1. What calculator(s) are you using to solve the problems on this paper? = e accurate to four significant digits

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Cameron Hood
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1 Math A Calculus Chapter Take-home problems Name Show your work for credit. Do ot copy other peoples work. Describe calculator use explicitly. What buttos did you push to get your results?. What calculator(s) are you usig to solve the problems o this paper?. Suppose a particle is movig alog the curve Approximate to four sigificat digits. y x e x /8 dx dy = so that =. Fid dt dt whe x =.. Fid the coordiates of the iflectio poits for. Estimate the left-most coordiates o the curve for these coordiates? y x x /8 = e accurate to four sigificat digits =, = + l. Ca you fid exact values x t t y t t 5. Use a graph to estimate the value of the limit. The use L Hospital s rule to fid the exact value sec( x) lim ta x. value: x π / 6. Fid the coordiates of the poit o curve y arcta ( x) = closest to the poit (0,). 7. Fid the value(s) of x o the iterval [0,] that satisfy the coclusios of the mea value theorem y = x x x. Approximate to sigificat digits. for 8. Ivestigate Newto s method for the family of cubic polyomials, f ( x) = ( x+ )( x + c) a. Ca you fid a two-cycle i the case where c =? b. For values of c betwee 0 ad 0., what dyamics do you observe from the iitial value x =?

2 Math A Calculus Chapter Take-home problems Solutios. What calculator(s) are you usig to solve the problems o this paper? SOLN: I ll be usig a variety icludig TI8, TI8, TI85, TI86, TI89, TI9. Should have a HP.. Suppose a particle is movig alog the curve y x e x /8 dx dy = so that =. Fid whe x =. dt dt Approximate to four sigificat digits. x /8 x /8 dy x /8 x x /8 e SOLN: y = x e = e e = ( x ) dx x /8 dy dx dy dx e x /8 so that = = ( x ) = e ( x ). Whe x =, dt dt dx dt dy 8/8 = e ( 8) = 79e 0.5. So who eeds a calculator? Well, you ca approximate this usig dt a TI8 to get dy/dt approximately Fid the coordiates of the iflectio poits x x for /8 y = e accurate to four sigificat digits dy x x /8 x x x /8 = e = e dx d y x x x /8 x x /8 SOLN: = e e dx x x x /8 = e x chages sig where = x = ±. Thus the coordiates of the iflectio poits are (, e ) /, e, 6, e / 6, e = =. Agai, who eeds a calculator? Oh, to estimate accurate to four sigificat digits, a calculator is hady. The TI8 yields so the poits are (,.8) ad (6,.8). To be sure, here s a graph as displayed o the TI9+: (baselie alog y=.). Estimate the left-most coordiates o the curve x = t t, y = t+ lt. Ca you fid exact values for these coordiates? SOLN: At the left-most coordiates we ll dx dx / dt t t t t = 0 = = have dy dy / dt + / t t + = t t = 0 ad

The use L Hospital s rule to fid the exact sec( x) lim ta x.

3 sice t = 0 is ot i the domai of y, it must be where t = so that (x, y) = (,). Well, dar who eeds these stikig calculators? They re good for a graph. Below we see the trace is early to the far left poit, which corroborates our previous result covicigly: 5. Use a graph to estimate the value of the limit. The use L Hospital s rule to fid the exact sec( x) lim ta x. value value: x π / SOLN: Graphig the fuctio o the iterval (0, π / ) ad tracig (it starts i the middle) we see the limit appears to be a local max ear (0.79, 0.68) This is a situatio, so we look at sec x l ta lim l y = lim sec( x) l ( ta x) = lim = lim = lim x π/ x π/ x π/ cos x x π/ si x x π/ si x si xcos x ( x) ta x = lim =, which meas that y = , sure eough! x π / si ( x) e 6. Fid the coordiates of the poit o curve y = arcta ( x) closest to the poit (0,). SOLN: The situatio is depicted at right. Evidetly the poit earest (0,) is where the ormal to the curve passes through (0,). The y = + a so slope of the taget at x=a is the slope of the ormal is y = ( + a ) so the equatio of the ormal lie through (0,) ad (a, aarcta(a)) is y ( a ) x the ormal actually passes through the poit of ormalcy, which is true iff = +. Requirig that aarcta a = + a a= a a+, aarcta a + a + a = 0. This is a equatio ot easily solvable by had. We could try Newto s method. You d iterate might fid it more user-friedly to use the solve feature o the TI89 as show at right, where we see x 0.85 with a omious warig about more solutios possibly existig (obviously there are t more.) Usig zoom square gives a more reasoable picture that this is, i fact, where the earest poit to (0,) o the curve is. x + = x ( x ) x x x x x arcta + + arcta x + x. Yikes. You

4 7. Fid the value(s) of x o the iterval [0,] that satisfy the coclusios of the mea value theorem y = x x x. Approximate to sigificat digits. for SOLN: Solve for x: f f 0 f ' x = x x = 5 x x=. Usig Tartaglia s 0 method we ote that, for all a ad b, a b + ab a b = a b so let x = a b ad set 6 ab = b = ad substitute ito a b = a = a a = to 6a 6a which we add to complete the square ad get a = + = = whece a = a= This meas that b = 6 6 whece x= a b= ad usig the TI9+, we see that x is approximately. Alteratively we could use Newto s method to fid a zero of x x 6= 0 by iteratig f ( x ) x x 6 8x + 6 x+ = x = x = f ' x x x To do this o the TI8, eter i the fuctio o the Vars page by hittig the Y= butto ad expressig the formula as show i the first scree shot at left. The quit ad o the home page store a iitial guess i x usig the STO key Ad the hit the Vars butto to brig up the meu to fid Y ad place it o the home page by arrowig over/dow to highlight ad pressig eter. The hit the Sto key ad store the output of Y back to the iput as show i the secod scree capture at right. The just keep hittig eter. As you ca see, Newto s method the coverges rapidly to the same value we got by the more itricate method above. Fially, if you like the Kraft Cheese method, you could use the solve feature o, say, the TI85. O the 85 you hit d +GRAPH to access the solver meu ad the type i the formula i the exp field ad set its value to zero as show i the scree capture at right. The put a iitial gues i the field for x ad, with the cursor still i that field, hit F5 (SOLVE). The result gives you two more digits tha the TI9+. GO 85!

5 8. Ivestigate Newto s method for the family of cubic polyomials, f ( x) = ( x+ )( x + c) a. Ca you fid a two-cycle i the case where c =? f x x + x + cx + c SOLN: Newto s method iterates x x x f x x + x + c x + x + c ( x + x c) + = = = ' Ok, have t used the 86 yet, so o the 86 eter the iteratio formula as y (first scree capture below.) The store - ito C usig the STO butto. You ca the start experimetig with various iitial values. If you start with x =, the you ll stay there (a cycle. ) Startig at x = 0.5 first pushes the iterates beyod ad the they decrease, approachig from above ( rd scree shot below.) To get some idea of the dyamics of this process, look at a graph of the fuctio whose zeros we seek. By ispectio, we see that the iitial value will be somewhere betwee the local extrema d ± 7 f '( x) = ( x + x x ) = x + x = 0 x= ad.59 dx So we start experimetig with iteratig Newto s formula with various iitial values. Below we see x = 0 immediately goes to the zero at (first scree capture below.) Nudgig the iitial value a bit to the left, x = 0. we get iterates that iitially oscillate but the settle i a path icreasig steadily towards (secod scree capture below.) Nudgig a otch further to the left, x = 0., the ext iterate eds up to the right of the local max ad the advaces steadily towards the zero at x= (this is the third scree capture below.) Experimetig with various values betwee 0. ad 0., we get various results covergig towards either or, but as we zero i o what seems to lead to a two-cycle we see that x = 0.0 actually eds up covergig to x =. Looks we could use some heavier gus. This could be a job for the TI9+! So for a two-cycle N N x = x. O the TI9+ we ca eter the formula for the Newto s method we wat ( ) iteratio ad put i c as we did o the TI86, but with the extra computer algebra features of N N x = x. This the TI9+, we ca actually use the calculator to approximate solutios to ( ) is show i the sequece of TI9+ scree captures which follow.

6 Evidetly, either x = or x =.5880 leads to the two-cycle oscillatig from the first back to the secod or vice-versa. Let s go back to the TI86 ad see if this actually works. Well, as the scree captures below show, it almost works: the trouble is the two-cycle is ustable, so ay slight deviatio from the perfect iitial value will evetually wobble off the cycle, as show i the sequece of values show below. Iitially it shows ice oscillatory behavior, but the errors compoud ad it eds up attractig to -. So there is a twocycle, but it s ustable ad sice our iitial value is ecessarily a approximatio, it will evetually wobble away. b. For values of c betwee 0 ad 0., what dyamics do you observe from the iitial value x =? SOLN: This is a ope eded questio ad a ivitatio to experimet. Let s start with c = 0.: Here we see quite a bit of roamig about before settlig i o the zero at. How about c = 0.5: With c = 0.5

7 With c = 0.9 So we experimet with various values util we discover this iterestig zoe close to c = 0. where we fid a attractive -cycle with c = 0.96: A attractive -cycle with c = 0.9: Ad, zouds! A attractive 8-cycle with c = 0.88

AP Calculus BC Review Applications of Derivatives (Chapter 4) and f,

AP Calculus BC Review Applications of Derivatives (Chapter 4) and f, AP alculus B Review Applicatios of Derivatives (hapter ) Thigs to Kow ad Be Able to Do Defiitios of the followig i terms of derivatives, ad how to fid them: critical poit, global miima/maima, local (relative)