$\newcommand{\R}{\mathbb R }$ $\newcommand{\N}{\mathbb N }$ $\newcommand{\Z}{\mathbb Z }$ $\newcommand{\bfa}{\mathbf a}$ $\newcommand{\bfb}{\mathbf b}$ $\newcommand{\bfc}{\mathbf c}$ $\newcommand{\bft}{\mathbf t}$ $\newcommand{\bff}{\mathbf f}$ $\newcommand{\bfF}{\mathbf F}$ $\newcommand{\bfk}{\mathbf k}$ $\newcommand{\bfg}{\mathbf g}$ $\newcommand{\bfG}{\mathbf G}$ $\newcommand{\bfh}{\mathbf h}$ $\newcommand{\bfu}{\mathbf u}$ $\newcommand{\bfv}{\mathbf v}$ $\newcommand{\bfx}{\mathbf x}$ $\newcommand{\bfp}{\mathbf p}$ $\newcommand{\bfy}{\mathbf y}$ $\newcommand{\ep}{\varepsilon}$
There are 3 natural ways to represent curve in $S\subset \R^2$:
as a graph: \begin{equation} \left. \begin{array}{c} S = \{ (x,y) \in \R^2: \ y = f(x) \mbox{ for }x\in I\}\\ \mbox{OR} \\ S = \{ (x,y) \in \R^2: \ x = f(y) \mbox{ for }y\in I\} \end{array}\right\} \label{graph} \end{equation} for some $f:I\to \R$, where $I\subset \R$ is an interval.
As a level set, that is, a set of the form
\begin{equation}
S = \{ (x,y)\in U : F(x,y) = c \}
\label{locus}\end{equation}
for some open $U\subset \R^2$, some $F:U\to \R$, and some $c\in \R$.
This is also called the zero locus
of $F$ when $c=0$.
Parametrically, that is, in the form \begin{equation} S = \{ \bff(t) : \ t\in I \} \label{parametrically}\end{equation} for some interval $I\subset \R$ and some $\bff:I\to \R^2$.
It is easy to see that $S$ is a curve in $\R^2$ that can be represented as the graph of a $C^1$ function $f:I\to \R$ for some open interval $I\subset \R$, then
$S$ can be represented as a zero locus of a function $F$ of class $C^1$, and
$S$ can also be represented parametrically, as the image of a function $\bff$ of class $C^1$.
These are true because, if $f:I\to \R$ is a $C^1$ function, then
for example
\begin{align}
\{ (x,y) \in \R^2: \ x\in I, y = f(x) \}
&=
\{(x,y)\in \R^2 : F(x,y)= 0\}
\nonumber \\
&=\{ \bfg(t): t\in I \} \nonumber
\end{align}
for $F(x,y) = y-f(x)$ and $\bfg(t) := (t, f(t))$.
Similar considerations apply if $S= \{ (x,y) \in \R^2: \ y\in I, x = f(y)\}$.
On the other hand, if a curve is represented as a zero locus, or parametrically, then it cannot always be represented as a graph.
Example 1. For example, the curve below shows the set $$ S = \{(x,y)\in \R^2 : y^2 = x^2(4-x^2) \}. $$ The curve can also be represented parametrically as $$ S = \{ \bff(t) : t\in \R\} ,\qquad\mbox{ for }\bff(t) = \binom{2\cos t}{2\sin 2t}. $$ But cannot be represnted as the graph of a $C^1$ function. Even worse, for every $r>0$, the set $S\cap B(r, \bf0)$, that is, the portion of the curve that is within distance $r$ from the origin, cannot be represented as a $C^1$ graph.
Example 2. Similarly, the curve below is (a part of) \begin{align} S &= \{(x,y) : x^5-y^2 = 0\} \nonumber \\ &= \{ \bff(t) : t\in \R\} \quad \mbox{ for }\ \ \bff(t) = \binom{t^2}{t^5}. \nonumber \end{align}
This can be written as $\{ (x,y) : x = |y|^{2/5}\}$, but the function $f(y) = |y|^{2/5}$ is not differentiable at $y=0$. So here too, for every $r>0$, the set $S\cap B(r, \bf0)$ cannot be represented as a $C^1$ graph.
Below we will say that a curve is a $C^1$ graph
as shorthand
for can be represented as a $C^1$ graph
.
Suppose that $S$ is a curve in $\R^2$ and $\bfa\in S$. The above examples suggest the following:
if, for every $r>0$, $S \cap B(r, \bfa)$ is not a $C^1$ graph, then $S$ is singular at $\bfa$.
if there exists some $r>0$ such that $S \cap B(r, \bfa)$ is a $C^1$ graph, then $S$ is regular (i.e. not singular) at $\bfa$.
We are using the words singular
and regular
as ordinary adjectives in English, not as precisely-defined mathematical terms. If we want to, we could take the above as definitions of regular and singular.
This is why we are interested in the question of when $B(r,\bfa)\cap S$ is or is not a $C^1$ graph; it gives important qualitative information about the behaviour of $S$ near $\bfa$.
We first consider a curve defined as a zero locus of a function $F$.
Theorem 1. Assume that $F:\R^2\to \R$ is $C^1$, and let $$ S := \{ \bfx\in \R^2 : F(\bfx) = 0\} $$ If $\bfa\in S$ and $\nabla F(\bfa)\ne \bf0$, then there exists some $r>0$ such that $B(r,\bfa)\cap S$ is a $C^1$ graph.
Proof.
If $\nabla F(\bfa)\ne 0$, then at least one partial derivative must be nonzero. If $\partial_y F(\bfa) \ne 0$, then the Implicit Function Theorem implies that there exists $r>0$ such that $B(r,\bfa)\cap S$ can be written in the form $\{ (x,y): y=f(x)\}$ for some $C^1$ function $f$ whose domain is a subset of $\R$. If $\partial_x F(\bfa)\ne 0$, then the Implicit Function Theorem implies that there exists $r>0$ such that $B(r,\bfa)\cap S$ can be written in the form $\{ (x,y): x=f(y)\}$ for some $C^1$ function $f$ whose domain is a subset of $\R$. In eithe case, $B(r,\bfa)\cap S$ is a $C^1$ graph.
As a result of Theorem 1, if $\nabla F$ does not vanish at any point of $S$, then $S$ is not singular anywhere. By contrast, if $\bfa$ is a point where $S$ is singular, then $\nabla F(\bfa) = 0$, that is, $\bfa$ is a critical point.
For example, you can easily check that for both Examples 1 and 2 above, the origin is a critical point of the function that appears in the definition of the curve $S$.
Example 3. (Question 3 on MAT237 Test 3, February 2018)
The picture below shows the set
$$
S := \{(x,y)\in \R^2 : y^2 - e^{2x} + 4e^x = \alpha \},
\qquad\mbox{ for a particular }\alpha\in \R.
$$
Determine the value of $\alpha$.
Solution. Please think about it before clicking.
Hey, I told you to think about it before clicking!
Let $\bfa$ denote the point where the two branches of $S$ cross. According to Theorem 1, $\bfa$ must be a critical point of $F(x,y) = y^2 - e^{2x} + 4e^x$. If you check, you will find that $F$ has exactly one critical point. So this point must be $\bfa$. You can then determine $\alpha$ by evaluating $F$ at $\bfa$.
We next consider a curve defined parametrically
Theorem 2. Assume that $\bff:(a,b)\to \R^2$ is $C^1$, and let $$ S := \{ \bff(t): t\in (a,b)\}. $$ If $\bff'(c) \ne\bf 0$ for some $c\in (a,b)$ then there exists some $r>0$ such that $\{ \bff(t) : |t-c|<r\}$ is a $C^1$ graph.
Proof (terse).
The assumption that $\bff'(c)\ne \bf0$ implies that at least one component $f_j'(c)\ne 0$, for $j=1,2$. Let us assume for concreteness that $f_1'(c)\ne 0$; the other case is essentially identical.
Since $f_1'(c)\ne 0$ and $f_1'$ is continuous, there is some $\ep>0$ such that $f_1'(t)\ne 0$ for $t\in (c-\ep, c+\ep)$. Thus $f_1$ is either strictly increasing or strictly decreasing in this interval, hence invertible.
Let $$ \bfF(t,x,y) := \left(\begin{array}{c} f_1(t) - x \\ f_2(t) - y \end{array}\right). $$ Since $f_1'(c)\ne 0$, the matrix $$ \left(\begin{array}{cc} \partial_t F_1& \partial_y F_1 \\ \partial_t F_2& \partial_y F_2 \end{array}\right) \ = \ \left(\begin{array}{cc} f_1'(t) &0 \\ f_2'(t) &-1 \end{array}\right) $$ is invertible at $(t,x,y) = (c, f_1(c), f_1(c))$, so the Implicit Function Theorem implies that there exists some $r_0, r_1>0$ and a function $\bfg: (f_1(c)-r_0,f_2(c)+r_0)\to \R^2$ such that \begin{multline}\label{iiift} \mbox{ For } (t,x,y)\mbox{ such that }|x - f_1(c)|<r_0\mbox{ and } |(t,y) - (c, f_2(c))| < r_1 , \\ \bfF(t,x,y) = {\bf 0} \mbox{ if and only if }(t,y) = \bfg(x) \end{multline}
Since $\bff$ is continuous, there exists some $r>0$ such that if $|t-c|<r$, then $(t,x,y) = (t, f_1(t), f_2(t))$ satisfies $$ |x - f_1(c)|<r_0\mbox{ and } |(t,y) - (c, f_2(c))| < r_1 . $$ Since $\bfF(t, f_1(t), f_2(t)) = \bf0$, it follows from \eqref{iiift} that the set $$ \{ (f_1(t), f_2(t)) : |t-c|<r \} = \{ (x, g_2(x)) : x\in I\} $$ where $I$ is some subset of $(f_1(c)- r_0, f_1(c)+r_0)$. This implies the conclusion of the theorem. $\quad \Box$
What this says is less intuitive than Theorem 1. If $\bff'(c)\ne \bf 0$, it does not say that $S$ is regular near $\bff(c)$. It says instead only that the parametrization is regular near $t=c$.
For example, consider again Example 1 above, where the curve can be written as $\{\bff(t) : t\in \R\}$ for $\bff(t) = \binom{2\cos t}{2\sin 2t}$. It is easy to see that
$\bff'(t)$ never vanishes. In particular, $\bff'(t)\ne \bf 0$ for $t = \pi/2, 3\pi/2,... $.
But $S$ looks singular near ${\bf0}$, which is $\bff(t)$ for $t = \pi/2, 3\pi/2,... $.
This is consistent with the theorem. To see this, consider the picture below, which shows the sets $$ \{ \bff(t) : |t-c|<\pi/4\} \quad\mbox{ for }\begin{cases} c = \pi/2 &\mbox{ in blue}\\ c = 3\pi/2 &\mbox{ in purple}. \end{cases} $$
It is easy to see that both the blue and the purple curves are $C^1$ graphs. This illustrates how the parametrization can be regular even if the curve is not.
By contrast, in Example 2 above, $\bff'(t)= 0$ only for $t=0$, and $\bff(0)$ is exactly the singular point of the curve, at the origin.
Example 4.
The picture below shows part of the set
$$
S := \{ \bff(t) : t \in \R \},
\quad\mbox { for } \bff(t) = \binom{t^3-4t^2+5t}{t^5+6t^4+4t^3-16t^2-9t}.
$$
If $\bfa$ denotes the point where the curve forms a cusp, then for which value of $t$ is it true that $\bfa = \bff(t)$?
Solution.
Here there appear to be no self-intersections
as in Example 1, so we are presumably justified in
assuming that the cusp is a point $\bff(c)$
for $c$ such that $\bff'(c)= \bf 0$.
This is actually true in this case, though it
might be hard to prove (making this
a bad question in some ways.) Try to solve for $c$.
The discussion here closely follows our discussion of curves in $\R^2$.
There are 3 natural ways to represent surface $S\subset \R^3$:
as a graph: $$ \left. \begin{array}{c} S = \{ (x,y,z) \in \R^3: \ z = f(x,y) \mbox{ for }(x,y)\in T\}\\ \mbox{OR} \\ S = \{ (x,y,z) \in \R^3: \ y = f(x,z) \mbox{ for }(x,z)\in T\}\\ \mbox{OR} \\ S = \{ (x,y,z) \in \R^3: \ x = f(y,z) \mbox{ for }(y,z)\in T\}\\ \end{array}\right\} $$ for some $f:T\to \R$, where $T$ is an open subset of $\R^2$.
As a level set, that is,a set of the form
$$
S \ = \ \{ (x,y,z)\in U : F(x,y,z) = c \}
$$
for some open $U\subset \R^3$, some $F:U\to \R$, and some $c\in \R$.
This is also called the zero locus
of $F$ when $c=0$.
Parametrically, that is, in the form $$ S \ = \ \{ \bff(\bft) : \ \bft\in T \} $$ for some open $T\subset \R^2$ and some $\bff:T\to \R^3$.
It is easy to see that $S$ is a surface in $\R^3$ that can be represented as the graph of a $C^1$ function $f:T\to \R$ for some open interval $T\subset \R^2$, then
These assertions can be proved by adapting the (easy) arguments from our discussion (above) of curves in $\R^2$.
On the other hand, if a curve is represented as a zero locus, or parametrically, then it cannot always be represented as a graph.
Examples 5. Below is a picture of the surface $$ S = \{ (x,y,z)\in \R^3 : y^2 + z^2 = x^2(4-x^2)\} $$ It can be represented parametrically as $$ \left\{ \left( \begin{array}{c}2\cos t \\ 2\cos s \, \sin 2t \\ 2\sin s\, \sin 2t \end{array} \right) : (s,t)\in \R^2 \right\} $$
It is intuitively clear that $S\cap B(r, {\bf 0})$ is not a $C^1$ graph for any $r>0$.
The same is true for the surface pictured below, which is $$ S = \{ (x,y,z) : x^5 - y^4-z^2=0 \} $$ It can also be written parametrically as (for example) $$ \left\{ \left( f_1(s,t) , s^5 , t^5 \right) : (s,t)\in \R^2 \right\} \quad \mbox{ for }f_1(s,t) := \begin{cases} (s^{20}+t^{10})^{1/5}&\mbox{ if }(s,t)\ne (0,0)\\ 0&\mbox{ if }(s,t) = (0,0). \end{cases} $$ It is not obvious that $f_1$ is $C^1$, but one can in fact check that it is.
Below we will say that a surface is a $C^1$ graph
as shorthand
for can be represented as a $C^1$ graph
.
Suppose that $S$ is a surface in $\R^3$ and $\bfa\in S$. As with curves:
This is again the reason that we are interested in the question of when $B(r,\bfa)\cap S$ is or is not a $C^1$ graph.
We first consider a surface defined as a zero locus of a function $F$.
Theorem 3. Assume that $U$ is an open subset of $\R^3$ and $F:U\to \R$ is $C^1$, and let $$ S := \{ \bfx\in U : F(\bfx) = 0\} $$ If $\bfa\in S$ and $\nabla F(\bfa)\ne \bf0$, then there exists some $r>0$ such that $B(r,\bfa)\cap S$ is a $C^1$ graph.
Proof.
As with Theorem 1, this is essentially a special case of the Implicit Function Theorem.
As a result of Theorem 3, if $\bfa$ is a point where $S$ is singular, then $\nabla F(\bfa) = 0$, that is, $\bfa$ is a critical point. And if there are no critical points of $F$ on $S$, then $S$ is regular everywhere.
For example, you can easily check that for both Examples 5 above, the origin is a critical point of the function that appears in the definition of the surface $S$.
Example 6.
The picture below shows the set
$$
S := \{(x,y,z)\in \R^3 : x^2-4x+y^4+4y^3+6y^2+4y-z^4 = \alpha \}
$$
for a particular $\alpha\in \R$. Determine the value of $\alpha$.
The idea of the solution is very similar to that of Example 3.
We next consider a parametrized surface.
Theorem 4. Assume that $T$ is an open subset of $\R^2$ and that $\bff:T\to \R^3$ is $C^1$, and let $$ S := \{ \bff(\bft): \bft\in T \}. $$ If $D\bff(\bfc)$ has rank $2$ at some $\bfc\in T$ (that is, if $\{ \partial_1 \bff , \partial_2\bff\}$ are linearly independent) then there exists some $r>0$ such that $\{ \bff(\bft) : \bft \in B(r,\bfc) \}$ is a $C^1$ graph.
The proof is similar in spirit to that of Theorem 2. Filling in the details is an exercise (probably a little on the hard side).
Like Theorem 2, the conclusion of the theorem is not that $S$ is regular near $\bff(\bfc)$, but only that the parametrization is regular near $\bfc$.
There are 3 natural ways to represent a curve $S\subset \R^3$:
as a graph: $$ \left. \begin{array}{c} S = \{ (x,y,z) \in \R^3: \ (x,y) = \bff(z) \mbox{ for }z\in (a,b)\}\\ \mbox{OR} \\ S = \{ (x,y,z) \in \R^3: \ (x,z) = \bff(y) \mbox{ for }y\in (a,b)\}\\ \mbox{OR} \\ S = \{ (x,y,z) \in \R^3: \ (y,z) = \bff(x) \mbox{ for }x\in (a,b)\}\\ \end{array}\right\} $$ for some $\bff:(a,b)\to \R^2$, where $(a,b)$ is some interval.
As a level set, that is,a set of the form
$$
S \ = \ \{ (x,y,z)\in U : \bfF(x,y,z) = \bfc \}
$$
for some open $U\subset \R^3$, some $\bfF:U\to \R^2$, and some $\bfc\in \R^2$.
This is also called the zero locus
of $\bfF$ when $\bfc=\bf0$.
Parametrically, that is, in the form $$ S \ = \ \{ \bff(t) : t\in(a,b) \} $$ for some interval $(a,b)\subset \R$ and some $\bff:(a,b)\to \R^3$.
It is easy to see that $S$ is a curve in $\R^3$ that can be represented as the graph of a $C^1$ function $\bff:I\to \R^2$ for some open interval $I\subset \R$, then
These assertions can be proved by adapting the (easy) arguments from our discussion (above) of curves in $\R^2$.
On the other hand, if a curve is represented as a zero locus, or parametrically, then it cannot always be represented as a graph.
Example 7. Below is a picture of the curve $$ S = \{ (x,y,z) : \bfF(x,y,z) = {\bf 0}\}\quad\mbox{ for } \bfF(x,y,z) = \binom{x^2-z}{y^3-z} $$ It can be represented parametrically as $$ \left\{ \left( \begin{array}{c}t^3 \\t^2 \\ t^6 \end{array} \right) : t\in \R \right\} $$
It is intuitively clear that $S\cap B(r, {\bf 0})$ is not a $C^1$ graph for any $r>0$.
Suppose that $S$ is a curve in $\R^3$ and $\bfa\in S$. As above:
This is again the reason that we are interested in the question of when $B(r,\bfa)\cap S$ is or is not a $C^1$ graph.
We first consider a curve defined as a zero locus of a function $\bfF$.
Theorem 5. Assume that $U$ is an open subset of $\R^3$ and $\bfF:U\to \R^2$ is $C^1$ with components denoted $(F_1,F_2)$, and let $$ S := \{ \bfx\in U : \bfF(\bfx) = {\bf 0} \} $$ If $\bfa\in S$ and rank $(D \bfF(\bfa))= 2$ (that is, if $\{ DF_1(\bfa), DF_2(\bfa)\}$ are linearly independent) then there exists some $r>0$ such that $B(r,\bfa)\cap S$ is a $C^1$ graph.
Proof.
As with Theorems 1 and 3, this is essentially a special case of the Implicit Function Theorem. Here, the point is that if rank$(D F(\bfa))= 2$, then at least one of $$ \left( \begin{array}{cc} \partial_x F_1 & \partial_y F_1 \\ \partial_x F_2 & \partial_y F_2 \end{array}\right), \quad \left( \begin{array}{cc} \partial_x F_1 & \partial_z F_1 \\ \partial_x F_2 & \partial_z F_2 \end{array}\right), \quad \left( \begin{array}{cc} \partial_y F_1 & \partial_z F_1 \\ \partial_y F_2 & \partial_z F_2 \end{array}\right), $$ must be invertible, and hence the Implicit Function Theorem implies that at least one pair of variables can be written as a function of the remaining variable near $\bfa$. When written out carefully, this implies the conclusion of the theorem.
As a result of Theorem 5, if $\bfa$ is a point where $S$ is singular, then $D \bfF(\bfa)$ has rank less than $2$.
For example, you can easily check that for Example 7 above, $D\bfF({\bf 0})$ has rank $1$.
We finally consider a parametrized curve.
Theorem 6. Assume that $I$ is an open interval and that $\bff:I\to \R^3$ is $C^1$, and let $$ S := \{ \bff(t): t\in I \}. $$ If $\bff'(c)\ne \bf 0$ at some $c\in I$ then there exists some $r>0$ such that $\{ \bff(t) : |t-c|<r \}$ is a $C^1$ graph.
The proof is similar in spirit to those of Theorems 2 and 4. Filling in the details is an exercise (probably a little on the hard side).
Like Theorems 2 and 4, the conclusion of the theorem is not that $S$ is regular near $\bff(c)$, but only that the parametrization is regular near $\bfc$.
Note that in Example 7, $\bff'(0) = \bf 0$, consistent with Theorem 6.
All the above examples fall into the category of a
$m$-dimensional object
in $\R^M$, where $1\le m < M$.
For completely general $m,M$, we can represent such an object as a level set, parametrically, or (sometimes) as a $C^1$ graph.
Theorems 1, 3, and 5 are special cases of a general theorem. Similarly, Theorems 2, 4, and 6 are special cases of a sloghtly different general theorem. If you are interested, it is possible to extrapolate these general theorems from the cases considered above.
This exercise is barely related to the Implicit Function Theorem but is useful for some of the later exercises.
Let $F_1$ and $F_2$ be $C^1$ functions defined on some open set $U\subset \R^2$. Further define
$$
F_3 = F_1 F_2\qquad F_4 = F_1^2 + F_2^2,
$$
and for $j=1,2,3,4$, let $S_j := \{(x,y)\in \R^2 : F_j(x,y)= 0\}$.
(a) Prove that $S_3 = S_1\cup S_2$.
(b) Prove that $S_4 = S_1\cap S_2$.
Whether or not you write out the proofs of these facts, feel free to use them below.
For some problems, it may be useful to review Tutorial 8.
For each of the following functions $F(x,y)$, define $S := \{ (x,y)\in \R^2 : F(x,y)= 0\}$. Then
The functions are:
(a) $F(x,y) = x^2+y^2 -1$.
(b) $F(x,y) = (y-4)(y-x)(y+x)$.
(c) $F(x,y) = (y-x^2+1)(y+x^2-1)$.
(d) $F(x,y) = (x^2+y^2-1)^2$.
(e) $F(x,y) = (x^2+y^2 -1)(x^2+y^2)$.
(f) $F(x,y) = y(y-x^2)$.
(g) $F(x,y) = (y-1)^3 - x^2$.
(h) $F(x,y) = (x^+)^2 + y^2$, where $x^+ := \max\{ x, 0\}$ .
You do not need to go through the exercise for all of these, but please do enough to be confident that you can answer this type of question.
Let
$$
F(x,y) = 2x^4 - 8xy +4y^2 +2.
$$
(a) Draw a contour plot, showing the level sets
$$
\{ (x,y) : F(x,y) = c\} \qquad \mbox{ for }c = -1, 0,1,2,3.
$$
To do this by hand, you can think of the equation $F(x,y)= c$
as a quadratic equation for $y$, by writing it as
$4 y^2 - (8x) y + (2x^4+2-c) = 0$. This can be solved for $y$, by using the quadratic formula. This may be time-consuming to plot by hand for several different choices of $c$, but if you rely entirely on a computer, you will probably get some details wrong. Perhaps it is best to rely on a combination of
computer and analysis.
(b) Find all critical points of $F$
and indicate them on your contour plot. Interpret what you see in terms of the implicit function theorem and/or theorems about curves in $\R^2$.
(c) If possible, determine the character of each critical point (local max, local min, or saddle point). Do the results of this exercise make sense in terms of the contour plot?
Let
$$
F(x,y) = 2(x^2 - 1)(y^2-1)
$$
(a) Draw a contour plot, showing the level sets
$$
\{ (x,y) : F(x,y) = c\} \qquad \mbox{ for }c = -1, 0,1,2,3.
$$
If you rely entirely on a computer, you will
probably get some details wrong. Also, for $c=0$ it is easy to plot the level
set, and for $c\ne 0$ it is easy to solve for $y$ as a function of $x$, or vice versa, although after solving it still may be hard to plot.
(b) Find all critical points of $f$
and indicate them on your contour plot. Interpret what you see in terms of the implicit function theorem and/or theorems about curves in $\R^2$.
(c) If possible, determine the character of each critical point (local max, local min, or saddle point). Do the results of this exercise make sense in terms of the contour plot?
Let
$$
F(x,y) = \frac 14(5x^2 + y^2-4)(x^2+5y^2-4)
$$
Below is a contour plot showing the level sets
$$
\{ (x,y) : F(x,y) = c\} \qquad\mbox{ for }c = -5, -4, -3, \cdots , 10.
$$
Unfortunately, the level sets are not labelled.
Let
$$
F(x,y) =(2y-x^2)(2x-y^2).
$$
Below is a contour plot showing the level sets
$$
\{ (x,y) : F(x,y) = c\} \qquad\mbox{ for }c = -5, -4.5, -4, \cdots ,9.5 , 10.
$$
Unfortunately, the level sets are not labelled.
Below is a contour plot showing the level sets
$$
\{ (x,y) : F(x,y) = c\} \qquad\mbox{ for }c = -45, -44,\ldots, 45
$$
for a $C^1$ function $F$. Unfortunately, the level sets are not labelled
and the formula for $F$ has been lost.
Given $\bff:(a,b) \to \R^2 $, we say that $ \bff $ is a regular parametrization of
$ S := \{ \bff(t) : a < t < b \}$ if $\bff'(t) \ne {\bf 0}$
for all $t\in (a,b)$.
For each $\bff:\R\to \R^2$ below,
The functions are:
(a) $\bff(t) = (t-1,t+1)$.
(b) $\bff(t) = (t^2-1,t^2+1)$.
(c) $\bff(t) = (t^3-1, t^3+1)$.
(d) $\bff(t) = (\cos^2 t, \sin^2 t)$.
(e) $\bff(t) = (\cos t^2, \sin t^2)$.
(f) $\bff(t) = (\cos^3 t, \sin^3 t)$.
(a) Answer the question from Example 6.
(b) A variant of the same question: The picture below shows the set
$$
S := \{(x,y,z)\in \R^3 : x^2-2x - y^2+2yz-z^2-z^3 = \alpha \}
$$
for a particular $\alpha\in \R$. Determine the value of $\alpha$.
(a) Assume that
$$
F(x,y,z) = x^4+y^4+ 2x^2y^2 - 2\alpha(x^2+y^2) +\alpha^2 - z^2
$$
for some $\alpha\in \R$, and let
$$
S := \{(x,y,z)\in \R^3 : F(x,y,z) = 0\}.
$$
Find all points in $S$ where $\nabla F = \bf 0$. (Your answer may depend on $\alpha$.)
Hint. $F(x,y,z) = [(x^2+y^2) -\alpha]^2 - z^2$
(b) Asssume that the picture below shows $S$. Using your answer to part (a), determine whether $\alpha$ is positive, negative, or zero. Justify your answer.
For each of the functions $F:\R^3\to \R$ below, let $$ S := \{(x,y,z)\in \R^3 : F(x,y,z) = \alpha \}, $$ for some $\alpha\in \R$. Determine for which values of $\alpha$ it is true that $S$ is regular (in the sense that for every $\bfx\in S$, there exists $r>0$ such that $S\cap B(r,\bfx)$ is a $C^1$ graph.)
(a) $F(x,y,z) = (x^2+(y-z)^2)z$
(b) $F(x,y,z) = x^2+2xy+ 2yz - z^2$.
(c) $F(x,y,z) = x^2+2xy+ 2xz - z^2$.
Determine which
of the following sets $S$ have the property that
$$
\forall \bfx\in S, \exists r>0\mbox{ such that }S\cap B(r,\bfx)\mbox{ is a }C^1\mbox{graph}.
$$
Prove that
your answer is correct.
(a) $ S = $ the empty set.
(b) $S$ ia a set consisting of a single point, say $S = \{(a,b)\}$ for some $(a,b)\in \R^2$.
(c)
$S = \{(x,y)\in \R^2 : |x|< 1, y=0 \}$.
(d)
$S = \{(x,y)\in \R^2 : |x|\le 1, y=0 \}$.
Let $$ F(x,y) = \frac{y^3-x^8y}{x^6+y^2}. $$ In the exercises in Section 2.1 of the online notes, you (may) have shown that $F$ is differentiable everywhere in $\R^2$, and that $\nabla F(0,0) = (0,1)$. Thus $\partial_y F(0,0)\ne 0$.
(a) Sketch the set $S:= \{(x,y)\in \R^2 : F(x,y) = 0\}$.
(b) Does there exist any $r>0$ such that $S\cap B(r,{\bf 0})$ is a $C^1$ graph?
(c) Is this consistent with the Implicit Function Theorem (that is, with Theorem 1 above)? is there any real or apparent contradiction? If so, how can this contradiction be resolved?
Write out the proof of Theorem 3, which is a modification of the proof of Theorem 1.
Write out the proof of Theorem 4, which is a modification of the proof of Theorem 2. (It may help to consult the proof of Theorem 5.)