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\begin{document}

\pagestyle{empty}

\noindent
\parbox{2in}{\bf Math 2130}
\hfill {\Large \bf Test \#2} \hfill
\parbox{2in}{\bf \hfill March ${\bf 16}^\mathrm{\bf th}$, 2021}

\vspace{0.1in}

\noindent {\large\bf Name:} \underline{\Large\sc\color{red}\quad Answer Key \quad}
\hfill {\bf Be sure to show your work!}

\vspace{0.1in}

\noindent {\bf \large 1. (12 points)} Three level curve plots are shown below. I have labeled the levels so you know which curves are higher and which are lower.

\force 
\hfill
\begin{tikzpicture}[scale=0.6]
  \begin{axis}[
      axis x line=middle, 
      axis y line=middle, 
      axis equal,
      xlabel = {},  ylabel = {},
      xtick={0}, ytick={0},
      restrict y to domain = -1:1,
      restrict x to domain = -1:1,
      color=black!30
    ]
    \foreach \C in {-5,...,5}
       \addplot [domain = -1:1, samples = 50, color=black] ({x}, {-2*x^3+\C/2-0.15});
    
    \node[fill=white] at (axis cs: -0.5,-0.85) {\color{red}\bf lower};
    \node[fill=white] at (axis cs: 0.5,0.85) {\color{red}\bf higher};
    
    \draw[red,fill=red,ultra thick] (axis cs:0.25,-0.175) circle [radius=3pt];
    \node[below left] at (axis cs:0.25,-0.175) {\color{red}\bf A};
    
    \draw[blue,ultra thick,->] (axis cs:0.25,-0.175)--(axis cs:0.425,0.375);

    \draw[blue,ultra thick,-] (axis cs:0.25,-0.175)--(axis cs:0.285,-0.065)--(axis cs:0.395,-0.1)--(axis cs:0.36,-0.21);
    \draw[blue,ultra thick,-] (axis cs:0.03,-0.105)--(axis cs:0.47,-0.245);

  \end{axis}
\end{tikzpicture}
\hfill
\begin{tikzpicture}[scale=0.6]
  \begin{axis}[
      axis x line=middle, 
      axis y line=middle, 
      axis equal,
      xlabel = {},  ylabel = {},
      xtick={0}, ytick={0},
      restrict y to domain = -1:1,
      restrict x to domain = -1:1,
      color=black!30
    ]
    \addplot [domain = -1:1, samples = 30, color=black] ({x},{x});
    \addplot [domain = -1:1, samples = 30, color=black] ({x},{-x});

    \foreach \C in {0,0.45,0.65,0.85} { 
       \addplot [domain = -3:3, smooth, samples=50, color=black] ({\C*cosh(x)}, {\C*sinh(x)});
       \addplot [domain = -3:3, smooth, samples=50, color=black] ({-\C*cosh(x)}, {\C*sinh(x)});
       \addplot [domain = -3:3, smooth, samples=50, color=black] ({\C*sinh(x)}, {\C*cosh(x)});
       \addplot [domain = -3:3, smooth, samples=50, color=black] ({\C*sinh(x)}, {-\C*cosh(x)});
    }
    \node[fill=white] at (axis cs: -0.25,-0.85) {\color{red}\bf lower};
    \node[fill=white] at (axis cs: 0.25,0.85) {\color{red}\bf lower};
    \node[fill=white] at (axis cs: 0.75,-0.15) {\color{red}\bf higher};
    \node[fill=white] at (axis cs: -0.75,0.15) {\color{red}\bf higher};
    
    \draw[red,fill=red,ultra thick] (axis cs:-0.15,-0.15) circle [radius=3pt];
    \node[below left] at (axis cs:0.05,-0.1) {\color{red}\bf B};
    
    \draw[blue,ultra thick,->] (axis cs:-0.15,-0.15)--(axis cs:-0.45,0.15);

    \draw[blue,ultra thick,-] (axis cs:-0.15,-0.15)--(axis cs:-0.22,-0.08)--(axis cs:-0.15,-0.01)--(axis cs:-0.08,-0.08);
      
    \draw[blue,ultra thick,-] (axis cs:-0.29,-0.29)--(axis cs:-0.01,-0.01);
    
  \end{axis}
\end{tikzpicture}
\hfill
\begin{tikzpicture}[scale=0.6]
  \begin{axis}[
      axis x line=middle, 
      axis y line=middle, 
      axis equal,
      xlabel = {},  ylabel = {},
      xtick={0}, ytick={0},
      restrict y to domain = -1:1,
      restrict x to domain = -1:1,
      color=black!30
    ]
    \foreach \C in {1,...,6}
       \addplot [domain = 0:360, samples = 50, color=black] ({\C/5*cos(x)}, {\C/5*sin(x)});
    
    \node[fill=white] at (axis cs: 0.25,-0.45) {\color{red}\bf lower};
    \node[fill=white] at (axis cs: 0.75,-0.85) {\color{red}\bf higher};
    
    \draw[red,fill=red,ultra thick] (axis cs:-0.205,0.564) circle [radius=3pt];
    \node[below right] at (axis cs:-0.205,0.564) {\color{red}\bf C};

    \draw[blue,ultra thick,->] (axis cs:-0.205,0.564)--(axis cs:-0.355,1.014);

    \draw[blue,ultra thick,-] (axis cs:-0.205,0.564)--(axis cs:-0.235,0.654)--(axis cs:-0.145,0.685)--(axis cs:-0.115,0.595);
    \draw[blue,ultra thick,-] (axis cs:-0.385,0.504)--(axis cs:-0.025,0.625);

  \end{axis}
\end{tikzpicture}
\hfill \force

\iffalse

\fi
  
\noindent  
\force \hfill 
\fbox{$f(x,y)=2x^3+y$} \quad 
\hfill \quad
\fbox{$f(x,y)=x^2-y^2$}
\hfill \quad\quad
\fbox{$f(x,y)=\sqrt{x^2+y^2}$}
\hfill \force
  
\begin{enumerate}[(a)]
   \item The plots above correspond to 3 of the 5 functions listed here: $f(x,y)=1-x^2-y^2$, $f(x,y)=\sqrt{x^2+y^2}$, $f(x,y)=2x^3+y$, $f(x,y)=2x^2+y$, and $f(x,y)=x^2-y^2$. Write the correct formula below each plot.
   \item Sketch a gradient vector at the points A, B, and C. If the vector is $\B 0$ or does not exist, draw an ``X'' on the point.\\ {}[Don't worry about having the correct length. I'm just looking for the correct direction.] 
\end{enumerate}

\vspace{-0.05in}

First, for part (b), keep in mind that the gradient vectors should be {\bf perpendicular} to the level curve and point towards {\bf higher} level curves since the gradient gives the direction of maximal {\it increase}.

Next, consider the formulas. Both $1-x^2-y^2=C$ (so $x^2+y^2=1-C$) and $\sqrt{x^2+y^2}=C$ (so $x^2+y^2=C^2$) are families of circles. However, the second formula's circles grow in radius as the level $C$ increases (also the radius equals the level $C$ so these curves should be equally spaced). Thus the third graph shows level curves of $f(x,y)=\sqrt{x^2+y^2}$ (a cone). Next, $2x^3+y=C$ implies $y=-2x^3+C$ (a family of cubics running downhill) which are pictured in the first plot. As opposed to $2x^2+y=C$ which implies $y=-2x^2+C$ (a family of parabolas opening downward) which are not pictured. Finally, $x^2-y^2=C$ are a family of hyperbolas which is the middle plot (or by process of elimination this must be the middle plot's formula).

\vspace{0.05in}

\noindent {\bf \large 2. (7 points)} Let $w=f(x,y,z)$, $x=g(t)$, $y=h(t)$ and $z=\ell(t)$. State the chain rule for the derivative of $w$ with respect to $t$. Clearly distinguish between regular derivatives (i.e., d's) and partial derivatives (i.e., $\partial$'s).

\[ \dfrac{dw}{dt} = \dfrac{\partial w}{\partial x}\dfrac{dx}{dt} + \dfrac{\partial w}{\partial y}\dfrac{dy}{dt} + \dfrac{\partial w}{\partial z}\dfrac{dz}{dt} \quad \text{OR (in other notation)} \quad \dfrac{df}{dt} = f_x\cdot g'(t)+f_y \cdot h'(t)+f_z \cdot \ell'(t) \]

\noindent {\bf \large 3. (9 points)} Consider some unknown function $f(x,y)$.

\vspace{-0.1in}

\begin{enumerate}[(a)]

\item It is possible to have a function where $f_{xy}(3,4)=5$ and $f_{yx}(3,4)=6$? \qquad {\Large\bf \fbox{YES} \quad / \quad NO}\\
\force \qquad If not, why not? If so, what does this tell us? 

%\vspace{0.05in}

Clairaut's theorem says that if $f_{xy}$ and $f_{yx}$ are {\it continuous} at $(x,y)=(3,4)$, then $f_{xy}(3,4)=f_{yx}(3,4)$. Thus, $f_{xy}(3,4)\not=f_{yx}(3,4)$, implies that the {\bf mixed second partials} of $f$ are {\bf not continuous} at $(x,y)=(3,4)$. 

%\vspace{0.05in}

\item If $\nabla f(x,y)$ exists, can I conclude that $f(x,y)$ is differentiable? \qquad {\Large\bf YES \quad / \quad \fbox{NO}}

%\vspace{0.05in}

Asserting $\nabla f(x,y)$ exists, just means that $f$'s first partials exist. We know that this is not sufficient to guarantee $f$'s differentiability. On the other hand, the converse holds: if $f$ is differentiable, we can conclude that $\nabla f$ exists. 

%\vspace{0.05in}

\item If $\nabla f(x,y)$ is continuous, can I conclude that $f$ is continuous? \qquad {\Large\bf \fbox{YES} \quad / \quad NO}

%\vspace{0.05in}

If $\nabla f$ is continuous, this means $f$'s first partials are continuous. Thus $f$ is differentiable. Since $f$ is differentiable, it must be a continuous function.

%\vspace{0.05in}

\end{enumerate}


\noindent {\bf\large 4. (10 points)} Limits and continuity.

\begin{enumerate}[(a)]
\item Show that $\displaystyle \lim\limits_{(x,y) \to (0,0)} \dfrac{3x^2+3y^2+xy^2}{x^2+y^2}$ exists and find this limit. \qquad [Switch to polar coordinates.]

\[ \lim\limits_{(x,y) \to (0,0)} \dfrac{3x^2+3y^2+xy^2}{x^2+y^2} = \lim\limits_{(r,\theta) \to (0,\theta)} \dfrac{3r^2+r\cos(\theta)\cdot(r\sin(\theta))^2}{r^2}
 = \lim\limits_{(r,\theta) \to (0,\theta)} \dfrac{r^2(3+r\cos(\theta)\sin^2(\theta))}{r^2} \]
 $= \lim\limits_{(r,\theta) \to (0,\theta)} 3+r\cos(\theta)\sin^2(\theta) = 3+0= \mbox{\fbox{3}}$ where we substituted $x=r\cos(\theta)$, $y=r\sin(\theta)$, and $x^2+y^2=r^2$. Note that $-r \leq r\cos(\theta)\sin^2(\theta) \leq r$ since $-1 \leq \cos(\theta) \leq 1$ and $-1 \leq \sin(\theta) \leq 1$. Thus since $\pm r \to 0$ as $r \to 0$, the squeeze theorem guarantees that $r\cos(\theta)\sin^2(\theta) \to 0$ too.

\vspace{-0.05in}

\item Show that $\displaystyle \lim\limits_{(x,y) \to (0,0)} \dfrac{10xy}{x^2+y^2}$ does not exist. 

\vspace{-0.05in}

If we approach along the $y$-axis (i.e., $x=0$), we get: \qquad $\displaystyle \lim\limits_{y\to 0} \dfrac{10(0)y}{0^2+y^2} = \lim\limits_{y \to 0} \dfrac{0}{y^2} =\lim\limits_{y \to 0} 0 =0.$\\
Approaching along the $x$-axis, also yields a $0$ limit. We need to try something different. For example, if we approach along the diagonal line $y=x$, we get: \qquad
$\displaystyle \lim\limits_{x\to 0} \dfrac{10x(x)}{x^2+x^2} = \lim\limits_{x\to 0} \dfrac{10x^2}{2x^2} = \lim\limits_{x\to 0} 5 = 5.$\\
Since approaching along these curves yields different answers, we must conclude that our limit does not exist.  

\end{enumerate}

\noindent {\bf \large 5. (15 points)} Let $F(x,y,z)=x^2+e^{yz}+\sin(xy^3z^2)+yz^2$.

\vspace{0.05in}

In each part, we need to know the gradient of $F$ at the point $(1,0,2)$, so we compute:

\vspace{-0.15in}

\[ \nabla F = \langle F_x, F_y, F_z \rangle = \left\langle 2x+y^3z^2\cos(xy^3z^2), ze^{yz}+3xy^2z^2\cos(xy^3z^2)+z^2, ye^{yz}+2xy^3z\cos(xy^3z^2)+2yz \right\rangle \]

\vspace{-0.05in}

Thus $\nabla F(1,0,2) = \langle 2(1)+0,2e^0+0+2^2,0+0+0 \rangle = \langle 2,6,0 \rangle$.

\begin{enumerate}[(a)]
\item Find an equation for the plane tangent to $x^2+e^{yz}+\sin(xy^3z^2)+yz^2=2$ at $(x,y,z)=(1,0,2)$

\vspace{0.05in}

We know that $\nabla F(1,0,2)=\langle 2,6,0\rangle$ is perpendicular to this level surface $F(x,y,z)=2$ at the point $(x,y,z)=(1,0,2)$. Therefore, an equation for the tangent plane is 
\fbox{$2(x-1)+6(y-0)+0(z-2)=0$} or equivalently $x+3y=1$.

\vspace{0.05in}

\item Find the directional derivative of $F$ at the point $(1,0,2)$ in the direction of the vector $\langle -1,2,-2 \rangle$.

\vspace{0.05in}

We must normalize $\B v = \langle -1,2,-2 \rangle$ so we have a direction vector. Notice that $|\B v| = \sqrt{(-1)^2+2^2+(-2)^2}=\sqrt{9}=3$. Thus $\B v$ points in the direction $\B u = \dfrac{1}{3}\langle -1,2,-2\rangle$. Our directional derivative is then $D_{\B u}F(1,0,2) = \nabla F(1,0,2) \bdot \B u = \langle 2,6,0 \rangle \bdot \dfrac{1}{3}\langle -1,2,-2 \rangle = \dfrac{2(-1)+6(2)+0(-2)}{3} =$ \fbox{$\dfrac{10}{3}$}.

\vspace{0.05in}

\item Is it possible to find a direction vector $\B u$ so that $D_{\B u}F(1,0,2)=-2$? Why or why not?

\vspace{0.05in}

Note that $|\nabla F(1,0,2)|=|2\langle 1,3,0 \rangle| = 2\sqrt{1^2+3^2+0^2}=2\sqrt{10}$. We know that the maximum and minimum values of $D_{\B u}F(1,0,2)$ as $\B u$ varies in every direction are $\pm |\nabla F(1,0,2)| = \pm 2\sqrt{10}$. In fact, the directional derivative takes on all values between this min and max. Therefore, since $-2\sqrt{10} < -2 < 2\sqrt{10}$, \fbox{yes}, it is possible to find such a direction vector.

\vspace{0.05in}

\end{enumerate}

\noindent {\bf\large 6. (8 points)} Suppose that $z=x^3y$ where $x$ is measured within 1\% accuracy and $y$ is measured within 2\% accuracy. Use a total derivative to estimate the maximum percent error in the corresponding $z$ measurement.

\vspace{0.05in}

The total derivative gives us a (linearization of the) error: $dz=z_x\,dx+z_y\,dy = 3x^2y\,dx+x^3\,dy$ if we think of $dx,dy,$ and $dz$ as the error in measured $x,y,$ and $z$ respectively. Then $dx/x,dy/y,$ and $dz/z$ measure percent error. We have $\displaystyle \dfrac{dz}{z} = \dfrac{3x^2y\,dx+x^3\,dy}{x^3y} = \dfrac{3x^2y\,dx}{x^3y}+\dfrac{x^3\,dy}{x^3y} = 3\dfrac{dx}{x}+\dfrac{dy}{y}$. So that $\left|\dfrac{dz}{z}\right| \leq 3 \left|\dfrac{dx}{x}\right|+\left|\dfrac{dy}{y}\right| \leq 3(1\%)+2\%=$ \fbox{5\%} max error in $z$.

\vspace{0.05in}

\noindent {\bf \large 7. (13 points)} Let $f(x,y) = -x^3+12x+y^3-3y$.
\begin{enumerate}[(a)]
   \item Compute the gradient and Hessian matrix for $f$.
   \item Find the quadratic approximation of $f$ at $(x,y)=(-1,2)$.

\vspace{-0.1in}

\[ \nabla f = \langle f_x,f_y \rangle = \langle -3x^2+12, 3y^2-3 \rangle \text{ and } H_f = \begin{bmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{bmatrix} = \begin{bmatrix} -6x & 0 \\ 0 & 6y \end{bmatrix} \]

Evaluating at our point: $f(-1,2) = -(-1)^3+12(-1)+2^3-3(2) = -9$, $\nabla f(-1,2) = \langle 9,9 \rangle$, and $H_f(-1,2) = \begin{bmatrix} 6 & 0 \\ 0 & 12 \end{bmatrix}$. Therefore,
\fbox{$Q(x,y)=-9+9(x+1)+9(y-2)+\dfrac{6}{2}(x+1)^2+\dfrac{12}{2}(y-2)^2$} or\\ \force \hfill \fbox{$Q(x,y)=-9+\langle 9,9 \rangle \bdot \langle x+1,y-2 \rangle + \dfrac{1}{2} \begin{bmatrix} x+1 & y-2 \end{bmatrix} \begin{bmatrix} 6 & 0 \\ 0 & 12 \end{bmatrix} \begin{bmatrix} x+1 \\ y-2 \end{bmatrix}$}.

\vspace{0.05in}
   
   \item Find and classify all of the critical points of $f$. {}[Use the ``$2^{\mathrm{nd}}$-derivative'' test to determine if critical points are relative max's, min's or saddle points.]
   
\vspace{0.05in}

Since $f$'s first partials are continuous, it is differentiable everywhere (we don't get any critical points from lack of differentiability). So we solve $\nabla f(x,y)=\B 0$. This means $-3x^2+12=0$ and $3y^2-3=0$ so that $x^2=4$ and $y^2=1$. Therefore, $x=\pm 2$ and $y = \pm 1$ (there are a total of 4 critical points). We now apply the second derivative test to classify these points: $H_f(-2,-1) = \begin{bmatrix} 12 & 0 \\ 0 & -6 \end{bmatrix}$  has a negative determinant, so this is a saddle point. Likewise, $H_f(2,1) = \begin{bmatrix} -12 & 0 \\ 0 & 6 \end{bmatrix}$ is also a saddle point. Next, $H_f(-2,1) = \begin{bmatrix} 12 & 0 \\ 0 & 6 \end{bmatrix}$ has a positive determinant and $f_{xx}(-2,1)=12>0$, so this is a relative minimum. Finally, $H_f(2,-1) = \begin{bmatrix} -12 & 0 \\ 0 & -6 \end{bmatrix}$ has a positive determinant and $f_{xx}(2,-1)=-12<0$, so this is a relative maximum. Therefore, \fbox{$(2,1)$ and $(-2,-1)$ are saddle points}, \fbox{$(-2,1)$ is a relative minimum}, and \fbox{$(2,-1)$ is a relative maximum}.

\vspace{0.05in}

\end{enumerate}

\noindent {\bf \large 8. (14 points)} Suppose $f(x,y)$ is a ``nice'' function (with continuous partials of all orders).

\begin{enumerate}[(a)]
\item $Q(x,y) = 13+{\color{red}0(x-3)+0(y+2)+}\dfrac{8}{2}(x-3)^2-3(x-3)(y+2)+\dfrac{4}{2}(y+2)^2$ is the quad. approx. at $(x,y)=(3,-2)$. 

\vspace{0.05in}

$\nabla f(3,-2) = \left\langle 0,0 \right\rangle$ \qquad $H_f(3,-2) =\begin{bmatrix} 8 & -3 \\ -3 & 4 \end{bmatrix}$

\vspace{0.05in}

Is $(x,y)=(3,-2)$ a critical point of $f(x,y)$?  {\large\bf \qquad \fbox{YES} \quad / \quad NO}

If not, why not? If so, what kind (relative min, relative max, saddle point or not enough information)?

\vspace{0.05in}

Since $\nabla f(3,-2)=\B 0$, this is a critical point. Notice that $\det H_f(3,-2) = 8(4)-(-3)^2>0$ and $f_{xx}(3,-2)=8>0$, so $(3,-2)$ is a \fbox{\bf relative minimum}.

\vspace{0.05in}

\item $Q(x,y) = 11+5(x+1)-y+\dfrac{8}{2}(x+1)^2-6(x+1)y$ is the quadratic approx. at $(x,y)=(-1,0)$. 

\vspace{0.05in}

$\nabla f(-1,0) = \left\langle 5,-1 \right\rangle$ \qquad $H_f(-1,0) =\begin{bmatrix} 8 & -6 \\ -6 & 0 \end{bmatrix}$

\vspace{0.05in}

Is $(x,y)=(-1,0)$ a critical point of $f(x,y)$?  {\large\bf \qquad YES \quad / \quad \fbox{NO}}

If not, why not? If so, what kind (relative min, relative max, saddle point or not enough information)?

\vspace{0.05in}

Since $\nabla f(-1,0) \not= \B 0$, this is not a critical point. To get the linearization (below) just ignore the quadratic terms.

\vspace{0.05in}

The linearization of $f(x,y)$ at $(x,y)=(-1,0)$ is $L(x,y)=$ \underline{\quad $11+5(x+1)-y$ \quad}\\
\force \hfill {}[If there is not enough information answer ``N/A''.]

\end{enumerate}

\noindent {\bf \large 9. (12 points)} Use the method of Lagrange multipliers to find the minimum and maximum values of 

\force \hfill $f(x,y)= xy^2$ constrained to $x^2+y^2=12$. 

\vspace{0.05in}

Let $g(x,y)=x^2+y^2$. At a max or min of $f(x,y)$ (subject to the constraint $g(x,y)=12$), we must have $\nabla f = \lambda \nabla g$. We have that $\langle y^2,2xy \rangle = \nabla f = \lambda \nabla g = \lambda \langle 2x,2y \rangle$. Thus we need to solve the equations: $y^2=2x\lambda$, $2xy=2y\lambda$, and $x^2+y^2=12$. 

If $y=0$, then $x^2+0^2=12$ so that $x = \pm 2\sqrt{3}$. Alternatively, suppose $y \not=0$. Then, after canceling off $2y$ from both sides of the second equation: $2xy=2y\lambda$, we have $x=\lambda$. Plugging this into the first equation: $y^2=2x\lambda$, gives us $y^2=2x^2$. Plugging this into the constraint equation gives $x^2+2x^2=12$ so that $3x^2=12$ so $x^2=4$. Therefore, $x= \pm 2$. This then tells us that $12=x^2+y^2=4+y^2$ so $y^2=8$ and thus $y =\pm 2 \sqrt{2}$. 

Finally, plugging our solutions into the objective function yields: $f(\pm 2\sqrt{3},0) = \pm 2\sqrt{3}\cdot 0^2=0$, $f(\pm 2,\pm 2\sqrt{2}) = \pm 2 \cdot (2\sqrt{2})^2 = \pm 16$. Therefore, the function $f(x,y)$ subject to the constraint $x^2+y^2=12$ attains a maximum value of \fbox{16} and a minimum value of \fbox{$-16$}.


\vspace{0.05in}



\end{document}