%
\def\whichexam{Exam I}
\def\examdate{September 25, 2001}

\long\def\newp #1.{
\par\noindent\vtop\bgroup\noindent{\bf #1.}
}
\long\def\problembreak{\par\egroup\vfill\eject\vtop\bgroup}
\def\endp{\par\noindent\vrule width\hsize depth 0pt height .1pt\egroup\par\bigskip}
\def\frac#1#2{{{#1}\over{#2}}}
\let\ds=\displaystyle
\font\sm=cmr7
\magnification1200
\centerline{\whichexam}
\centerline{\examdate}
\def\Int{\displaystyle\int}
\def\Lim{\displaystyle\lim}
\def\frac#1#2{{{#1}\over{#2}}}
\bigskip
\newp 11.
To find the tangent line we need to know the slope, which we find via implicit differentiation:
$$2x+y^\prime=3\cos(x+y)(1+y^\prime)\leqno(\ast)$$ 
so at the point $(1,-1)$ we get the equation
$2\cdot 1+y^\prime\vert_{(1,-1)}=3\cos(0)(1+y^\prime\vert_{(1,-1)})=3(1+y^\prime\vert_{(1,-1)})$, so
$y^\prime\vert_{(1,-1)}=-\frac12$.
Hence an equation for the tangent line is $y-(-1)=-\frac12(x-1)$.

The curve lies above the tangent line in a neighborhood of the point $(1,-1)$ provided
$y^{\prime\prime}\vert_{(1,-1)}>0$.
Likewise it lies below the tangent line if $y^{\prime\prime}\vert_{(1,-1)}<0$.
We compute implicitly by differentiating $(\ast)$:
$$2+y^{\prime\prime}=-3\sin(x+y)(1+y^\prime)^2+3\cos(x+y)(y{\prime\prime})\ .$$
At the point $(1,-1)$ we get 
$2+y^{\prime\prime}\vert_{(1,-1)}=
-3\sin(0)(1+y^\prime)^2+3\cos(0)(y^{\prime\prime}\vert_{(1,-1)})=
3(y^{\prime\prime}\vert_{(1,-1)})$, so $y^{\prime\prime}\vert_{(1,-1)}=1>0$
and the curve lies above the tangent line in a neighborhood of the point.
\endp
\newp 12.
You are being asked to find $\frac{dV}{dt}$ at a certain moment in time.
You are given that $V=\frac{\pi}{3}r^2h$ so
$\frac{dV}{dt}=\frac\pi3 2r\frac{dr}{dt} h+\frac\pi3 r^2\frac{dh}{dt}$.
The time being asked for is a time when $h=r=2$, $\frac{dr}{dt}=4$ and
$\frac{dh}{dt}=2$, so
$\frac{dV}{dt}=\frac\pi3 2\cdot 2\cdot 4\cdot 2 +\frac\pi3 2^2\cdot2=(32+8)\frac\pi3=\frac{40}{3}\pi$.
\endp
\newp 13.
By the Mean Value Theorem,
$\sin(a)-\sin(b)=\cos(c)(b-a)$ for some $c$ between $a$ and $b$.
If we take $a=2.674$ and $b=2.670$ we see $b-a=0.004$ and we know that $-1\leq\cos(c)\leq1$
and it follows that
$-0.004\leq \sin(2.674)-\sin(2.670)\leq 0.004$ which is equivalent to our absolute value inequality.
Since $\frac\pi2<2.670<2.674<\pi$, we actually know $\frac\pi2<c<\pi$ and for any $c$ in this
range, $\cos(c)<0$, so we see $\sin(2.674)-\sin(2.670)<0$.
\endp
\newp 14.
First we locate the critical points: $y^\prime$ is defined and continuous everywhere since
$x^2+1>0$ so we are never dividing by $0$ or taking the square root of a negative number.
Therefore $x=0$ is the only critical point.
Similar reasoning shows $y^{\prime\prime}$ is defined and continuous everywhere and is never $0$.
Therefore,
\itemitem{a.} $f$ is increasing on the interval $(0,\infty)$ and decreasing on the interval $(-\infty,0)$.
\itemitem{b.} $f$ is concave up everywhere and concave down nowhere.
\itemitem{c.}Since $\sqrt{x^2+1}$ is defined and continuous everywhere, 
the graph can have no vertical asymptotes.

To check the slant asymptote assertions we need to show
$\Lim_{x\to\infty} \sqrt{x^2+1}-x=0$ and $\Lim_{x\to\infty} \sqrt{x^2+1}-(-x)=0$.

Compute
$\Lim_{x\to\infty} \sqrt{x^2+1}-x=
\Lim_{x\to\infty} (\sqrt{x^2+1}-x)\cdot\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}+x}=$ \par\noindent
$\Lim_{x\to\infty} \frac{(\sqrt{x^2+1})^2-x^2)}{\sqrt{x^2+1}+x}=\Lim_{x\to\infty} \frac{1}{\sqrt{x^2+1}+x}=
\Lim_{x\to\infty}\frac1x\cdot\frac{1}{\sqrt{1+\frac{1}{x^2}}+1}=$\par\noindent$
\Lim_{x\to\infty}0\cdot\frac{1}{\sqrt{1+0}+1}=0
$.

Compute
$\Lim_{x\to-\infty} \sqrt{x^2+1}-(-x)=
\Lim_{x\to-\infty} (\sqrt{x^2+1}+x)\cdot\frac{\sqrt{x^2+1}-x}{\sqrt{x^2+1}-x}=$ \par\noindent
$\Lim_{x\to-\infty} \frac{(\sqrt{x^2+1})^2-x^2)}{\sqrt{x^2+1}-x}=
\Lim_{x\to-\infty} \frac{1}{\sqrt{x^2+1}-x}=
\Lim_{x\to-\infty}\frac1x\cdot\frac{1}{-\sqrt{1+\frac{1}{x^2}}-1}=$\par\noindent$
\Lim_{x\to-\infty}0\cdot\frac{1}{-\sqrt{1+0}-1}=0
$.

\endp
\end