\section{Graphs and Surfaces}

\begin{enumerate}[1.]
\item
{\it 
Let $f(x,y)=\sqrt{y-x^2}$.
Find the domain of $f$, and find the equations of its level curves.}

\bigskip
The domain is $\{(x,y)\in{\mathbb R}^2:\ y\geq x^2\}$. The level curves $f(x,y)=c$ for $c\geq 0$ are 
parabolas $y=x^2+c^2$.

\bigskip

\item {\it 
Find the function $f(x,y,z)$ if for each constant $D$ the level surface $f(x,y,z)=D$ is the plane  perpendicular to $\mathbf{n}= -3 \ii + \jj + 2\kk$ and passing through the point $(0,0,D^5 )$.}

\bigskip
First we find the equation of the plane described above.  Since the plane is normal to $\mathbf{n}$ and passes through the point $(0,0,D^5)$, we know that any point $Q =( x, y,z )$ on $\mathcal{P}$ necessarily satisfies the equation
$$
0=\langle x,y,z-D^5\rangle  \cdot \mathbf{n} = -3 x + y + 2(z-D^5),
$$
which simplifies to $-3x + y + 2z = 2 D^5$.  

Now, if $(x,y,z)$ is such that $f(x,y,z) = D$, then for that $(x,y,z)$ we have
 $$-3x + y + 2z = 2 D^5=2(f(x,y,z))^5.
 $$
We solve this to find
$$
f(x,y,z) = 2^{-1/5} (-3x + y + 2z)^{1/5}.
$$
\end{enumerate}
