TEN OF MY FAVORITE PRECALCULUS AND
CALCULUS PROBLEMS
MMC Conference of Workshops
Steven Condie
Illinois Mathematics and Science Academy
scondie@imsa.edu
1. Find the sum:
1 1 1 1
   
2 4 8 16
Solutions:




Paper cutting
Geometric Series
2S-S =
Lim(Sn)
Comments:
1. When does 2S-S work.
2. Idea of finding formula for Sn is important.
3. .3333333…….=1/3
4. 1 1  .5  .25  .125  .0625 
1 0 
Assume that students know: The transformation matrix 
 reflects points in the plane over
0 1
the x-axis. Suppose further that they know the transformation matrices that reflect points over the
y-axis, the line y  x and the line y   x . Finally, suppose they know the rotation matrix
cos  sin  
 sin  cos  which rotates the plane counter-clockwise through an angle  about the origin.


Problems:
a. Find a transformation matrix that reflects points over the line y  2 x .
b. Find a transformation matrix that reflects points over the line y  mx .
Solution: We look at the general case - y  mx . The strategy here is to rotate the line y  mx so that it is
so that it is coincident with the x-axis, reflect the plane over the x-axis, then rotate the plane so
that the x-axis is again coincident with y  mx . To this end, let  be the angle that y  mx
makes with the positive x-axis as shown.
cos( )  sin     cos( ) sin  
Now, 

 rotates the
 sin( ) cos( )    sin( ) cos( ) 
plane clockwise about the origin through an angle  . Then
cos( )  sin  

 rotates the plane counter-clockwise
 sin( ) cos( ) 
through the angle  . Now, being careful with the order we do
the transformations (right to left), the transformation matrix we
are looking for is:

cos( )  sin   1 0   cos( ) sin  


.

 sin( ) cos( )  0 1   sin( ) cos( ) 
cos2 ( )  sin 2 ( ) 2sin   cos( )  cos(2 ) sin  2 
Simplifying, we get 

.
2
2
 2sin( )cos( ) cos ( )  sin ( )   sin(2 ) cos(2 ) 

1
m
Looking at the triangle at right, we can see that
m
1
sin  
and cos  
. So our transformation
2
1  m2
1 m
becomes:
1  m 2
2m 
2
2

2
cos ( )  sin ( ) 2sin   cos( )  1  m 1  m2 



2
2
1  m2 
 2sin( )cos( ) cos ( )  sin ( )   2m
1  m2 1  m2 
 3
5
For m = 2. 
4
 5
4
5

3 
5 
3. In a classroom, 34 students are seated in 5 rows of 7 chairs. The place at the center of the room is
unoccupied. A teacher decides to reassign the seats such that each student will occupy a chair adjacent
next to his/her present one (i.e. move one seat forward, back, left, or right). In how many different ways
can this assignment be done? (From 2010 Havard/MIT Math Tournament; written by Tarimn Dragner I
suppose).
Solution: Think of the seating arrangement as a 5  7 chessboard as shown. Initially, 34 squares are
occupied: 18 gray and 16 white. When a student moves to an adjacent chair (square), they will
necessarily change color (from gray to white or white to gray). Since initially, there are 18 gray squares
occupied, after the rearrangement there must be 18 white squares occupied. Oops, there are only 17
white squares, so there are no such rearrangements.
Martin Gardner’s mutilated chess board.
A standard chess board has 64 squares (8 x 8). Suppose that we have 32 dominoes that each cover
exactly two squares of the chess board; then the entire board can be covered with the 32 dominoes. Now,
cut the two opposite corners from the chess board as shown. Can this mutilated chess board be covered
with 31 dominoes?

Nice example of a elegant proof.
4. A man on a dock is pulling a rope fastened to the bow of a small boat. The man’s hands are 12 feet
higher than the point where the rope is attached to the boat and he is retrieving the rope at a constant
rate of 2 feet per second. What is the average speed of the boat as it moves from 9 feet from the dock
to 5 feet from the dock? What is the average speed of the boat as it moves from 5 feet from the dock
to 1 foot from the dock?
Solution: The three situations look like:
15
12
9
12
13
5
12
12.04
1
Note: 122  12  12.04
As the boat moves from 9 feet away from the dock to 5 feet from the dock, the man must pull in 2 feet of
rope (15 – 13 = 2). Since this takes 1 second, the boat is moving at an average rate of 4 feet per second.
As the boat moves from 5 feet away from the dock to 1 foot from the dock, the man must pull in
approximately 1 foot of rope. Since this takes ½ second, the boat is moving at an average rate of 8 feet
per second.
What is the average speed of the boat as it moves from 5 inches from the dock to 1 inch from the
dock??? Explain.



Return to this in calculus
What really happens when the boat gets close to the dock?
Mechanical advantage
5. Find the equation of the line through (2, 1) that is parallel to 3x  2 y  4 .
3(2)  2(1)  8
So the equation we want is 3x  2 y  8
Notes:



My way is not always the best way
a b
 ax  by  e
has a unique solution if and only if
0

c d
cx  dy  f
If you’re no learning from your students, you’re not listening.
6. How we find the area of the region pictured below: (graph of y  x 2 )
rn
r4 r3 r2 r
Subdivide the interval [0,1] using the points
1 > r > r2 > r3 > …………> 0.
Find the sum, A(r), of the areas of the rectangles whose base is along the x-axis between r n and r n 1 and
top is at y   r n   r 2 n .
2
A(r )  1  r  r 2   r  r 2  r 4   r 2  r 3  r 6   r 3  r 4  r 8 
 1  r  r 2  1  r  r 5  1  r  r 8  1  r  r11 

 1  r   r 2  r 5  r 8 
 r2 

 1  r   
3
 1  r  
r2

1  r  r 2 

 1
r2

Now, lim  A(r )  lim 
2
r 1
r 1
 1  r  r   3
7. Let R be the region in the plane bounded by the graphs of y  x 2 , y  6  x, and the y  axis.
a. Carefully and accurately sketch the graph of the region R.
b. Find the volume of the solid obtained when R is rotated about the line y  6  x .
Solution: Rotate a vertical slice ( 0  x  2 ) about the line y  6  x as shown. This creates a
“conical shell”. The volume of such a shell is given by:
dV  ( S . A)  (thickness)
  r  slant ht  dx
2

((6  x)  x 2 )  (6  x)  x 2  dx
2
2

((6  x)  x 2 ) 2 dx
2
2

Then the volume is 
0
2
248 2
((6  x)  x 2 ) 2 dx 
2
15
This solution is by Lael Costa in my BC 2-3 class in the fall of 2012.
8. Evaluate
e x (1  x)
 x2 dx
Solution :
i.Let
u
1
,
x
dv  e  x
.
1
du   2
x
Then
v  e
x
e x (1  x)
 x2 dx   vdu  udv
 uv

ii.Or just notice that
e  x
x
e  x (1  x)
d  f ( x) 
looks like

 and figure out f ( x ) .
2
dx  x 
x
9. Jeopardy: What is 2 n ?






The sum of the numbers in the nth row of Pascal’s Triangle.
n n n
 n  n

   
 
 0 1  2
 n  1  n 
The number of base two whole numbers with n digits or less.
The number of different possible outcomes if a coin is tossed n times.
 2n 
For every positive integer M, there is an n and m such that  m 
10 
Not a sum of (more than one) consecutive positive integers. (iff)
p
1
 if x  is a rational number in (0,1)
q
10. Let f ( x)   q
.
 0 if x is an irrational number in (0,1)
