\magnification1200\footline{}

\def\frac#1#2{{{#1}\over{#2}}}
\def\equiz{\par\noindent\vrule width\hsize height 0pt depth .1 pt
\par}
\def\leaderfill{\leaders\hbox to .5em{\hss.\hss}\hfil}

\def\dquiz{\par\noindent\hbox to \hsize{\leaderfill}
\par\medskip}

\long\def\stp\centerline#1 \centerline#2 \vskip10pt \noindent{%
\par\bigskip\noindent{#1} \qquad{ #2}\par\medskip\noindent}

\input epsf.def
\gdef\localbox#1{\epsfscale=1600\vtop to1.5in{\epsfbox{#1.eps}\vss}}


%%%%%%%%%%%%%%%
%################
\def\frac#1#2{{{#1}\over{#2}}}
\def\sL#1{\displaystyle\lim_{#1\to\infty}}
\def\isum#1{{\displaystyle{\sum_{n={#1}}^{\infty}}}\>}

\newcount\cc\cc=1
\def\np{%
\par\noindent{\bf\number\cc.}\ %
\global\advance\cc by1%
}



\stp
\centerline{\bf Math.~125 Quiz \#1}
\centerline{September 3, 2002}
\vskip10pt
\noindent
If $$(f\circ g)(x)=f\bigl(g(x)\bigr)=(\sin x)^2 + 3\sin x +7$$
what is a possibility for $f$ and $g$?

\noindent
If $h(u)=\cos u$  and $m(s)=2s+3$ what is
$(h\circ m)(y)=h\bigl(m(y)\bigr)$?

\noindent
If the domain of $f$ is the interval $(3,7)$, 
what is the domain of $f(x+2)$?


\dquiz
$g(x)=\sin x$ and $f(x)=x^2+3x+7$.


$(h\circ m)(y)=h(m(y))=h(2y+3)=\cos (2y+3)$.

The domain of $f(x+2)$ is in the interval $(1,5)$.
\equiz

%################
\stp
\centerline{\bf Math.~125 Quiz \#2}
\centerline{September 10, 2002}
\vskip10pt
\noindent

\np
What is 
$\displaystyle\lim_{x\to 3} \frac{3 x - 1}{1-x}$\ ?

\np
What is 
$\displaystyle\lim_{x\to 2^-}\frac 1{2-x}$\ ?

\np
Suppose $\displaystyle\lim_{x\to 1}f(x)=4$ and
$\displaystyle\lim_{x\to 1}g(x)=-2$.
What is the value of the following limit?
$$\displaystyle\lim_{x\to 1}\frac{f(x) \cdot g(x)}
{\bigl(4+g(x)\bigr)^2}$$

\dquiz\cc=1
\np
$\displaystyle\lim_{x\to 3} \frac{3 x - 1}{1-x} = \frac{3\cdot 3
-1}{1-3} = \frac 8{-2} = -4$
 
\np
$\displaystyle\lim_{x\to 2^-}\frac 1{2-x} = +\infty $
\np
$$\displaystyle\lim_{x\to 1}\frac{f(x) \cdot g(x)}
{\bigl(4+g(x)\bigr)^2} = \frac{\lim_{x\to 1}f(x)\cdot \lim_{x\to
1}g(x)}{\bigl(4+\lim_{x\to 1}g(x)\bigr)^2} = \frac{4 \cdot
(-2)}{(4-2)^2} = -2$$ 
\equiz

%################
\stp
\centerline{\bf Math.~125 Quiz \#3}
\centerline{September 17, 2002}
\vskip10pt
\noindent
\newcount\cc\cc=1

\noindent
For each of the functions below, select one of
the properties A-D.
\itemitem{A.} Is continuous for all $x$.
\itemitem{B.} Has one or more vertical asymptotes.
\itemitem{C.} Has one or more removable singularities.
\itemitem{D.} None of the above.
\def\\{\par\noindent}
\def\mathbb#1{{\bf #1}}
\cc=1

\np
The function $[[x]]+1$.

\np 
The function $\displaystyle\frac{x^2+3x+2}{x+1}$.

\np 
The function $\frac{(x-3)\sin x}{x^2-6x+9}$.

\np
The function $x+\root 3\of{1-x}$.
\dquiz
\cc=1
\np
The function $[[x]]+1$.\\
Answer:  D\\
There is a jump discontinuity at each integer value.

\np
The function $\displaystyle\frac{x^2+3x+2}{x+1}$.\\
Answer:  C\\
\noindent
$$
\frac{x^2+3x+2}{x+1}=\frac{(x+2)(x+1)}{x+1}=x+2
$$
Therefore, we have a removable singularity at $x=-1.$


\np
The function $\frac{(x-3)\sin x}{x^2-6x+9}$.\\
Answer:  B\\
\noindent
$$
\frac{(x-3)\sin x}{x^2-6x+9}=\frac{(x-3)\sin
x}{(x-3)(x-3)}=\frac{\sin
x}{x-3}.
$$
Therefore, we have a vertical asymptote at $x=3$.

\np
The function $x+\root 3\of{1-x}$.\\
Answer:  A\\
\noindent
Root functions are continuous everywhere in their domain.  So
$\root 3\of{1-x}$ is continuous on $\mathbb{R}$.  $g(x)=x$ is
continuous on $\mathbb{R}$.  Hence, $x+\root 3\of{1-x}$ is
continuous for all $x$.
\equiz

%################
\stp
\centerline{\bf Math.~125 Quiz \#4}
\centerline{October 1, 2002}
\vskip10pt
\noindent
\newcount\cc\cc=1
\def\np{%
\vskip20pt\noindent{\bf\number\cc.}\ %
\global\advance\cc by1%
}
\def\frac#1#2{{{\textstyle #1}\over{\textstyle #2}}}

Find a formula for the derivative of 
$f(x)=\frac{\tan(x)+\sin(x)}{x^2+4x-3}$.

\dquiz
\cc=1
We have a quotient so the answer is 

\noindent
(derivative of the numerator) $\times$ (denominator) - 
(numerator) $\times$ (derivative of the denominator)
divided by (the denominator) squared
\vskip10pt

$f^\prime(x)=\frac{(\sec^2(x)+\cos(x))(x^2+4x-3)-
(\tan(x)+\sin(x))(2x+4)}{(x^2+4x-3)^2}$.
\vskip4pt
\equiz

%################
\stp
\centerline{\bf Math.~125 Quiz \#5}
\centerline{October 8, 2002}
\vskip10pt
\noindent
\newcount\cc\cc=1
\def\np{%
\vskip20pt\noindent{\bf\number\cc.}\ %
\global\advance\cc by1%
}
\def\frac#1#2{{{\textstyle #1}\over{\textstyle #2}}}

\itemitem{1.a)}Find a formula for $\frac{dy}{dx}$ if
$$x^4+5xy^2+y^5=-5\ .$$
\itemitem{1.b)}Find an equation for the tangent line to the curve
at the point $(-1,-1)$.
\vskip4pt
\itemitem{2\hphantom{.c})} Suppose that, when $x=0$,
$f^\prime(0)=-0.5253$. If $h(x)=f(x^2+10x)$, find $h^\prime(-10)$.
\dquiz
1a)
Using implicit differentiation:
$$4 x^3 +\bigl(5y^2+10xyy^\prime\bigr)+5y^4 y^\prime=0$$
or
$$y^\prime=-\frac{4x^3 + 5y^2}{10 x y + 5 y^4}$$
\vskip2pt
1b)
Line goes thru $(-1,-1)$ and has slope 
$\frac{4(-1)^3+5(-1)^2}{10(-1)(-1)+5(-1)^4}=
-\frac{5-4}{10+5}=-\frac{1}{15}$, so an equation for the 
tangent line is
$y-(-1)=-\frac{1}{15}\bigl((x-(-1)\bigr)$, or
$15y+15=-(x+1)$.

\vskip4pt
2) By the Chain Rule, $h^\prime(x)=f^\prime(x^2+10x) (2x+10)$,
so \par\noindent
$h^\prime(-10)=f^\prime(0)\cdot(2(-10)+10)=-0.5253\cdot(-10)=5.253$.
\equiz


%################
\stp
\centerline{\bf Math.~125 Quiz \#6}
\centerline{October 15, 2002}
\vskip10pt
\noindent
\newcount\cc\cc=1
\def\np{%
\vskip20pt\noindent{\bf\number\cc.}\ %
\global\advance\cc by1%
}
\def\frac#1#2{{{\textstyle #1}\over{\textstyle #2}}}

Consider a clock whose hour hand has length $a$ and whose
minute hand has length $b$.
At $3$ o'clock, how fast is the distance between the
tip of the hour hand and the tip of the minute hand changing?

Let $\theta$ be the angle between the minute and hour
hand.
The distance, $c$, between the tips is given by the law of cosines
$$c^2=a^2+b^2-2ab\cos(\theta)\ .$$
\vskip-.75in
\localbox{clock}
\vskip -.5in
\noindent
Step 1.
It is true that $\frac{d \theta}{dt}$ is a constant.
What constant is it?
\par\noindent{\bf Hint:} $\theta=\theta_{\rm hr}-\theta_{\rm min}$
where $\theta_{\rm hr}$ is the angle the hour hand makes with the vertical
and  $\theta_{\rm min}$ is the angle the minute hand makes 
with the vertical.
Both $\frac{d \theta_{\rm hr}}{dt}$ and 
$\frac{d \theta_{\rm min}}{dt}$ are constants. What are they?


\vskip 1in
\noindent Step 2.
Use this fact and your related rates techniques to answer
the original question.
If the distance is {\sl decreasing\/} be sure your formula
gives a negative value; if increasing, a positive one.
\dquiz
$$\eqalign{%
c^2=&a^2+b^2-2ab\cos(\theta)\ \Rightarrow
2c \frac{dc}{dt}=
-2ab(-\sin(\theta)) \frac{d\theta}{dt}
=2ab\sin(\theta) \frac{d\theta}{dt}
\cr&\Rightarrow \frac{dc}{dt}=\frac{2ab\sin(\theta)
\frac{d\theta}{dt}}{2c}=\frac{ab\sin(\theta)
(-\frac{11\pi}{6})}{c}\cr
}$$
At 3:00, $\theta=\frac {\pi}{2}$ and $\sin(\theta)=
\sin (\frac {\pi}{2})=1.$  Therefore, at 3:00,
$$
  \frac{dc}{dt}=\frac{ab\sin(\frac {\pi}{2})
(-\frac{11\pi}{6})}{c}=\frac{ab
(-\frac{11\pi}{6})}{c}=-\frac{11ab\pi}{6c}.
$$
\equiz

\stp
\centerline{\bf Math.~125 Quiz \#7}
\centerline{November 5, 2002}
\vskip10pt
\noindent
\item{1.}Find all the critical points for the function below.
At which critical points is the function
value a local maximum? Why?
At which critical points is the function
value a local minimum? Why?
\item{2.}The existence theorem for global max/min does not apply to
this function restricted to the interval
$[0,\infty)$ since the interval must be a closed interval for
the theorem to apply.
Nevertheless, this function does assume a minimum value on this interval.
Where is it and why?
Why does this function assume no maximum value on this interval?

$${{(x-3)^6}\over{(x+1)^{2}}}$$

\dquiz
\noindent 1.
Let $r(x) = \frac{(x-3)^6}{(x+1)^{2}}$. 
Therefore,
$$\eqalign{
r^\prime(x)=&
\frac{(x+1)^2\cdot 6\cdot(x-3)^5-(x-3)^6\cdot 2 \cdot (x+1)}{(x+1)^4}
\cr=&
\frac{(x+1)(x-3)^5[6(x+1)-2(x-3)]}{(x+1)^4}
=\frac{(x-3)^5(4x+12)}{(x+1)^3} =\frac{4(x-3)^5(x+3)}{(x+1)^3}\ .\cr
}$$

\noindent
$r^\prime(x)=0 \Leftrightarrow (x-3)^5(x+3)=0 \Leftrightarrow x=-3, 3$.
$r^\prime(x)$ does not exist 
$\Leftrightarrow (x+1)^3=0 \Leftrightarrow x=-1$. 
So the critical numbers of $r(x)$ are $x=-3, -1 ,3$.  
We have $x\in (-\infty, -3) \Rightarrow r^\prime(x)<0$,  
$x \in (-3,-1) \Rightarrow r^\prime(x)>0$, 
$x \in (-1,3) \Rightarrow r^\prime(x)<0$ 
and $x \in (3,\infty) \Rightarrow r^\prime(x)>0$.  
Hence, local minima occur at $x=-3, 3$ because 
the derivative changes from negative to positive at these points. 
There is no local minimum or maximum at $x=-1$ since 
$r(x)$ is not defined there.  


\noindent 2.  
$\displaystyle\lim_{x \rightarrow \infty} r(x)=\infty$ and $r(0)=729$. 
We {\sl suspect\/} that the global minimum occurs at $x=3$ 
since this point is in the interval of interest and 
a local minimum occurs here.
By the first derivative test, the value of $r(x)$ is increasing 
on $(3,\infty)$ so $r(3)\leq r(x)$ for $x>3$.
Likewise, the value of $r(x)$ is deceasing on $[0,3)$ so
$r(x)\geq r(3)$ for $x$ less than $3$ and greater than or equal to $0$.
Hence $r(3)\leq r(x)$ for all $x\in [0,\infty)$, which says that the value
at $x=3$ is a global minimum for the function on the interval $[0,\infty)$.

The idea in the last paragraph is a general result.
Suppose
\itemitem{a)}
a function $f(x)$ is differentiable on an interval (closed, open,
or half open), 
\itemitem{b)}
$x=c$ is the {\sl only\/} critical number
in the interval, and 
\itemitem{c)} the value of $f$ 
is a local maximum (respectively minimum) at $x=c$. 

\noindent
Then
$f(c)$ is a global maximum (respectively minimum) on that interval.

\bigskip
No maximum value occurs in $[0, \infty)$ since 
$\displaystyle\lim_{x \rightarrow \infty} r(x)=\infty$.
\equiz

\stp
\centerline{\bf Math.~125 Quiz \#8}
\centerline{November 12, 2002}
\vskip10pt
\noindent
We wish to use Newton's method to find a solution to the equation
$$\cot x=2\tan x\ .$$
If we make a first approximation of $x=\frac{\pi}{4}$, 
what does Newton's
method give for a second approximation?

\bigskip
Leave your answer in terms of sums of fractions and rational
multiples of $\pi$.

\dquiz
To apply Newton's method, observe that our original 
equation is equivalent to $f(x)=0$ where $f(x)=\cot x-2\tan x$.

Since $f^\prime(x)=-\csc^2x -2\sec^2 x$, we get
$$x_{n+1}=x_n-\frac{\cot x-2\tan x}{-\csc^2x -2\sec^2 x}=
x_n+\frac{\cot x-2\tan x}{\csc^2x +2\sec^2 x}\ .$$

Taking $x_0=\frac\pi4$, we get
$$x_1=\frac\pi4-\frac{f(\frac\pi4)}{f^\prime(\frac\pi4)}$$

$f(\frac\pi4)=\cot(\frac\pi4)-2\tan(\frac\pi4)=1-2=-1$, 
$f^\prime(\frac\pi4)=-\csc^2(\frac\pi4) -2\sec^2 (\frac\pi4)=
-2-2\cdot 2=-6$
so $x_1=\frac\pi4 -\frac {-1}{-6}=\frac\pi4 -\frac16$.



\equiz

\stp
\centerline{\bf Math.~125 Quiz \#9: Solutions}
\centerline{November 19, 2002}
\vskip10pt
\noindent
1. Consider the region in the plane below the graph
$y=\csc x$; above the $x${--}axis and between the vertical
lines $x=0$ and $x=2$.

\item{a.}Divide the interval into $6$ equal parts and write down the
Riemann sum using the mid{--}point rule.
Do not evaluate the sum or even try to simplify it.
\bigskip

\dquiz
Midpoints are $\frac{13}{12}$, $\frac{15}{12}$, \dots $\frac{23}{12}$ :
The sum is 
$\displaystyle\sum_{i=1}^{6}\frac{1}{6} 
\csc \Bigl(\frac{13}{12}+\frac{i-1}{6}\Bigr)
=\sum_{i=1}^{6} \frac{1}{6} 
\csc \Bigl(\frac{11}{12}+\frac{i}{6}\Bigr) =
\frac{1}{6} \biggl(
\csc \Bigl(\frac{13}{12}\Bigr) + \csc \Bigl(\frac{15}{12}\Bigr) + 
\ldots + 
\csc \Bigl(\frac{23}{12}\Bigr)\biggr)$.

\dquiz
\vskip10pt

\item{b.}
Write a definite integral which evaluates the area precisely.
Do not evaluate the integral.
\bigskip
\dquiz
$\int_{1}^{2} \csc x dx.$
\dquiz
\vskip12pt

\noindent
2. What number is represented by the sum
$$\sum_{q=3}^6 (q+2)\ ?$$

\dquiz

$\sum_{q=3}^{6} (q+2) = (3+2)+(4+2)+(5+2)+(6+2) = 26.$
\equiz


\stp
\centerline{\bf Math.~125 Quiz \#10}
\centerline{Decsmber 10, 2002}
\vskip10pt
\noindent
\itemitem{1.}
Write a definite integral that evaluates to the area of the
shaded region below. Do not evaluate the integral.
\itemitem{2.}
Rotate the shaded region about the line $y=2$ and write
a definite integral that evaluates to the volume of the resulting
solid of revolution. Do not evaluate the integral.
\itemitem{3.}
Rotate the shaded region about the line $x=-3$ and write
a definite integral that evaluates to the volume of the resulting
solid of revolution. Do not evaluate the integral.
\vskip0pt
\epsfscale=800\centerline{\epsfbox{graph10.eps}}

\dquiz
\def\smfrac#1#2{{{#1}\over{#2}}}

Because the curves are given as $y$ equals a function of $x$, the 
answers will be integrals with respect to $dx$ and so
the first issue is to find the points of intersection of the two curves,
or solve $\sin x = \cos x$ or $\tan x =1$.
$x = \frac{\pi}{4}$ is one solution and $x = \frac{5\pi}{4}$ is the 
next and these are all the solutions in the relevant interval.
\smallskip\noindent
The area is given by the integral
$\displaystyle\int_{\smfrac{\pi}{4}}^{\smfrac{5\pi}{4}}
\Bigl(\sin x - \cos x \Bigr)\ dx$: top curve minus bottom curve.


\smallskip\noindent
The volume  integral when rotated about $y = 2$ is given by
$$\pi\int_{\smfrac{\pi}{4}}^{\smfrac{5\pi}{4}}
\Bigl( (2 - \cos x)^2 - (2 - \sin x)^2 \Bigr)\ dx\ .$$
With this set up, we {\sl must\/} use the washer method.
The outer radius is the distance between $y=2$ and $y = \cos x$
because from the graphs, the distance from $y = 2$ to $y = \sin x$
is smaller.

\smallskip\noindent
The volume  integral when rotated about $x = -3$ is given by
$$2\pi\int_{\smfrac{\pi}{4}}^{\smfrac{5\pi}{4}}
\Bigl(x - (-2)\Bigr)\cdot\Bigl(\sin x - \cos x \Bigr)\ dx\ .$$
With this set up, we {\sl must\/} use the shell method.
\equiz
\end