CALCULUS CLASSROOM CAPSULES

CALCULUS CLASSROOM CAPSULES SESSION S86 Dr. Sham Alfred Rartan Valley Communty College salfred@rartanval.edu 38th AMATYC Annual Conference Jacksonvlle, Florda November 8-, 202

2 Calculus Classroom Capsules Sesson S86 Overvew I. Fnte Dfferences of Polynomals and Exponental Functons II. Logarthmc Dfferentaton: Two Wrongs Make a Rght III. Product and Quotent Rules for Dervatves: Usng Logarthmc Dfferentaton IV. Crtcal Ponts of Polynomal Functons V. An Overlooked Calculus Queston

3 Calculus Classroom Capsules Sesson S86 I. Fnte Dfferences of Polynomals and Exponental Functons Before you begn to teach the dervatve of an exponental functon consder a motvatonal exercse to compute fnte dfferences of polynomals and exponental functons.. Consder the set of data gven n the frst two columns of Table () The data s non-lnear. In fact t s quadratc: u(x) = ax 2 + bx + c Its regresson equaton s u(x) = 0.5 x 2 + 0.5x wth R 2 =

4 Calculus Classroom Capsules Sesson S86 Table () x y st Fnte Dfferences 2 nd Fnte Dfferences 3 rd Fnte Dfferences u = y+ y + u u v + = + w = v+ v + =,...,9 =,...,8 =,...,7 3 = 2 2 3 2 = 2 2 3 3 6 4 0 4 0 5 0 5 5 6 0

5 Calculus Classroom Capsules Sesson S867 60 50 y = 0.5x 2 + 0.5x - 3E-3 R² = 40 30 20 0 Seres Graph of data n Table () Poly. (Seres) 0 0 2 4 6 8 0 2 The column of frst fnte dfference s lnear: 2 2 ( ) 2 2 ( ax ) ( ) a x x b( x x) c c + bx+ + c ax + bx + c + + + + = + + ( x x)( a( x + x) + b) + + = = ( + + ) + x+ x a x x b

6 Calculus Classroom Capsules Sesson S86 The column of second fnte dfferences s constant: mx+ + b ( mx + b) m( x+ x) + b b = = + + The column of thrd fnte dfferences s 0: m m m + = 0 Fnte dfferences are dfference quotents and are a good motvaton for the dervatve of polynomal functons and for analyzng data sets n general

7 2. Consder the second set of data n Table (2) below Table (2) x y st Fnte Dfferences u y+ y = + 2 nd Fnte Dfferences u+ u v = + 3 rd Fnte Dfferences =,...,9 =,...,8 =,...,7 v v + w = x + x 4 th Fnte Dfferences z w+ w = + 4 = 3 2 6 3 = 3 2 4 3 = 2 = 0 2 2 4 0 4 = 6 3 2 0 6 = 4 3 2 5 4 = 3 2 = 0 3 2 3 0 0 5 0 4 20 5 6 0 5 35 2 7 0

8 A regresson on the frst two columns wll result n a cubc model As n the frst example we can show that the frst fnte dfference wll be quadratc, the second wll be lnear, the thrd wll be constant and the fourth wll be zero 250 200 y x x x wth R 6 2 3 3 2 2 = + + = y = 0.667x 3 + 0.5x 2 + 0.3333x + 7E-4 R² = 50 00 Seres Poly. (Seres) 50 Graph of data n Table 2 0 0 2 4 6 8 0 2

3. Consder the thrd set of data {(x, y )}, = 0 n Table (3) below 9 x y st Fnte Dfferences 2 nd Fnte Dfferences 3 rd F4nte Dfferences 4 th Fnte Dfferences 5th Fnte Dfferences Table (3) u = y+ y + u+ u v = + w = v+ v + w+ w z = + =,...,9 =,...,7 =,...,6 =,...,5 5 = 4 2 2 5 3 5 4 35 5 70 20 5 6 0 35 2 7 0 56 28 8 0 6 26 84 36 9 0

0 The regresson equaton on the frst two columns results n a fourth degree equaton shown below. 4 3 2 800 700 600 y = x + x + x + x 24 4 24 4 y = 0.047x 4 + 0.25x 3 + 0.4583x 2 + 0.25x + 6E-0 R² = 500 400 300 200 00 Seres Poly. (Seres) Poly. (Seres) Graph of data n Table (3) 0 0 2 4 6 8 0 2 A further analyss, as was done n the frst example, wll show that the column of frst fnte dfferences represents cubc data; the second quadratc data; the thrd lnear data; the fourth constant data and the and the ffth represents the 0 functon. The three examples above motvate average rate of change and the dervatve of polynomals.

Now consder the exponental data n Table (4) Table (4) x y st Fnte Dfferences 2 nd Fnte Dfferences 3 rd Fnte Dfferences 4 th Fnte Dfferences 5th Fnte Dfferences y y u + = + =,...,9 u u v + = + v+ v w = + =,...,8 =,...,7 w+ w z = + =,...,6 =,...,5 2 4 2 = 2 2 4 2 = 2 2 4 2 = 2 2 4 2 = 2 2 4 2 = 2 2 2 4 8 4 = 4 3 2 8 4 = 4 3 2 8 4 = 4 3 2 8 4 = 4 3 2 8 4 = 4 3 2 3 8 8 8 8 8 8 4 6 6 6 6 6 6 5 32 32 32 32 32 32 6 64 64 64 64 64 64 Unlke polynomal functons, fnte dfferences of exponental functons reman exponental 2 + 2 2(2 ) = = 2 2

2 Consder the followng exponental data of powers of 3 n Table (5) Table (5) x y st Fnte Dfferences y y u + = + 2 nd Fnte Dfferences u u v + = + 3 rd Fnte Dfferences v+ v w = + 4 th Fnte Dfferences w+ w z = + 5th Fnte Dfferences =,...,9 =,...,8 =,...,7 =,...,6 =,...,5 3 9 3 = 2 3 2 2 2 3 2 3 2 = 2 2 3 4 2 4 2 = 5 2 3 3 2 3 2 = 4 2 3 4 2 4 2 = 5 2 3 2 9 23 2 2 2 3 2 4 2 5 2 3 27 2 3 3 2 3 3 3 4 3 5 3 4 8 5 243 4 2 3 5 5 2 4 2 5 3 4 3 3 4 4 4 5 5 4 5 5 6 729 23 6 2 6 3 6 4 6 5 6

3 Exponental functons reman exponental when computng ther fnte dfferences. In ths case, a constant tmes an exponental functon remans an exponental functon. Compare wth the dervatve of an exponental functon y = e x whch after ths exercse comes wth no surprse for the student as y = e x Compare outcome of Table (5) wth repeated dervatve of. In both cases the exponental functon s multpled by a constant. Fnte dfferences are a good motvaton for the dervatve of exponental functons.

4 II. Logarthmc Dfferentaton: Two Wrongs Make a Rght Brannen and Ford wrte about ther experence n Calculus I class teachng the dervatve of y = x x and contrbute an mportant nsght from that experence. Some students vew y = x x as y = x n, a power functon and dfferentate t as y = xx x- (Wrong!) Other students vew t as an exponental functon such as y = a x snce the exponent s a varable and dfferentate t as y = x x lnx (Also wrong!) So y = x x s nether a power functon nor an exponental functon. As teachers we yearn for the student who wll wrte t as

5 At ths tme, Brannon and Ford are about to ntroduce logarthmc dfferentaton when a student rases her hand and suggests that y = x x s both a power functon and an exponental functon so ts dervatve must be the sum of the two namely, y = xx x- + x x lnx () At ths pont they use the correct method logarthmc dfferentaton to get lny = ln x x = x lnx y' = ()lnx + x = lnx+ y x y' = y(lnx+ ) = x x (lnx+ ) (2) But equaton () and equaton (2) turned up to be exactly the same. That was not the expected outcome!!!!. They assured the class that that was just a concdence!

6 Usng the student s suggested method of the dervatve of y = f(x) g(x) n general, they got: Whch can also be wrtten as: Usng logarthmc dfferentaton they got (3) Whch when the y = f(x) g(x) s dstrbuted s exactly the same as the student s method (4)

7 So t was not a concdence. Two wrongs ndeed make a rght but why???? Why can one treat g(x) as a constant and then treat f(x) as a constant? Ths s precsely what one does wth partal dfferentaton namely, holdng one varable constant whle dfferentatng the other. y = f(x) g(x) s a functon of two varables (f, g) each of whch s functon of x. So usng the multvarable chan rule we get dy δydf δydg = + dx δf dx δgdx dy dx = g x f x + g( x) g( x) ( ) ( ) f'( x) f( x) g'( x)ln( f( x)) (5) Equaton (5) s the same as (3) and (4). The student s vndcated. But contnue to teach logarthmc dfferentaton.

8 III. Alternatve Proof of the Product and the Quotent Rule for Dervatves Most calculus books use the formal defnton of the dervatve to prove the product and quotent rules. In some cases the product and quotent rules can be proved usng logarthmc dfferentaton The Product Rule such that for all x n the doman of f, and u and v are dfferentable functons of x, then ln f( x) = ln[ u( xv ) ( x)] = ln u( x) + ln v( x) Takng the dervatve of both sdes yelds:

9 The Quotent Rule u( x) If f( x) = whereu( x) andv( x) are defned as above, v( x) u( x) Thenln f( x) = ln = ln u( x) ln v( x) Dfferentatng both sdes, v( x) f'( x) u'( x) v'( x) v( xu ) '( x) u( xv ) '( x) = = Multply both sdes by to get f( x) u( x) v( x) u( xv ) ( x) v( xu ) '( x) u( xv ) '( x) u( x) v( xu ) '( x) u( xv ) '( x) = u( xv ) ( x) v( x) u( xv ) ( x) f'( x) = f( x) f'( x) = vxu ( ) '( x) u( xv ) '( x) 2 [ vx ( )] Fndng the dervatve of a product and quotent functons under the condtons stated gves an alternatve useful method for renforcng the product and quotent rules for dervatves and provdes good practce and ncentve for usng logarthmc dfferentaton.

20 IV. Crtcal Ponts of Polynomal Functons [4] Nowadays we use computers and graphng calculators more often to provde opportuntes for expermentaton. In ths actvty, sutable for a student n a calculus I class, crtcal ponts of a one parameter famly of polynomals wll be nvestgated and a general proof of how to obtan the locus of the { ft( x)} crtcal ponts for each famly wll be shown. Example. If (, ( )) s a crtcal pont of Fnd the crtcal ponts of the famly of polynomals 4 2 f ( x ) = tx x + 6 4 x f t x f '( ) 0 t x = then mples that Then In addton mples that Therefore t Ths shows that every crtcal pont ( x, ft( x)) ( 2 6) 4 2 les on the graph of the polynomal p( x ) = x +

2 Conversely, If s an arbtrary pont on the graph of ( 2 2 6) p( x ) = x + 4 and f x = 0 then (x, p(x)) = (0, 4) whch s also a crtcal pont of for every t. If then (x, p(x)) s a crtcal pont of when as derved above. Therefore the locus of crtcal ponts of the famly s gven by the graph of

22 Fgure () shows the graph of the functon and the locus of crtcal ponts Ths result can be generalzed. Fgure for Example

23 Theorem: If f( x) s a one parameter famly of polynomals t m of the form ft( x) = g( x) + tx where and g(x) s a nonzero polynomal wth no m th degree term, then the locus of crtcal ponts of s the graph of the polynomal p( x) together wth the pont (0, g(0)) n case g(x) has no frst degree term where

24 Proof: Suppose m =. If ( x, ft( x)) s a crtcal pont of then and Therefore f s a crtcal pont of then as requred to show for m =. Conversely, for every x, the pont s a crtcal pont of when Therefore the locus of crtcal ponts s gven by the equaton for m =

25 m 2 Suppose. If s a crtcal pont of Then and Multply both sdes of last equaton by x to get Therefore, f s a crtcal pont of then Conversely, consder an arbtrary pont xg, ( x) g'( xx ) m g'( xx ) xg, ( x) m

26 Three cases should be observed:. If x = 0 and g has no frst degree term then and s a crtcal pont of for all t 2. If x = 0 and g has a frst degree term then g'(0) 0 and (0, g(0)) s not a crtcal pont of any 3. If then s a crtcal pont of when Ths completes the proof. t= g'( x) mx m

27 m From the frst example and the theorem ft( x) = g( x) + tx m and g(x) are gven; t s requred to fnd Note that: Frst, f m=0 n the theorem then g(x) s a polynomal wth no constant term. So the graph of s a vertcal translaton of the graph of g(x). The locus of crtcal ponts of the famly conssts of vertcal lnes passng through the crtcal ponts of g. Second, f the degree of s n, then the locus of crtcal ponts p(x) s a polynomal wth and

28 Example 2 Gven Requred to Fnd t=0,,2,...,0 m=? m=3 g x x x ( ) = 4 6 2+ 4 p( x) = g( x) g'( x) x m p x 3 4 2 ( ) = ( x 6x + 2)

Fgure for Example 2 29

30 Example 3 Gven Requred to Fnd f ( ) ( ) m t x = g x + tx f = + + ( ) 4 2 t x x tx 4 t=, 2,..., 0 m=? m=2

Fgure for Example 3 3

32 Example 4 Gven Requred to Fnd m=? g( x) m=2 g x x x ( ) = 2 3+ + 3

Fgure for Example 4 33

34 V. An Overlooked Calculus Queston Eugene Couch of the Unversty of Calgary, n Alberta, Canada, notes that when ntroducng and teachng exponental functons of the form f(x) = a x and ther nverse functons g(x) = log a x the graphs that are drawn n most Precalculus and Calculus texts gve the mpresson that these two functons do not ntersect. Let us then consder the two questons: ) Do the graphs of f(x) = a x and g(x) = log a x for a >, ever ntersect? 2) If they do ntersect, for what bases a and for what values of x does ths happen? Clearly they must ntersect for some value of a >. If a s suffcently close to, such as a =., the graph of f(x) = a x stays close to for x as large as we lke. So t must be ntersected by y = ln x and also by g(x) = log a x because g(x) = log a x les above y = ln x for x > when a < e.

35 Ths graph shows f(x) = a x =. x ntersectng g(x) = log. x at x =. In addton, t shows y = ln x ntersectng f(x) = a x =. x and g(x) = log. x. Note that g(x) = log. x s above y = ln x for a =. < e. Fgure 5: a =., a < e

36 Observe algebracally that a x > x mples a x > x > log a x, and a x < x mples a x < x < log a x Furthermore, f a x = x then log a x = x and the graphs of f(x) = a x and g(x) = log a x ntersect at x Thus the graphs of f(x) = a x and g(x) = log a x ntersect at x f and only f a x = x whch s equvalent to a = x /x

37 The three graphs shown below are: y = x /x, y = a =.25 and y = where < a < e /e =.4446 Fgure 6 From the graph, t s evdent that for < a < e /e =.4446 f(x) = a x and g(x) = log a x ntersect exactly twce and exactly once at x = e for a = e /e

38 The above analyss gves rse to the followng three cases for a Case : < a < e /e there are two Intersecton ponts. Fgure 7: Graphs of y = a x y = log a x and y = x for a =.25 Intersecton ponts are: (x=.3522, y=.3522) and (x = 0.56527, y =0.56527)

39 Case 2. For a = e /e =.4446 there s one Intersecton pont. Fgure 8: Graphs of y = a x y = log a x and y = x for a =.4446 Intersecton pont s at (x = e, y = e) Note that n cases and 2 all ntersecton ponts le on the y = x lne.

40 Case 3: For a > e /e (a =.6 >.4446 ) there are no ntersectons Fgure 9: Graphs of y = a x y = log a x and y = x for a =.6

4 Calculus Classroom Capsules Sesson S86 References. Data for Tables (), (2) and (3) are from Pascal s Trangle 2. Noah Samuel Brannen and Ben Ford, The College Mathematcs Journal, Vol. 35. No. 5, November (2004), p. 388-390. 3. Jorge Sarmento, MATYCNJ Presentaton, Aprl 202, County College of Morrs 4. Elas Y. Deeba, Denns Rodrguez, Ibrahm Wazr, The College Mathematcs Journal, Vol.27, No. 4 September 996 5. Eugene Couch, Classroom Capsules, The College Mathematcs Journal, Vol. 33. No. 5, November 2002