\magnification1200\footline{}
\def\frac#1#2{{{\textstyle #1}\over{\textstyle #2}}}
\def\np{%
\par\noindent{\bf\number\cc.}\ %
\global\advance\cc by1%
}

\input epsf.def

\long\def\ans#1\ll{\cc=1 %
\par\noindent {****************** Answer(s) ******************}
\par
#1\ll}

\newcount\cc\cc=1

\long\def\nf#1#2{\vskip20pt\noindent
\hbox to2in{{\bf Math.~125 Quiz \#{#1}}\hfil}#2\par\noindent
}
\def\ll{\par\noindent\vrule width \hsize depth .1pt height 0pt\par\cc=1}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\nf{1}{September 4, 2001}
\par\noindent
\np If $$(f\circ g)(x)=f\bigl(g(x)\bigr)=\sin(x^2+1)$$
what what is a possibility for $f$ and $g$?
\np If $h(t)=\frac{t}{t+1}$ and $m(s)=2s+3$ what is
$(h\circ m)(y)=h\bigl(m(y)\bigr)$?
\ans
\np One choice is $f(x)=\sin(x)$ and $g(x)=x^2+1$. 
Another option is $f(x)=\sin(x+1)$ and $g(x)=x^2$.
\np 
$(h\circ m)(y)=\frac{2y+3}{2y+4}
$.
\ll
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\nf{2}{September 11, 2001}
\np
What is 
$\displaystyle\lim_{x\to-1} \frac{2+x}{1-x}$\ ?

\np
What is 
$\displaystyle\lim_{x\to 0^+}\frac 1x$\ ?

\np
Suppose $\displaystyle\lim_{x\to 1}f(x)=4$ and
$\displaystyle\lim_{x\to 1}g(x)=-2$.
If the following limit exists, what is it?
$$\displaystyle\lim_{x\to 1}\frac{2\cdot f(x)+3\cdot g(x)}
{\bigl(8+f(x)\cdot g(x)\bigr)^2}$$

\np
What is
$\displaystyle\lim_{x\to 3} x^2-2x-3$\ ?
\ans
\np$\frac{2+(-1)}{1-(-1)}=\frac{1}{2}$ (just plug in).
\np$+\infty$. This is a basic limit you should know.
\np Using various theorems we get
$\frac{2\cdot 4+3\cdot(-2)}{\bigl(8+4\cdot(-2)\bigr)^2}=\frac{2}{0}$
so our theorems are not immediately applicable.
However, the top of this fraction is close to $2$ where as the bottom
is close to $0$. Because of the square in the denominator, not only is
the denominator close to $0$, but it is also positive so the
answer is $+\infty$.
\np Just plug in again $3^3-2\cdot 3-3=0$.
\ll
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\nf{3}{September 18, 2001}
\np
Show $f^\prime(x)=3x^2+3$ using the definition of the derivative where
$$f(x)=x^3+3x+2\ .$$ 
\np
Find all the vertical tangents of $f$.
\ans
\np$$\eqalign{%
f^\prime(x)=&\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=
\lim_{h\to0}\frac{\bigl((x+h)^3+3(x+h)+2\bigr)-
\bigl(x^3+3x+2\bigr)}h\cr
=&\lim_{h\to0}\frac{\bigl(x^3+3x^2h+3xh^2+h^3+3x+3h+2\bigr)-
\bigl(x^3+3x+2\bigr)}h\cr
=&\lim_{h\to0}\frac{3x^2h+3xh^2+h^3+3h}h=
3x^2+3xh+h^2+3=x^2+3\ .\cr
}$$
\np The only place where a vertical tangent can occur is when 
$\displaystyle\lim_{x\to a^\pm} f^\prime(x)=\pm\infty$.
But $f^\prime(x)$ is a polynomial and hence continuous everywhere
so there are no vertical tangents.
\ll

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\nf{4}{October 2, 2001}
\cc=1
\np
Find the derivative of
$$f(x)=\sin\bigl(\cos(x^2)\bigr)$$ 

\np
Find all values of $x$ for which the tangent line is horizontal.

\noindent{\bf Remark:}$\frac{\pi}{2}>1$.
\ans
\np By the chain rule
$f^\prime(x)=\cos\bigl(\cos(x^2)\bigr)\cdot\bigl(-\sin(x^2)\bigr)
\cdot(2x)$.
\np We need to solve $f^\prime(x)=0$, so first ``simplify'' our expression
for $f^\prime(x)$:
$f^\prime(x)=(-2x)\cdot\bigl(\sin(x^2)\bigr)\cdot\cos\bigl(\cos(x^2)\bigr)$.
The only way a product can vanish is for one of the factors to vanish
so we need to solve
$(-2x)=0$, $\sin(x^2)=0$ and $\cos\bigl(\cos(x^2)\bigr)=0$.
The frist equation has the unique solution $x=0$ and the hint plus the
facts that $-1\leq \cos(x)\leq 1$ and $\cos(x)=0$ only when
$x=\frac{\pi}2+k\pi$ for integers $k$, shows that the third equation
has no solutions.
This leaves $\sin(x^2)=0$.
Well, $\sin(x)=0$ if and only if $x=k\pi$, $k$ any integer, so we need
to solve $x^2=k\pi$, and therefore $x=\pm\sqrt{k\pi}$ for $k$ a non{--}negative
integer.
The solution $x=0$ is included in the set $x=\pm\sqrt{k\pi}$ 
for $k$ a non{--}negative integer so these are all the solutions
and hence all the horizontal tangents.
\ll

\nf{5}{October 9, 2001}
\np Car $A$ and car $B$ are approaching the intersection ``$C$''
of two streets intersecting at a right angle.
Car $A$ is going South at 45 mph, car $B$ is heading West at 30
mph. We denote the angle $\angle (C,B,A)$ by $\theta$, the distance 
from $C$ to $B$ by $x$, and the distance from $C$ to $A$ by $y$.
Then, $\tan \theta = {y\over x}$.
At what rate is the angle $\theta$ changing when car $A$ and
car $B$ are both 1 mile from the intersection?

\epsfbox{q5.eps}
\ans
We are asked to compute $\frac{d\theta}{dt}$ when $x=1$ and $y=1$.
We are given a relation $\tan\theta=\frac{y}{x}$ which holds all the time,
so differentiating 
$$\bigl(\sec^2\theta\bigr)\frac{d\theta}{dt}=
\frac{x\frac{dy}{dt}-y\frac{dx}{dt}}{x^2}\leqno(\ast)$$
We are further given 
$\frac{dy}{dt}=-45$ and $\frac{dx}{dt}=-30$ and we can plug onto $(\ast)$
provided we can compute $\sec^2\theta$ when $x=y=1$.
When $x=y=1$, $\tan\theta=1$ so $\sec^2\theta=1+\tan^2\theta=2$
and we get
$2\frac{d\theta}{dt}=(-45)-(-30)=-15$
so
$\frac{d\theta}{dt}=-\frac{15}2$.

\ll
\nf{6}{October 16, 2001}

\noindent Find all the critical points for the function below.
Which ones are local maxima? Why?
Which ones are local minima? Why?

\noindent The existence theorem for global max/min does not apply to 
this function restricted to the interval
$(-\infty,-2)$ since the interval must be a closed interval for 
the theorem to apply.
Nevertheless, this function does have a minimum value on this interval. 
Where is it and why is it where you say?
Why does this function have no maximum value on this interval?

$$\frac{(x-5)^8}{(x+2)^{4}}$$
\ans
First we need to compute the derivative of 
$r(x)=\frac{(x-5)^8}{(x+2)^{4}}$:
$$\eqalign{%
r^\prime(x)=&\frac{8(x-5)^7(x+2)^4-(x-5)^8 4(x+2)^3}{(x+2)^8}=
(x-5)^7(x+2)^3\frac{8(x+2)-(x-5) 4}{(x+2)^8}\cr
=&
\frac{(x-5)^7\bigl(8x+16-4x+20\bigr)}{(x+2)^5}=
\frac{(x-5)^7(4x+36)}{(x+2)^5}=
\frac{4(x-5)^7(x+9)}{(x+2)^5}\ .\cr
}$$
To find the critical points, note that
$r^\prime(x)$ is continuous everywhere it exists
so we only need to solve $r^\prime(x)=0$ and find all
the points where $r^\prime$ does not exist.
Well, $r^\prime(x)=0$ only when $(x-5)^7(x+9)=0$ so
$x=5$ and $x=-9$: $r^\prime(x)$ does not exist precisely
when $(x+2)^5=0$ so $x=-2$.

\hskip.15in$-$\hskip1in$+$ \hskip1in$-$
\hskip.75in$+$ 
\par
\vrule width 4true in depth .1pt height 0pt
\par\hskip.5in$-9$\hskip 1in$-2$\hskip1in $5$

Local minima occur at $x=-9$ and $x=5$ because the
derivative changes from negative to positive at these points.
The value $x=-2$ is not a local extremum since the function
is not defined there.
To understand $r(x)$ on the interval $(-\infty,-2)$ it helps to
compute $\displaystyle\lim_{x\to\infty} r(x)=+\infty$
and $\displaystyle\lim_{x\to-2^-} r(x)=+\infty$.
Hence $r(x)$ starts out very big far to the left and decreases
until $x=-9$.
Thereafter, $r(x)$ steadily increases, getting larger and larger.
Hence, if $x<-9$, $f(x)>f(-9)$ and if $x>9$, $f(x)>f(9)$ which
is the definition of an absolute minimum.
There can be no maximum on the interval $(-\infty,-2)$
since $r(x)$ gets larger and larger as $x$ gets closer and closer to $-2$.

\ll
\nf{7}{November 6, 2001}
\noindent
Find the rectangle of largest area that can be inscribed in the
ellipse
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$.

\epsfbox{q7.eps}
\ans
This is a max.-min. problem: specifically, we are being asked to maximize
the area of a rectangle.
The sides of the rectangle are $2x$ and $2y$, so the area is $A=4xy$.
This is typical of many max.-min. problems in that $A$ is a function
of two variables, so we must eliminate one of them.
The required equation comes from the fact that the point $(x,y)$ lies
on the ellipse, so $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.
Solve for $y=b\sqrt{1-\frac{x^2}{a^2}}$.
(Since our point is in the first quadrant, we want the solution for which
$y\geq 0$.)
Hence $A=4b x\sqrt{1-\frac{x^2}{a^2}}$ is now a function of one variable.
Additionally, we have set things up so that $0\leq x\leq a$.

The next step is to locate the critical points of $A(x)$ so
calculate
$$A^\prime(x)=4b\biggl(\sqrt{1-\frac{x^2}{a^2}}+
x\frac{\frac{-2x}{a^2}}{2\sqrt{1-\frac{x^2}{a^2}}}\biggr)
=4b\biggl(\frac{1-\frac{x^2}{a^2}-\frac{x^2}{a^2}}
{\sqrt{1-\frac{x^2}{a^2}}}\biggr)=
4b\Biggl(\frac{1-2\frac{x^2}{a^2}}{\sqrt{1-\frac{x^2}{a^2}}}
\Biggr)
$$
The critical points occur where $A^\prime$ is not defined ($x=a$)
or where $A^\prime=0$ (or $x=\frac{a}{\sqrt2}$).
To locate the maximum, note $A(0)=A(a)=0$ and 
$A(\frac{a}{\sqrt2})=2ab$ so the largest area rectangle occurs
when $(x,y)=\biggl(\frac{a}{\sqrt2},\frac{b}{\sqrt2}\biggr)$.
\ll

\nf{8}{November 13, 2001}
We wish to use Newton's method to find a solution to the equation
$$\sec x=\tan x\ .$$
Of course this is equivalent to solving the equation
$\sec x-\tan x=0$.
If we make a first approximation of $x=0$, what does Newton's
method give for a second approximation?
\ans
Since $f(x)=\sec x-\tan x$,
$f^\prime(x)=\sec x \tan x -\sec^2 x$,
so we get 
$$x_1=0-\frac{f(0)}{f^\prime(0)}\ .$$
Since $f(0)=\sec(0)-\tan(0)=1$, 
$f^\prime(0)=\sec(0) \tan(0) -\sec^2 (0)=1\cdot 0-1^2=-1$
so $x_1=1$.
\ll

\nf{9}{November 20, 2001}
1. Consider the region in the plane below the graph
$y=\sin x$; 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 left{--}hand end points of each interval.
Do not evaluate the sum or even try to simplify it.

\item{b.}
Write a definite integral which evaluates the area precisely.
Do not evaluate the integral.
\vfill
2. What number is represented by the sum
$$\sum_{q=-1}^2 q(q^2-1)\ ?$$
\ans
\item{1a.}
The width of each interval is $\frac{2-0}{6}=\frac{1}{3}$;
the left hand end points are $0$, $\frac{1}{3}$,
$\frac{2}{3}$,  $\frac{3}{3}=1$,  $\frac{4}{3}$ and  $\frac{5}{3}$.
Hence the asked for Riemann sum is\par
$\sin(0)\cdot\frac13+\sin(\frac13)\cdot\frac13+
\sin(\frac23)\cdot\frac13+\sin(1)\cdot\frac13+
\sin(\frac43)\cdot\frac13+\sin(\frac53)\cdot\frac13$.
\item{1b.}
$\displaystyle\int_0^2 \sin x\> dx$.
\item{2.}
The sum is $(-1)\bigl((-1)^2-1\bigr)+
(0)\bigl((0)^2-1\bigr)+
(1)\bigl((1)^2-1\bigr)+
(2)\bigl((2)^2-1\bigr)$.
All the terms but the last vanish so the number is $2(4-1)=6$.
\ll
\nf{10}{November 27, 2001}
Find the derivative of the function
$$G(x)=\int_0^{\int_1^x\frac1t dt} \csc\theta\ d\theta$$
with respect to $x$.
\ans
By the Chain Rule and the Fundamental Theorem of Calculus,
$$G^\prime(x)=\csc\biggl(\int_1^x\frac1t dt\biggr)
\cdot\frac{d \int_1^x\frac1t dt}{dx}\ .$$
We can do nothing with the integral inside the $\csc$,
but we can apply the Fundamental Theorem again to
compute $\frac{d \int_1^x\frac1t dt}{dx}=\frac 1x$,
so
$$G^\prime(x)=\csc\biggl(\int_1^x\frac1t dt\biggr)
\cdot\frac 1x\ .$$
\ll
\nf{11}{November 11, 2001}
\itemitem{1.}
Write a definite integral that evaluates to the area of the
shaded region below. Do not evaluate the integral.
\vskip 2cm\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.
\vskip 2cm\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.

\vskip 2cm
\centerline{\epsfbox{graph11.eps}}
\ans
\item{1.} It is easy to describe the two curves as graphs of $y$
as a function of $x$.
Hence\par
${\rm Area}=\displaystyle\int_{-1}^1\bigl( x^2-x^4\bigr)\>dx$.

Having done the area, the remaining two problems are practically done.

\item{2.} Since we are rotating around $y=2$ which is perpendicular
to the $x${--}axis, we get washers so\par
${\rm Volume}_{_{y=2}}=
\pi\displaystyle\int_{-1}^1\bigl( (2-x^4)^2-(2-x^2)^2\bigr)\>dx$.


\item{3.}
Since we are rotating around the line $x=3$ which is parallel to
the $x${--}axis, we get shells so\par
${\rm Volume}_{_{x=3}}=
2\pi\displaystyle\int_{-1}^1\bigl(3-x\bigr)\bigl( x^2-x^4\bigr)\>dx$.
\ll

\end