```Using the TI-89 Calculator
in
LTC Troy Siemers
Virginia Military Institute
1
Course Outline
• Calculator basics
• Algebra
• Graphing
• Calculus
• Matrices and Vectors
• Differential Equations
• Statistics
• Assorted Topics
2
Important Information!!!
The manual that came with the calculator is your
friend. Read it and bring it to every class!!
Manual Organization
Front cover: Short Cuts
Pg 388-391 : Condensed list of all functions
Pg 392-513 : Function descriptions (with examples)
Pg 537 : Reserved variable names
Online manual
http://education.ti.com/us/product/tech/89/guide/89guideus.html
3
Screen Layout
History Area
Entry line
(1)
(2) (3)
1)
2)
3)
4)
5)
(4)
(5)
Current folder
Exact (vs approximate, auto)
Graph type
Number of pairs in history area
4
Important Keys
F1
F2
F4
F5
ESC
2nd
HOME
F3
alpha
APPS
MODE
CATALOG
CLEAR
ENTER
5
Important Keys (cont)
When fed up :
HOME
To clear history area :
CATALOG
and
or
or ESC or
F1
2nd
ESC
(QUIT)
8
Lists all functions and syntax on their use
(to scroll :
or press beginning letter)
6
Basic Operations
MODE
+
F1
F2
or
F3
(or up/down)
Most of the entries in mode are self explanatory. Make sure
that the Angle is in radians, the Base is decimal,and the
Exact/Approx is exact.
Some of the things in mode that we will look at later include
changing the Graph type, and splitting the screen.
Other parts deal with how information is displayed. Try changing
the Pretty Print to OFF and see how it affects things.
NOTE!!! You must press ENTER twice to save any changes!!!!!
(This is true on many popup screens as well)
7
Basic Operations (cont)
F1
to
F8
These function keys bring up menus that depend on
the screen you are currently on.
Note: they can be used in conjunction with the
2nd
and
keys to bring up different screens and menus.
8
Basic Operations (cont)
2nd
and
These are the 2nd and 3rd buttons to access the functions
in orange and green on the keyboard. Important ones to
keep in mind are:
2nd
+
+
5
6
(–)
ENTER
ENTER
Memory access
Last entry
Gives approx. numerical value
9
Tips
1) To view all variables: 2nd
From here, one can also delete, copy, rename, etc.
variables.
This is also used when transmitting data between TI89s.
2) If something in the history area is too big (pg 91),
either press
or
or
2nd
10
Tips (cont)
3) Use
2nd
alpha
when entering letters.
Press alpha again to get out alpha mode.
4) To use copy/paste/cut, you need to highlight
the object. Hold
and use the arrow keys.
5) When selecting an item from a menu, you can either
scroll down/up and press enter, or you can press the
number or letter next to the item.
11
Algebra
(from home:
F2
)
One of the main difficulties that people have in using
a calculator to do mathematics is entering the
information properly. It should appear as it does on
your piece of paper. Don’t forget rules of operations,
and don’t forget your PARENTHESES !!!
The TI89 is a great tool to check your algebra even
if it includes variable names. Let’s look at some
examples.
12
Questions:
1) Type in both xy*x*y and x*y*x*y. Why are the
2) Factor the polynomial y = x5 – 1.
3) Solve the equation x5 – 1 = 0.
4) Find the common denominator for 1/3465 + 1/8085
5) Enter the expression:
 c  a  1 


1/ 3 
 bc

c
2/3




c  a 1
b
1/ 2

c
c  a 1 b
1 / 2
13
1) The TI89 thinks xy is a variable name.
2) factor(x^5 – 1, x) gives
 2 ( 5  1)  x   2 ( 5  1)  x 
( x  1)   x 
 1   x 
 1
2
2

 

2) cfactor(x^5 – 1, x) gives (complex roots)

 ( 5  1)

( x  1)  x 

4


x 

5 1
4

2

4
2
55

55
  i    x   (
4
  i    x 
 
 
5 1
4
5  1)

2
4

2

55
4

4
55
  i  

  i 

14
3) solve(x^5 – 1 = 0, x) gives x  1 (real roots).
3) csolve(x^5 – 1 = 0, x) gives (complex roots)
x 
 ( 5  1)

2
4
x 
5
1 4 
4
or

55
i
or
x 
4
2

55
4
 ( 5  1)
2


4
i
or
x 
5
4
1 4 
55
i
4
2

55
i
4
x 1
4) comDenom(1/3465 + 1/8085) gives 2/4851.
5) Work it out by hand. The expression equals 1.
The TI89 doesn’t give 1. Why?
15
Homework Assignment #1
1) Find the partial fraction decomposition for
x  3x  1
5
x  3x  3x  3x  2
4
3
2
2) Find all of the zeros of the function
f ( x )  x  6 x  49 x  64 x  230 x  300
5
3) Expand
4
3
sin( 2  x )  cos( 5  x )
2
in terms of
sin( x )
and
cos( x )
16
Homework Assignment #1 (cont)
2
2
4) In the expression A x  B xy  C y  D x  E y  F  0
make the substitutions
x  x  cos   y  sin 
y  x  sin   y  cos 
and simplify into the form
A x   B x y   C y   D x   E y   F   0
2
2
What are
A , B , C , D , E , F  in terms of A , B , C , D , E , F
?
17
HW Assignment #1 Solutions
1)
5


x  3x  1

expand  4
, x 
3
2
 x  3x  3x  3x  2 
x
10 ( x  1)
2
2) czeros(

7
10 ( x  1)
2

1
2 ( x  1)

gives
27
5( x  2 )
 x3
x  6 x  49 x  64 x  230 x  300 , x )
5
4
3
2
gives
x   5 , 3, 10 ,  1  i ,  1  i
3)
texpand( sin( 2  x )  cos( 5  x ), x )
sin( 2  x )  cos( 5  x )   32 sin
3
gives
x cos x  8 sin x cos x  10 sin x cos x
4
4
2
18
Homework Assignment #1 Solutions (cont)
4) Use xx instead of x (otherwise circular definition error).
Use the following sequence of steps (press enter between each):
x ' cos(  )  y ' sin(  )
STO
x ' sin(  )  y ' cos(  )
xx
STO
yy
A  xx  B  xx  yy  C  yy  D  xx  E  yy  F
2
2
gives
A   Acos ( )  ( B cos(  )  C sin(  )) sin(  ),
2
B    ( 2 A sin(  ) cos(  )  B ( 2 cos ( )  1)  2 C sin(  ) cos(  )),
2
C   A sin ( )  ( B sin(  )  C cos(  )) cos(  ),
2
D   D cos(  )  E sin(  ),
E    ( D sin(  )  E cos(  )),
F  F
19
Graphing
The TI83-TI92 make up the “graphing calculator”part
of the TI calculator lineup.
Displaying pictures of graphs of functions, differential
equation vector fields, statistical data, sequences of
points, etc. add to understanding of information.
The TI89 separates itself from its predecessors with
its ability to create 3D graphs as well.
20
Overview:
1) Accessing graphs and related operations:
with
F1
to
F5
(note: menus will change depending on current screen)
2)
MODE
: Graph type select, split screen to display
function with its graph.
3) Up to 100 functions of each type (2D, 3D, etc.)
can be stored simultaneously.
4) Halting the graphing process : press ON
21
“y =’’ screen: (from anywhere
F1
)
Enter the functions (up to 100)
F2
- zoom possibilities (in, out, standard, trig, etc.)
F4
- (de)select (functions to be graphed have check mark)
F6
F1
and
9
- Gives format possibilities (including
simultaneous/sequential creation of graphs)
22
Graphing Example
y 1( x ) 
1
2 
x
e
2
2
Window:
F2
x: -3 to 3, y: 0 to 1/2
To graph,
F3
(look familiar?)
The F5 button in “graph’’ mode gives the math menu.
It includes the function’s: value, zeros, minimum,
maximum, intersection, (numerical) derivatives and
integrals, inflection points, distance, tangent line, and
arc length.
23
Questions:
Find the following for y1(x):
1) The inflection point in the second quadrant.
2) The derivative at that point.
3) The tangent line at that point.
4) The length of the curve from x = –1 to 1.
5) The total area under the curve.
24
Answers: (from the graph screen and
F5
1) 8: Inflection (enter range from x = -3 to x = 0) .
ans.: x = –1, y=.24191
2) 6: Derivatives, select dy/dx (enter x = -1).
ans.: dy/dx=.24197
3) A: Tangent (enter x = -1).
ans.: y=.24191x+.483
4) B: Arc (enter x = -1 to x = 1).
ans.: Length = 2.02983
3
5) 7: 
f ( x ) dx
(enter x= - 3 to x = 3).
ans.: 
f ( x ) dx  . 9973
3
(look familiar?)
25
3D graphing
(In MODE change “Graph” entry to “3D”. “y=” shows z1, z2, ...)
x y y x
3
Example:
z1 
3
360
type
F1
z1=(x^3*y - y^3*x)/360
F3
to graph
26
x y y x
3
z1 
3
360
27
3D graphing (cont)
Options:
F1
- type of axes, coords.
9
Gives expanded view
Changes style (wire frame, contour, shaded, etc.)
X
Y
Z
Gives projected views
Rotate (hold to animate)
(Note:
0
returns to original view)
28
Other types of graphing
(In MODE change the “Graph” entry)
Parametric : “y=” screen shows x1(t)= and y1(t)=
(Note:
F5
button now gives dx/dt and dy/dt)
Polar: “y=” screen shows r1=
(Note:
F5
button now gives dr/dq)
Sequence: “y=” screen shows u1=
(Note: Sequence can be recursive, formulaic, etc.)
29
Homework Assignment #2
1) Graph
y 
x 
x
and
y 
x 
x
Where do they intersect?
Find the area between the curves from x = 1 to x = 3.
Find the tangent line of each at x = 1.
2) Graph
y ( t )  0 . 78540  0 . 63662 cos 2 t  0 . 07074 cos 6 t
 0 . 02546 cos 10 t  0 . 01299 cos 14 t
and
y ( t )  1 . 2732 sin 2 t  0 . 4244 sin 6 t
 0 . 2546 sin 10 t  0 . 18186 sin 14 t
How are the two functions related?
(note: these are the first few terms in the Fourier series for
some common functions)
30
Homework Assignment #2 (cont)
3) Create a graph where the area under the function
y  1 . 25 x  cos( x ) and above the x-axis is shaded.
4) Multipart Function:
Define a “step” function as
0,
u c (t )  
 1,
Let
tc
tc
 15 t   15   t / 4
1
t / 4

e
h ( t )  1  e
cos 
sin




2 
 4   15 
 15 t  


 4 

 
31
Homework Assignment #2 (cont)
4) Multipart Function: (cont)
Using the definitions of uc(t) and h(t), graph the function
y ( t )  u 5 ( t )  h ( t  5 )  u 20 ( t )  h ( t  20 )
Note: Have a look at the “Multi-statement” functions, pg. 195.
Also, this is the graph of the solution to the differential equation
2y’’ + y’ + 2y = u5(t) – u20(t)
which models a spring-mass system with variable mass.
32
HW Assignment #2 Solutions
1) Graph both functions and use the calculus menu
to get the intersection, integration and tangent line
commands.
F5
Intersection: specify the two functions and range.
(the graphs intersect at each non-zero integer)
Integration: Find each integral from 1 to 3 and subtract.
Areas: 2.60269 – 1.50408 = 1.09861 (approx)
Tangent line: specify the function and x=1.
No solution found since y’(1) doesn’t exist for each.
33
Homework Assignment #2 Solutions (cont)
2) The derivative of the first equals the second.
34
Homework Assignment #2 Solutions (cont)
It will prompt you for shading above/below the axis.
A possible graph is:
35
Homework Assignment #2 Solutions (cont)
4) This is most easily done by defining our function as the product
of a couple others. Two of these are multipart functions defined
by the “when” command.
y1( x )  when ( x  5 , 0 , 1)
y 2 ( x )  when ( x  20 , 0 , 1)
gives
0, x  5
y1( x )  
 1, else
gives
 0 , x  20
y 2( x)  
 1, else
 15 t   15   t / 4
1
t / 4

e
y 3 ( t )  1  e
cos 
sin




2 
 4   15 
 15 t  


 4 

 
y 4 ( x )  y1( x )  y 3 ( x  5 )  y 2 ( x )  y 3 ( x  20 )
36
Homework Assignment #2 Solutions (cont)
4) (cont) The graph looks like: (use x: 0..40, y=(-.3)..(.8))
37
Calculus
(from home:
F3
)
Limits
Symbolic and numerical (partial) Derivatives, Integrals
Sums, Products, Max/Min
Arc Length, Tangent lines
Taylor Polynomials
38
Questions:
1) Find the derivative of y = x*cos(x).
2) Find the integral of x*ln(x).
b
3) Compute 
2
n
n 1

4) Compute  1
n
2
n 1
5) Find the Taylor polynomial of order 4 centered at x=0
for y=cos(x). Now graph both the function together
with this polynomial.
39
1) d(x*cos(x), x) gives cos(x) – x*sin(x)
2)  ( x * ln( x ) , x ) gives
3)  ( n ^ 2 , n ,1, b ) gives
4)
 (1 / n ^ 2 , n ,1,  )
x  ln( x )
2

2
x
2
4
b  ( b  1)  ( 2  b  1)
gives
6

2
6
40
5) taylor(cos(x),x,3,4) gives
cos( 3 )  ( x  3 )
4

24
sin( 3 )  ( x  3 )
3

6
cos( 3 )  ( x  3 )
2
 sin( 3 )  ( x  3 )  cos( 3 )
2
or (approx)
 . 04125  ( x  3 )  . 02352  ( x  3 )  . 4949  ( x  3 )  . 1411  ( x  3 )  . 9899
4
3
2
41
Homework Assignment #3
1) Compute
2) Compute
1 
1
 3
lim  2  sin    1  cos 
x 
x 
x
x
lim 
x1
( x  1)
2
x ln x  x  x cos  x
(how would you do it by hand?)
3) Compute the derivative of
y  x 
4) Compute the following


0

e
x
2
dx
and
x3
 ( x  1)  ( x
2
2
 1)
dx
42
Homework Assignment #3 (cont)
5) Compute
2x  5x  8x  4
3

2
( x  2 x  2)
2
2
dx
6) Using nested functions, find dy/dx for
0
y 

cos( x )
1
1 t
2
dt
7) In one line, using repeated derivatives, find
 f
3
x z y
for f ( x , y , z )  tan
1
(x  y  z)
43
Homework Assignment #3 (cont)
8) Find the area under y  1 . 25 x  cos(
the x-axis from x = – 2 to x = .
x)
and above
9) Find the arc length of the curve
y 
4
4x
2
9
from x = – 3 to x = 3.
Note: This is the perimeter of the upper half of an ellipse.
10) Find the maximum and minimum of the function
y  1 . 25 x  cos( x )
on the interval  4 , 1
44
HW Assignment #3 Solutions
1)
 3

1 
1
limit   2  sin    1  cos  , x ,    0
x 
x
x

2) By hand, you use l’Hopitals rule. By calculator, it’s
2


( x  1)
limit 
, x ,1, 0   0
 x ln x  x  x cos  x

3)
y   x   y '  d ( ceiling(
x ), x )  
d
( floor(  x ))
dx
45
Homework Assignment #3 Solutions (cont)
4)  ( e ^ (  x ^ 2 ), x , 0 ,  )  . 886227
(in exact mode, the integral will be returned. You must
force it to approximate)

(( x  3 ) /(( x  1)  ( x
2
 1)) , x , 2 ,  )  ln( 5 )  tan
1
(1 / 2 )
(approximate mode is not exact)
5)  (( 2 x ^ 3  5 x ^ 2  8 x  4 ) /( x ^ 2  2 x  2 )^ 2 , x )
 ln(| x  2 x  2 |)  tan
2
1
( x  1) 
1
x  2x  2
2
46
Homework Assignment #3 Solutions (cont)
6)
d (  (1 /( 1  t ^ 2 ), t , cos( x ), 0 ), x ) 
7) d ( d ( d (tan
1
1
x
(cos( x )  1)  tan  
2
( x  y  z ), y ), z ), x )
2 ( 3 x  6 x ( y  z )  3 y  6 yz  3 z  1)
2

2
2
( x  2 x ( y  z )  y  2 yz  z  1)
2
8)   1 . 25 x  cos( x ), x ,

2
 3
2
,
2
3
 
 


1
.
25
x

cos(
x
),
x
,
0
,
 
  8 . 56748
2 
2

(or do it from the graph screen with these bounds)
47
Homework Assignment #3 Solutions (cont)
9) Graph the function
y 
4
4x
2
9
from x = – 3 to x = 3. Then use the Arc command on
the math menu with parameters – 3 and 3.
10) Graph the function y  1 . 25 x  cos( x ) for x = -4 to 1 and
use the maximum/minimum commands on the math menu.
Maximum: (-3.42562, 4.11046) (approx)
Minimum: (-.860334, -.70137)(approx)
48
Matrices and Vectors
(from home: APPS
6
)
Matrix and vector manipulations with the TI89
are very useful and fast. Vectors are treated as row or
column matrices so the computations are the same as
for matrices. As with many applications, the bulk of the
time is spent in entering the information.
Solve the system
of equations:
 25 x  61 y  12 z  25 


 18 x  12 y  7 z  10 
 3x  4 y  z  4 


49
We create the coefficient matrix and use the reduced
row echelon form to read off the solutions.
APPS
6
3
Data/Matrix editor.
Type : Matrix
Folder : main
Variable : m
Rows : 3
Columns : 4
50
 25 x  61 y  12 z  25 


 18 x  12 y  7 z  10 
 3x  4 y  z  4 


51
Computation: rref(m)
(either type this in manually, get the function
from the catalog or follow the procedure below)
From the home screen:
2nd
5
4
4
rref function
Solution: x = -2.0523 , y = 3.878 , z = 13.356
(approx)
52
Questions:
1) Find the inverse, transpose, determinant, eigenvalues,
eigenvectors, and LU decomposition of the matrix
6

m  5

 3
12
14
8
18 

31

18 
2) Find the dot and cross products of the vectors

 



u 1  i  2 j u 2  3i  4 j
53
Questions: (cont)
3) Three armored cars, A, B, and C, are engaged in a
three-way battle. Armored cat A had probability 1/3
of destroying its target, B has probability 1/2 of destroying its
target, and C has probability 1/6 of destroying its target. The
armored cats fire at the same time and each fires at the strongest
opponent not yet destroyed. Using as states the surviving cats at
any round, set up a Markov chain and answer the following
questions:
a) How many of the 8 states are absorbing?
b) Find the expected number of rounds fired.
c) Find the probability that A survives the battle.
54
Questions: (cont)
4) The method of least squares (for 4 data points). Given a set of
data points (xi,yi), i=1,2,3,4, we wish to find the “line of best fit.”
This line is given by y = mx + b where the m and b are found
by solving the equation
ATAX
= ATY
for
x1

x2

A 
x 3

x 4
Find the line of best fit to the following
data that describes the concentration of
a certain drug in a persons body after a
certain number of hours. Use the line to
estimate the amount of drug present
after 5 hours.
1

1

1

1
m 
X   
b
 y1 
 
y2


Y 
y3 
 
y 4 
Hours
Conc.
(ppm)
2
2.1
4
1.6
6
1.4
8
1.0
55
1) Use the method in the example to enter the matrix m.
m^-1, mT, det(m), eigVl(m), eigVc(m), and
LU m, m1, m2, m3 give
m
1
 1/6


1/8

  1 / 12
3
9/4
1/2
eigVl( m )  {1.92818
 .971

eigVc( m )  - .227

 - .068
 1

m1  5 / 6

 1 / 2
0
1
1/2
5 

4

1 
T
35.7234
- .547
- .725
- .420
0

0

1 
m
6

 12

18
5
14
31
3

8

18 
det( m )  24
.348426}
.824 

- .556

.112 
6

m2  0

 0
12
4
0
18 

16

1 
1

m3  0

 0
0
1
0
0

0

1 
56
2) Use the method in the example to enter u1, u2 (as rows).
dotP(u 1 , u 2 )   5
crossP(u
1
, u 2 )  0
0
 10 
3) In order to solve this problem, we need the possible states that
the system can be in after each shot is fired. Since each car can
be either dead or alive after a shot, there are 2*2*2=8 states.
These are:
none, A, B, C, AB, AC, BC, ABC
We create a transition matrix with the (i,j) entry giving the
probability of moving from state i to state j in one shot. (Of
course, state AB can never be achieved, but we need it for the
calculations)
57
3) (cont) The transition matrix P is given below.


A
B
P 
C
AB
AC
BC
ABC
 1

0

 0

 0
. 17

 . 06
 . 08

 0
A
0
B
C
AB
0
0
AC
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
. 17
. 33
0
. 33
0
. 28
0
. 11
0
. 55
0
. 42
. 06
0
0
0
0
. 22
0
. 22
BC
0
ABC
0 

0
0

0
0 

0
0 
0
0 

0
0 
. 42
0 

. 28 . 28 
Note: P can be blocked off into four 4X4 pieces as
I 4
P  
S
0

Q
58
3) (cont) From the theory of absorbing Markov chains, we define
the matrix
T  (I 4  Q)
1
The information we need is included in the matrices T and T*S.
AB
T  AC
BC
ABC
AB AC BC
ABC
0
0
0 
1 . 5


0
2 . 25
0
0


 0
0
1 .7
0 


0
.
69
.
66
1
.
4



AB
T  S  AC
BC
ABC
 . 25

. 13

. 14

. 09
A
. 25
B
.5
. 63
0
0
. 71
. 19
. 27
C
0 

. 25

. 14 

. 44 
59
3) (cont) The sum of the entries in the last row of T gives
the expected number of shots fired
0  . 69  . 66  1 . 4  2 . 75
The last row of T*S gives the probabilities of the battle ending
in the state associated with that column.
P ( None survive)
9%
P ( A survives)
 19 %
P ( B survives)
 27 %
P ( C survives)
 44 %
60
4) Enter the matrices A, X, and Y as before. The solution to
ATAX = ATY is X = (ATA)-1ATY
Computations:
120
A A  
 20
T
20 
,
4 
T
( A A)
-1
1/20
 
 - 1/4
- 1/4 

3/2 
 2.1 


1.6

Y  
1.4 


1.0


m 
  . 175 
T
-1
T

X

(
A
A)
A
Y

 


b
 2 .4 
So, our least squares line is C = –.175 t + 2.4 where C is the
concentration (in ppm) and t is the time (in hours).
When t = 5, C = 1.525.
61
Homework Assignment #4
1) Solve the system of equations
x





 2 y  3z  2 w  5

y  4 z  2 w  1 

3 x  y  4 z  10 
2 x  y  3 w  5 
2) Find the inverse, transpose, determinant, eigenvalues, and
eigenvectors for
 25

m  18

 3
61
 12
4
 12 

7

 1 
Use 4 point decimal
approximation
62
Homework Assignment #4 (cont)
3) Find a unit vector that points in the same direction as u1 and
the dot and cross products of the vectors

 

u1  i  j  2 k



u 2  3i  3 j
4) Find the (Householder) QR factorization of the matrix
6

m  5

 3
12
14
8
18 

31

18 
(use 4 point decimal approximations)
63
Homework Assignment #4 (cont)
5) Using the least squares example above as a guide, find the
line of best fit for the data in the table below.
Rainfall
(inches)
Yield of Wheat
(bushels per acre)
12.9
62.5
7.2
28.7
11.3
52.2
18.6
80.6
8.8
41.6
10.3
44.5
15.9
71.3
13.1
54.4
Use the line to make a guess as
to the yield of wheat if 22
inches of rain falls.
64
HW Assignment #4 Solutions
1) Enter the matrix using the matrix editor, then use the
rref command.
Answer: x = 683/185, y = -21/37, z = -76/185, w = -112/185
(approx: x = 3.6919, y = -.5676, z = -.4108, w = -.6054)
2) Enter the matrix and perform the calculations:
  . 0234

m ^ (  1)  . 0571

 . 1581
. 019
. 0161
. 1215

 25


T
 . 5725 , m  61


  12
 2 . 0469 
. 4143
18
 12
7
3 

4

 1
det( m )  683 , eigVl ( m )  {  31 . 6486 , 44 . 1376 ,  . 488942 }
 .7342

eigVc( m )  - .6786

 .0166
- .9449
- .3143
- .0906
- .1958 

.2674

.9434 
65
Homework Assignment #4 Solutions (cont)
3) Input the vectors as a single row matrix and compute.
 6
unitV( u 1 )  
 6

6

3 
6
6
dotP(u 1 , u 2 )  6
crossP(u
1
, u 2 )  6
6
0
4) Input the matrix into the editor
QR m , m1, m 2
 . 7171

m 1  . 5976

. 3586
 . 693
. 6663
. 2754
gives
 . 0743 
 8 . 3667


 . 446 , m 2 
0


 0
. 892 
19 . 8408
3 . 2160
0
37 . 8887 

13 . 1395

. 892 
66
Homework Assignment #4 Solutions (cont)
5) Input
Solution:
12 . 9

7 .2

11 . 3

18 . 6
A  
 8 .8

10 . 3
15 . 9

13 . 1
1

1

1

1
1

1
1

1
 62 . 5 


28 . 7


 52 . 2 


80
.
6

Y  
 41 . 6 


 44 . 5 
 71 . 3 


 54 . 4 
m 
 4 . 42372 
T
-1
T
   X  ( A A) A Y  

b
.
22919
 


The line is y = 4.42372 x + .22919.
If x = 22(inches), y = 97.55(bushels/acre) (approx.)
67
Differential Equations
The TI89 can solve first and second order differential
equations using the deSolve() function.
It can plot the solutions of higher order differential
equations by transforming an nth order differential
equation into a system of n 1st order equations.
Initial conditions can be entered as well to give a specific
solution (instead of a general solution).
68
2
Example: Solve: y ' y  x
Solution: deSolve(y’ + y = x^2, x, y) :
y  c1  e
Note: deSolve() is under the Calc menu, prime is
x
 x 2x  2
2
2nd
We can also give the vector field plot for our solution.
Change the
MODE
Graph setting to DIFF EQUATIONS.
On the y= screen, enter
y1’= t^2 - y1, yi1 = {0,3}
F2
- x, y min/max: -5 to 5, fldres to 20
F3
- graphs vector field and particular
solutions
69
This graph displays the slope field along with two solutions
given by the initial contitions y(0)=0 and y(0)=3.
Note: other initial conditions can be specified with
2nd
F8
70
Example: Solve: y ' ' '  2 y ' '  2 y '  y  sin( x ),
y ( 0 )  0 , y ' ( 0 )  1, y ' ' ( 0 )  1
Since this is not 1st or 2nd order, the best we can do is
graph the solution. This is always useful to get an idea
of the behavior of the solution.
First, we must transform the equation into a system of four 1st order
equations by introducing the intermediate variables y1, y2, y3 as
y1  y ,
y 2  y1 '  y ' ,
y 3  y 2 '  y1 ' '  y ' '
transforming
the equation
to the system
y1 '  y 2 ,
y2 '  y3,
y 3 '  sin( t )  2  y 3  2  y 2  y 1
Now we can enter the equation in the calculator.
71
Example:(cont)
In the y= screen, enter the system of equations with
initial conditions yi1=0, yi2=1, yi3=1.
72
Example:(cont)
On the y= screen, make sure y1’ is the only one checked.
Enter
and set
Axes=ON, Labels=ON, Solution Method=RK,
Fields=FLDOFF (important!)
In the y= screen, enter
2nd
F2
and set Axes=TIME
In the window screen enter
t0=0, tmax=10, tstep=.1, tplot=0
xmin=-1, xmax=10, xscl=1, ymin=-3, ymax=3, yscl=1
ncurves=0, diftol=.001
73
Example:(cont)
Now graph.
F3
A solution (using MA311 tools) can be found to be
y  . 554 e
 1 . 54 x
 . 554 e
 2 . 28 x
cos( 1 . 12 x )  . 367 e
 2 . 28 x
sin( 1 . 12 x )  sin( x )
74
Homework Assignment #5
1) Solve: y ' '  2 y '  y  sin( x )  cos( x )
2) Solve: ( 2 x  3 ) y '  y  ( 2 x  3 )
1/ 2
3) In the solution to number 2), find the value of the constant
when y=0 and x= –1. (Hint: See pg 184)
75
Homework Assignment #5 (cont)
4) Follow our example for third order differential equations
to plot the solution to
y ' ' '  3 y ' '  9 y '  13 y  0 ,
y ( 0 )  1, y ' ( 0 )  2 , y ' ' ( 0 )  3
Use the window screen settings
t0=0, tmax=3, tstep=.1, tplot=0,
xmin=-1, xmax=2, xscl=1, ymin=-15, ymax=20, yscl=1, ncurves=0, diftol=.001
Note: the actual solution is
y(x) 
4e
x
9

4e
2x
cos( 3 x )
9

4e
2x
sin( 3 x )
9
76
HW Assignment #5 Solutions
1) deSolve( y ' ' 2 y ' y  sin( x )  cos( x ), x , y )
gives
 2 cos( 2 x )
25

3 sin( 2 x )
 ( c1  x  c 2 ) e
x
50
2) deSolve( ( 2 x  3 ) y '  y  ( 2 x  3 )1 / 2 , x , y )
gives
2 x  3  ln( 2 x  3 )
2
 c1 
2x  3
77
Homework Assignment #5 Solutions (cont)
3) Define y 
2 x  3  ln( 2 x  3 )
 c1 
2x  3
2
then solve(y  0, c1) | x  -1 gives c1  0
4) Follow the example to get the graph:
78
Statistics
The TI89 statistical capabilities include finding one
and two variable descriptive statistics (mean,
median, variance, etc.), regressions (linear, quadratic,
cubic, logistic, etc.), correlations, and plots of data
along with their regression curves.
As with any set of data, the bulk of time is spent in
entering the data. Computations are very fast.
Note: from the home screen,
2nd
5
6
79
Example:
Input is similar to method for Matrices.
APPS
6
3
- now on data entry screen (“New”)
Keep as “data” and input Variable name:
Example: set1 ( alpha or
- not case sensitive)
80
In column c1,
type 1, 4, 7, 7, 10, 39
From
F5
screen
Calculation Type : OneVar
X : c1
ENTER
gives mean, x, x2, Sx, # entries, minX, quartiles, maxX
81
Example:
APPS
6
Input data name: set2
Press
3
C1:1,2,3,4,5,6,10
C2:1,8,27,64,125,216,1000
F5
Change Calculation Type: LinReg
X: C1
Y:C2
Store ReGEQ to y1(x)
82
The regression line is now stored in y1(x).
(Change the MODE Graph to FUNCTION to see it)
Do same for QuadReg in y2(x) and CubicReg in y3(x).
We now plot the data with the regression curves.
83
Statistical Plots
From the worksheet
F1
F2
F2
X: C1
Y:C2
- brings up the “y=” screen
(note: push
F3
F1
to see the Plots)
- graphs it all
- zoom to the appropriate fit (ZoomFit) or,
F2
- specify range of x and y
84
Here, the data is plotted as squares together with
the linear, quadratic, and cubic regression curves.
85
Homework Assignment #6
1) This is a continuation of problem 5) from assignment #4.
a)
b)
c)
d)
Enter the rainfall vs. yield information as a data set.
Find the linear regression between the two variables.
Plot the regression line together with a scatter plot of the data.
Use the value command in the Math submenu to find the
yield for 22 inches of rain.
86
Homework Assignment #6 (cont)
2) The following 150 data points are scores from a recent
government achievement test.
62
43
58
55
46
45
81
57
68
46
37
59
77
44
50
60
72
40
56
60
49
56
74
63
35
47
54
44
47
45
56
70
63
47
56
72
57
55
86
69
89
64
37
28
43
87
92
36
52
74
52
55
68
83
61
54
61
55
70
42
41
62
41
46
76
67
42
44
59
55
70
79
52
55
63
45
30
40
40
46
80
48
60
53
66
76
57
57
71
50
28
26
69
72
42
52
58
28
56
53
54
61
58
54
50
57
62
63
34
77
45
56
73
83
65
32
86
45
62
70
95
62
14
70
41
55
45
86
81
49
52
49
60
61
62
70
63
61
58
58
66
71
84
36
74
44
28
51
43
63
a) Find the one variable descriptive statistics.
b) Create a histogram of the data using classes 10-19, 20-29, …, 90-99.
87
Homework Assignment #6 (cont)
3) The following table holds the scores obtained by 44 cadets firing at a target
from a kneeling position, X and from a standing position, Y.
X
Y
X
Y
X
Y
X
Y
81
93
76
86
99
98
82
92
95
98
91
83
88
78
83
94
87
77
94
94
84
83
81
96
86
91
90
87
90
98
94
75
88
76
81
91
76
81
85
89
91
94
76
88
94
86
91
85
93
83
83
99
90
96
85
86
76
90
87
84
87
81
97
96
86
84
77
97
83
86
98
93
88
90
97
89
88
83
86
78
89
91
82
78
93
92
87
92
Create a scatter plot and describe the relationship between the scores
in the two positions.
88
HW Assignment #6 Solutions
1) APPS
6
3
Create data set “s”. Input the data.
F5
Change calculation type to LinReg. X : c1, Y : c2.
Store RegEQ to y1(x). Gives y=4.4424 x + .2292
F2
F1
X : c1, Y : c2.
F2
X : 0 to 23, Y : 0 to 100.
F3
Graphs it all.
89
Homework Assignment #6 Solutions (cont)
1) (cont)
F5
Select value and input x=22. Result y=97.55
90
Homework Assignment #6 Solutions (cont)
2) Again, use the data editor to create a data set and
input the values.
F5
Change calculation type to OneVar. X : c1.
Mean = 57.05, Standard Deviation = 15.02
Min = 14, Q1 = 46, Median = 56.5, Q3 = 67, Max = 95
91
Homework Assignment #6 Solutions (cont)
2) (cont)
F2
(Highlight Plot 2) F1
Change Plot Type to Histogram, X : c1, Hist. Width = 10
F2
X : 0 to 100, Y : 0 to 50.
F3
Graphs it all.
92
Homework Assignment #6 Solutions (cont)
3) Again, use the data editor to create a data set and
input the values.
F2
(Highlight Plot 2) F1
Change Plot Type to Scatter, X : c1, Y : c2
F2
X : 60 to 100,
Y : 60 to 100.
F3
Graphs it all.
(Shown with y=x to indicate standing scores
93
aren’t as good as kneeling)
Functions, Programming, and
Numeric Solver
User-defined functions:
• Expand existing TI89 functions.
• Useful in evaluating the same expression with
different values.
• Can graph or store resulting values.
Numeric Solver:
• Provides fast solutions to expressions or equations.
94
Programming:
• Similar syntax to common programming languages
(e.g. If…EndIf, loops, etc.)
• Can call other programs as subroutines.
• Can change the TI89’s configuration inside a program
(e.g. setMode command)
• Can prompt user for input.
• Can get or create Assembly-Language programs.
Now, some examples…
95
Example: A Millionaire in the Making
Under what saving conditions can you become a
millionaire?
We answer this question using three methods:
1) A user-defined function.
2) A program.
3) The numeric solver.
96
Assumptions and Variables:
• Time horizon. Variable name: t
(the amount of time until \$1,000,000 is achieved).
• Number of interest compounding periods per year.
Variable name: n.
• The annual (nominal) interest rate expressed as a
decimal. Variable name: r.
• Amount invested per interest period (equal per period).
Variable name: P.
• We assume that the interest rate is constant over
the time horizon.
• No inflation is assumed.
97
Formulas:
The basic formula for the future value of a one time
investment P, at rate r, for t years, with n compounding
periods is:
r

F  P  1  
n

t n
To get the formula we are interested in, we use
a finite sum over the entire time horizon. (next page)
98
Formulas: (cont)
2
r
r
r



P  P  1    P  1      P  1  
n
n
n



t n
Using a simple formula on partial sums of geometric
series, we have
t n 1


P n 
r
Value 
  1  
 1
r
n
 

We now use this in our function, program, and the
numeric solver.
99
User-defined Function: (pg. 85)
We create a function called “value1” which takes the
variables, P, n, r, and t, and returns the future value.
There are three ways to do this (see pg 85). We use
the store command here.
Type:
p*n/r*((1+r/n)^(t*n+1)-1)
STO
value1(p,n,r,t)
press
ENTER
100
User-defined Function: (cont)
To use this function, you can either type for example:
value1(1000, 4, 0.08, 10)
or,
2nd
and select the function from there. On the entry line of
the home screen, it will place “value1(”. Input the data
above and press enter.
101
Program: (ch. 17)
We use the program editor to enter our program
which prompts the user for values of the variables
and returns the value of the investment.
APPS
7
3
- opens the program editor
Type: Program,
Folder: main,
Variable: value2
102
Program: (cont)
Using the CATALOG
key,enter the
commands on the
line under “Prgm”.
value2()
Prgm
ClrIO
Disp "Enter P":Prompt p
Disp "Enter n":Prompt n
Disp "Enter r":Prompt r
Disp "Enter t":Prompt t
p*n/r*((1+r/n)^(t*n+1)-1)->val
Disp "Value is"
Disp val
EndPrgm
103
Program: (cont)
To run the program, from the home screen either type
value2()
(no input here)
or,
2nd
and select the function from there. On the entry line of
the home screen, it will place “value2(”. Close the
parenthesis and press enter.
104
Numeric Solver: (ch. 19)
At this point you may be wondering about the whole
millionaire part. The trouble is that there are four
where the numeric solver can help since you get to
choose which variable to solve for.
APPS
9
opens the
numeric solver
105
Numeric Solver: (cont.)
Enter the equation:
1000000=p*n/r*((1+r/n)^(t*n+1)-1)
press
ENTER
Input values for all but one of the variables, move the
cursor to the remaining variable, and press F2
Example: n = 4, r = .08, t = 10. Move cursor to
p = and press F2
p = 15971 (note: this is dollars per quarter)
106
Final Problem Set
Instructions: The solutions to these problems MUST include
printouts. I can supply the software for installation on your
computer.

1) Calculate 
n 1
1
n
4

1
and  n .
6
n 1

1
What does this say about  n ? (Note: The exact value
is unknown)
5
n 1
107
Final Problem Set (cont)
2a) Compute
2b) Plot
2



0
sin( y )
dy
y
4
sin( 3 x ) sin( 5 x )
sin( 13 x ) 
y   sin( x ) 

 

 
3
5
13

(Note: This problem is related to something called
Gibb’s phenomenon in signal processing)
108
Final Problem Set (cont)
3a) Find the determinant of the following matrices.
These special matrices are called Vandermond matrices.
1

2
2

2
1

2

 3
2
2
3
1

2

3

4
3

3

3 
2
3
2
3
3
3
4
4
4

4

4

4
3b) Find the formula for the determinant of
1

2

3


n

2
3

2
3

3
3




n
n
n
n

n

n

n
n 
109
Final Problem Set (cont)
4) The Hessian of a function f ( x , y ) is defined to
be the determinant
f xx
f xy
f yx
f yy
 f
  f 



yx
y  x 
2
(Note: f xy means
)
Find the Hessian of the function
f ( x , y )  x  y  2 x  ln( x y )
2
110
Final Problem Set (cont)
5) The given chart represents mile run times (in seconds)
by world class runners in the given year (after 1900).
Year
Time
Year
Time
Year
Time
Year
Time
54
54
56
56
58
239.4
238.0
238.1
238.5
234.5
58
60
60
62
62
236.2
235.3
234.8
235.1
234.4
64
64
66
66
68
234.1
234.9
231.3
232.7
231.4
68
70
70
72
72
231.8
232.0
231.9
231.4
231.5
a) Find the linear, cubic, and logarithmic regression curves
for the data.
b) Use each curve to predict a time for the year 2002.
c) The current world record of 223.1 was set in 1999 by
Moroccan runner Hicham El-Guerrouj. Which of the
curves in part a) best approximate this?
111
Final Problem Set (cont)
6) When a tractor trailer turns into a cross street or
driveway, its rear wheels follow a curve called a tractrix.
The function that traces this curve is the function y  f ( x )
that is a solution to the differential equation
y'
1
x 1 x

2
x
1 x
2
a) Solve the differential equation.
b) The solution provided by the calculator is wrong.
Explain what is wrong with this solution.
c) The actual solution is given below. Plot this function.
y  cosh
1
(1 / x )  1  x
2
112
Final Problem Set (cont)
7) A random variable X is said to have an Erlang
distribution (with parameters λ and r) if the associated
probability distribution function is given by
f (x) 

r
( r  1)!
x
r 1
e
 x
,
x0
The Erlang distribution is a special case of the gamma
distribution and is appropriate for queuing theory
applications including loss and waiting times in telephone
calls.
(cont next page)
113
Final Problem Set (cont)
7) (cont) The mean μ and variance σ2 formulas
for any distribution are

 
 x  f ( x ) dx ,
0


2

x
2
 f ( x ) dx  
2
0
Compute the mean and standard deviation for the
Erlang distribution in the case that λ=3 and r=2.
114
Final Problem Set (cont)
8) The center of mass of an object is
Mx My Mz


 M , M , M 


b f2 ( x) g2 ( x, y )
where
M 
 
  ( x , y , z ) dzdydx
a f1 ( x ) g 1 ( x , y )
b f2 ( x) g2 ( x, y )
b f2 ( x) g2 ( x, y )
M
x

 
 x   ( x , y , z ) dzdydx
a f1 ( x ) g 1 ( x , y )
M
y

 
 y   ( x , y , z ) dzdydx
a f1 ( x ) g 1 ( x , y )
b f2 ( x) g2 ( x, y )
M
z

 
 z   ( x , y , z ) dzdydx
a f1 ( x ) g 1 ( x , y )
and δ(x,y,z) is the density of the object.
115
Final Problem Set (cont)
8)(cont) Find the centroid of the object defined by
 ( x, y, z )  1
(a  2, b  2)
2 x2

4x
2
 y
2
0 z  2x
4x
2
2

 f (x)   
 1

4x
2
2
, f2 ( x) 

2
4  x 

2

( g1 ( x, y )  0, g 2 ( x, y )  2  x )
116
Final Problem Set (cont)
9) For what values of λ does the following system of
equations have a solution? For those λ, give the solution.
  x  61 y  12 z  25 


 18 x   y  7 z  10 
 3x  4 y  z  4 


117
Final Problem Set (cont)
10) One of the roots of x5 – 1 = 0 is 1. Find the other
(complex) roots of the equation x5 – 1 = 0. Lable the
roots 1, r2, r3, r4, r5 and complete the chart below
(where the entry in the ith row and jth column is ri * rj).
1
r2
r3
r4
r5
1
r2
r3
r4
r5
118
Final Problem Set Solutions
1)


n 1


n 1
1
4
n
1
n
So,
6


 (1 / n ^ 4 , n ,1,  ) 
 (1 / n ^ 6 , n ,1,  ) 


4
90


n 1
1
n
5




4
90
6
945
6
945
119
Final Problem Set Solutions (cont)
2) ( 2 /  ) *  (sin( y ) / y , y , 0 ,  )  1 . 17898
Store
4
sin( 3 x ) sin( 5 x )
sin( 13 x ) 
y1( x )   sin( x ) 

 

 
3
5
13

Plot with x between –2π and 2π, y between –2 and 2
120
Final Problem Set Solutions (cont)
3a) det([[1,2][2,2]]) = –2
det([[1,2,3][2,2,3][3,3,3]]) = 3
det([[1,2,3,4][2,2,3,4][3,3,3,4][4,4,4,4]]) = –4
3b) The pattern is (-1)n*n, where n is the number of
rows in the matrix.
121
Final Problem Set Solutions (cont)
4) Define
f ( x , y )  x  y  2  x  ln( x  y )
2
To get the Hessian, use the command
det([[ d ( d ( f ( x , y ), x ), x ), d ( d ( f ( x , y ), x ), y )] 
[ d ( d ( f ( x , y ), y ), x ), d ( d ( f ( x , y ), y ), y )]])
(the three dots indicate that this is entered on one line)
f xx
f yx
f xy
f yy

2
2
x y
2
1
122
Final Problem Set Solutions (cont)
5a) Store the information in the data editor as “mile”.
Using the calc menu, compute the linreg, cubicreg
and lnreg with x:c1, y:c2.
linreg :
T   . 4190 * y  260 . 7218
cubicreg
:
lnreg :
T  ( 5 . 0742 E  4 ) * y  . 0791 * y  3 . 4711 * y  201 . 9988
3
2
T  343 . 7471  26 . 4370 * ln( y )
5b) Using the value command on the graph (or direct calc)
with y = 102 (for year 2002).
linreg :
T (102 )  217 . 981
cubicreg
:
lnreg :
T (102 )  272 . 024
T (102 )  221 . 579
Best approximation
123
Final Problem Set Solutions (cont)
6)

1
deSolve  y ' 


2
x
1

x

gives
y  ln( x )  ln

x
1 x

1 x 1 
2
2

x, y 


1  x  c1
2
This cannot be true since the domain of this function
is empty. Look at the second term’s domain.
Plot
y 2 ( x )  cosh
1
(1 / x ) 
1 x
2
x:0  1
y :0  5
124
Final Problem Set Solutions (cont)
7) Mean :
 3 2 1 3 x

    x   x e , x , 0 ,    2 / 3
1!


Variance

2
:
 2 3 2 1 3 x

2
   x 
 x e , x , 0 ,      2 / 9
1!


Standard
 
Deviation
2 /3
125
Final Problem Set Solutions (cont)
8)
M
M
M
M
2
2 
 


4

x
4

x
 , x ,  2 , 2   12 . 5664
      1, z , 0 , 2  x , y ,
,

 

2
2

 

2
2 
 


4

x
4

x
 , x ,  2 , 2    6 . 28319
       x , z , 0 , 2  x , y ,
,

 

2
2

 

2
2 
 


4

x
4

x
, x , 2,2   0
       y , z , 0 , 2  x , y ,
,

 

2
2

 

2
2 
 


4

x
4

x
 , x ,  2 , 2   15 . 708
       z , z , 0 , 2  x , y ,
,

 

2
2

 

So, center of mass is (-1/2,0,5/4)
126
Final Problem Set Solutions (cont)
9) rref ([[  , 61 ,  12 , 25 ][ 18 ,   , 7 ,10 ][ 3 , 4 ,  1,  4 ]])
gives
1

0

 0
0
0
1
0
0
1
A

B

C 
A
with
B 
73   2278
p ( )
3 ( 6   733 )
p ( )
4   35   802
2
C 
p ( )
p (  )    64   1515
2
czeros(  ^2 – 64   1515,  ) gives   32 
491 i
So the solution is x = A, y = B, z = C if   32  491 i
127
Final Problem Set Solutions (cont)
10) In approximate mode, czeros ( x 5  1, x )
gives 1 and
rr 2   . 8090  . 5868 i, rr 3   . 8090  . 5868 i
rr 4  . 30917  . 9511 i, rr 5  . 30917  . 9511 i
(Note, you can’t use the system variables r2 to r5)
1
rr2
rr3
rr4
rr5
1
1
rr2
rr3
rr4
rr5
rr2
rr2
rr5
1
rr3
rr4
rr3
rr3
1
rr4
rr5
rr2
rr4
rr4
rr3
rr5
rr2
1
rr5
rr5
rr4
rr2
1
rr3
128
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