Using the TI-89 Calculator in Undergraduate Mathematics Courses LTC Troy Siemers Virginia Military Institute © by Troy Siemers, 2006 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 Menus History Area Entry line (1) (2) (3) 1) 2) 3) 4) 5) (4) (5) Current folder Radians (vs degrees) 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) To adjust brightness: 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 Math menus 6 (–) Last answer ENTER ENTER Memory access Last entry Gives approx. numerical value 9 Tips 1) To view all variables: 2nd (var-link) 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 answers different? 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 bc c 2/3 c a 1 b 1/ 2 c c a 1 b 1 / 2 13 Answers: 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 55 55 i x ( 4 i x 5 1 4 5 1) 2 4 2 55 4 4 55 i i 14 Answers: (cont) 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 55 i or x 4 2 55 4 ( 5 1) 2 4 i or x 5 4 1 4 55 i 4 2 55 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 ) x3 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 - drawing/shading (line, dot, animated, shade above/below) 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 menu) 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 tc tc 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) 3) Plot the graph. Under the calculus menu select shade. 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 Answers: 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 Answers:(cont) 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 x1 ( 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 x3 ( 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. Answer = 7.93272 (approx.) 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. Let’s start with an example. 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 Enter coefficient matrix in spreadsheet. 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 Math menus 4 Matrix submenu 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 Answers: 1) Use the method in the example to enter the matrix m. From the math menu (matrix submenu), or catalog, 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 Answers: 2) Use the method in the example to enter u1, u2 (as rows). From the math menu (matrix submenu), or catalog, 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 Answers: 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 Answers: 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 Answers: 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 Answers: 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 2x 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. A few adjustments have to be made before graphing the solution. 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 (MATH menu) 6 (Statistics submenu) 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. Answers: 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 (var-link) 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 (var-link) 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 variables that you can adjust to meet your goal. That’s 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 details about how the calculator was used in addition to your final answers. Graph link software may be used for 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 yx 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 , x0 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 x2 4x 2 y 2 0 z 2x 4x 2 2 f (x) 1 4x 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) Answer: 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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# TI-89 Calculator Seminar