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\begin{document}

\begin{center}
  {\bf \huge Class Activities
}
   \end{center}
\begin{center}
  {\bf \Large
Math 108, Spring 1999
}
   \end{center}
{\bf  Introduction.} The activities below were used during the
spring semester of 1999 and their main purpose was to encourage
participation and active learning in class.  The first thing I did when I
walked in the first class was to distribute Activity 1, which introduces
antidifferentiation, and asked the students to start working on
it.  Also, I told them that 
\begin{itemize}
\item  They should work in groups of 2-4
\item Raise hand if their group needs help
\item One paper with all names listed on it will be collected from
each group and it will be graded.
\end{itemize}
While the students were working on the first activity, I finished
distributing other handouts (general information about the
course,  and the first assignment), and walking around the room I
provided guidance to groups that asked for help.  Some groups
finished fast, others took longer time than I expected.  In any
case, after about 15 minutes all groups finished and then I
collected a paper from each group.
\vskip .05 in
This first interaction with the students gave me valuable
information about the class which I used immediately in my
lecture about antidifferentiation.  The rest of the class went the
traditional lecturing style interrupted by a few questions from
students and a few questions from the teacher.
At the end of the class I informed the students that there will be
activities similar to the one we had today in most classes of this
course.  In addition, it was stressed that
\begin{itemize}
\item  They should read the material before  is discussed in
class with the objective to obtain an initial understanding of the concepts
involved, and to recognize possible difficulties needed to be resolved in
class.
\end{itemize}
Although many students at the beginning did follow the
suggestion of reading before the material was considered, only a
few continued to read till the end of the semester.  It seems that
pure encouragement alone does not work.  Giving a simple quiz at the
beginning of each class, testing reading and whose contribution in
the course grade is significant, could work.  However this was not 
an option since Math 108 is a multisection course having a
common grading scheme, and the other seven  sections were not
using activities.  The activities' grade was part of the homework
grade.  And the homework contributed only 50 points in the total
of 500 points.
\vskip .05 in
Taking into consideration that most students did not read
the material to be discussed before class, the activities
given fall into the  following three kinds.
\vskip .05 in
\noindent
{\bf 1.}  Activities based mostly on previous knowledge and on the new
concepts, which could be easily described in the activity.  Such examples
are Activities 1, 2, 14.
\vskip .05 in
\noindent
{\bf 2.}  Activities introducing the concept(s) to be considered in the
lecture to follow in concrete situations.  Such examples are Activities 1,
4, 18.
\vskip .05 in
\noindent
{\bf 3.}  Activities applying the concepts and methods introduced in
the lecture, and also demonstrated afterwards with at least an
example by the teacher.  Typically the lectured material  and the
solved examples were displayed on the blackboard before the
activity started.  Most of the activities were of this type.   Typical
examples are 9, 19, 20.
\vskip .05 in
Obviously, there is plenty of room for improvements.  However,
these activities provided a good opportunity for students to
participate and learn mathematics by doing it.  Also they gave
students the opportunity to contribute in a collaborative effort
of their group and class.  At the same time they helped the teacher 
to be more in tune with the class.
\vskip .05 in
\noindent
{\bf Covering the Material.}  The activities did not present any
serious problem in covering the syllabus.  My section went at the
same pace as the other sections of Math 108.  The reason is that
they were well integrated in the lectures.  For example, the
students were doing the concrete examples leading to the
general theory to be discussed as an activity instead of the
teacher.  And that a second application of the theory already
discussed was done by  the students instead of the teacher.  Also,
when sometimes the class time was not enough to complete the
activity, then the groups were asked to do it after class and
return their paper in the next class.  Finally there were days with
where no activities were given.
\vskip .05 in
\noindent
{\bf Grading.}  7 out of the 22 activities given were collected and
graded.  The scale used was 0-3.  the total, 21 points, was entered
into the homework, which contributed 50 points in the total of
500 points for the course.


\pagebreak

\begin{center}
  {\bf \huge Rules for Activities}
   \end{center}

{\Large\bf

\noindent
{$\bullet$} Read material before is discussed in class. (most difficult)
 
\medskip
\noindent
{$\bullet$} Work in groups of 2-4.

\medskip
\noindent
{$\bullet$} Raise hand if group needs help.

\medskip
\noindent
{$\bullet$} Some of the activities will be collected and graded. Grade will
form a part of homework grade. (7 out of 22 were graded, and almost all got
perfect score)
}  


\pagebreak



\begin{center}
  {\bf \huge Conclusions}
   \end{center}

{\Large\bf

\noindent
{$\bullet$} Activities promoted active learning.

\medskip
\noindent
{$\bullet$} Students helped each other learn. (their way)

\medskip
\noindent
{$\bullet$} Students helped teacher too!

\medskip
\noindent
{$\bullet$} Promoted asking questions. (good ones)


\medskip
\noindent
{$\bullet$} The questions asked by the students provided a valuable and immediate
feedback, which resulted to immediate adjustments in teaching.

\medskip
\noindent
{$\bullet$} Created a sense that everyone learned something in each class.

\medskip
\noindent
{$\bullet$} Enhanced attendance.

\medskip
\noindent
{$\bullet$} No sleeping students. (at 8:30 am)

}
\medskip
\medskip
\begin{center}
  {\bf \huge Remarks}
   \end{center}

{\Large\bf

\medskip
\noindent
{$\bullet$} Not easy to make students read material before class.

\medskip
\noindent
{$\bullet$} Could write better activities.
}



\pagebreak

\begin{center}
  {\bf \Large Activity  1}
   \end{center}


Today we begin by  reviewing  integration, and the basic integration
techniques of substitution and integration by parts.
\vskip 0.3in
{\bf A} Recall that the antiderivative or indefinite  integral of  a
function $f(x)$ is a function
$F(x)$ such that $F'(x)=f(x)$ i.e. the derivative of $F(x)$ is the
original function $f(x)$. This is written 
$\int f(x)dx=F(x)+C$ where $C$ is a constant. 
Calculate the following indefinite integrals.


\vskip 0.3 in

1.\quad   $\int x^3-x^2+4 dx$

\vskip 0.3 in

2.\quad    $\int e^{2t} dt $


\vskip 0.3 in

3.\quad    $\int t^{-1} dt$

\vskip 0.3 in

4. \quad   $\int y^{1/2}dy$




\vskip 0.3 in 


{\bf B} Recall that integration by Substitution is  based on the {\bf
chain rule }  for differentiation. For practice compute 

\bigskip
5. \quad $\int (x^2-3x)^8(2x-3)dx=$

\vskip 0.3in



{\bf C} {\bf Integration by Parts.} This method
is based on the product rule for differentiation.  It uses the formula
$$ 
\int u dv = uv - \int v du.
$$
where
$dv = v' dx$ and  $du = u' dx$. Now use this method to compute the 
indefinite integral

\vskip 0.3in 

6 \quad $\int te^{3t}dt=$
\pagebreak

\begin{center}
  {\bf \Large Activity  2}
   \end{center}


{\bf Partial Fractions.} To compute the indefinite integral 
 $\displaystyle \int\frac{dx}{ax^2 + bx + c}$, in the case that the
quadratic in the denominator has real roots $r_1$, $r_2$, first
 we find the numbers $A$ and $B$ such that 
$$
\frac{1}{ax^2 + bx + c}= \frac{A}{x-r_1}+\frac{B}{x-r_2}.
$$
And then we integrate the two simple fractions to compute the
desired integral in ters of  logarithmic functions. Practice 
this method by computing the integral

\vskip 0.3in
7 \quad $
\displaystyle \int\frac{dx}{x^2 +3x -4}=$

\pagebreak

\begin{center}
  {\bf \Large Activity  3}
   \end{center}

\noindent
Today we shall discuss  the  {\bf definite (Riemann) integral } of a nonnegative
function. It is the area of the region 
under its graph and above the $x$-axis. To
compute this area we use  the following simple idea.

\medskip
\noindent
{\large \bf The Idea.}  We cover the region under the graph with very thin
non-overlapping rectangles. Each rectangle has its base on the $x$-axis,
 and its height is equal to the value of the function at a point in the base.
 If the rectangles are very thin, the total area they contain will approximate the area of
the region under the graph, and by making the rectangles thin enough we can get as good an
approximation as we want.


\medskip
\noindent
{\large \bf Example.} Use this idea to approximate  the definite integral
of the function shown below.  

\bigskip
$$\epsfxsize=4.2in\epsfbox{approx.area.eps}$$

\pagebreak

\begin{center}
  {\bf \Large Activity 4}
   \end{center}


\noindent
{\bf A.}
The method of {\bf substitution for indefinite integrals}
together with the {\bf Fundamental Theorem of Calculus} 
give the following method for computing
definite integrals
\begin{equation}
\int^b_a f(g(x)) g^\prime (x) dx \stackrel{(u = g(x))}{=}
\int^{g(b)}_{g(a)} f(u) du=F(g(b)) - F(g(a)),
\end{equation}
where $F$ is an antiderivative of $f$. Use formula (1)
to compute the integral

\bigskip
\noindent
 $ 
\int_1^2 (2x-5)e^{-x^2+5x-6} dx=
$


\vskip 2.5 in
\noindent
{\bf B.} 
Using the {\bf the integration by parts formula for 
indefinite integrals} and the {\bf fundamental theorem of calculus}
 we obtain
\begin{equation}\label{eq72}
\int^b_a u dv = uv |^b_a - \int^b_a v du,
\end{equation} which is the integration by parts formula for definite
integrals. Use formula (2) to compute the integral

\bigskip
\noindent
$ \int^e_1 t^4 \ln t dt=$


\pagebreak


\begin{center}
  {\bf \Large Activity 5}
   \end{center}



\noindent
The area between two curves is 
equal to the area under the  {\bf upper curve} {\large\bf minus} the area under the 
{\bf lower curve}.

\bigskip
\noindent
{\bf Problem:}
Compute the area of the region enclosed by the curve
 $y= 3-x^2$ and the curve $y= x^2-4x+3$. Sketch the region first.


\pagebreak

\begin{center}
  {\bf \Large Activity 6}
   \end{center}


\noindent
Today we shall see that the {\bf
average value} of a function $ f$ over an interval $[a,b]$ is given by the formula:
\[
\fbox{average value of $f$ over $[a,b]$ \ =  \ $\displaystyle \frac{1}{b-a}
\int^b_a f(x) dx $}.
\]
Use this formula to compute the average value of the function
$$f(x) =\frac{1}{ x^2 +5x + 4},\, 0 \le x \le 4.$$


\pagebreak

\begin{center}
  {\bf \Large Activity 7}
   \end{center}

\noindent
Today we shall discuss how we can go 
from Marginal Function to Total Function by using
the Fundamental Theorem of Calculus (FTC):
\[
\int^b_a F^\prime (x) dx = F(b) - F(a), 
\] 
which can be stated as the {\bf Fundamental Principle}:
\begin{center}
\fbox{ The definite integral of a rate of change equals the total change.}
\end{center}

\bigskip
\noindent
{\bf Problem.}  The marginal profit function for producing and selling
$x$ units of a certain product is given by $MP = -0.6x + 420$. 
The company operates at the production level of 500 units.  Is it 
profitable for  the company to increase production? If yes,
 find the production level which yields maximum  profit, as well as
the additional profit gained by raising production to that level.


\pagebreak

\begin{center}
  {\bf \Large Activity  8}
   \end{center}


\noindent
{\bf Problem.}  Find the equilibrium quantity, price, the consumer surplus, and
the producer surplus for  the following demand and supply
curves.

 $$\displaystyle{D (q) = \frac{25}{q+2}},\quad S (q) = q+2.$$



\pagebreak

\begin{center}
  {\bf \Large Activity 9}
   \end{center}

\noindent
{\bf Problem.}  Suppose that an amount of money $M_0$ is deposited in the
bank which pays interest at an annual rate $r$ compounded
continuously.

\bigskip

a.  Find the time $T$ needed for $M_0$ to double as a function of
$r$.

b.  Find $T$ if $r$ is $5\%$
   

\pagebreak

\begin{center}
  {\bf \Large Activity 10}
   \end{center}

\noindent
{\bf Problem.}  An employee of a company is offered a choice between the
following two retirement plans.  For the first plan the company
will open an IRA with the initial amount of $\$5,000$ and
thereafter for the next 25 years an income stream of $\$15,000$
will be deposited in the IRA with an annual rate of $8\%$
compounded continuously.  For the second plan the company will
pay the employee $\$1,100,000$ at the end of 25 years.  Which
one is the more beneficial plan for the employee.

\pagebreak




\begin{center}
  {\bf \Large Activity 11}
   \end{center}

\noindent
{\bf Problem.}  Using the trapezoidal rule, estimate the integral
$$ \hskip .10 in
\displaystyle{\int^1_0   \sqrt{1-x^2} \ dx;\, n=5. }$$


\pagebreak


\begin{center}
  {\bf \Large Activity 12}
   \end{center}

\noindent
{\bf Problem.} Check that the function  $ y(x) = a+ce^{kx}$, 
$a,c,k \in \mathbb{R}$, is a solution  to
 $\displaystyle{\frac{dy}{dx} = k(y-a)}$.


\pagebreak


\begin{center}
  {\bf \Large Activity 13}
   \end{center}

\noindent
Today we shall consider separable differential equations.
A DE is {\bf separable} if the independent and  dependent
variables can be separated. More precisely if it can
be written in the form
 
\begin{equation} \label{eq:separablede}
\fbox{$\displaystyle{f(y)\frac{dy}{dt}=g(t)\quad \mbox{ or }
\quad f(y)dy=g(t)dt,}$}
\end{equation}
where $f$ and $g$ are known functions. To solve such equations,
we integrate both sides and we obtain
$\int f(y)\frac{dy}{dt} dt=\int g(t) dt$,
which is the same as
\[
\int f(y)dy=\int g(t) dt.
\]
If these integrals are computed then the DE (1) is solved.

\vskip 0.2 in
\noindent
{\bf Problem A.}  Solve the differential equation $\displaystyle{\frac{dy}{dt} = 4-y}$.
Check your answer.

 \vskip 2.75 in

\noindent
{\bf Problem B.}  Solve the initial value problem 
$
\displaystyle{\frac{dy}{dx} = \frac{x-1}{y},\, y (3) = 8 }.
$
Check your answer.

\pagebreak


\begin{center}
  {\bf \Large Activity 14}
   \end{center}

\noindent
A constant solution to a  differential equation is called an {\bf
equilibrium} solution or {\bf steady-state} solution.
An equilibrium solution  is called {\bf stable} if any solution arising from a small change in the
initial condition tends to the equilibrium solution
as  the independent variable tends to infinity. It is  called
{\bf  unstable} if a small change in the initial
condition gives rise to a solution that moves away from the equilibrium solution.

\bigskip
\noindent
{\bf Problem A.}  Find the equilibrium solution of each of
the differential equations
\[
{\bf (a)}  \ \frac{dy}{dt} = y - 20,  \hspace{25mm} {\bf (b)} \ \ \frac{dy}{dt} = 3 -
0.1y,
\]
and decide whether they are stable or unstable.





\vskip 2 in
\noindent
{\bf Problem B.}
 Solve these
differential equations and then look at the solutions to draw the
same conclusions.
 

\pagebreak


\begin{center}
  {\bf \Large Activity 15}
   \end{center}

To solve a linear system of equations we apply Gaussian
elimination. This method   consists of
performing a sequence of the following  

\vspace{2mm}
\noindent {\bf Elementary Operations}
\begin{itemize}
 \item[1. ] Interchange equations $E_i$ and $E_j$.
\item[2.  ] Replace equation $E_i$ by $cE_i$, where $c$ is a nonzero
constant.
\item[3.  ] Replace equation $E_i$ by the combination $E_i+dE_j$,
where $E_j$ is another equation and $d$ is a constant.
\end{itemize}

\vspace{2mm}
\noindent In practice frequently we combine  operations (2) and
(3); i.e. we replace an equation $E_i$ by a combination
$cE_i+dE_j$, where $c,\ d$ are constants with $c\ne 0$ and $E_j$ is
another equation.

\vspace{2mm}
\noindent  Each time we apply an elementary operation to our
system we obtain a new system with the same solution as the old
one; i.e. we obtain a new system that is {\bf equivalent} to the original 
one. This is the reason that this method works.

\vspace{2mm}
\noindent By applying Gaussian elimination we either arrive at an equation
of the form 
$ 0=b,\ \mbox{with}\ b\ne 0,
$ 
which tells us that the system  is {\bf inconsistent};
i.e. it has no solution, or we obtain a simple system in the
{\bf echelon, or triangular form}, which is solved by back-substitution.

\bigskip
\noindent
{\bf Problem.} Apply Gaussian elimination to solve the system

\vspace{2mm}
$
\begin{array}{cl}  &x_1 - 3x_2 + 4x_3 -2x_4 = -1\\
&3x_1 - 8x_2 + 10x_3 -5x_4 =2\\
&-2x_1 + 7x_2 - 9x_3 + 5x_4 =6\\
&2x_1 - 5x_2 + 7x_3 -2x_4 =4. 
\end{array}
$
 


\pagebreak

\begin{center}
  {\bf \Large Activity 16}
   \end{center}


\bigskip
\noindent
{\bf Problem A.} Find the equation of the plane passing
through the  point $(-4,-2,\frac{9}{2})$ and having
slope in $x$-direction
$- \frac{5}{2}$ and slope in $y$-direction $\frac{7}{2}$.

 \vskip 3 in

\bigskip
\noindent
{\bf Problem B.} Find the equation of the planes
passing through the three  points
$(1,0,-1)$, $(0,1,3), (1,-1,-2)$.


\pagebreak

\begin{center}
  {\bf \Large Activity 17}
   \end{center}

\noindent
Given a function $z = f (x, y)$ and $z_o$  a  number 
then the {\bf level curve of
height $\mathbf z_o$} is the set of all points in the $xy$-plane satisfying
equation $f (x, y) = z_o.$
Geometrically, the level curve of height $z_o$, of a given
function $z = f (x, y)$ is obtained by slicing its graph by a
horizontal plane at height $z_o$ and then projecting the intersection
of the plane with the graph   onto the $xy$-plane. In
topographical maps,  level curves are drawn with the
corresponding altitude to help readers visualize the 3D-picture
of the ground surface.

\bigskip
\noindent
{\bf Problem A.} Sketch the level curve of the function 
$f(x,y)=(x-3)^{2}+(y-5)^{2}$ of height  $z_{0}=4$,and the level curve of the function 
$f(x,y)=y-x^{2}$ of height  $z_{0}=1$.


 \vskip 0.1 in
\vspace{2mm}
\noindent
If $f(x,y)$ is a function of the two variables $(x,y)$ and if we let only one of the
 variable be free and the other fixed, then $f(x,y)$ becomes a function of one variable.  The
ordinary derivative of this function with respect to the free variable ($x$, or $y$) 
is called  the {\bf partial derivative} of $f(x,y)$ for that variable.

\medskip
\noindent
{\bf Problem B.} Find  the first order partial
derivatives  the given functions:

\vskip .3 in
{\bf 1.} $f(x,y)=5-8x+5y $

\vskip .4 in
{\bf 2.}  $f(x,y)=x^3 y^2 -2x^2 y+y$


\vskip .4 in
{\bf 3.} $f(x,y) = y^3 e^{2xy} - x^5$

\vskip .4 in
{\bf 4.} $M(r,t)=1000 e^{rt}$

\vskip .4 in
{\bf 5.} $\displaystyle{R(s,t)= \frac{5 t^{2}}{s}}$
 



\pagebreak

\begin{center}
  {\bf \Large Activity 18}
   \end{center}


\noindent
{\bf Problem A.} Use the first and second derivative
tests to find the local minima, maxima, and saddle points of the
function  $g(x,y)=x^2 -xy+y^2 +3y-1$.  If the second derivative test is inconclusive, so
state.



 \vskip 4.0 in

\medskip
\noindent
{\bf Problem B.} Find the equation of
the tangent plane to the graph
 of the function $z=f(x,y)=x^{1/3}y^{2/3}$  at the point $(1,8)$, as well as its
linear aproximation at the same point. Then use its
linear aproximation to estimate $f(1.1, 8.2)$.




\pagebreak



\begin{center}
  {\bf \Large Activity 19}
   \end{center}


\noindent
{\bf Problem}.  The quantity, $Q$, of a product manufactured by a company is
given by
$Q(K,L) = 40K^{0.35} L^{0.65},$
where  $K$ is the quantity of capital and $L$ is the
quantity of labor used.  Capital costs are \$70 per unit, labor costs
are  \$65 per unit, and the company wants to keep costs for capital and
labor combined  equal to $\$4,000,000$.Find the combination of capital and
labor which gives the maximum output



\pagebreak


\begin{center}
  {\bf \Large Activity 20}
   \end{center}


\noindent
{\bf Problem}.  Determine which of the given functions is
a probability density of a random variable $X$.

\medskip

a)\quad $f(x) = \frac{4}{15} x^3 , -1 \leq x \leq 2$
 \qquad b)$f (x) = \frac{3}{4} x (2 - x), 0 \leq  x \leq 2$ 


\medskip
\noindent
Then compute the probability of $ 1<X<2$.



\pagebreak

\begin{center}
  {\bf \Large Activity 21}
   \end{center}


\noindent
{\bf Problem}.   Suppose that the  amount of milk in a gallon container
is a normal random  variable $X$ with
$\mu =128.2$ ounces and $\sigma =0.2$ ounce. Then find the probability 
that a bottle of milk chosen at random  contains  more than 128.6 onces.




\end{document}
 
                                                                                                                                                                                

                                                           
