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%  Homework for the course "Calculus II", 
%  Spring semester, 2024, Wei-Cheng Wang.

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\begin{document}

%          The title and header.

\noindent
{\scriptsize Calculus II, Spring 2024 
(http://www.math.nthu.edu.tw/\~\,wangwc/)} \hfill 

\begin{center}
\large
Homework 09   
\normalsize
\end{center}

\vspace{.3in}

%           The questions!

\begin{enumerate}

% 14.4: The Chain Rule.
\item
Section 14.4: 
Problems 1, 7, 10, 21, 24, 29, 31, 43, 51.

% Same as s14.10, problem 9
% 14.4: The Chain Rule.
% \item
% Section 14.4: 
% Suppose that $F(x,y,z) = 0$ can implicitly define
% $x = f(y,z)$, or 
% $y = g(z,x)$, or 
% $z = h(x,y)$ near some point $(x_0,y_0,z_0)$ with
% $F(x_0,y_0,z_0) = 0$.
% (for example, $F(x,y,z) = x + 2y + 3z -4$ can).
% Show that, for any such
% point $(x_0,y_0,z_0)$, we have
% \begin{equation} \label{m1}
% {\partial f \over \partial y}
% {\partial g \over \partial z}
% {\partial h \over \partial x}
% =
% {\partial f \over \partial z}
% {\partial g \over \partial x}
% {\partial h \over \partial y}
% = -1
% \end{equation}
%%% % Hint: Read Example 5 and Example 6.
% 
% Remark: Sometimes the identity \eqref{m1}
% may be written as or another notation (s14.10 problem 9)
% \begin{equation} \label{m2} 
% \left({\partial x \over \partial y}\right)_z
% \left({\partial y \over \partial z}\right)_x
% \left({\partial z \over \partial x}\right)_y
% =
% \left({\partial x \over \partial z}\right)_y
% \left({\partial y \over \partial x}\right)_z
% \left({\partial z \over \partial y}\right)_x
% = -1
% \end{equation}
% or simply (and confusingly)
% \begin{equation} \label{m3} 
% \left({\partial x \over \partial y}\right)
% \left({\partial y \over \partial z}\right)
% \left({\partial z \over \partial x}\right)
% =
% \left({\partial x \over \partial z}\right)
% \left({\partial y \over \partial x}\right)
% \left({\partial z \over \partial y}\right)
% = -1
% \end{equation}
% The identities \eqref{m2} or \eqref{m3}
% are exactly the same as \eqref{m1} with
% a somewhat confusing notation.
% 
% 


% % 14.5: Directional Derivatives and Gradient Vectors.
\item
Section 14.5: 
Problems 9, 15, 19, 25, 27, 29, 35, 36, 40 (See page 850).

% 14.5: Directional Derivatives and Gradient Vectors.
\item Section 14.5:
Let $f_1(x,y) = \sqrt{x^2+y^2}^{1\over 2} = (x^2+y^2)^{1\over 4}$, \\ 
$f_2(x,y) = 2x+ 3y + 4 +\sqrt{x^2+y^2}^{3\over 2}
          = 2x+ 3y + 4 + (x^2+y^2)^{3\over 4} $  and \\ 
$f_3(x,y) = \ds{x^3 \over x^2+y^2}$ for $(x,y) \neq (0,0)$,
$f_3(0,0) = 0$.
\begin{enumerate}
\item Are $f_i$ continuous at $(0,0)$?
\item Do $\partial_x f_i$ and $\partial_y f_i$ exist at $(0,0)$?
\item Use the definition of directional derivative
to evaluate $\ds{df_i \over ds}_{(0,0), (\cos\theta,\sin\theta)}$,
i.e.  the directional derivative
of $f_i$ at $(x_0, y_0) = (0,0)$ in the direction $(\cos\theta,\sin\theta)$,
if it exists.
\item Are $f_i$ differentiable at $(0,0)$?

Hint:
\begin{enumerate}
\item
If you know or guess that $f$ is differentiable at $(x_0,y_0)$,
you can try to prove it using Theorem 3 (Section 14.3, page 832). It may or may not work.
\item
If you know or guess that $f$ is NOT differentiable at $(x_0,y_0)$,
you can try to prove it using Theorem 4 (Section 14.3, page 832)
or Theorem 9 (Section 14.5, page 847). It may or may not work.
\item
Note: You definitely CANNOT use Theorem 3 in (ii) or 
Theorem 4, Theorem 9 in (i).
If you don't know whether 
$f$ is differentiable at $(x_0,y_0)$ or not,
it is always safe to follow the definition as outlined in page 7 of Lecture 15.
It almost always works.
\end{enumerate}

% How to prove $f$ is not differentiable at $(x_0,y_0)$?

% If you don't know the answer yet, it is always safe to follow the definition (as the method in page 7 of Lecture 15.
% If you already know (or guess) the answer is "No", you can apply 

\end{enumerate}
\end{enumerate}
 
\end{document}

% 14.3: 
\item
Section 14.3: 
(Optional, try it if time permits)

Consider the function
\begin{equation} \label{f1}
f(x,y) = \left \{ \begin{array}{rc}
                \ds{xy \over \sqrt{x^2+y^2}}, & (x,y) \neq (0,0) \\
                0, & (x,y) = (0,0) 
                \end{array}
             \right. 
\end{equation}
\begin{enumerate}
\item
Show that
\begin{equation} \label{f2}
\partial_x f(x,y) = \left \{ \begin{array}{rc}
                \ds{y^3 \over (\sqrt{x^2+y^2})^3}, & (x,y) \neq (0,0) \\
                0, & (x,y) = (0,0) 
                \end{array}
             \right. 
\end{equation}
Hint:
 For $(x,y) \neq (0,0)$, compute the derivative directly.
 For $(x,y) = (0,0)$, find the partial derivative from definition.
\item
Use Two Path Test to show that $\partial_x f$ is not continuous at $(0,0)$.

Since 
$\partial_x f$ is not continuous at $(0,0)$, Theorem 3 is not
applicable and therefore the differentiability of $f$ at
$(0,0)$ is still inconclusive. To find out whether
$f$ is differentiable at
$(0,0)$ or not, we continue with the following steps.
 
\item
Find the linear function 
$L(x,y) = f(0,0) + \partial_x f(0,0)(x-0) + \partial_y f(0,0)(y-0)$.

\item
Check if this $L(x,y)$ satisfies the requirement "(ii)" on
page 1 of Lecture 14.
In this example, 
the $\varepsilon \cdot \sqrt{(x-0)^2+(y-0)^2}$ version is easier
to check (see page 3 of Lecture 14 and problem 2 above). 
In the end, you should reach the conclusion
that $f$ is not differentiable at $(0,0)$.

\end{enumerate}
