\documentclass[12pt]{math23a}
\problemset{10}
\year1999\month12\day16
\author{Li-Chung Chen}

\DeclareMathOperator{\Img}{Img}
\DeclareMathOperator{\Mat}{Mat}
\newcommand{\R}{{\mathbb R}}
\def\theenumii{\Alph{enumii}}

\begin{document}

\item[1]
\begin{enumerate}
\item % (a)
See part (b) below.
\item % (b)
$DF(x, y) = \begin{bmatrix}
3x^2 - 3y & 3y^2 - 3x
\end{bmatrix}$.  Thus $DF(x,y) = 0 \implies
x^2 = y$ and $y^2 = x \implies y = x^2 = (y^2)^2 = y^4
\implies y = 0$ or $1$ (since $y \in \R$) $\implies
(x,y) = (0,0)$ or $(1,1)$.  One can check that these pairs do
make $DF(x, y) = 0$.  Therefore, the set of real critical points is
$\{(0,0),(1,1)\}$.  This set consists of isolated points, so the
complex level curves containing $(0,0)$ or $(1,1)$ are singular at
these points.  For the purposes of this course, you can say that
this implies that the \emph{real} level curves are singular, though this
conclusion doesn't really follow.

The answer: $C_{F(0,0)} = C_0$ is smooth except at $(0,0)$,
$C_{F(1,1)} = C_{-1}$ is smooth except at $(1,1)$, and $C_r$ is
smooth for any $r \in \R-\{0,-1\}$.

The later parts of the problem will show that $C_0$ really is singular
at $(0,0)$.  As for $C_{-1}$, one can show that $(1,1)$ is an isolated
point in the curve.  (All that follows is \emph{not} required for
your solution.)  Say $\epsilon > 0$ is small.  We claim that
the equation $F(1+\epsilon,y) = -1$ has no solution near $-1$.
The equation is equivalent to
$y^3 - 3(1+\epsilon)y + (1+\epsilon)^3+1 = 0$.  Let
$f_1(y) = y^3 - 3(1+\epsilon)y + (1+\epsilon)^3+1$.  Then
$f_1'(y) = 3y^2 - 3(1+\epsilon)$, so around $1$, $f_1(y)$ has a minimum
value at $y = \sqrt{1+\epsilon}$.  But this yields
$f_1(\sqrt{1+\epsilon}) = -2(1+\epsilon)^{\frac{3}{2}}+(1+\epsilon)^3+1
= \left((1+\epsilon)^{\frac{3}{2}}-1\right)^2 > 0$.  Hence $f_1$ is
positive around $y=1$, and $C_{-1}$ has no point of the form
$(1+\epsilon,y)$ near $(1,1)$.  As for $1-\epsilon$, one can show that
near $1$, the function $f_2(y)=F(1-\epsilon,y)+1$ achieves its minimum
at $\sqrt{1-\epsilon}$, and that $f_2(\sqrt{1-\epsilon})
= \left((1-\epsilon)^{\frac{3}{2}}-1\right)^2 > 0$.  Lastly, we note
that the same argument will work for $(x,1+\epsilon)$ and
$(x,1-\epsilon)$ by symmetry.  Therefore, $C_{-1}$ does indeed have
an isolated point at $(1,1)$.

\item % (c)
$DF(x,y)$ is still $\begin{bmatrix}
3x^2 - 3y & 3y^2 - 3x
\end{bmatrix}$.  Thus $DF(x,y) = 0 \implies
x^2 = y$ and $y^2 = x \implies y = x^2 = (y^2)^2 = y^4$.  However,
$y$ is now allowed to be complex, so $y$ can be $0$, $1$, 
$\frac{-1+\sqrt{3}i}{2}$, or $\frac{-1-\sqrt{3}i}{2}$.  The relation
$x = y^2$ implies that $(x,y)$ must be $(0,0)$, $(1,1)$,
$(\frac{-1+\sqrt{3}i}{2}, \frac{-1-\sqrt{3}i}{2})$, or
$(\frac{-1-\sqrt{3}i}{2}, \frac{-1+\sqrt{3}i}{2})$.  One can check
that all four of these pairs satisfy $DF(x,y) = 0$.
Hence this is the set of critical points.  Because it consists of
isolated points, by a theorem given in class the level curves containing
these points are singular at these points.

\item % (d)
See the attached Mathematica worksheet.

\item % (e)
The problem is equivalent to showing that given $t \in \R-\{0,-1\}$, there
exists unique $(x,y) \in \R^2$ such that $F(x,y)=0$, $(x,y)\neq (0,0)$,
and $y=tx$.  This amounts to solving two equations.  If $(x,y)$ is a solution,
then $0=x^3+y^3-3xy=x^3+(tx)^3-3x(tx)=(t^3+1)x^3-3tx^2 \implies
x = 0$ or $\frac{3t}{t^3+1}$ (the latter well-defined because $t\neq -1$).
But if $x$ were $0$, then $y=tx=0$, violating the condition that $(x,y)\neq
(0,0)$.  Therefore, $x = \frac{3t}{t^3+1}$ and $y = tx = \frac{3t^2}{t^3+1}$.
This is the only possible solution, and it's easy to check that it works.
(Note that it's nonzero because $t \neq 0$.)

Note that because such $(x,y)$ is unique for each $t \in \R-\{0,-1\}$, we
can denote as $(x(t),y(t))$ the $(x,y)$ obtained for a particular $t$.
Hence $x(t) = \frac{3t}{t^3+1}$ and $y(t) = \frac{3t^2}{t^3+1}$.
These will be used in the later parts.

\item % (f)
First we show that as $t$ ranges in $\R-\{0,-1\}$, $(x(t),y(t))$ covers all
of $C_0-\{(0,0)\}$.  Suppose $(a,b)\in C_0-\{(0,0)\}$.  Then $a$ and $b$
are nonzero
(because $a = 0 \implies 0=F(a,b)=b^3 \implies b = 0$; similar for
$b = 0$).  Further,
$b/a \neq -1$ (because $b=-a \implies 0=F(a,b)=3a^2 \implies a=b=0$).
Hence $(a,b)$ satisfies the conditions $F(a,b)=0$, $(a,b)\neq(0,0)$, and
$b=(b/a)a$ for $b/a \in \R-\{0,-1\}$.  By part (e), $(a,b)$ is the unique point
that satisfies all these conditions for $t=b/a$.  Hence $(a,b)=(x(b/a),y(b/a))$.
But as $t$ ranges in $\R-\{0,1\}$, $-1+1/t$ ranges in $\R-\{0,-1\}$ (check
this).  Therefore, $(x(-1+1/t),y(-1+1/t))$ ranges in $C_0-\{(0,0)\}$, as
claimed.

Now
\begin{align*}
\phi(t) & = \left(\frac{3(-1+1/t)}{(-1+1/t)^3+1},
\frac{3(-1+1/t)^2}{(-1+1/t)^3+1}\right) & \\
& = \left(\frac{-3t^3+3t^2}{(-t+1)^3+t^3}, 
\frac{3t(-t+1)^2}{(-t+1)^3+t^3}\right)
& (\text{multiplying top and bottom by }t^3) \\
& = \left(\frac{3t^2(1-t)}{3t^2-3t+1}, \frac{3t(1-t)^2}{3t^2-3t+1}\right). &
\end{align*}
Since $3t^2-3t+1$ is always positive (why?), the above fractions'
denominators are nonzero for any $t$.  Therefore, we can \emph{define}
a function $\bar{\phi}\from\R\to\R^2$ by letting
$\phi(t) = \left(\frac{3t^2(1-t)}{3t^2-3t+1}, 
\frac{3t(1-t)^2}{3t^2-3t+1}\right)$.
This way, $\bar{\phi}$ is certainly continuously differentiable (being a quotient
of two continuously differentiable functions where the denominator is
never zero), \emph{plus} it has the property of being an extension of
the original $\phi\from\R-\{0,1\}\to\R^2$. 

The original $\phi$ covers all of $C_0-\{(0,0)\}$ as noted above, so
$\bar{\phi}$ does as well.  But in addition, $\bar{\phi}(0) = (0,0)
= \bar{\phi}(1)$.  Hence the image of $\bar{\phi}$ is exactly $C_0$,
and $\bar{\phi}$ is a parametrization for $C_0$.

Note: It is a common technique to define something first, then to prove it
has the desired properties.  This is often useful for proving existence.

\item % (g)
We've shown in (f) that $\bar{\phi}$ parametrizes $C_0$.  We claim that
it is one-to-one except for $\bar{\phi}(0)=\bar{\phi}(1)
=(0,0)$.  The key is to show that $\phi\from\R-\{0,1\}\to\R^2$ is 
one-to-one: the
function $t \mapsto -1+1/t$ is one-to-one, and the function
$u \mapsto (x(u),y(u))$ is one-to-one as well (since $(x(u),y(u))=
(x(u'),y(u')) \implies u=y(u)/x(u)=y(u')/x(u')=u'$).  Therefore,
the composition $t \mapsto \phi(t) = (x(-1+1/t),y(-1+1/t))$ is also
one-to-one.  $\bar{\phi}$ is just $\phi$ defined at two extra values of $t$,
so our claim is proved.

Next we claim that $D\bar{\phi}$ is nonzero near $0$ and near $1$.
Calculations show that $D\bar{\phi}(t)
= \begin{bmatrix}
\frac{3t(-3t^2+6t^2-6t+2)}{(3t^2-3t+1)^2}
&
\frac{3(1-t)(3t^3-3t^2+3t-1)}{(3t^2-3t+1)^2}
\end{bmatrix}$.  So in particular $D\bar{\phi}(0)$ and
$D\bar{\phi}(1)$ are nonzero.  By the continuity of $D\bar{\phi}$,
it is nonzero near $0$ and $1$.  Our second claim is proved.

Now we know that the image of $\bar{\phi}$ is $C_0$.  Let $\epsilon > 0$
be small.  
Let $\bar{x}(t) = \frac{3t^2(1-t)}{3t^2-3t+1}$ and
$\bar{y}(t) =
\frac{3t(1-t)^2}{3t^2-3t+1}$.  By one-variable calculus techniques,
one can show that
$\{t: \bar{x}(t)\in(-\epsilon,\epsilon)\}$ is of the form
$(-\delta_1,\delta_2) \cup
(1-\delta_3,1+\delta_4)$ for some small positive $\delta_1,\delta_2,
\delta_3,\delta_4$, and that
$\{t: \bar{y}(t)\in(-\epsilon,\epsilon)\}$ is of the form
$(-\delta_5,\delta_6)\cup (1-\delta_7,1+\delta_8)$.  Hence
\begin{align*}
\{t:\bar{\phi}(t) & = (\bar{x}(t),\bar{y}(t))\in
U_{\epsilon}\} \\
& = ((-\delta_1,\delta_2)\cup(1-\delta_3,1+\delta_4))\cap
((-\delta_5,\delta_6)\cup(1-\delta_7,1+\delta_8)) \\
& = (-\delta_9,\delta_{10})\cup(1-\delta_{11},1+\delta_{12})
\end{align*}
for some small positive $\delta_9,\dots,\delta_{12}$.
Let $I_1 = (-\delta_9,\delta_{10})$ and $I_2 = (1-\delta_{11},1+\delta_{12})$.
Recall that $D\bar{\phi}$ is nonzero close to $0$ and $1$.  We can force
$I_1$ and $I_2$ to be sufficiently close to $0$ and $1$ by making
the original $\epsilon$ sufficiently small.  Hence there exists
$\epsilon > 0$ such that $D\bar{\phi}$ would be nonzero in
$I_1$ and in $I_2$; we pick such an $\epsilon$ at the very beginning.

Therefore, $C_0\cap U_{\epsilon}
= \bar{\phi}(I_1 \cup I_2)
= \bar{\phi}(I_1) \cup \bar{\phi}(I_2)$.
But $\bar{\phi}\mid_{I_1}$ and $\bar{\phi}\mid_{I_2}$ are one-to-one, and
$D\bar{\phi} \neq 0$ in $I_1$ and $I_2$.  Hence
by the parameterization definition of a smooth curve,
the image $C'$ of $\bar{\phi}\mid_{I_1}$ and the image $C''$ of
$\bar{\phi}\mid_{I_2}$ are smooth curves.  The above discussion shows that
$C_0\cap U_{\epsilon} = C' \cup C''$ and that $C' \cap C'' = \{(0,0)\}$.

\item % (h)
The tangent line to $C'$ at $(0,0)$ (corresponding to $t = 0$) is given by
the parameterization $\left(D_1\bar{\phi}(0)t+0,
D_2\bar{\phi}(0)t+0\right)
= \left(0t+0,-3t+0\right) = \left(0,-3t\right)$.  This corresponds
to the line $x = 0$.

The tangent line to $C''$ at $(0,0)$ (corresponding to $t = 1$) is given by
the parameterization $\left(D_1\bar{\phi}(1)t+0,
D_2\bar{\phi}(1)t+0\right)
= \left(-3t+0,0t+0\right) = \left(-3t,0\right)$.  This corresponds
to the line $y = 0$.

\end{enumerate}

\item[2]
\begin{enumerate}
\item % (a)
Let $f = x^2+y^2-z^2$.  Then $\Sigma$ is a level set of $f$.  Now
$Df(x,y,z) =
\begin{bmatrix}
2x & 2y & -2z
\end{bmatrix}$, which is zero iff $x=y=z=0$.  However, $(0,0,0)$
is not in $\Sigma$.  Hence $Df$ is nonzero in $\Sigma$, implying that
$\Sigma$ is a smooth surface by the level
set definition of a smooth surface in $\R^3$.

The surface looks like an hourglass (an apt description provided by
some of you).

\item % (b)
Define $F\from\R^3\to\R^2$ by $F(x,y,z) =
\begin{bmatrix}
f(x,y,z) \\ x+y+z
\end{bmatrix}
=
\begin{bmatrix}
x^2 + y^2 - z^2 \\ x+y+z
\end{bmatrix}$.  Then $C_r$ is a level set of $F$:
$C_r = \left\{(x,y,z):F(x,y,z) = \begin{bmatrix}1 \\ r \end{bmatrix}
\right\}$.  Hence if for a given $r$, $DF(x,y,z)$ is onto for any 
$(x,y,z)\in C_r$, then $C_r$ is a smooth curve by the level set
definition of a smooth curve in $\R^3$.  Now $DF(x,y,z) =
\begin{bmatrix}
2x & 2y & -2z \\
1 & 1 & 1
\end{bmatrix}$.  If any two of $2x$, $2y$, and $-2z$ are nonequal,
then the two corresponding rows are linearly independent (why?) and
$DF(x,y,z)$ is onto.  The only time this doesn't happen is when
$2x=2y=-2z \implies 1=f(x,y,z)=x^2+x^2-(-x)^2 = x^2 \implies
x = 1$ or $-1 \implies r = x+x+(-x) = 1$ or $-1$.  Hence for any $r$
not equal to $1$ or $-1$, $DF(x,y,z)$ is onto in all of $C_r$, and
$C_r$ is a smooth curve.

But we have yet to settle the cases $r = 1$ or $-1$.  First
suppose $r = 1$.  Then $x+y+z=1$ and $x^2+y^2-z^2=1 \implies
1=x^2+y^2-(1-x-y)^2 = -2xy+2x+2y-1 \implies 0=xy-x-y+1 = (x-1)(y-1)
\implies x = 1$ or $y = 1$.  But once $x=1$, the two equations
reduce to $y+z=0$ and $y^2-z^2=0$, so that $y$ can be anything and
$z$ must be $-y$.  Similarly, if $y=1$, then $x$ can be anything and
$z$ must be $-x$.  Thus $C_1 =
\{(1,y,-y):y\in\R\}\cup\{(x,1,-x):x\in\R\}$.  This is just a pair
of lines in $\R^3$ that intersect at $(1,1,-1)$.  Hence $C_1$ is not
smooth and has a sole singular point $(1,1,-1)$.

Lastly suppose $r = -1$.  Then $x+y+z=-1$ and $x^2+y^2-z^2=1 \implies
1=x^2+y^2-(-1-x-y)^2 = -2xy-2x-2y-1 \implies 0=xy+x+y+1 = (x+1)(y+1)
\implies x = -1$ or $y = -1$.  From this we can deduce that
$C_{-1} = 
\{(-1,y,-y):y\in\R\}\cup\{(x,-1,-x):x\in\R\}$.  This is just a pair
of lines in $\R^3$ that intersect at $(-1,-1,1)$.  Hence $C_{-1}$ is
not smooth and has a sole singular point $(-1,-1,1)$.

\item % (c)
Let $g = x^2+y^2-z^2$.  Then $\Sigma'$ is a level set of $g$.  Now
$Dg(x,y,z) =
\begin{bmatrix}
2x & 2y & 2z
\end{bmatrix}$, which is zero iff $x=y=z=0$.  However, $(0,0,0)$
is not in $\Sigma'$.  Hence $Dg$ is nonzero in $\Sigma'$, implying that
$\Sigma'$ is a smooth surface by the level
set definition of a smooth surface in $\R^3$.

The surface looks like a hollow sphere, of radius $1$ around the origin.

Now define $G\from\R^3\to\R^2$ by $G(x,y,z) =
\begin{bmatrix}
g(x,y,z) \\ x+y+z
\end{bmatrix}
=
\begin{bmatrix}
x^2 + y^2 + z^2 \\ x+y+z
\end{bmatrix}$.  Then $C'_r = \{(x,y,z)\in\Sigma':x+y+z=r\}$ 
is a level set of $G$:
$C'_r = \left\{(x,y,z):G(x,y,z) = \begin{bmatrix}1 \\ r \end{bmatrix}
\right\}$.  Hence if for a given $r$, $DG(x,y,z)$ is onto for any 
$(x,y,z)\in C'_r$, then $C'_r$ is a smooth curve by the level set
definition of a smooth curve in $\R^3$.  Now $DG(x,y,z) =
\begin{bmatrix}
2x & 2y & 2z \\
1 & 1 & 1
\end{bmatrix}$.  If any two of $2x$, $2y$, and $2z$ are nonequal,
then the two corresponding rows are linearly independent (why?) and
$DF(x,y,z)$ is onto.  The only time this doesn't happen is when
$2x=2y=2z \implies 1=f(x,y,z)=x^2+x^2+x^2 = 3x^2 \implies
x = \frac{\sqrt{3}}{3}$ or $-\frac{\sqrt{3}}{3} \implies 
r = x+x+x = \sqrt{3}$ or $-\sqrt{3}$.  Hence for any $r$
not equal to $\sqrt{3}$ or $-\sqrt{3}$, $DG(x,y,z)$ is onto in all 
of $C'_r$, and
$C'_r$ is a smooth curve.

Now we take care of the cases $r = \sqrt{3}$ or $-\sqrt{3}$.
First suppose $r = \sqrt{3}$.  Then $(x,y,z)$ is in $C'_r$ iff it
satisfies $x+y+z=\sqrt{3}$ and
$x^2+y^2+z^2=1$.  However, the only solution is
$x=y=z=\frac{\sqrt{3}}{3}$.
(One way to see this is to note that
$0 = 1-2+1 = (x^2+y^2+z^2)-\frac{2\sqrt{3}}{3}(x+y+z)+1
= (x-\frac{\sqrt{3}}{3})^2 +
(y-\frac{\sqrt{3}}{3})^2 +
(z-\frac{\sqrt{3}}{3})^2$, so that each square must be $0$.)
Hence $C'_{\sqrt{3}} = \left\{(\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3},
\frac{\sqrt{3}}{3})\right\}$.  It is now clear that $C'_{\sqrt{3}}$
is not smooth and has the sole singularity point
$(\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3})$.

The case $r = -\sqrt{3}$ is similar and yields
$C'_{-\sqrt{3}} = \left\{(-\frac{\sqrt{3}}{3},-\frac{\sqrt{3}}{3},
-\frac{\sqrt{3}}{3})\right\}$.  Thus $C'_{-\sqrt{3}}$
is not smooth and has the sole singularity point
$(-\frac{\sqrt{3}}{3},-\frac{\sqrt{3}}{3},-\frac{\sqrt{3}}{3})$.
\end{enumerate}

\item[3]
\begin{enumerate}
\item % (a)
One direction of this problem is actually false: $C_f$ being smooth
at $(0,0,0)$ does not imply that $f_{\Sigma} \neq 0$.
Here is an example courtesy of Nils:
Let $g(x,y,z)=x^2+(y-1)^2$ and let $\Sigma=\{(x,y,z):g(x,y,z)=1\}$.
Let $f(x,y,z) = x$.  Then $C_f$ is the intersection of $\Sigma$ with
$\ker f$ (the $yz$ plane); it turns out to be the union of the lines
$\{(0,0,z): z\in\R\}$ and $\{(0,2,z): z\in\R\}$, hence a smooth curve.
However,
the tangent plane $T_{\Sigma}(0,0,0)$ is $y=0$, on which $f$ is
\emph{not} identically $0$.

The other direction is true.  Because $\Sigma$ is a smooth surface,
there exists a continuously differentiable function $g\from\R^3\to\R$
such that 
\begin{itemize}
\item
$Dg(0,0,0)\neq 0$ and
\item
near $(0,0,0)$, $\Sigma$ coincides with the level set
$\{(x,y,z):g(x,y,z)=0\}$.
\end{itemize}
Define $F\from\R^3\to\R^2$ by $F(x,y,z)
= \begin{bmatrix}
g(x,y,z) \\ f(x,y,z)
\end{bmatrix}$.  Then near $(0,0,0)$, $C_f$ coincides with the level
set $\left\{(x,y,z):F(x,y,z)=\begin{bmatrix}0 \\ 0\end{bmatrix}\right\}$.
Hence to show that $C_f$ is smooth at $(0,0,0)$, it suffices to show
that $DF(0,0,0)$ is onto.

Let $f(x,y,z)=ax+by+cz$.  Then $$DF(0,0,0) = 
\begin{bmatrix}
D_1g(0,0,0) & D_2g(0,0,0) & D_3g(0,0,0) \\
a & b & c
\end{bmatrix}.$$  Thus $DF(0,0,0)
\begin{bmatrix} x \\ y \\ z \end{bmatrix} = 
\begin{bmatrix}
D_1g(0,0,0)x + D_2g(0,0,0)y + D_3g(0,0,0)z \\
ax+by+cz
\end{bmatrix} \\
=
\begin{bmatrix}
D_1g(0,0,0)x + D_2g(0,0,0)y + D_3g(0,0,0)z \\
f(x,y,z)
\end{bmatrix}$.  Now the tangent plane $T_{\Sigma}(0,0,0)$ is exactly
the set $\{(x,y,z): D_1g(0,0,0)x + D_2g(0,0,0)y + D_3g(0,0,0)z = 0\}$,
so because $f$ is not identically $0$ on $T_{\Sigma}(0,0,0)$,
there exists $(x_1,y_1,z_1)$ such that
\begin{itemize}
\item
$D_1g(0,0,0)x_1 + D_2g(0,0,0)y_1 + D_3g(0,0,0)z_1 = 0$ and
\item
$f(x_1,y_1,z_1) = r \neq 0$.
\end{itemize}
Hence $DF(0,0,0)
\begin{bmatrix} x_1 \\ y_1 \\ z_1 \end{bmatrix} =
\begin{bmatrix} 0 \\ r \end{bmatrix}$.

Now, because $Dg(0,0,0) \neq 0$,
there exists $(x_2,y_2,z_2)$ such that
$D_1g(0,0,0)x_2 + D_2g(0,0,0)y_2 + D_3g(0,0,0)z_2 = s \neq 0$.  Thus
$\begin{bmatrix} s \\ f(x_2,y_2,z_2) \end{bmatrix} = DF(0,0,0)
\begin{bmatrix} x_2 \\ y_2 \\ z_2 \end{bmatrix}$ and
$\begin{bmatrix} 0 \\ r \end{bmatrix} = DF(0,0,0)
\begin{bmatrix} x_1 \\ y_1 \\ z_1 \end{bmatrix}$ are in the image of
$DF(0,0,0)$.  But these two vectors form a basis for $\R^2$ because
$s\neq 0$ and $t\neq 0$.  Therefore, the image of $DF(0,0,0)$ is all
of $\R^2$, i.e. $DF(0,0,0)$ is onto.  As discussed above, this implies
that $C_f$ is smooth at $(0,0,0)$.

\item % (b)
As discussed in lecture, the tangent line to $C_f$ at $(0,0,0)$ is
\begin{align*}
\ker DF(0,0,0) & =
\left\{(x,y,z): D_1g(0,0,0)x+D_2g(0,0,0)y+D_3g(0,0,0)z
\text{ and } f(x,y,z)=0\right\} \\
& = \left\{(x,y,z): (x,y,z)\in T_{\Sigma}(0,0,0) \text{ and }
f(x,y,z)=0\right\} \\
& = T_{\Sigma}(0,0,0) \cap \ker f \\
& = \ker f_{\Sigma}.
\end{align*}
\end{enumerate}

\end{document}
