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\newcommand{\Title}{Math 234, Final Exam, Tuesday December 18, 2001}
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\def\Points{ Points}

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\def\Qii{50}    \advance\total by \Qii  %
\def\Qiii{50}    \advance\total by \Qiii  %
\def\Qiv{60}     \advance\total by \Qiv   %
\def\Qv{50}    \advance\total by \Qv  %
\def\Qvi{50}   \advance\total by \Qvi %
\def\Qvii{50}  \advance\total by \Qvii %
\def\Qviii{50}  \advance\total by \Qviii %
%\def\Qvii{30}  \advance\total by \Qvii %
%\def\Qviii{30}    \advance\total by \Qviii  %
%\def\Qix{15}     \advance\total by \Qx   %


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\newcommand{\prob}[2]{\pagebreak[3]\noindent{\bf #1.} (#2 points.) }
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\begin{document}

\vspace*{-0.5in}

\ifanswer
\else
Name \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_
\fi
\medskip

\centerline{\large \Title}

\medskip

\ifanswer
 \centerline{\large Answers}
\bigskip

(The page references below are to Thomas and Finney fifth edition.)

\bigskip

\else



\medskip

\fi

\ifanswer\else
\begin{center}

{\em Circle your section.}

\begin{tabular}{|llr|llr|}
\hline
321  &  Cho &  7:45 T  &
322  &  Cho &  7:45 R  \\
323  &  Cho &  8:50 T  &
324  &  Cho &  8:50 R   \\
325  &  Raichev &  9:55 T  &
326  &  Raichev &  9:55 R  \\
329  &  Russell & 12:05 T  &
330  &  Russell & 12:05 R   \\
331  &  Russell &  1:20 T  &
332  &  Russell &  1:20 R    \\
333  &  Raichev &  2:25 T  &
334  &  Raichev &  2:25 R    \\ \hline
 \end{tabular}

\large
\renewcommand{\arraystretch}{1.8}

\bigskip


\begin{tabular}{|l|l|@{\qquad\qquad}c|} \hline\hline
  I      & \Qi \Points    &     \\ \hline
  II     & \Qii \Points   &     \\ \hline
  III    & \Qiii \Points  &    \\ \hline
  IV     & \Qiv \Points   &     \\ \hline
  V      & \Qv \Points    &      \\ \hline
  VI     & \Qvi \Points   &      \\ \hline
  VII    & \Qvii \Points  &     \\ \hline
  VIII   & \Qviii \Points  &     \\ \hline
 % IX    & \Qix \Points  &     \\ \hline
 % X     & \Qx \Points  &     \\ \hline
\hline \hline
Total & \the\total \Points &   \\ \hline \hline  % Total = 250
\end{tabular}
\end{center}
\bigskip

{\sc You may leave iterated integrals unevaluated so long as you
correctly indicate the limits of integration.
Show your reasoning. Your grade will be based in part
on the clarity of your writing.
YOU MUST EXPLAIN HOW YOU GOT YOUR ANSWER.}


\newpage
\fi


\prob{I}{\Qi} Find an equation for the plane that is
tangent to the surface $z=x^2-xy-y^2$ at the point $P_0(1,1,-1)$.

\ifanswer \Answer{The equation for the surface has form
$z=f(x,y)$ so the answer is given by the linearization
$$
    z=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0),
$$
i.e.
$$
    z= -1+ (x-1)-3(y-1).
$$
Alternatively the surface has equation $F(x,y,z)=0$ where
$F(x,y,z)=z-x^2+xy+y^2$ so the equation for the tangent plane is
$$
 0=\nabla F(x_0,y_0,z_0)\cdot\overrightarrow{P_0P}
=-(x-1)+3(y-1)+(z+1).
$$
(This is problem~3 on page~583.)
\PageBreak{1}
}\else
\vfill\par\newpage
\fi

\prob{II}{\Qii} {\bf(1)}
Find the linear function $L(x,y)$ which best approximates
the function
$$
f(x,y)=\frac{1}{1+x-y}
$$
near the point $(2,1)$.

\ifanswer \Answer{This is the linearization
$$
    L(x,y)=f(2,1)+f_x(2,1)(x-2)+f_y(2,1)(y-1),
$$
i.e.
$$
    L(x,y)=\frac{1}{2}-\frac{x-2}{4}+\frac{y-1}{4}.
$$
(This is Example~3 on page~589.)
}\else
\vfill\par
\fi

\medskipnoindent {\bf(2)} Find the quadratic function $Q(x,y)$ which best approximates
the function  $f(x,y)$ of part~(1) near the point $(2,1)$.


\ifanswer \Answer{This is the quadratic approximation
$$
    Q(x,y)=L(x,y)+\frac{f_{xx}(2,1)(x-2)^2+2f_{xy}(2,1)(x-2)(y-1)+f_{yy}(2,1)(y-1)^2}{2},
$$
i.e.
$$
  Q(x,y)=
\frac{1}{2}-\frac{x-2}{4}+\frac{y-1}{4}+\frac{(x-2)^2}{8}-\frac{(x-2)(y-1)}{4}+\frac{(y-1)^2}{8}.
$$
(This was  explained in the lecture and in section 16-11.)

\PageBreak{1}
}\else
\vfill\par\newpage
\fi


\prob{III}{\Qiii} {\bf(1)} Find the total differential $dw$ of the function
$$
w=e^{2x+3y}\cos4z.
$$

\ifanswer \Answer{$\ds dw=\frac{\p w}{\p x}\,dx+\frac{\p w}{\p y}\,dy+\frac{\p w}{\p z}\,dz$
so
$$
dw
=2e^{2x+3y}\cos4z\,dx+3e^{2x+3y}\cos4z\,dy-4e^{2x+3y}\sin4z\,dz.
$$
}\else
\vfill
\fi

\medskipnoindent {\bf(2)} Find the derivative $\ds\frac{dw}{dt}$ of the
function $w$ of part~(1) along the curve given by the parametric equations
$$
   x=\ln t, \qquad y=\ln(t^2+1),\qquad z=t.
$$

\ifanswer \Answer{$\ds
\frac{dw}{dt}=
\frac{\p w}{\p x}\,\frac{dx}{dt}+
\frac{\p w}{\p y}\,\frac{dy}{dt}+
\frac{\p w}{\p z}\,\frac{dz}{dt}$
so
$$
\frac{dw}{dt}
=2e^{2\ln t+3\ln(t^2+1)}\cos4t\,\frac{1}{t}\,+
 3e^{2\ln t+3\ln(t^2+1)}\cos4t\,\frac{2t}{t^2+1}\,-
4e^{2\ln t+3\ln(t^2+1)}\sin4t.
$$
Alternatively, along the curve we have
$$
   w=e^{2\ln t+3\ln(t^2+1)}\cos4t=t^2(t^2+1)^3\cos4t
$$
so, by the product rule,
$$
\frac{dw}{dt}= 2t(t^2+1)^3\cos4t+t^2(t^2+1)^26t\cos4t-4t^2(t^2+1)^3\sin4t.
$$
(This is exercise~3 on page~604.)
\PageBreak{1}
}\else
\vfill\par\newpage
\fi


\prob{IV}{\Qiv} The unit tangent vector and curvature vector of a curve
whose position vector is $\bfR$ are  defined by
$$
  \bfT=\frac{d\bfR}{ds},\qquad \kappa\bfN=\frac{d\bfT}{ds}=\frac{d^2\bfR}{ds^2}
$$
where $s$ is the arclength. The osculating circle  to a curve
at a point $P$ on the curve is that circle through $P$
which has the same unit tangent vector and curvature vector at $P$ as the curve.


\medskipnoindent{\bf(1)} Find $\bfT$ and $\kappa\bfN$ for the curve $y=e^x$.
Express your answers as functions of the $x$ coordinate.
If either (or both) of your answers depend on the direction
of parameterization so indicate by placing a $\pm$ in front of your answer.

\ifanswer \Answer{ $\bfR=x\bfi+e^x\bfj$ so
$$
\bfT=\frac{d\bfR}{dx}\frac{dx}{ds}= \pm(\bfi+e^x\bfj)(1+e^{2x})^{-1/2}
$$
and
$$
\kappa\bfN=\frac{d\bfT}{dx}\frac{dx}{ds}=
\left(e^x\bfj(1+e^{2x})^{-1/2}-(\bfi+e^x\bfj)(1+e^{2x})^{-3/2}e^{2x}\right)(1+e^{2x})^{-1/2}.
$$
}
\else\par\vfill\par \centerline{(Problem IV continues on next page)} \newpage
\centerline{(Problem IV continued from previous page)}\par\medskip
\fi

\medskipnoindent{\bf(2)} Find an equation for the osculating circle to the  curve
of part~(1) at the point $P(0,1)$.

\ifanswer \Answer{At the point $P(0,1)$ we have
$$
 \bfT=\pm\frac{\bfi+\bfj}{\sqrt{2}}, \qquad
\kappa\bfN=\frac{\bfj}{2}-\frac{\bfi+\bfj}{4}=-\frac{\bfi}{4}+\frac{\bfj}{4}.
$$
The osculating circle has curvature $\kappa=\frac{\sqrt{2}}{4}$
and hence radius $\frac{4}{\sqrt{2}}$ and thus has an equation of form $(x-a)^2+(y-b)^2=8$.
The  center $Q(a,b)$ of this circle lies on the concave side of the curve $y=e^x$
and the line $\overline{PQ}$ from the center $Q$ to the point $P$ on the curve
is perpendicular to the tangent line $y=x+1$ to the curve at $P$.
Hence the line $\overline{PQ}$ has slope  $\ds\frac{b-1}{a-0}=-1$
so $b=1-a$. Since the distance from $Q$ to $P$ is
$\frac{4}{\sqrt{2}}=\sqrt{(a-0)^2+(b-1)^2}$
we have $8=a^2+(b-1)^2=2a^2$ so $a=-2$ and $b=3$. Thus an equation for the circle is
$$
  (x+2)^2+(y-3)^2=8.
$$
(This is problem~13 on page~557.)
\PageBreak{1}
}\else
\vfill\par\newpage
\fi


\prob{V}{\Qv} {\bf(1)} If the   vector field $\bfv=M\bfi+N\bfj$
represents the velocity vector field  of a fluid in the plane
then the flux across a curve $C$ measures the rate at which
the fluid flows across the curve. Find the flux of the field
$$
\bfv=2x\bfi-3y\bfj
$$
outward across the ellipse
$$
 16x^2+y^2=16.
$$


\ifanswer \Answer{Let $R$ denote the interior of the ellipse,
$\bfn$ denote the unit outward normal to the ellipse,
and $ds$ denote the arclength element of the ellipse.
Orient the ellipse counter clockwise.
The unit tangent and the outward normal are
$$
\bfT=\frac{dx}{ds}\bfi+\frac{dy}{ds}\bfj,\qquad
\bfn=\frac{dy}{ds}\bfi-\frac{dx}{ds}\bfj.
$$
By the Divergence Theorem
the  flux of $\bfv=M\bfi+N\bfj$ across the ellipse is
$$
\oint_{\p R}\bfv\cdot\bfn\, ds =
\int\!\!\!\int_R\left(\frac{\p M}{\p x}+\frac{\p N}{\p y}\right)\, dA
=\int_{x=-1}^1\int_{y=-\sqrt{16-16x^2}}^{\sqrt{16-16x^2}} (2-3)\,dy\,dx.
$$
The value of the integral is $-4\pi$ since
the area of the ellipse $\ds\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $\pi ab$.
(This is example~3 on page~708.)
\PageBreak{1}
}
\else\par\vfill\par \centerline{(Problem V continues on next page)} \newpage
\centerline{(Problem V continued from previous page)}\par\medskip
\fi


\medskipnoindent{\bf(2)}
Find the work done by the force field
$$
     \bfF= 3y\bfi+2x\bfj,
$$
in moving a particle once\ifanswer\footnote{The wording of the question
in  Thomas Finney fifth would be
``$\ldots$ when the point of application moves once $\ldots$''.} \fi
counter clockwise  around the ellipse in part~(1).
Hint: The vector field $\bfv$ arises from  $\bfF$ by rotation through ninety degrees.

\ifanswer \Answer{The force field is $\bfF=-N\bfi+M\bfj$ where $\bfv=M\bfi+N\bfj$
is the field in part~(1). The work is
$$
  \oint_{\p R}\bfF\cdot\bfT\, ds= \oint_{\p R}\bfv\cdot\bfn\, ds
$$
so the answer is the same as for part~(1).
(This is example~4 on page~709.)
\PageBreak{1}
}\else
\vfill\par\newpage
\fi



\prob{VI}{\Qvi} Find the area cut off the surface $y^2+z^2=2x$
by the plane $x=1$.


\ifanswer \Answer{This is the same as the area of the surface
$x^2+y^2=2z$ cut off  by the plane $z=1$ and the area is
$$
\int\!\!\!\int_{x^2+y^2\le 2}\sqrt{1+
\left(\frac{\p z}{\p x}\right)^2+\left(\frac{\p z}{\p y}\right)^2
}\, dx\,dy =\int\!\!\!\int_{x^2+y^2\le 2}\sqrt{1+x^2+y^2}\,dx\,dy.
$$
In cylindrical coordinates the surface is $r^2=2z$ and $z\le 1$ when
$r\le\sqrt{2}$ the area is
$$
\int\!\!\!\int_{x^2+y^2\le 2}\sqrt{1+x^2+y^2}\,dx\,dy = \int_0^{2\pi}\int_0^{\sqrt2}
\sqrt{1+r^2}\, r\, dr\,d\theta.
$$
The integral can be evaluated with the substitutions
$u=1+r^2$,  $du=2r\,dr$ so $u=1$ when $r=0$ and $u=3$ when $r=\sqrt2$ and
$$
\int_0^{\sqrt2}\sqrt{1+r^2}\, r\, dr=\frac12\int_1^3\sqrt{u}\,du=
\left.\frac{1}{2}\cdot\frac{2}{3}u^{3/2}\right|_1^3=\frac{\sqrt{27}-1}{3}.
$$
(This is problem~48 on page 678.)
}\else
\vfill\par\newpage
\fi




\prob{VII}{\Qvii} Let  $A$, $B$, $C$ be twice continuously differentiable functions of $(x,y,z)$
and let $\bfF$ be the vector field defined by
$$
\bfF=A\bfi+B\bfj+C\bfk.
$$
The curl of $\bfF$ is denoted by $\nabla\times\bfF$ and the divergence of $\bfF$
is denoted by $\nabla\cdot\bfF$.

\medskipnoindent{\bf(1)}
Write formulas for the
divergence  and
curl  of the vector field $\bfF$
and show that the divergence of the curl is zero.

\ifanswer \Answer{
$$
\nabla\cdot\bfF=\frac{\p A}{\p x}+\frac{\p B}{\p y}+\frac{\p C}{\p z},
$$
$$
\nabla\times\bfF=
\left(\frac{\p C}{\p y}-\frac{\p B}{\p z}\right)\bfi+
\left(\frac{\p A}{\p z}-\frac{\p C}{\p x}\right)\bfj+
\left(\frac{\p B}{\p x}-\frac{\p A}{\p y}\right)\bfk,
$$
so
$$
\nabla\cdot\left(\nabla\times\bfF\right)
=(C_y-B_z)_x+(A_z-C_x)_y+(B_x-A_y)_z=0.
$$
(This is problem~24 on page~702.)
\PageBreak{1}
}
\else\par\vfill\par \centerline{(Problem VII continues on next page)} \newpage
\centerline{(Problem VII continued from previous page)}\par\medskip
\fi


\medskipnoindent{\bf(2)} State Stokes' Theorem and the Divergence Theorem
for the vector field $\bfF$. Be sure to define any notation you use.

\ifanswer \Answer{
For any three dimensional region $R$ with boundary $\p R$ the Divergence Theorem
(see page~722)
says that
$$
  \int\!\!\!\int\!\!\!\int_R \nabla\cdot\bfF\,dV = \int\!\!\!\int_{\p R} (\bfF\cdot\bfn)\,d\sigma
$$
where $dV$ is the volume element, $\bfn$ is the outward unit normal, and $d\sigma$ is the area element.
For any surface $S$ with boundary $\p S$ Stokes' Theorem (see page~729)
says that
$$
  \int\!\!\!\int_S (\nabla\times\bfF)\cdot\bfn\,d\sigma = \int_{\p S} \bfF\cdot d\bfR
$$
where  $d\bfR=dx\,\bfi+dy\,\bfj+dz\,\bfk=\bfT\,ds$; here $\bfT$ is the unit tangent vector
to the boundary $\bfR$. The relation between the direction of the unit normal $\bfn$
and the direction of the unit tangent $\bfT$ is such that if the normal is the thumb of
your right hand, the forefinger of your right hand indicates the direction around the boundary.
\PageBreak{1}
}\else
\vfill\par\newpage
\fi


\prob{VIII}{\Qviii} Let $\bfF$ denote the force field in the Kepler
problem, i.e.
$$
   \bfF= -\frac{x\bfi}{\rho^3}- \frac{y\bfj}{\rho^3}-\frac{z\bfk}{\rho^3},
 \qquad \rho=\sqrt{x^2+y^2+z^2}.
$$

\medskipnoindent{\bf(1)} Is there a  function $W$
such  $\nabla W=\bfF$?
If so, find it; if not, say why not.



\ifanswer \Answer{As we learned in our study of the Kepler problem,
$\bfF=\nabla W$ where $W=\rho^{-1}$.
}\else
\vfill
\fi


\medskipnoindent{\bf(2)} Write
an iterated integral (you need not evaluate it) for the outward flux
$$
     \int\!\!\!\!\int_S \bfF\cdot\bfn \,d\sigma
$$
over the ellipsoid $S$ defined by
$$
\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1.
$$
Here $d\sigma$ denotes the area element on the ellipsoid
and $\bfn$ denotes the outward unit normal to the ellipsoid.
If you don't see how to do this,
do the special case of the sphere ($a=b=c$)  for partial credit.\footnote{
In grading this question, it was sometimes difficult to decide if the student
was doing the special case or the general case.}

\ifanswer \Answer{The problem is easy when $a=b=c$; then the outward unit normal
is $\bfn=a^{-1}\bfR$ and $\bfF=-a^{-3}\bfR=-a^{-2}\bfn$ so $\bfF\cdot\bfn=-a^{-2}$.
As this is constant, the integral is this constant times the area of the sphere,
i.e.
$$
     \int\!\!\!\!\int_S \bfF\cdot\bfn \,d\sigma = -a^{-2}(4\pi a^2)=-4\pi.
$$
For the general case,
parameterize the ellipsoid by the equations
$$
  x=a\cos\theta\sin\phi,\qquad y=b\sin\theta\sin\phi,\qquad z=c\cos\phi.
$$
Let $\bfR=x\bfi+y\bfj+z\bfk$ be the position vector. Then
$$
\frac{\p\bfR}{\p\phi}=a\cos\theta\cos\phi\bfi+b\sin\theta\cos\phi\bfj-c\sin\phi\bfk
$$
and
$$
\frac{\p\bfR}{\p\theta}=-a\sin\theta\sin\phi\bfi+b\cos\theta\sin\phi\bfj
$$
so
$$
  \frac{\p\bfR}{\p\phi}\times \frac{\p\bfR}{\p\theta} =
 bc\sin\theta\sin^2\phi\bfi+ac\cos\theta\sin^2\phi\bfj+ab\cos\phi\sin\phi\bfk.
$$
Now
$\ds\bfn \,d\sigma=\frac{\p\bfR}{\p\phi}\times \frac{\p\bfR}{\p\theta}d\phi\,d\theta$
and
$$
   \bfF= -\frac{a\cos\theta\sin\phi\bfi}{\rho^3}- \frac{b\sin\theta\sin\phi\bfj}{\rho^3}-\frac{c\cos\phi\bfk}{\rho^3},
$$
where
$$
 \rho=\sqrt{x^2+y^2+z^2}=(a^2\cos^2\theta\sin^2\phi+b^2\sin^2\theta\sin^2\phi+c^2\cos^2\phi)^{1/2}
$$
so
$$
     \int\!\!\!\!\int_S \bfF\cdot\bfn \,d\sigma =
-abc\int_0^{2\pi}\int_0^\pi
\frac{2\cos\theta\sin\theta\sin^3\phi+\cos^2\phi}{
(a^2\cos^2\theta\sin^2\phi+b^2\sin^2\theta\sin^2\phi+c^2\cos^2\phi)^{3/2}
}
\,d\phi\,d\theta
$$
{\bf Remark.} Some of you may have learned in a course in electrostatics
that the integral is $4\pi$ times the total charge inside. This is actually
a consequence of the Divergence Theorem as follows. A direct calculation
(see problem~11 on page~632) shows that for $W=\rho^{-1}$ we have
$$
   \nabla\cdot\bfF=\nabla\cdot\nabla W=
\frac{\p^2 W}{\p x^2}+\frac{\p^2 W}{\p y^2}+\frac{\p^2 W}{\p z^2}=0.
$$
Let $R$ be the region outside the tiny sphere $\rho=h$ and inside the
ellipsoid $S$. Then
$$
   0=\int\!\!\!\!\int\!\!\!\!\int_R\nabla\cdot\bfF\,dV =\int\!\!\!\!\int_{\p R} \bfF\cdot\bfn \,d\sigma
=\int\!\!\!\!\int_S \bfF\cdot\bfn \,d\sigma-\int\!\!\!\!\int_{\rho=h} \bfF\cdot\bfn \,d\sigma.
$$
The integral over the sphere $\rho=h$ is  $-4\pi$. If you used this
method to answer the question you will get full credit provided you indicated
that  the divergence of $\bfF$ vanishes and that you are using the Divergence Theorem.
}\else
\vfill
% last page
\centerline{(Continue on reverse side if you need more space.)}
\fi

\ifanswer
\begin{verbatim}
____ Thu Dec 20 17:43:32 2001
  

--
There are  253 scores
grade      range       count      percent
   A     320...400     31         12.3%
  AB     300...319     15          5.9%
   B     230...299     65         25.7%
  BC     180...229     60         23.7%
   C     150...179     30         11.9%
   D     100...149     32         12.6%
   F       0... 99     20          7.9%
Mean score = 217.1. Mean grade = 2.42. 
--

\end{verbatim}


The cutoffs for the toal score (out of 1000) were

A 865, AB 785, B 705, BC 603, C 500, D 385.

\fi

\end{document}
