Using the TI-92+ : examples

Transcription

1 Using the TI-9+ : eamples Michel Beaudin École de technologie supérieure 1100, rue Notre-Dame Ouest Montréal (Québec) Canada, H3C 1K3 mbeaudin@seg.etsmtl.ca 1- Introduction We incorporated the use of the TI-9/9+/89 in order to help our engineering students to use Computer Algebra in the classroom, not only in the labs. Students are often asked to use the «rule of four» : where appropriate, topics should be presented geometrically, numerically, analytically and verbally. With a tool like the TI-9, we can present and perform eamples that will use that rule. We decided, here, to present eamples from our first calculus course, one from our multiple variable course and some from our ordinary differential equations course. - Eamples Eample 1 : students learn that every eponential growth function eventually dominates every power function. Let see what is going on when we want to solve the equation 16. There is no close formula for the solutions of the above equation (Lambert W-functions are not seen in the first year of engineering studies!). A plot, in the same window, of the sides shows intersections and the «solve» command gives a warning that there might be more solutions (figure 1)! FIGURE 1

2 But there is, in fact, a third intersection! A table shows that it is between 100 and 110 and using «nsolve» with appropriate guess allows us to find this third intersection (figure on the right). FIGURE Because Newton s method is the one used by students as an application of the derivative, we find the first positive intersection (figure 3) simply by asking the TI to perform the iteration f ( ) f ( ) where f ( ) 16 FIGURE 3 As we can see, applying Newton s method is fast and quite easy with the TI! Finally, because every power function eventually dominates every logarithmic function, it would be easy to find the largest (positive) intersection by looking at the equation ln ln 16. Eample : the concept of limit is best understood by students if again we use the rule of four. Suppose we want to make a conjecture about the value of e lim 1 1 1

3 (this problem is encountered before studying l Hopital s rule). We are going to make a table of values of the epression for = 1.1, 1.01, 1.001,..., 0.9, 0.99, 0.999,..., (figure 4) as well as we will look at the graph of the epression. FIGURE 4 Now we will find a value of that will work for a value of, say = 0.01 : this means, that we will find an interval containing 1 such that the difference between our conjecture and the value of the epression is less than 0.01 on that interval (we will have to zoom in order to find a window of height 0.0 such that the graph eits the sides of the window and not the top or bottom of the window). We set the window parameters for y between 1.99 and.01 and try for the value 0.05 (so we fi between 0.95 and 1.05). Figure 5 (on the left) shows that the value of is not small enough! Zooming in for only yields a correspondig value of (figure 5 on the right), in fact will work here for. FIGURE 5 Of course, l Hopital s rule or a Taylor epansion of the numerator s epression would be the «analytic» part of the «rule of four» (figure 6). 3

4 FIGURE 6 Eample 3 : find the minimum distance from the point (1,, 10) to the paraboloid given by the equation z y. It is funny to solve this problem with and wihout Lagrange multiplier. If we minimize the square of the distance, we have to minimize the function f ( 1) ( y ) ( z 10) under the constraint g y z non linear system R S T f g : g 0 R S T ( 1) ( y ) y ( z 10) y z 0. That means to solve the Solving the first three equations for, y and z and combining with the forth one yields the cubic equation 5 ( 1 ) 0. This former equation has 3 real roots and we find that the minimum distance is , at the point (1.4040,.8080, ). The following screens show complete solution of this problem. We can note the use of the new «zeros» function of the TI-9+/89. 4

5 FIGURE 7 Another approach will be to substitute the constraint into f and to consider the function that yields the distance from the point (1,, 10) to an arbitrary point on the paraboloid : Here is the graph of the last epression : ( 1) ( y ) y 10 c h FIGURE 8 Eample 4 : in differential equations, students solve ODEs without looking at the slope field and without finding where the solution is defined! Let s consider, for eample, the first order separable equation dy dt 3t 4t, ya0f. ( y 1) 5

6 3 Of course, it is easy to solve by hand this DE and we find y y 8 t t t. If we want to solve for y, there will be two answers, depending on wich branch we want to select (because y = 1 is a singularity and the slope field allowed us to see this). If we apply Euler s method starting at t = 0 and y =, with a negative step, we will encounter problems... (figure 9 on the right ; note that the RK method used by the TI-9+/89 «recognize» that there will be a problem and stops drawing when we approach y = 1 : figure 9 on the left) FIGURE 9 Eemple 5 : a linear first order ODE that leads to the concept of steady-state solution. Let us consider, for a a real number, the following diffential equation dy dt y 3 t y 0 a cos( ), a f FIGURE 10 The interactive IC menu F8 allows us to choose the initial conditions and a solution is drawn by RK ou Euler methods. (figure 10 on the left). This can be done before solving eactly the first order linear equation dy dt b g p( t) y q( t), y t0 y0. And, using the «desolve» function, we see 6

7 easily that the answer is the sum of a transient solution and a steady-state solution whose amplitude seems to be and phase angle 0 ; using, for eample, Trigcollect (figure 10 on the right). Eemple 6 : finding the inverse Laplace transform of an epression using partial fraction and/or convolution! Suppose we have a table of Laplace transforms and we want to inverse the transform F( s) 1 ( s ) s 6s 13 better first to set s + 3 = w and to epand after : c h. Because a completion of the square has to be done, it is FIGURE 11 A table of transforms contains the following associations : s a ( s a) b e cos bt, b ( s a) b e sin bt, 1 1 e and te. So, s a ( s a) at at at at the inverse is simply (because w = s + 3) e t cost 1e t sin t 10e t te t Another approach that our students are encouraged to use is the convolution. We have 7

8 F( s) ACA 99 Session : Computer Algebra meets Education F I HG K J F H G I K J ( s ) s 6s 13 ( s ) ( s 3) 4 c h t 1 3t te e sin t where * means the convolution of signals. The TI can do it for us if we simply use the definition of the convolution : ( t) h( t) z t ( ) h( t ) d. 0 FIGURE 1 We find important that our students perform these operations... before downloading from the WEB a powerful inverse Laplace transform function! Eemple 7 : it is very simple to calculate a Fourier epansion with a TI-9! Let f be the periodic function of period P = 3, defined by R f ( ) ST si 0 si 3 P 3 We define this function using the «when» and etend it periodically by using the modulo function (same as in DERIVE). We note that the TI connects the discontinuity (figure 13 on the right) : 8

9 FIGURE 13 The TI knows that a number n is an integer if we use for eample the arbitrary We calculate the Fourier coefficients : We will plot the partial sum a 3 0 a n 1 same window (figure 15 on the right). n FIGURE 14 F H I K F H I K n n cos bn sin and the periodic signal in the 3 3 FIGURE 15 9

10 We conclude this eample and this paper by looking at the Energy Theorem (Parseval identity) that states that z 0 P f d a ( ) an bn P What fraction of the energy of f is contained in the constant term and the first three harmonics? About % : n 1 FIGURE Conclusion Teaching mathematics with a CAS like DERIVE is a great eperience for us since the last 5 years. Being able to see our students in the classroom performing calculations and eamples like those presented in the precedent section, on their own «portable CAS», is fantastic. Thanks to those who worked so hard in order to «take out» computer algebra from big computers and labs! Long life to these machines! 10

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