Math 34A. Final Review

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Benjamin Shepherd
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1 Math A Final Rviw

1) Us th graph of y10 to find approimat valus: a) 50 0. b) y (0.65) solution for part a) first writ an quation: 50 0. now tak th logarithm of both sids: log() log(50 0.

2 1) Us th graph of y10 to find approimat valus: a) b) y (0.65) solution for part a) first writ an quation: now tak th logarithm of both sids: log() log(50 0. ) pand th right sid using log ruls: log() 0.log(50) 0.[log(5)log(10)] 0.log(5)0.(1) now look up log(5) on th graph it is about 0.7 log() 0.(0.7) finally w tak th antilog of both sids: antilog(0.51) Look this up on th graph: solution for part b) us th graph to find points on th tangnt lin at 0.65 thn calculat th slop as usual th slop should b about 10. (10 is clos nough)

3 ) Intgrat: a) c) b) C C d d π π π π π π ( ) ) ( 6 6 b a k a b ka a kb b k k d b a b a ( ) ( ) ( ) ( ) 0 (.5) (.5) ) (0.5 ) 1 ( ln(.5) ln(.5) ln() ln() ln() ln(.5) ln() ln(0.5) dt t t t t

4 ) Th population of a crtain town is incrasing at a rat of 5 popl pr month ( rprsnts th numbr of months). If th population is currntly,000 popl, by how much will th population incras during th nt 8 months? Option 1: Us a dfinit intgral Option : Us an indfinit intgral Dfin p() population of town at tim months W ar givn p () 5. Intgrat this btwn 0 and 8: 8 ( 5 ) 0 d [ 5(8) (8) ] [ 5(0) (0) ] Th population will incras by 10 popl Dfin p() population of town at tim months W ar givn p () 5 and p(0),000 Intgrat to find a formula for p(): p( ) p'( ) d 5 d ( ) p( ) 5 C Now w hav to find th constant C by plugging in th initial valu: p(0),000 5(0) (0),000 C p( ) 5,000 C Now plug in 8: p(8) 5(8)(8),000,10 Th incras in th population is 10 popl.

5 ) Find valus of whr th function f() a) is incrasing b) is concav down c) has a minimum point W will nd drivativs of f(): f '( ) f ''( ) 1 ( ) f() will b incrasing whr f () is positiv st f () 0 to dtrmin whr f() might switch from incrasing to dcrasing, or vic-vrsa: or or 0 f ngativ f ngativ f positiv f() will b concav down whr f () is ngativ st f () 0 to dtrmin whr f() might switch concavity from up to down, or vic-vrsa: ( ) or 0 f positiv f ngativ f positiv 0 Again, chcking points in ach rgion shows that f() is concav down for 0<< 0 Th minimum point occurs whn. At this point, f ()0 and f () is positiv. A numbr lin can hlp chcking points in ach rgion shows us that f() is incrasing whn >

6 5) A thatr currntly slls tickts at a pric of $8, and slls 1000 tickts. It is stimatd that for ach 10-cnt incras in th pric, 0 fwr tickts will b sold. What is th optimal pric for th thatr to charg (to maimiz thir incom)? W want to maimiz rvnu, so w nd a formula for rvnu in trms of pric. Th basic formula is somthing lik Rvnu (Pric)(#of tickts sold) Dfin variabls: f()rvnu whn pric is $ pr tickt W will nd a formula for #of tickts sold in trms of pric, so dfin y# of tickts sold. Th ky to this typ of problm is ralizing that th formula for # of tickts will b linar. W alrady know on point on th lin: (8,1000) this is th information givn in th first sntnc. W can find anothr point on th lin by using th givn info about th pric incrass: for ampl, if th pric is $9, 00 fwr tickts will b sold, so (9,800) should b anothr point on th lin. Now w just nd to us ymb to find th quation of th lin. Th slop is using (8,1000) w gt: (8) b b 600 y This is our quation for th # of tickts. Putting this into our rvnu quation: f ( ) ( )( ) f ( ) To find th maimum, w st f () 0: f '( ) Th optimal pric to charg is $6.50

7 6) Givn th two functions f() 8 and g() - 5 a) Find th quation of th lin tangnt to f() at b) For which valu of is th slop of f() qual to th slop of g()? answr for a) Th slop of th tangnt lin will b th drivativ at : f '( ) 6 8 f '() 6() W nd a point on th lin as wll. Using, w find f()() -8()-. Thus (,-) is on th lin. Now just us point-slop form for th quation of th lin: y ( ) ( ) y 8 y 1 answr for b) W nd th first drivativ of ach function to b qual: f '( ) 6 8 g'( )

8 7) A rock is thrown from th top of a 00-mtr high cliff. Its vlocity (in mtrs pr scond) is givn by th formula v(t) 0 10t, whr t is th tim in sconds aftr th rock is thrown. a) Whn dos th rock hit th ground? b) What is th spd of th rock at th tim found in part a)? W can us an indfinit intgral, along with th initial hight of 00 mtrs, to find a formula for position: ( 0 10t) f ( t) dt f ( t) 0t 5t C f (0) 00 0 f ( t) 5t 0t 00 ( 0) 5( 0) C C 00 Th rock will hit th ground whn f(t) 0: 5t 0t 00 0 t t 60 0 ( t 10)( t 6) 0 or t 10 0 t 6 0 or t 10 t 6 Th rock hits th ground at t10 sconds (t-6 sconds dosn t mak sns in this contt) To gt th spd, just plug into th original vlocity formula: v(10) 0-10(10) -80 Th spd is 80 mtrs pr scond

9 8) Th tabl givs th position of a car at various tims during an aftrnoon driv. a) Find th avrag spd of th car btwn pm and pm. b) During what tim intrval was th avrag spd gratst? Tim 1 pm 1:0 pm pm pm pm :0 pm Position 0 mi. 0 mi. 5 mi. 70 mi. 105 mi. 10 mi. a) This is just a slop problm: mi. Avg. Spd 0 hr. mi hr b) For this part, just find th avrags for ach intrval (sam mthod as part a) Th largst valu is from 1:0 to pm mi. Avg. Spd 50 : 00 1:0 1 hr. mi hr

10 9) A cylindrical containr with no top is to b mad to hold a volum of 1π cm. Matrial for th sid costs cnts pr cm and th matrial for th bottom costs cnts pr cm. What ar th dimnsions of th last pnsiv containr? W nd formulas for th Volum and th Cost: πr r Volum π(r )(h) 1π (w will us this latr) Cost ()(ara of sid) ()(ara of bottom) C ()(πrh) ()(πr ) h sid bottom r h This has variabls, so w nd to rplac on of thm using th volum quation: 1 solving for h: h r 1 8π now substitut into th cost quation: C( r) π r πr πr r r W nd to st th drivativ of this cost function qual to zro to find th minimum: 8π C'( r) 6πr 0 r 8π 6πr r 8 r r plug this back in to find h: h 1 Radius cm Hight cm

11 10) Th rat (in kilowatts) at which powr is usd in a houshold is shown in th graph blow. Find th total nrgy usd btwn 1pm and 6pm. A1 A A Sinc th graph givs th rat, w just nd to find th ara undr th graph btwn 1pm and 6pm. Brak it up into as many pics as you nd I will us rgions (labld A1, A and A): Ara of A1 (hr)(1.5kw kw)/.5 Kwh Ara of A (hr)(kw) Kwh Ara of A (hr)(kw kw)/ 5 Kwh Answr: 1.5 kilowatt-hours

12 11) Givn m(t) 5t k π t - a) Find m (t) b) Find m () m'( t) 15 5t t notic that k π is constant, so its drivativ is 0 m''( t) 75 5t m''() t 6()

1973 AP Calculus AB: Section I

1973 AP Calculus AB: Section I 97 AP Calculus AB: Sction I 9 Minuts No Calculator Not: In this amination, ln dnots th natural logarithm of (that is, logarithm to th bas ).. ( ) d= + C 6 + C + C + C + C. If f ( ) = + + + and ( ), g=