\section{Gradients and Directional Derivatives}
We now focus on the following definition.

\begin{definition}
The \underline{gradient} of a function $f(x_1,...,x_n)$ is given by
\begin{eqnarray*}
\nabla f (x_1,...,x_n) & = & f_1 (x_1,...,x_n) \mathbf{e}_1 + \cdots + f_n (x_1,...,x_n) \mathbf{e}_n \\[1ex]
\quad & = & \frac{\partial f}{\partial x_1} \mathbf{e}_1 + \cdots + \frac{\partial f}{\partial x_n} \mathbf{e}_n.
\end{eqnarray*}
If $f$ is a function $f: \mathbb{R}^n \rightarrow \mathbb{R}$, the gradient $\nabla f$ is a vector-valued function that maps points in $\mathbb{R}^n$ to vectors in $\mathbb{R}^n$.  
We occasionally use the notation $\hbox{\rm grad}\, f$ to denote the gradient.
\end{definition}


\medskip
For a function $f(x,y)$ of two variables, the gradient is written as
$$
\nabla f(x,y) = f_1 (x,y) \mathbf{i} + f_2 (x,y) \mathbf{j} = \frac{\partial f}{\partial x}\, \mathbf{i} + \frac{\partial f}{\partial y}\, \mathbf{j}.
$$
We can also consider the gradient as an \textit{operation} on functions, similar to taking derivatives like $\frac{d}{dt}$ or $\frac{\partial}{\partial x_j}$. We write:
\begin{eqnarray*}
(\textrm{two dimensions}) & \nabla : = & \mathbf{i} \frac{\partial}{\partial x} + \mathbf{j} \frac{\partial}{\partial y} \\[1ex]
(n \textrm{ dimensions}) & \nabla : = & \mathbf{e}_1 \frac{\partial}{\partial x_1} + \cdots + \mathbf{e}_n \frac{\partial}{\partial x_n}
\end{eqnarray*}
The notation here means that we apply $\nabla$ to $f$ by evaluating the partial derivatives of $f$ and then multiplying them by the corresponding basis vectors. Usually, when multiplying a vector by a scalar function, we write the scalar function first (for instance, $g(x)\ii$ rather that $\ii g(x)$). Here, we reverse this convention to prioritize clarity: we want to make sure that $\nabla f=( \mathbf{i} \frac{\partial}{\partial x} + \mathbf{j} \frac{\partial}{\partial y})f$ is understood as $ \mathbf{i} \frac{\partial f}{\partial x} + \mathbf{j} \frac{\partial f}{\partial y}$, as opposed to taking the partial derivatives of the $\ii$ and $\jj$ vectors and then multiplying by $f$.



Using the gradient operation allows us to:
\begin{enumerate}

\item Find directional derivatives;


\smallskip

\item Find directions of steepest ascent or descent;

\smallskip

\item Find normal vectors to level surfaces or curves of functions.


\end{enumerate}

We now examine each of these three ideas. Examples appear in the Worked Problems below.


\subsection{Directional Derivatives for Functions of Two Variables}
Recall that a \underline{unit vector} $\mathbf{u} = u_1 \mathbf{i} + u_2 \mathbf{j}$ in $\mathbb{R}^2$ is any vector which has length one, so that $u_1^2 + u_2^2 = 1$.  For any unit vector $\mathbf{u}$, we define the \textbf{directional derivative} in the direction of $\mathbf{u}$ at $(a,b)$ as
the one-sided limit
\begin{equation}\label{eq:33eq2}
D_{\mathbf{u}} f(a,b) : = \lim_{h \rightarrow 0^+} \frac{1}{h} \big[f(a + h u _1, b + hu_2) - f(a,b) \big],
\end{equation}
assuming that this limit exists. This can be interpreted as a single-variable derivative in the following sense: if the function $f(a + t u_1, b + t u_2)$ is differentiable in $t$ at $t=0$, then
\begin{equation}\label{eq:33eq1}
D_{\mathbf{u}} f (a,b) = \frac{d}{dt} f(a + t u_1, b + t u_2)\Big\vert_{t = 0}.
\end{equation}
(In fact, many sources use (\ref{eq:33eq1}) as the definition of the directional derivative.) However, the one-sided limit in (\ref{eq:33eq2}) may exist even when the full derivative in (\ref{eq:33eq1}) does not. This happens for example when $f(x,y)=\sqrt{x^2+y^2}$ and $(a,b)=(0,0)$ (exercise: check this!).

  If $f$ happens to be differentiable at $(a,b)$, we have a particularly simple expression for $D_{\mathbf{u}}$.

\begin{theorem}\label{thm:33thm2}
Suppose that $f(x,y)$ is differentiable at $(a,b)$. Then, for every unit vector $\mathbf{u} = \langle u_1, u_2 \rangle$, one has
$$
D_{\mathbf{u}} f (a,b) = f_1 (a,b) u_1 + f_2 (a,b) u_2 = \nabla f(a,b) \cdot \mathbf{u}.
$$
This generalizes to differentiable functions of several variables: if $f(x_1,\dots,x_n)$ is differentiable at $(a_1,\dots,a_n)$, then
$$
 D_{\mathbf{u}} f (a_1,\dots,a_n) =  \nabla f(a_1,\dots,a_n) \cdot \mathbf{u}.
$$
\end{theorem}


\begin{proof}
Evaluate the derivative \eqref{eq:33eq1} using the Chain Rule.
\end{proof}

Two special cases are of particular interest. First, we will look at what happens when we try to maximize or minimize the directional derivative. Next, we will consider the relationship between directional derivatives and the level curves or surfaces of the function.


\subsection{Direction of Steepest Ascent or Descent}
Let $f(x,y)$ be differentiable at some point $(a,b)$, and let $\mathbf{u}$ (a unit vector in $\mathbb{R}^2$) represent a direction. Then
$$
D_{\mathbf{u}} f (a,b) = \nabla f (a,b) \cdot \mathbf{u} = |\nabla f (a,b)| | \mathbf{u}|\cos \theta,
$$
where $\theta$ is the angle between the unit vector $\mathbf{u}$ and the gradient vector $\nabla f (a,b)$. Since $|{\bf u}|=1$, we get
\begin{equation}\label{eq:33eq3}
D_{\mathbf{u}} f (a,b) = |\nabla f (a,b)| \cos \theta.
\end{equation}

Suppose that we fix $(a,b)$, and try to choose the direction ${\bf u}$ so that the directional derivative is either maximized or minimized.
Let us think about what the identity \eqref{eq:33eq3} tells us here.

\begin{itemize}

\item The direction of \underline{steepest ascent} is the direction in which the rate of increase of $f$ is maximized. In other words, we are looking for a unit vector $\mathbf{u}$ that maximizes $D_{\mathbf{u}} f (a,b)$.  
Since $|\cos \theta| \leq 1$, the directional derivative $D_{\mathbf{u}} f(a,b)$ is maximized when $\cos \theta = 1$.  This means that $\theta = 0$, so that $\mathbf{u}$ points in the same direction as $\nabla f (a,b)$.  In other words,  
$$\mathbf{u} : = \frac{\nabla f}{|\nabla f|}$$
 always gives the direction of \underline{maximal increase}. The \underline{rate of steepest ascent} is given by
$$
D_{\frac{\nabla f}{|\nabla f|}} f : = |\nabla f|.
$$


\medskip

\item Similarly, $D_{\mathbf{u}} f (a,b)$ is \underline{minimized} when $\cos \theta = -1$. This implies that $\theta = \pi$, so that $\mathbf{u}$ and $\nabla f (a,b)$ point in opposite directions.  Thus, the direction of \underline{steepest descent} is always given by $\mathbf{u} = - \frac{\nabla f}{|\nabla f|}$, the reverse direction of $\nabla f (a,b)$!
\end{itemize}


\begin{figure}[H]\label{33fig3}
\centering
\includegraphics[width= .4\textwidth]{33fig3.pdf}
\caption{The level curves of some surface $z = f(x,y)$. The arrows in the picture indicate the directions of {steepest ascent}, and so point in the same direction as $\nabla f$.  }
\end{figure}


To illustrate this graphically, we assume that $f(x,y)$ is a $C^1$ function on an open domain, and represent $f(x,y)$ as a ``topographical map" of the surface $z = f(x,y)$ in $\mathbb{R}^3$. 
The arrows indicate the directions where, intuitively, we would expect the climb to be the steepest. The above computations show that they should also match the direction of the gradient. It is not a coincidence that they are perpendicular to the level curves; we will investigate this in the next subsection.








\subsection{Normal Vectors and Gradient}



\begin{theorem}\label{thm:33thm1}
Suppose that $f (\mathbf{x})$ is $C^1$ on some neighbourhood of $\mathbf{a}$, and that $\nabla f (\mathbf{a}) \neq \mathbf{0}$.  Then the vector $\nabla f (\mathbf{a})$ is normal to the level surface of $f(\mathbf{x})$ which passes through $\mathbf{a}$.
\end{theorem}



\begin{proof}[Sketch of Proof for Functions of Two Variables]
We start with some $f(x,y)$ satisfying the assumptions of the theorem at $\mathbf{a} =( a,b)\in\RR^2$, so that $f$ is $C^1$ on some neighbourhood of $(a,b)$ and $\nabla f (a,b) \neq \langle 0,0 \rangle$.  We also let $c = f(a,b)$, so that the level curve of $f(x,y)$ passing through $(a,b)$ is given by the equation $f(x,y) = c$. 

\smallskip
We need to show that, at $(a,b)$, the vector $\nabla f (a,b)$ is normal to the level curve $f(x,y) = c$.
To accomplish this,  we parametrize the level curve $f(x,y) = c$ by a differentiable vector-valued function
$$
\mathbf{r} (t) = x(t) \mathbf{i} + y(t) \mathbf{j},
$$
where $x(0) = a$ and $y(0) = b$, so that $r(0) = a \mathbf{i} + b \mathbf{j}$. 
(In this sketch, we will just {\it assume} that this can be done. See Remark \ref{33-remark1} below.)
Then the vector ${\bf v}=\langle x' (0), y' (0) \rangle$ is tangent to that level curve. Thinking about ${\bf r}(t)$ as the position vector of a point moving along the level curve, the vector ${\bf v}$ represents the velocity of that object at $t=0$. 


 

\begin{center}
\begin{minipage}{.8\textwidth}
\begin{figure}[H]\label{33fig1}
\centering
\includegraphics[width= .6\textwidth]{33fig1.png}
\caption{The two vectors appearing in the proof sketch.}
\end{figure}
\end{minipage}
\end{center}


Since $\mathbf{r}(t)$ parametrizes the level curve $f(x,y) = c$, we know that $f(x(t), y(t)) = c$ for all values of $t$ in some neighbourhood of $t_0 = 0$. Differentiating and using the Chain Rule in such a neighbourhood, we get
\begin{eqnarray*}
0 & = & \frac{d}{dt} f(x(t), y(t))\vert_{t = 0} = \frac{\partial f}{\partial x} (a,b) \,x' (0) + \frac{\partial f}{\partial y} (a,b) \, y' (0) \\[1ex]
\quad & = &  \left\langle \frac{\partial f}{\partial x} (a,b), \frac{\partial f}{\partial y}(a,b) \right\rangle  \cdot \langle x' (0), y' (0) \rangle  \\[1ex]
&= & \nabla f(a,b)\cdot {\bf v} .
\end{eqnarray*}
This means that $\nabla f(a,b)$ is perpendicular to ${\bf v}$. Therefore 
$\nabla f (a,b)$ is also normal to the level curve $f(x,y) = c$ at the point $(a,b)$.

Alternatively, we can invoke directional derivatives. Thinking again about ${\bf r}(t)$ as a trajectory of a point, and considering that the value of $f({\bf r}(t))$ remains constant along that trajectory, we expect that $D_{\mathbf{u}} f (a,b) = 0$ if $\mathbf{u}$ is a unit vector tangent to the curve. But 
$$D_{\mathbf{u}} f (a,b) = 0\hbox{ if }\cos \theta = 0,$$
where $\theta$ is the angle between $\nabla f (a,b)$ and $\mathbf{u}$. This implies that $\theta = \pi/2$, so that we again see that $\nabla f (a,b)$ is perpendicular to the level curve $f(x,y) = c$ at $(a,b)$.
\end{proof}



\begin{remark}\label{33-remark1}
The fact the level curve $f(x,y) = c$ has a nice differentiable parametrization $\mathbf{r}(t) = x(t) \mathbf{i} + y(t) \mathbf{j}$ is \underline{not a given}! In fact, at this point, we do not even have any justification for saying that the set
$$
\{(x,y) \in \mathbb{R}^2 : f(x,y) = c\}
$$
is a curve.  Yet, as it happens, the hypotheses of Theorem \ref{thm:33thm1} are indeed strong enough to guarantee that the graph of $f(x,y) = c$ is in fact a curve and also admits a suitable parametrization near $(x,y)=(a,b)$. Here,  \underline{suitable} means a $C^1$ parametrization so that one can apply the Chain Rule as in the previous proof.  We will address this question a little bit later in connection with the Implicit Function Theorem.
\end{remark}

In dimensions $n \geq 3$, the proof is similar if we consider curves lying on the level surface $f(\mathbf{x}) = f(\mathbf{a})$ and their tangent vectors. We give a figure below, but omit the computations.

\begin{center}
\begin{minipage}{\textwidth}
\begin{figure}[H]\label{33fig2}
\includegraphics[width= .8\textwidth]{33fig2.png}
\centering
\caption{The surface defined by the equation $z = \sqrt{1 - x^2 - y^2}$ overlain with the curve obtained by taking the intersection with the plane $y = \frac{1}{2}$. Several vectors normal to the surface and along the curve are shown in red.}
\end{figure}
\end{minipage}
\end{center}



\subsection{Non-differentiable Functions} We conclude by reminding the reader that, while formula for the directional derivative $D_{\mathbf{u}} f(a,b)$ simplifies greatly if we assume that $f(x,y)$ is differentiable at $(a,b)$,  the directional derivative \underline{may still exist} even if $f$ fails to be differentiable. In this situation, we must appeal directly to the definition of the directional derivative.  So, without the assumption of differentiability at $(a,b)$, the formula $D_{\mathbf{u}} f(a,b) = \nabla f (a,b) \cdot \mathbf{u}$ may or may not hold. Several examples are given in the Worked Problems below.

\section*{Worked Problems}
\begin{enumerate}[1.]
\item {\it Let $f(x,y) = x^2 y - xe^y$.  Find $D_{\mathbf{u}} f (1,0)$ if $\mathbf{u}$ is the unit vector in the direction of $\mathbf{v} = \langle 3, -1 \rangle$.}

\medskip
The function $f(x,y)$ is differentiable (why?) for all $(x,y)$ in the plane. So, we first calculate the gradient,
$$
\nabla f (x,y) = \langle 2 xy - e^y, x^2 - x e^y \rangle,
$$
so that $\nabla f (1,0 ) = \langle 0 - 1, 1 - 1 \rangle = \langle -1, 0 \rangle$.  Next, we normalize $\mathbf{v}$ to a unit vector by writing,
$$
\mathbf{u} = \frac{\mathbf{v}}{|\mathbf{v}|} = \frac{1}{\sqrt{10}} \langle 3, -1 \rangle.
$$
Finally, we can use the formula in Theorem \ref{thm:33thm2} to write
$$
D_{\mathbf{u}} f (1,0) = \langle -1, 0 \rangle \cdot \frac{1}{\sqrt{10}} \langle 3, -1 \rangle = - \frac{3}{\sqrt{10}}
$$

\bigskip

\item (In this problem, we show that if $f$ is a differentiable function, then two directional derivatives in non-parallel directions determine the gradient of $f$.)

\medskip

\noindent
{\it Suppose $f$ is differentiable at $(a,b)$ and that $D_{\mathbf{u}} f (a,b) = 3$, $D_{\mathbf{v}} f (a,b) = 2$, where
$$
\mathbf{u} = \frac{1}{\sqrt{5}} \langle 2, 1\rangle \textrm{ and } \mathbf{v} =\langle -1,0 \rangle.
$$
Determine $\nabla f (a,b)$.
 }
 
 \medskip
 
 We are given that
 \begin{eqnarray*}
 3= D_{\mathbf{u}} f(a,b) & = & \frac{2}{\sqrt{5}} f_x (a,b) + \frac{1}{\sqrt{5}} f_y (a,b)  \\
 2= D_{\mathbf{v}} f (a,b)&  = & - f_x (a,b) + 0 
 \end{eqnarray*}
 which is a system of two linear equations in the unknown variables $f_x (a,b)$ and $f_y (a,b)$.  Solving this system of equations gives
 $$
 f_x (a,b) = -2, \, f_y (a,b) = 4 + 3 \sqrt{5},
 $$
 so that $\nabla f (a,b) = \langle -2, 4 + 3 \sqrt{5} \rangle$.  
 
 \bigskip
 
 \item (In this problem, we find the equation of a \underline{tangent plane to a surface} by calculating the gradient)
 
 \medskip
 \noindent
 {\it The point $P=(2,3,1)$ lies on the surface $x^2 - y^2 + 4z^2 = -1$.  Find the equation of the tangent plane at that point.}
 
 \medskip Let $f(x,y,z) = x^2 - y^2 + 4 z^2$. Then
 $$
 \nabla f (x,y,z) = \langle 2x, -2y, 8z\rangle\  \Rightarrow \ 
 \nabla f (2,3,1)= \langle 4, -6, 8 \rangle.
 $$
We use this vector as our normal vector at $P$.  The equation of the tangent plane is then
$$
4(x-2) - 6(y-3) + 8(z-1) = 0
$$
 which simplifies to
 $$
 2x - 3y + 4z = -1.
 $$
 
 \bigskip
 
 \item (In this question, we show that the formula for computing directional derivatives from the gradient does not necessarily hold if the function is not differentiable.)
 
 \medskip
 
 {\it Show that, for the function
 $$
 f(x,y) =
 \begin{cases}
 \frac{x^2 y}{x^4 + y^2}, (x,y) \neq (0,0) \\
 0, \quad \, \, \, \; (x,y) = (0,0)
 \end{cases}
 $$
 the formula $D_{\mathbf{u}} f = \nabla f \cdot \mathbf{u}$ fails to hold at the origin.
 }
 
\medskip
 
 Let $\mathbf{u} = u_1 \mathbf{i} + u_2 \mathbf{j}$ be some unit vector.  Using the definition of the directional derivative, we calculate
 \begin{eqnarray*}
 D_{\mathbf{u}} f (0,0) &=& \lim_{h \rightarrow 0^+} \frac{1}{h} \bigg[f(h u_1, h u_2) - f(0,0) \bigg] \\[1ex]
 \quad & = & \lim_{h \rightarrow 0^+} \frac{1}{h}\bigg[\frac{h^3 u_1^2 u_2}{h^4 u_1^4 + h^2 u_2^2} - 0 \bigg] \\[1ex]
 \quad & = & \lim_{h \rightarrow 0^+} \frac{u_1^2 u_2}{h^2 u_1^4 + u_2^2}.
 \end{eqnarray*}
 Evaluating the final limit as $h \to 0^+$, we have
\begin{equation}\label{eq:33eq4}
 D_{\mathbf{u}} f (0,0) =
 \begin{cases}
 \frac{u_1^2}{u_2},  \quad u_2 \neq 0 \\
 0, \quad \, \, \, u_2 = 0
 \end{cases}
\end{equation}
Let's now see what happens if we try to calculate $D_{\mathbf{u}} f (0,0)$ using the gradient formula. Since $f(x,0) = f(0,y) = 0$ for all $x$ and $y$, we know that $\nabla f (0,0) = \mathbf{0}$.  Hence,  for \underline{all} unit vectors $\mathbf{u}$, we have $\nabla f (0,0) \cdot \mathbf{u} = 0$, which disagrees with our calculated value of $D_{\mathbf{u}} f (0,0)$ in (\ref{eq:33eq4})!
 
 In this example, the formula in (\ref{eq:33eq4}) is correct, and the gradient formula is not. The reason for it is that $f$ is not differentiable at $(0,0)$, so that the assumptions needed for the gradient formula are not satisfied.
 
  
 
\end{enumerate}
\section*{Practice Problems}
\begin{enumerate}[1.]
\item {
Find all points $(x,y,z)$ on the surface $xy+z^2=20$ where the tangent plane to that surface is 
perpendicular to the vector $2\ii + 3\jj +4\kk$.}

\bigskip

\item { Let $f(x,y)=x\sin(xy^2)$.

\bigskip

(a)  Find the equation of the tangent plane to the surface $z=f(x,y)$ at the point
where $x=\pi$, $y=1$.}

\bigskip

{ (b) Find the directional derivative of $f$ at $(\pi,1)$ in the direction of the vector 
${\bf v}={\bf i}- 3{\bf j}$.}

\bigskip

{(c)  Find a vector {\bf w} in the $xy$-plane tangent to the level curve of $f$ at $(\pi,1)$. (The answer might not be unique. Any non-zero tangent vector will suffice.)}

\bigskip

\item {  Let $f(x,y)=xe^{x-2y}$.

\bigskip

(a)  Find the directional derivative of $f$ at $(x,y)=(2,1)$ in the direction of the vector $3{\bf i}-{\bf j}$.}

\bigskip

{ (b) Find a unit vector ${\bf u}$ such that $D_{\bf u}f(2,1)=0$.}

\bigskip

\item {\it 
Let $f(x,y)$ be a differentiable function such that $f(2,5)=3$, $f_x(2,5)=4$, and
$D_{\bf u} f(2,5) = 1$, where ${\bf u}$ is the unit vector in the direction of $\ii-\sqrt{3}\jj$.
Find the equation of the plane tangent to the graph of $z=f(x,y)$ at $(x,y)=(2,5)$.}

\bigskip

\item { 
The tangent plane to the graph of $z=f(x,y)$ at $(x,y)=(3,4)$ has the equation
$x-6y-2z=1$. 

\bigskip

(a) Find the directional derivative $D_{\bf u} f(3,4)$ if $\displaystyle \uu=\frac{\sqrt{3}}{2}\,\ii
-\frac{1}{2}\, \jj$. }

\bigskip

{ (b) Is there a unit vector ${\bf v}$ such that $D_{\bf v} f(3,4)=4$?
If yes, find it. If no, explain why.}

\bigskip

\item { 
The surfaces $4x^2+y^2=z^2$ and $xy=z+1$ intersect in a curve that contains the point
$P=(2,3,5)$. Find the equation of the line tangent to that curve at $P$.}

\bigskip

\item { 
Let 
$$g(x,y)=\left\{\begin{matrix}
\displaystyle{\frac{x^2y}{x^6+ y^2}}&\hbox{ if }(x,y)\neq (0,0)\\
0&\hbox{ if }(x,y)=(0,0)\\
\end{matrix}
\right.
$$
Find the directional derivative $D_{\bf u} g(0,0)$, if ${\bf u}$ is the unit vector in the
direction of $3\ii +4\jj$. (This function is not differentiable or even continuous at $(0,0)$, so you have to 
use the definition of the directional derivative.) }

\bigskip

\item 
{Let
$$f(x,y)=\begin{cases}
\displaystyle{\frac{3xy^2-y^3}{x^2 + y^2}}&\hbox{ if }(x,y)\neq (0,0)\\
0&\hbox{ if }(x,y)=(0,0)\\
\end{cases}
$$

(a) Is $f$ differentiable at $(0,0)$? Prove your answer. }




\medskip

{(b)  Find the directional derivative $D_{\bf u}f(0,0)$ for a general unit vector $\displaystyle {\bf u}=u_1{\bf i}+u_2{\bf j}$. If it does not exist, explain why.}






\bigskip



\item { 
Let 
$$f(x,y)=\left\{\begin{matrix}
\displaystyle{\frac{x^5}{x^4+y^4}}&\hbox{ if }(x,y)\neq (0,0)\\
0&\hbox{ if }(x,y)=(0,0)\\
\end{matrix}
\right.
$$

\bigskip

Find the general formula for  $D_{\bf u}f(0,0)$, where ${\bf u}=(u_1,u_2)$ is a 
unit vector, in terms of $u_1,u_2$.}

\bigskip








\item 
{   Suppose that $f(x,y)$ is a function defined for all $(x,y)$ in a neighbourhood of $(a,b)$. 
Suppose also that
$$
D_{{\bf u}}f(a,b) = 2,\ \ D_{{\bf v}}f(a,b) = 3,\ \
$$
where 
$$ {\bf u}= \frac{\sqrt{3}}{2}\,{\bf i} +  \frac{1}{2}\,{\bf j},\ \ \ 
 {\bf v}= \frac{\sqrt{3}}{2}\,{\bf i} -  \frac{1}{2}\,{\bf j},
 $$ 
 and that $f_1(a,b)=-1$. Prove that $f$ is not differentiable at $(a,b)$.}

\bigskip

\item { Is there a differentiable function $f:\rr^2\to \rr$ such that 
 $D_{\bf u}f(0,0)=u_1^2-u_2^2$ for every unit vector ${\bf u}=(u_1,u_2)$?
If yes, find it. If no, explain why.}

\bigskip

\end{enumerate}





