1. (16 points) Answer the following derivative-related questions. dx tan sec x. dx tan u = du d. dx du tan u. du tan u d v.

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1 Exam #2 Soluions. (6 poins) Answer he following eriaie-relae quesions. (a) (8 poins) If y an sec x, fin. This is an applicaion of he chain rule in wo sages, in which we shall le u sec x, an sec x: an sec x an u u u an u ( ) ( u an u ) u an u (sec x an x) 2 sec2 u sec x an x 2 sec x sec2 sec x (b) (4 poins) Fin ((ex + 2x) arcsin x). This is an applicaion of he prouc rule: ((ex + 2x) arcsin x) (ex + 2x) arcsin x + (e x + 2x) arcsin x (e x + 2) arcsin x+)e x + 2x) x 2 (c) (4 poins) If g() ln 3, fin g (). This is an applicaion of he quoien rule: g () ln 3 3 ( ln ) ln /3 ( 3 ) 2 3 ln ( 3 ) 2 3 2/3 This problem coul alernaiely be sole as a prouc-rule inocaion on g () ln /3. 2. (2 poins) Calculae cos(x2 e x ). Page of Karika 6

2 Exam #2 Soluions Le u x 2 e x ; hen his is a sraighforwar chain rule inocaion, alhough wih a prouc rule embee wihin i: cos(x2 e x ) cos(u) u u cos(u) ( x 2 e x) ( sin u) ( 2xe x + x 2 e x) ( sin u) ( 2xe x + x 2 e x) ( sin(x 2 e x ) ) 3. (0 poins) Fin an equaion of he angen line o he cure y e x (x 2 3x + ) a (0, ). Using he prouc rule, ex (x 2 3x + ) + e x (2x 3) an specifically when x 0, e0 + e 0 ( 3) 2, so he esire line has slope 2, an hus equaion y 2x + b. Since i passes hrough he poin (0, ) we may specifically eermine ha b, an hus ha b, for an equaion y 2x + of he angen line. 4. (5 poins) The cissoi of Diocles is a cure saisfying he equaion x(x 2 + y 2 ) 4y 2. (a) (2 poins) Fin a formula for on his cure. Taking he eriaie of each sie, we ge ( x(x 2 + y 2 ) ) ( ) 4y 2 (x 2 + y 2 ) + x (x2 + y 2 ) ( ) 4y 2 (x 2 + y 2 ) + 2x 2 + x y2 8y 3x 2 + y 2 + x y2 8y 3x 2 + y 2 + 2xy 8y 2xy 8y 3x2 y 2 (2xy 8y) 3x2 y 2 3x2 y 2 2xy 8y 3x2 + y 2 8y 2xy (b) (3 poins) Fin he equaion of he angen line o he cure a (2, 2). Page 2 of Karika 6

3 Exam #2 Soluions A he specific alues x 2, y 2, we see from he aboe ha ( 2) 2 8( 2) 2 2( 2) Thus we hae a line of he form y 2x + b; plugging in (2, 2) we can sole for b: so y 2x + 2 is he angen line b 2 b 5. (9 poins) Fin he absolue maxima an minima of he funcion f(x) x 3 2x 2 4x + 3 on he ineral [0, 3]. We know ha f (x) 3x 2 4x 4 We hen seek ou criical poins. f (x) is a polynomial, so poins where f (x) oes no exis are no uner consieraion. We also look for alues where f (x) 0. This occurs when 3x 2 4x 4 0, which, eiher by facorizaion or by he quaraic formula can be foun o be he case when x 2 or x 2. x 2 is ousie our ineral, so he only criical poin 3 3 wihin he ineral is x 2. Thus, he only caniaes for minimum an maximum are he criical poin x 2 an he enpoins x 0 an x 3. f(0) , f(2) , an f(3) Of hese, we may hus say wih cerainy ha he highes alue (he absolue maximum) is a x 0, an he lowes alue (he absolue minimum) a x (6 poins) Amy is saning moionless 50 meers eas of a norh-souh roa wih a raar gun, while Bob, who is 20 meers o he norh, is riing souh. The raar gun repors how quickly he isance beween Amy an Bob is changing (which may no be Bob s acual spee). Our seup for his problem inoles a righ riangle beween Bob s car, he place on he roa parallel o Amy, an Amy s posiion well off he roa. The eas-wes leg beween Amy an he roa is a consan lengh of 50, while he lengh of he norh-souh leg an he lengh of he hypoenuse are changing. We may enoe hese respeciely by y an s, an noe ha, by he Pyhagorean Theorem, hey hae he relaionship y s 2. We also know ha y is currenly 20, an can compue in aion ha s is currenly y (a) (2 poins) If Bob is riing souh a 30 meers per secon, wha will he raar repor as he rae of change of he isance beween Bob an Amy? The informaion gien o us here is ha 30 (since y is ecreasing a a rae of 30 meers per secon), an ha we wish o fin s. Using he relaionship escribe in he Page 3 of Karika 6

4 Exam #2 Soluions genral problem, we iffereniae boh sies wih respec o : y s 2 ( y ) s2 ( y ) s s 2y 2s y s s s s2 an hus s (b) (4 poins) Conersely, if he raar repore a change-rae of 25 meers per secon, wha woul Bob s acual spee be? In his case we are gien s ±25, an aske o fin. We may sar wih he secon-o las line of he eriaion seen in he preious secion: an sole for s s ±25 30 y 20 ± 325 s 2y 2s 7. (9 poins) Esimae he following alues using appropriae linear approximaions. 2. (a) (4 poins) (.993) 4. We consier he funcion f(x) x 4, whose eriaie is f (x) 4x 3. For x close o 2 (as.993 is), we can use he linear approximaion: f(x) f( 2) + (x + 2)f ( 2) Since f( 2) 6 an f ( 2) 32, i follows ha f(.993) ( 32) For purposes of comparison, he acual alue of (.993) 4 is aroun (b) (5 poins) We consier he funcion f(x) x, whose eriaie is 2. For x close o 25 (as x is), we can use he linear approximaion: f(x) f(25) + (x 25)f (25) Since f(25) 5 an f (25) 2, i follows ha 25 0 f(25.07) 5 + (0.07) For purposes of comparison, he acual alue of is aroun Page 4 of Karika 6

5 Exam #2 Soluions 8. (3 poins) If f(x) arcan 3x x 4 +2, hen fin f (x). We shall nee he quoien rule immeiaely, an will eenually nee recourse o he chain rule; preempiely we shall efine u 3x an x 4 + 2, which we will nee laer. Then: f (x) arcan u ( arcan u) arcan u ( ) 2 ( u arcan u) arcan u u ( 3x) arcan u ( +u 2 x4 + 2 ) 2 x (3x) 2 arcan(3x) 4x 3 2 x 4 +2 x x x 2 4x3 arcan(3x) 2 x 4 +2 x The las line of he aboe calculaion is cleanup an is opional. 9. (6 poin bonus) Currenly Yee is 0 miles norh of he Library of Babel, walking souh a 3mph, while Zachary is mile eas of he Library, walking eas a 5mph. How soon will i be he case ha he isance beween hem is (if only momenarily) unchanging? This can be sole eiher purely as a eriaie calculaion, or as a relae raes problem. The key obseraion is ha we are inerese in arious properies a imes oher han he presen, so i woul be erroneous o work uner he presumpion ha Yee an Zachary s isances from he library are specifically 0 an mile respeciely; insea, in hours, heir isances will be 0 3 an + 5 respeciely. To sole his purely as a eriaie, one coul hen compue he isance beween hem a ime o be s (0 3) 2 + ( + 5) an hen using he chain rule, s an since we wan o know when s is zero, we se his expression o zero an sole for, geing Alernaiely, we coul le Yee s an Zachary s isances from he library be gien by he ariables y 0 3 an z + 5, an by he furher known eriaies 3 an Page 5 of Karika 6

6 Exam #2 Soluions z 5. We know y2 + z 2 s 2, so iffereniaing boh sies wih respec o : ( y 2 + z 2) s2 z z2 s y2 + z s s2 z s 2y + 2z 2s 2 ( 3)(0 3) + 2 5( + 5) s 2s an since we are inerese in when s wih soluion , his equaion can be simplifie o , Page 6 of Karika 6