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

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\begin{document}

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

\begin{center}
\large
Homework 08   
\normalsize
\end{center}

\vspace{.3in}

%           The questions!

\begin{enumerate}

\item % 14.3: Partial Derivatives.
Section 14.3: Problems 19, 21, 53, 60, 65, 67, 69, 72, 81, 91.

Hint for problem 72: 

% Since $(0,0)$ is a special point, we need 
To compute % $f_{xy}(0,0)$ from definition:
$$
f_{xy}(0,0) = \lim_{y\to 0} \ds{ f_x(0,y) - f_x(0,0) \over y - 0 },
$$
we need $f_x(0,y)$ for $y\neq 0$ (direct differentiation)
and
$$
f_{x}(0,0) = \lim_{x\to 0} \ds{ f(x,0) - f(0,0) \over x - 0 }.
$$
The procedure for computing $f_{yx}(0,0)$ is similar.

% 14.3: Partial Derivatives.
\item
Section 14.3: 
Further explanation on definition of differentiability on Lecture 14, page 6:
% Some details about the definition of differentiability at $(x_0,y_0)$: 

Show that 
\begin{equation} \label{e1}
h(x,y) = \varepsilon_1 \cdot (x-x_0) 
+ \varepsilon_2 \cdot (y-y_0)
\text{ with } \displaystyle{\lim_{(x,y) \to (x_0,y_0)} (\varepsilon_1, \varepsilon_2) = (0,0)} 
\end{equation}
is the same as %if and only if
\begin{equation} \label{e2}
h(x,y) = \varepsilon \cdot \sqrt{ (x-x_0)^2 + (y-y_0)^2 }
\text{ with } \displaystyle{\lim_{(x,y) \to (x_0,y_0)} \varepsilon = 0}.
\end{equation}
That is, if $h(x,y)$ is given by \eqref{e1}, then it can be rewritten (combined)
as \eqref{e2}.
Also, 
if $h(x,y)$ is given by \eqref{e2}, then it can be rewritten (split)
as \eqref{e1}.

\noindent{\bf Hint}: Use 
$$ \sqrt{ (\Delta x)^2 + (\Delta y)^2} 
= \displaystyle{ \Delta x \over \sqrt{ (\Delta x)^2 + (\Delta y)^2} } \Delta x
+ \displaystyle{ \Delta y \over \sqrt{ (\Delta x)^2 + (\Delta y)^2} } \Delta y
$$
or
$$
|\Delta x| \leq \sqrt{ (\Delta x)^2 + (\Delta y)^2 }, \quad
|\Delta y| \leq \sqrt{ (\Delta x)^2 + (\Delta y)^2 }, 
$$
$$
\sqrt{ (\Delta x)^2 + (\Delta y)^2 }\; \leq |\Delta x| + |\Delta y|
$$

\noindent{\bf Remark}:
% If you still remember the 'Big O' and 'small o' notation from last semester, 
In terms of the notation
`$O$' ('Big O') and `$o$' ('Big o') in Section 7.4 (page 464), 
the above statement can be recast as:
\begin{equation} \label{o1}
o(1) \cdot \Delta x + o(1) \cdot \Delta y = o(1) \cdot \sqrt{ (\Delta x)^2 + (\Delta y)^2 },
\end{equation}
or
\begin{equation} \label{o2}
o(\Delta x) + o(\Delta y) = o(\sqrt{ (\Delta x)^2 + (\Delta y)^2 } ),
\end{equation}
where the small $o(\cdot)$'s refer to $2D$ limit 
$\displaystyle{\lim_{(\Delta x,\Delta y) \to (0,0)}}$, same as
$\displaystyle{\lim_{(x,y) \to (x_0,y_0)}}$.

The statements \eqref{o1} or \eqref{o2} are more concise and widely used.
% If you forgot the the 'Big O' and 'small o' notation, just stick to the 
% original statements \eqref{e1} and \eqref{e2}.

% 14.3: Partial Derivatives.
% move to note
% \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}

% 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.
% 
% 
\end{enumerate}
 
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