Let’s go over some basics for using the TI-89 to perform various calculus “calculations” on a specific x

1 function, say

( x ) = x

2
. To begin, we will have the TI-89 define this function so that we can repeatedly use it without having to type it over and over. This can be done using
Define
command, which can be found under the
F4
(
Other
) menu.
Next, let’s examine the limit of 
( x ) as x approaches 2. We will use the
CATALOG
to access the limit(
command, rather than get it from the
F3
(
Calculus
) menu, so that we can see the correct syntax. According to the
CATALOG
, we need to enter an expression (
EXPR
) for which we wish to evaluate the limit, the variable (
VAR
) that is changing in this limit, and the point (
POINT
) at which we are taking the limit (more on
[DIRECTION]
later). When this calculation is performed, the result is undefined ( undef
), which is correct since the limit as x approaches 2 of

( x ) does not exist.
But what if we want to examine the one-sided limits at x = 2, which do happen to exist? This is where the
[DIRECTION]
part of the limit(
command comes into play. In part of a command enclosed in
[ ]
is an optional part of the command. In the case of the limit(
command, it is included when you want to calculate one-sided limits. Include any positive number (like 1) as a
DIRECTION
and the limit from above the number will be calculated. Include any negative number (like –3) as a
DIRECTION
and the limit from below the number will be calculated.
Dr. Don Porzio, IMSA Page 1 USACAS 2007, TI-89 Workshop

The limit(
command can also be used to determine the limit of

( x ) as x approaches +

or –

.
The d( differentiate
command found under the
F3
(
Calculus
) menu can be used to determine the derivative of

( x ). Use the d( differentiate
command and the “with” or “such that” (
|
) key together to evaluate the derivative of

( x ) at a specific point (say x = 4).
The
∫( integrate
command found under the
F3
(
Calculus
) menu can be used to determine an antiderivative of

( x ) (the one where C = 0) or evaluate the definite integral of

( x ) over a specific interval
(say from x = 4 to x = 8).
The form of the antiderivative of

( x ), F ( x ) = 3 ln ( | x
– 2 | ), may seem a bit strange to some students. To see where it came from, apply the propFrac(
command found under the
F2
(
Algebra
) menu to

(x) x

1 to see that x

2

3 x

2

1 . When

( x ) is viewed in this form, the antiderivative provided by the TI-89 makes much more sense (and yes, the propFrac(
command did really determine the partial fraction x

1 decomposition for x

2
).
Dr. Don Porzio, IMSA Page 2 USACAS 2007, TI-89 Workshop

The TI-89 can be a powerful tool for exploring slopefields and Euler’s Method of approximating antiderivatives. To begin, the TI-89 needs to be set to Differential Equations Mode. To do this, Press the
MODE key, then press the key and select
6:DIFF EQUATIONS
. Next access the
Y=
Editor, press F1 and select
9:FORMAT
. Scroll down to
Solution Method
, press the key and select
2:EULER
. Finally, scroll down to
Fields
, press the key and select
1:SLPFLD
. In the
Y=
Editor, enter the differential equation in y1  =
(or y2  =
, y3  =
, etc.). Be sure to use “ t ” for your independent variable and “ y1 ” for your dependent variable. For example, the equation y

= 3 x + y would be entered in y1  =
as
3t+y1
.
After entering your differential equation, access the
WINDOW
Editor to enter your desired window then press F3 (
GRAPH
). Note, in the
WINDOW
Editor, tstep
is the usual step size in the Euler program. This will produce the slopefield corresponding to your differential equation. If you now want to plot the approximation to a specific solution to the differential equation using Euler’s Method, you need to return to the
Y=
Editor and enter the initial condition that defines the specific solution. To do this, enter the x-coordinate of your initial condition in t0
and enter the y-coordinate of your initial condition in yi1
.
For example, given the differential equation y
´ = 0.2
y (4 – y ), to plot the specific solution with initial condition (0, 0.3), you would have: t0=0 y1  =.2
 y1  (4-y1) yi1=.3
Before graphing, be sure tstep
in the
WINDOW
Editor is the usual step size.
Once the specific solution is graphed, press F3
TRACE
and then use the and keys to scroll through the coordinates at each step. Note that the initial condition can also be entered interactively in the
GRAPH
screen after a slopefield is graphed (though you cannot trace the solution graph if you do this).
To do this, enter a differential equation and graph the slopefield as before. (Ignore t0
, leave yi1
blank, and be sure tstep
in the
WINDOW
Editor is the usual step size.) Once the slopefield is graphed, press
2nd F3 (
F8 IC
). Now you have two options:
(1) Use the and keys to move to the desired point ( approximate ) and press ENTER .
(2) Type in the x -coordinate of your initial condition and press ENTER . Then type in the y -coordinate of your initial condition and press ENTER .
Let’s look at ways in which the TI-89 should and should not be used when solving AP Exam problems.
2001 BC Exam #1d: An object moving along a curve in the xy -plane has position
     
at time t with dx dt
 cos and dy dt

3sin
Find the position of the object at time t = 3.
for 0 ≤ t ≤ 3. At tine t = 2, the object is at position (4, 5).
Dr. Don Porzio, IMSA Page 3 USACAS 2007, TI-89 Workshop

To solve this problem, students need to recognize the relationship between the position of the object and the integral of the object’s rate function. In this case,

2
3 dx dt dt x or x
 
2
3 cos
 
  x
 
and

2
3 dy dt dt y or y
 
2
3
3sin
 
  y
 
The important point here that students need to recognize (and often don’t) is that even thought the integrals
 cos
 
 dt and

3sin
 
 dt have no closed form solutions, the definite integrals

2
3 cos
 
 dt and

2
3
3sin
 
 dt can be calculated. Students often forget the TI-89 is expected to be used for its numerical integration capabilities and not its symbolic manipulation capabilities.
This issue came up on the 2002 AB/BC Exam #2. On this problem, students calculated the closed form of an indefinite integral when they did not have to, since they were only required to evaluate certain definite integrals, which can be done without knowing the closed form of the indefinite integral. This points out how critical it is that students recognize when and how best to use their calculators.
2002 AB Exam #3d
An object moves along the x -axis with initial position x (0) = 2. The velocity of the object at time t ≥ 0 is given by
   sin


3 t  . What is the total distance traveled by the object over the time interval 0 ≤ t ≤ 4?
This is another case where understanding the capabilities of the calculator plays a key role in solving the problem in a time efficient manner. Assuming students recognize that the total distance traveled over the time interval 0 ≤ t ≤ 4 is given by the integral

0
4    dt , then the question becomes how best to evaluate this integral.
One way is to determine when v ( t ) is positive and negative, determine an antiderivative of v ( t ), and then evaluate the antiderivative using the appropriate bounds. The easier way is to use the numerical integration capabilities of the calculator to evaluate the integral. As in the previous example, the key for students is understanding how best to use the capabilities of the calculator to answer the questions being asked
Dr. Don Porzio, IMSA Page 4 USACAS 2007, TI-89 Workshop

Problem #5 in the 2000 AB and BC Free Response Sections is as follows.
Consider the curve given by
xy
2  x
3 y

6. a. Show that dy dx

3
2 x
2 xy y

 x y
3
2
. b. Find all points on the curve whose the x -coordinate is 1, and write an equation for the tangent line at each of these points. c. Find the x -coordinate of each point on the curve where the tangent tine is vertical.
Part a presents difficulties for the TI-89’s d( differentiate
command if you use it to find dy dx implicitly. The difficulty here is that the TI-89 does not recognize that y is a function of x . This difficulty can be avoided by using the variable y ( x ), which the TI-89 does recognize as a function of x , instead of y .
The TI-89 cannot solve the latter equation for d dx
 
since it does not recognize this expression as a variable. To handle this difficulty, d set dx
 
equal to some other variable that the
TI-89 will recognize (say yp ) and solve the equation for this new variable.
That’s almost what they asked for, but not quite. However, using the expand(
command doesn’t help clear things up unless it is followed by the comDenom(
command and then all is well.
Dr. Don Porzio, IMSA Page 5 USACAS 2007, TI-89 Workshop

Dr. Don Porzio, IMSA Page 6 USACAS 2007, TI-89 Workshop

Of course, with the newest operating system for the
TI-89, this problem is very easily solved by first using the impDif(
command and then using the expand(
command to get an expression in the requested form.
For part b, the solve(
command quickly finds all the points whose x -coordinate is 1 In part c, this command can also be used to determine for what values of x and y the tangent line is vertical (assuming dy the student knows this can occur when the denominator of dx
is equal to zero).
For the solution y
 x
2
, substituting for y in the original equation and then solving for x should produce
2 the desired answers. For the solution x = 0, hopefully students recognize the need to substitute x = 0 back into the original equation to verify that this solution is valid. Doing so shows such is not the case.
So what is the point of all this work with the TI-89 considering that many of the uses made of the TI-89 in this activity are “illegal” on the AP Exam and that Problem #5 was in the NON -calculator portion of the
2000 AB and BC Free Response Sections? The point is that doing such work allows students to focus on coming up with the correct set-up, equations, and justifications without having to worry about doing the symbol manipulation correctly. The necessary symbol manipulation skills can be handled separately. It also points out that use of the TI-89 to do some of the symbol manipulation can occasionally be more difficult or time-consuming than doing the symbol manipulation by-hand.
Dr. Don Porzio, IMSA Page 7 USACAS 2007, TI-89 Workshop

Many formulas for derivatives and antiderivatives can be explored and developed through pattern recognition. For example, use the TI-89 to find the derivative of y = x + sin x ; of y = x
2
+ sin x ; and of y = x 3 + sin x . Based of your observations, predict the derivative of y = x 4 + sin x ; of y =

( x ) + g ( x ).
(You can do the same with subtraction, multiplication, division and composition of functions!) .
Here’s another example. Consider the Fundamental Theorem of Calculus, d
 
  dx
 
Students often have difficulty recognizing how to handle problems such as d

 a x

0 sin
 x t
2 e dt
 dx
.
where the upper bound of the integral, “ x
,” is replaced by a function of x. Have the students use the TI-89 to do a variety of these problems and then have them predict the solution to the general case, d dx

 a


. Have the students check their prediction using the TI-89. The result is quite interesting.
Prove or Disprove: If a cubic polynomial has three real roots, then the tangent line to the graph of this polynomial at a point whose x-coordinate is the average of two of the roots will intersect the graph at the third root. [See if it works for a specific case such as

( x ) = ( x – 2)( x + 3)( x – 4).]
Prove or Disprove: If P is any point on the graph of

( x ) = x
3
such that the tangent to the graph of

( x ) at
P intersects the graph at another point Q, then slope of the graph of

( x ) at Q is four times the slope of the graph at P. [Again, see if it works for a specific case such as P being the point (2,8).]
Draw a tangent line to any cubic polynomial at (almost) any point (say x = a ). Determine the second point
(say x = b ) where this tangent intersects the cubic. Determine the area of the region bounded by the tangent and the cubic. Next, draw the tangent line to the cubic at x = b and determine second point (say x = c ) where this tangent intersects the cubic. Determine the area of the region bounded by this second tangent and the cubic. Determine the ratio of the larger area to the smaller area? Prove this result.
Find the volume of the largest right circular cone inscribed in a sphere of radius r .
Dr. Don Porzio, IMSA Page 8 USACAS 2007, TI-89 Workshop