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%%%%%%  DEFINITIONS  %%%%
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 \magnification1200

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\def\pih{\sfrac\pi2}
\def\Lim{\displaystyle\lim}
\def\Sum{\displaystyle\sum}

\newcount\error\newcount\errorans

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\else\LL{#1}\fi}}%%
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\settabs\+  01\quad&184\quad&Nieuwland \quad
&11:00 - 11:50\quad&Instructor\cr

\long\def\hide#1{}
\hide{
%%%%%%%%%%%%%%%%%%%%
%%%  HEADER PAGE  %%%%%%
%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%

\null\vskip-.5true in
\centerline{\bigbold Math 126}
\centerline{\bigbold Fall 1997}
\centerline{\bigbold Final Examination}
\centerline{\bigbold Dec. 16, 1997}
\vskip 20true pt
{\big
{{\parindent=1true in
\itemitem{$\bullet$}
This test runs for two hours.
\itemitem{$\bullet$}
All work is to be your own.
\itemitem{$\bullet$}
Calculators are not to be used.
}}

\vbox to 1true in{
\vskip .5true in
\noindent{\bigbold Name:}\quad$\underline{\hskip 3true in}$%
\hskip.75true in
{\bf Professor:}\ $\underline{\hbox{\Prof}}$\vfill}

\big
If \Prof\ is not
the name of your professor,
please raise your hand.
\vskip20pt
\centerline{\bigbold Grading on this examination.}
\vskip8pt
\centerline{\bigbold READ CAREFULLY BEFORE BEGINNING.}
\vskip 10true pt
The first three questions will require you to demonstrate 
knowledge of a correct procedure for arriving at the answer.
Please circle the answer you want us to consider.
The correct answer is not enough for full credit, 
you must indicate how you arrived at it.
Please mark out any work on a problem that you do not want us
to consider.
Questions 4 through 24 are multiple choice and 
are either correct or wrong.
Please put an {\bf X} through your answer to each multiple choice
question in the 
space provided on this page as in the {\bf Sample} below.
}

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$\underline{\hskip\hsize}$\vskip10pt%
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%%%%%%%%%%%%%%%%%%%%
%%%%%  PROBLEMS  %%%%%%%%%
%%%%%%%%%%%%%%%%%%%%
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\problem(15pt) 
Find the area inside the 
3{--}leafed rose, $r=\sin(3\theta)$.\par
\vskip.75in
\hskip3in\hbox to0pt{\scaledpicture 100pt by 100pt (roseB scaled 500)\hss}

{\bf Area}=$\frac12\Int^\beta_\alpha r^2\> d\theta$.
From the graph check that the rose is swept out once as $\theta$ runs from
$0$ to $\pi$:  $0$ to $\frac\pi3$ is the leaf in the first quadrant;
$\frac\pi3$ to $\frac{2\pi}3$ is the leaf centered on the negative $y${--}axis;
and $\frac{2\pi}3$ to $\pi$ is the leaf in the second quadrant.

Hence {\bf Area} $= \frac12\Int^\pi_0 \sin^2(3\theta)\> d\theta=
\frac12\Int^\pi_0 \Frac{1-\cos(6\theta)}{2}\>d\theta=
\frac14\biggl( \theta\Big\vert^\pi_0-\frac16\sin(6\theta)\Big\vert^\pi_0\biggr)=
\Frac{\pi}4
$.
\def\nL{\par\vrule width\hsize depth .1pt height0pt\par}
\nL
\problem(15pt) 
Find the arclength of the curve $r=\sec\theta$ 
with $0\leq\theta\leq\Frac\pi4$.

There are two solutions.
The first is to note $r=\sec\theta$ is equivalent to $r\cos\theta=1$, or $x=1$,
so our graph is a part of the vertical line $x=1$.
The $y${--}coordinate is given by 
$y=r\sin\theta=\frac{\sin\theta}{\cos\theta}=\tan\theta$.
When $\theta=0$, $y=0$ and when $\theta=\frac\pi4$, $y=1$.
Hence the length is $1$.

The overwhelming majority of you proceeded as follows.
{\bf Length}$=\Int^\beta_\alpha\sqrt{ r^2+(r^\prime)^2\ }\> d\theta$.
$\frac{d\>r}{d\theta}=\sec\theta\tan\theta$;
$r^2+(r^\prime)^2=\sec^2\theta+\sec^2\theta\tan^2\theta=
\sec^\theta\>(1+\tan^2\theta)=\sec^2\theta\sec^2\theta$.
Hence
{\bf Length}$=\Int^{\frac\pi4}_0\sqrt{ \sec^4\theta\ }\> d\theta=
\Int^{\frac\pi4}_0\sec^2\theta\>d\theta=
\tan\theta\Big\vert^{\frac\pi4}_0=1-0=1
$.
\nL
\problem(15pt)
Find the surface area of the surface obtained
by rotating the piece of
$r^2=1+\cos(2\theta)$ in the first quadrant
around the $x${--}axis.
(The graph is that of the entire curve.)
\vskip.75in
\hskip3.5in\hbox to0pt{\scaledpicture 100pt by 100pt (lem scaled 500)\hss}

By studying the graph, see that we need to rotate the part of the curve
$r=\sqrt{1+\cos(2\theta)\>}$ for
$\theta$ between $0$ and $\pi$.
{\bf Surface Area} $=2\pi\Int^\beta_\alpha r\sin\theta
\sqrt{ r^2+(r^\prime)^2\ }\> d\theta$.

$\frac{d\>r}{d\theta}=\Frac{-2\sin(2\theta)}{2\sqrt{1+\cos(2\theta)\>}}=
\Frac{-\sin(2\theta)}{\sqrt{1+\cos(2\theta)\>}}$, so

$ r^2+(r^\prime)^2=1+\cos(2\theta)+\Frac{\sin^2(2\theta)}{1+\cos(2\theta)}=
\Frac{1+2\cos(2\theta)+\cos^2(2\theta)+\sin^2(2\theta)}{1+\cos(2\theta)}=$

$\Frac{2+2\cos(2\theta)}{1+\cos(2\theta)}=2
$.
Hence 
{\bf Surface Area} $=2\pi\Int^{\frac\pi2}_0 
\sqrt{1+\cos(2\theta)\ }\
(\sin\theta)\sqrt{2 }\> d\theta=$

$2\pi\Int^{\frac\pi2}_0 
\sqrt{2\cos^2\theta\ }\
(\sin\theta)\sqrt{2 }\> d\theta=
4\pi\Int^{\frac\pi2}_0 \cos\theta\sin\theta\> d\theta=
2\pi\sin^2\theta\Big\vert^{\frac\pi2}_0=2\pi
$.

A second approach proceeds as follows.
Rewrite $r=\sqrt{1+\cos(2\theta)\>}$ as $r=\sqrt{2}\>\cos\theta$ using the
half-angle formula.
Then $r^\prime=-\sqrt{2}\>\sin\theta$ so $r^2+(r^\prime)^2=2$ and
{\bf Surface Area} $=2\pi\Int^{\frac\pi2}_0 
(\sqrt{2}\>\cos\theta)\sin\theta\>\sqrt{2}
\> d\theta$ and finish as above.

\nL

%%%%%%%%%%%%%%%%%%%%
%%%  MULTIPLE CHOICE  %%%%
%%%%%%%%%%%%%%%%%%%%
\np(5pt) %#4
Which function below is the inverse function to
$f(x)= e^{2x}$?
\ans{54312}
{$\bullet\Frac12\ln x$}
{$\ln(x/2)$}
{$\sqrt{\ln x}$}
{$e^{-x/2}$}
{$e^{-x}+\ln\vert x\vert$}@
Since $f(\frac12\ln x)=e^{2\cdot\sfrac12\ln x}=e^{\ln x}=x$, 
$\frac12 \ln x$ is the inverse function.


\nL
\np(5pt) %#5
Which substitution reduces the integral
$\Int \Frac{dx}{\sqrt{4-x^2}}$ to the integral $\Int du$\ ?
\ans{13425}
{$\bullet x=2\sin u$}
{$u=2\sin x$}
{$x=\Frac12\tan u$}
{$x=\sin u + \cos u$}
{$x=\sqrt 2\cos u$}@
If $x=2\sin u$, $dx=2\cos u\>du$ and $\sqrt{4-x^2\>}=\sqrt{4-4\sin^2u\>}=
2\sqrt{cos^2u\>}=2\cos u$, so
$\Int \Frac{dx}{\sqrt{4-x^2}}=\Int \Frac{2\cos u\>du}{2\cos u}$.
\nL

\np(5pt) %#6
Use Integration by Parts to show $\Int_0^\pih \cos^{10}x\>dx$ equals
one of the numbers below.
\ans{34125}
{$\bullet\Frac{9}{10}\Int_0^\pih\cos^{8}x\>dx$}
{$0$}
{$\Frac\pi2$}
{$(-10)\Int^\pih_0\cos^{9}x \sin x\>dx$}
{$\Int^\pih_0\cos^{8}x\>dx$}@
$u=\cos^9x$ $dv=\cos x\> dx$ so $du=-9\cos^8x\sin x\>dx$, $v=\sin x$
and
$\Int_0^\pih \cos^{10}x\>dx=\cos^9x \sin x\Big\vert^\pih_0+
9\Int^\pih_0\cos^8x\sin^2x\>dx=
9\Int^\pih_0\cos^8x(1-\cos^2x)dx=
9\Int^\pih_0\cos^8x\> dx-9\Int^\pih_0\cos^{10}x\> dx$.
Now solve for $\Int^\pih_0\cos^{10}x\> dx$.

\nL

\np(5pt) %#7
The value of $\Lim_{t\to\infty}t\cdot\sin\biggl(\frac1{2t}\biggr)=$
\ans{42531}
{$\bullet\Frac12$}
{$2$}
{$\Frac14$}
{$4$}
{$1$}@
By L'H\^opital's Rule,
$\Lim_{t\to\infty}t\cdot\sin\biggl(\frac1{2t}\biggr)=
\Lim_{t\to\infty}\Frac{\sin\biggl(\frac1{2t}\biggr)}{\Frac1t}=
\Lim_{t\to\infty}\Frac{\cos\biggl(\frac1{2t}\biggr)\Frac12\Frac{-1}{t^2}}
{\Frac{-1}{t^2}}=$\par$
\frac 12\Lim_{t\to\infty}\cos\biggl(\frac1{2t}\biggr)=\frac12
$.
\nL
\np(5pt) %#8
Which function below is the solution to the initial value
problem $y^\prime=3y$, $y(1)=1$?
\ans{21354}
{$\bullet y(t)=\Frac{e^{3t}}{e^3}$}
{$y(x)=\Frac{e^3}{e^{3x}}$}
{$y(s)=e^{3s}-e^3+1$}
{$y(t)=e^{3t}$}
{$y(x)=x^3$}
@
This is a growth/decay differential equation so $y=Ce^{3t}$.
Since $y(1)=1$, $C=\frac1{e^3}$.
\nL
\snp(5pt) %#9
Indicate which one of the statements below is true.
The series $\Sum_{n=0}^\infty\Frac1{n^2+3}$
\ans{123}
{$\bullet$absolutely converges}
{conditionally converges}
{diverges}{}{}
@
Compare to the convergent $p${--}series $\Sum{n=1}\Frac{11}{n^2}$.

\nL
\snp(5pt) %#10
Indicate which one of the statements below is true.
The series $\Sum_{n=1}^\infty\Frac{(-1)^n}{\root3\of{n^2+n}}$
\ans{312}
{$\bullet$conditionally converges}
{diverges}
{absolutely converges}{}{}
@
Alternating series.
Compare to $p${--}series $\Sum{n=1}\Frac{1}{n^{\frac23}}$ which diverges.
Terms of the alternating series decrease to $0$, so converges conditionally.
\nL
\snp(5pt) %#11
Indicate which one of the statements below is true.
The series $\Sum_{n=3}^\infty\Frac{1}{n\ln n}$
\ans{231}
{$\bullet$diverges}
{absolutely converges}
{conditionally converges}{}{}
@
The Integral Test applies and we need to evaluate 
$\Int^\infty_2\Frac{dx}{x(\ln x)^{\sfrac12}}=2(\ln x)^{\sfrac12}\Big\vert^\infty_2=
$\par$2\Lim_{x\to\infty}(\ln x)^{\sfrac12}-2\sqrt{\ln 2}
$ which diverges since $\ln x$ goes to $\infty$ as $x$ does.
\nL

\np(5pt) %#12
Indicate which one of the statements below is true.
The series \par
\centerline{$\Sum_{n=0}^\infty\Frac{2n+1}{(n^2+1)(n^2+2n+2)}$}
\ans{21453}
{$\bullet$\hbox{has value $1$}}
{\hbox{has value $2$}}
{\hbox{has value $3$}}
{\hbox{has value $4$}}
{diverges}
@
This is a telescoping series since
$\Frac{2n+1}{(n^2+1)(n^2+2n+2)}=
\Frac{1}{n^2+1}-\Frac{1}{n^2+2n+2}=\Frac{1}{n^2+1}-\Frac{1}{(n+1)^2+1}$.
Since $\Lim_{n\to\infty}\Frac{1}{n^2+1}=0$, the series sums to
$\Frac{1}{0^2+1}$.
\nL

\np(5pt) %#13
Which series below is the Taylor series for the function $\ln x$ at $2$?
\ans{24153}
{$\bullet\ln 2+\Sum_{n=1}^\infty(-1)^{n+1}\Frac{(x-2)^n}{n2^n}$}
{$\Sum_{n=1}^\infty(-1)^n\Frac{x^n}n$}
{$\Sum_{n=0}^\infty\Frac{(x-2)^n}{n!}$}
{$\Sum_{n=1}^\infty(-1)^n\Frac{(x-2)^{2n}}{(2n)!}$}
{$\Sum_{n=2}^\infty(-1)^{n-1}\Frac{(x-2)^{n-3}}{\ln2}$}
@
(c) is the only series which has the correct constant term.
\nL

\np(5pt) %#14
The MacLaurin Series for $(1+x^3)^{\sfrac13}$ starts out
\ans{23451}
{$\bullet 1+\mfrac13x^3-\mfrac19x^6+\mfrac{5}{81}x^9\cdots $}
{$1+\mfrac13x-\mfrac19x^2+\mfrac{5}{81}x^3\cdots $}
{$\mfrac13x^3-\mfrac19x^6+\mfrac{5}{81}x^9\cdots $}
{$\mfrac13x-\mfrac19x^2+\mfrac{5}{81}x^3\cdots $}
{$\root3\of{1+x^3}+\cdots$}
@
Plug into the Binomial Theorem:
$(1+x^3)^{\sfrac13}=\Sum_{n=0}{\sfrac13\choose n}(x^3)^n=
1+\Frac13 x^3+\Frac{\frac13(\frac13-1)}{2}x^6+\cdots=
1+\frac13 x^3-\frac19 x^6+\cdots$

\nL
\np(5pt) %#15
The difference
$\Sum_{n=1}^\infty(-1)^{n+1}\Frac{n^2}{2^n}-
\Sum_{n=1}^4(-1)^{n+1}\Frac{n^2}{2^n}$ is
\ans{14253}
{$\bullet$\hbox{positive and less than $\Frac{25}{32}$}}
{\hbox{negative and greater than $-\Frac{25}{32}$}}
{\hbox{positive and less than $\Frac{49}{1024}$}}
{\hbox{negative and greater than $-\Frac{49}{1024}$}}
{\hbox{greater than $-\Frac{1}{1024}$ and less than $\Frac{1}{1024}$}}
@
$\Sum_{n=1}^\infty(-1)^{n+1}\Frac{n^2}{2^n}$ is an alternating series to which
the Alternating Series Test applies.
Hence the difference is bounded by the absolute value of the $5$th term,
$\frac{25}{32}$, and
is positive.

\nL
\np(5pt) %#16
The partial fraction decomposition for
$\Frac{x^5+4x^3-4x^2+2x-3}{x^2(x^2+1)}$ is
\ans{25413}
{$\bullet x+\frac2x-\frac3{x^2}+\frac{x-1}{x^2+1}$}
{$x+\frac3{x^2}+\frac{x-1}{x^2+1}$}
{$\frac2x-\frac5{x^2}+\frac{x-1}{x^2+1}$}
{$x-\frac3{x^2}+\frac{x-1}{x^2+1}$}
{$x^2+\frac2x-\frac5{x^2}+\frac{x-1}{x^2+1}$}
@
By polynomial long division,
$\Frac{x^5+4x^3-4x^2+2x-3}{x^2(x^2+1)}=
x+\Frac{3x^3-4x^2+2x-3}{x^2(x^2+1)}$ and
$\Frac{3x^3-4x^2+2x-3}{x^2(x^2+1)}=\Frac{A}{x}+\Frac{B}{x^2}+\Frac{Cx+D}{x^2+1}$.
Hence $3x^3-4x^2+2x-3=Ax(x^2+1)+B(x^2+1)+(Cx+D)x^2$.
Plug in $x=0$, $B=-3$.
Hence $Ax(X^2+1)+(Cx+D)x^2=3x^3-x^2+2x$ or
$A(x^2+1)+(Cx+D)x=3x^2-x+2$.
Plug in $x=0$, $A=2$.
Hence $(Cx+D)x=x^2-x$ so $Cx+D=x-1$.

\nL
\np(5pt) %#17
$\Int^1_0 xe^xdx=$
\ans{15342}
{$\bullet$1}
{2}
{3}
{4}
{0}
@
By parts: $u=x$ $dv=e^xdx$; $du=dx$, $v=e^x$.
$\Int^1_0 xe^xdx=xe^x\Big\vert^1_0-\Int^1_0 e^xdx=
xe^x\Big\vert^1_0-e^x\Big\vert^1_0=(e-0)-(e-1)=1$.
\nL

\np(5pt) %#18
The radius of convergence of
$\Sum_{n=0}^\infty \Frac{1\cdot 3\cdots(2n-1)}
{2\cdot 4\cdots(2n)}x^n$ is
\ans{53421}
%\ans{14523}
{$\bullet$1}
{2}
{8}
{3}
{$\infty$}
@
There is a typo - the sum should start at $n=1$.
Compute
\par$\Frac{\Frac{1\cdot 3\cdots(2n+1)}
{2\cdot 4\cdots(2n+2)}x^{n+1}}{\Frac{1\cdot 3\cdots(2n-1)}
{2\cdot 4\cdots(2n)}x^n}=\Frac{2n+1}{2n+2}\cdot x
$.
As $n$ goes to $\infty$ this quantity goes to $x$, so the radius of
Convergence is $1$.
\nL

\np(5pt) %#19
$\Frac{d\>3^x}{dx}=$
\ans{45132}
{$\bullet(\ln3)3^x$}
{$3^x$}
{$\Frac{3^x}{\ln3}$}
{$x3^{x-1}$}
{$\ln(3^x)$}
@
$\Frac{d\>3^x}{dx}=\Frac{d\>(e^{\ln 3})^x}{dx}=
\Frac{d\>e^{x\ln 3}}{dx}=(\ln 3)e^{x\ln 3}$.
\nL

\np(5pt) %#20
$\Int^\infty_0\Frac{dx}{x^2+1}=$
\ans{21543}
{$\bullet\Frac\pi2$}
{$\Frac\pi3$}
{$0$}
{$\arctan 4$}
{diverges}
@
$\Int^\infty_0\Frac{dx}{x^2+1}=\Lim_{x\to\infty}\arctan x - \arctan 0=
\Frac\pi2-0$.
\nL

\np(5pt) %#21
$\Int_0^2\Frac{dx}{x-1}=$
\ans{42315}
{$\bullet$diverges}
{$\ln3$}
{$\ln2$}
{$2$}
{$0$}
@
The integral is improper because $x-1$ vanishes at $x=1$.
$\Int\Frac{dx}{x-1}=\ln\vert x-1\vert+C$.
Since $\Lim_{x\to 1^+}\ln\vert x-1\vert=-\infty$, the integral diverges.

\nL
\np(5pt) %#22
Which of the functions below grows the most slowly?
\ans{34215}
{$\bullet x\ln x$}
{$x^6+3x^5+4$}
{$x^3-3x$}
{$\Frac{x^2}{4}$}
{$\ln\Bigl(e^{x^6}\Bigr)$}
@
(e) is a fancy way to write the polynomial $x^6$.
(d) grows faster than $x$ and more slowly than $x^2$.
\nL

\np(5pt) %#23
$\Frac{d\>\arc{\sec}(x^2)}{dx}=$
\ans{43251}
{$\bullet\Frac{2}{x\sqrt{x^4-1}}$}
{$\Frac{1}{x^2\sqrt{x^4-1}}$}
{$\Frac{1}{\sqrt{1-x^4}}$}
{$\Frac{2x}{\sqrt{1-x^4}}$}
{$\Frac{x^2}{\sqrt{1-x^4}}$}
@
$\Frac{d\>\arc{\sec}(x)}{dx}=\Frac{1}{x\sqrt{x^2-1\>}}$ so by the Chain Rule,
$\Frac{d\>\arc{\sec}(x^2)}{dx}=\Frac{2x}{x^2\sqrt{(x^2)^2-1\> }}$

\nL\np(5pt) %#24
Which function below is a solution to the differential equation
$y^\prime-x^2y=0$?
\ans{25134}
{$\bullet y=2e^{\sfrac{x^3}3}$}
{$y=3+e^{\sfrac{x^3}3}$}
{$y=3e^{\sfrac{x^2}2}$}
{$y=2+e^{\sfrac{x^2}2}$}
{$y=\frac32e^{\sfrac{x^3}2}$}
@
This is a linear equation in standard form.
The integrating factor is $v=e^{^{\int -x^2dx}}=e^{\sfrac{-x^3}{3}}$.
Hence
$\biggl(e^{\sfrac{-x^3}{3}}y\biggr)^\prime=0$ so
$e^{\sfrac{-x^3}{3}}y=C$, or
$y=Ce^{\sfrac{x^3}{3}}$.
(c) is the only function of this form among the choices.

\end