2} + {y^2} - 12x + 12} $
$\frac{{\partial z}}{{\partial x}} = \frac{{x - 6}}{{\sqrt {{x^2} + {y^2} - 12x + 12} }}$
For $z = f\left( {x,y} \right) = - 4 - \sqrt {{x^2} + {y^2} - 12x + 12} $
$\frac{{\partial z}}{{\partial x}} = - \frac{{x - 6}}{{\sqrt {{x^2} + {y^2} - 12x + 12} }}$
2. Calculate $\frac{{\partial z}}{{\partial x}}$ via implicit differentiation
Let $z$ be defined implicitly as a function of $x$ and $y$ by the equation
$F\left( {x,y,z} \right) = - {z^2} - 8z + {x^2} + {y^2} - 12x - 4 = 0$
The partial derivatives are
$\frac{{\partial F}}{{\partial x}} = 2x - 12$, ${\ \ }$ $\frac{{\partial F}}{{\partial y}} = 2y$, ${\ \ }$ $\frac{{\partial F}}{{\partial z}} = - 2z - 8$
Using the Chain Rule we get
$\frac{{\partial F}}{{\partial x}}\frac{{\partial x}}{{\partial x}} + \frac{{\partial F}}{{\partial y}}\frac{{\partial y}}{{\partial x}} + \frac{{\partial F}}{{\partial z}}\frac{{\partial z}}{{\partial x}} = 0$
Since $\frac{{\partial x}}{{\partial x}} = 1$, ${\ \ }$ $\frac{{\partial y}}{{\partial x}} = 0$, so
$\frac{{\partial F}}{{\partial x}} + \frac{{\partial F}}{{\partial z}}\frac{{\partial z}}{{\partial x}} = 0$
$\frac{{\partial z}}{{\partial x}} = - \frac{{\partial F}}{{\partial x}}/\frac{{\partial F}}{{\partial z}}$
Substituting $\frac{{\partial F}}{{\partial x}}$ and $\frac{{\partial F}}{{\partial z}}$ in the equation above gives
$\frac{{\partial z}}{{\partial x}} = - \frac{{2x - 12}}{{ - 2z - 8}} = \frac{{x - 6}}{{z + 4}}$
For $z = f\left( {x,y} \right) = - 4 + \sqrt {{x^2} + {y^2} - 12x + 12} $
$\frac{{\partial z}}{{\partial x}} = \frac{{x - 6}}{{\sqrt {{x^2} + {y^2} - 12x + 12} }}$
For $z = f\left( {x,y} \right) = - 4 - \sqrt {{x^2} + {y^2} - 12x + 12} $
$\frac{{\partial z}}{{\partial x}} = - \frac{{x - 6}}{{\sqrt {{x^2} + {y^2} - 12x + 12} }}$
We conclude that the two methods give the same results. Hence, the answers agree.
