Example 3 Find the tangent line to the following function at.

Transcription

1 Emple Given n fin eh of the following. () () () () Emple Given fin. Emple Fin the tngent line to the following funtion t.

2 Now ll tht we nee is the funtion vlue n erivtive (for the slope) t. The tngent line is then, Emple 4 Differentite eh of the following funtions. () () () () Emple 5 Evlute the following inefinite integrl. Emple 6 Integrte.

3 Emple 7 Evlute eh of the following integrls. () () () () Emple 8 Determine the verge vlue of eh of the following funtions on the given intervl. on

4 Emple 9 Determine the numer tht stisfies the Men Vlue Theorem for Integrls for the funtion on the intervl [,4] Clerly the seon numer is not in the intervl n so tht isn t the one tht we re fter. The first however is in the intervl n so tht s the numer we wnt. Emple 0 Determine the re of the region enlose y,,, n the y-is.

5 Emple evlute the inite limits, if they eist. () () ()

6 Emple Ans 0 E Determine the seon erivtive of Ans E 4 Determine the fourth erivtive of E 5) Fin y impliit ifferentition for

7 E 6) Critil Points Determine the ritil points of. EX 7 Determine the ritil points of Rell tht ritil points re simply where the erivtive is zero n/or oesn t eist. In this se the erivtive is rtionl epression. Therefore we know tht the erivtive will e zero if the numertor is zero (n the enomintor is lso not zero for the sme vlues of ourse). We lso know tht the erivtive won t eist if we get ivision y zero. So, ll we nee to o is set the numertor n enomintor equl to zero n solve. Note s well tht the - we ftore out of the numertor will not ffet where it is zero n so n e ignore. Likewise, the eponent on the whole enomintor will not ffet where it is zero n so n lso e ignore. This mens we nee to solve the following two equtions. As we n see in this se we neee to use the qurti formul oth of the qurti equtions. Rememer tht not ll qurtis will ftor so on t forget out the qurti formul!

8 Now, rell tht we on t use omple numers in this lss n so the solutions from where the enomintor is zero (i.e. the erivtive oesn t eist) won t e ritil points. Therefore, the only ritil points of this funtion re, EX 8. Use L Hospitl s Rule to evlute n so this is form tht llows the use of L Hospitl s Rule. So, t this point let s just pply L Hospitl s Rule. At this point ll we nee to o is try the limit n see if it n e one. So, the limit n e one n we one with the prolem! The limit is then, E 9:

9

10 Clulus Chet Sheet Integrls Definitions Definite Integrl: Suppose f ( ) is ontinuous Anti-Derivtive : An nti-erivtive of f ( ) on [, ]. Divie [, ] into n suintervls of is funtion, F( ), suh tht F f. * with D n hoose i from eh intervl. Inefinite Integrl : f F + * Then f limâ f ( i ) D. where F( ) is n nti-erivtive of f ( ). næ i Funmentl Theorem of Clulus Prt I : If f ( ) is ontinuous on [, ] then Vrints of Prt I : u g f () t t is lso ontinuous on [, ] f () t t u f u È Î n g f () t t f. f () t t v f v - È v Î Prt II : f is ontinuous on[, ], F( ) is u f () t t u f u v f v - v F f ) n nti-erivtive of f ( ) (i.e. f F F. then - f ± g f ± g ± ± f g f g 0 f - f f f ( ) f ( ) f ( ) + for ny vlue of. If f g on then If f 0 on then f 0 Properties f f [ ] [ ], is onstnt f f, is onstnt f g - f f If m f M on then m( - ) f M ( -) Clulus Chet Sheet Stnr Integrtion Tehniques Note tht t mny shools ll ut the Sustitution Rule ten to e tught in Clulus II lss. g u Sustitution : The sustitution u g will onvert f ( g ) g f ( u) u using g u g. For inefinite integrls rop the limits of integrtion. E. 5 os u fi u fi u :: 8 fi u fi u Integrtion y Prts : uv uv- vu n sin( u) ( sin( 8) -sin() ) os os u u uv uv - vu. Choose u n v from integrl n ompute u y ifferentiting u n ompute v using v v. - E. e u v e - fi u v -e - e - - e - + e - -e - - e - + E. 5 ln u ln v fi u v ( ) ln ln - ln - 5ln 5 -ln - Prouts n (some) Quotients of Trig Funtions n m n m For sin os we hve the following : For tn se we hve the following :. n o. Strip sine out n onvert rest to osines using sin - os, then use the sustitution u os.. m o. Strip osine out n onvert rest to sines using os - sin, then use the sustitution u sin.. n n m oth o. Use either. or. 4. n n m oth even. Use oule ngle n/or hlf ngle formuls to reue the integrl into form tht n e integrte.. n o. Strip tngent n sent out n onvert the rest to sents using tn se -, then use the sustitution u se.. m even. Strip sents out n onvert rest to tngents using se + tn, then use the sustitution u tn.. n o n m even. Use either. or. 4. n even n m o. Eh integrl will e elt with ifferently. sin sin os os os sin - os Trig Formuls :, ( + ), ( ) k k+ n n+ n n+ - ln ln + lnuu uln ( u) - u+ +, - + u u e u e Common Integrls osuu sinu+ Visit for omplete set of Clulus notes. sinuu - osu+ se uu tn u+ seutnuu seu+ suotuu - su+ s uu - ot u+ tnuu ln seu + seuu ln seu+ tn u + u tn + u - u + u - u sin u Pul Dwkins E. 5 tn se 4 ( se se ) tnse 4 ( u ) uu ( u se ) 5 4 tn se tn se tn se - - se - se Visit for omplete set of Clulus notes. E. sin5 os 5 4 sin sin sin (sin ) sin os os os (- os ) sin os (-u) u u4 - u u u ( os ) u u se + ln os - os Pul Dwkins

11 Clulus Chet Sheet Trig Sustitutions : If the integrl ontins the following root use the given sustitution n formul to onvert into n integrl involving trig funtions. - fi sinq os q - sin q 6 E. 4-9 sin q fi osq q 4 4sin 4os q q os q Rell. Beuse we hve n inefinite integrl we ll ssume positive n rop solute vlue rs. If we h efinite integrl we nee to ompute q s n remove solute vlue rs se on tht n, Ï if 0 Ì Ó - if < 0 In this se we hve 4-9 osq. - fi seq tn se P Q q q - Û ı ( os ) ( os ) + fi tnq se q + tn q 6 4 sin q q sin q 9 q q q s q - otq + Use Right Tringle Trig to go k to s. From sustitution we hve sinq so, From this we see tht ot 4-9 q. So, Prtil Frtions : If integrting where the egree of P( ) is smller thn the egree of Q( ). Ftor enomintor s ompletely s possile n fin the prtil frtion eomposition of the rtionl epression. Integrte the prtil frtion eomposition (P.F.D.). For eh ftor in the enomintor we get term(s) in the eomposition oring to the following tle. Ftor in Q( ) Term in P.F.D Ftor in Q( ) + A + A+ B ( + ) k ( + + ) k Term in P.F.D A A A k + + L A + B A k + Bk + L k k Clulus Chet Sheet Applitions of Integrls Net Are : represents the net re etween f f n the -is with re ove -is positive n re elow -is negtive. Are Between Curves : The generl formuls for the two min ses for eh re, Èupper funtion Èlower funtion Î Î & fi Îright funtion - Îleft funtion y f fi A - È È f y A y If the urves interset then the re of eh portion must e foun iniviully. Here re some skethes of ouple possile situtions n formuls for ouple of possile ses. A f ( y) -g( y) y A f -g A f g g f Volumes of Revolution : The two min formuls re V A n V A y y. Here is some generl informtion out eh metho of omputing n some emples. Rings Cyliners A p ( ) outer rius - inner rius A p ( rius) ( with / height) Limits: /y of right/ot ring to /y of left/top ring Limits : /y of inner yl. to /y of outer yl. f, f y, f y, f, Horz. Ais use g( ), A( ) n. Vert. Ais use g( y ), A( y ) n y. Horz. Ais use g( y ), A( y ) n y. Vert. Ais use g( ), A( ) n. E. Ais : y > 0 E. Ais : y 0 E. Ais : y > 0 E. Ais : y 0 7+ ( - )( + 4) E ( - )( + 4) ( ) 4ln - + ln tn Here is prtil frtion form n reomine. A( 7+ A B+ C + 4) + ( B+ C)( -) + ( - )( + 4) ( - )( + 4) Set numertors equl n ollet like terms. 7 + A+ B + C - B + 4A- C Set oeffiients equl to get system n solve to get onstnts. A+ B 7 C- B 4A- C 0 A 4 B C 6 outer rius : - f inner rius : - g outer rius: + g inner rius: + f rius : - y with : f ( y) - g( y) rius : + y with : f ( y) - g( y) An lternte metho tht sometimes works to fin onstnts. Strt with setting numertors equl in 7 + A B+ C -. Chose nie vlues of n plug in. previous emple : For emple if we get 0 5A whih gives A 4. This won t lwys work esily. These re only few ses for horizontl is of rottion. If is of rottion is the -is use the y 0 se with 0. For vertil is of rottion ( > 0 n 0 ) interhnge n y to get pproprite formuls. Visit for omplete set of Clulus notes. 005 Pul Dwkins Visit for omplete set of Clulus notes. 005 Pul Dwkins

12 Work : If fore of F moves n ojet in, the work one is W Clulus Chet Sheet F Averge Funtion Vlue : The verge vlue of f ( ) on is f vg f ( ) - Ar Length Surfe Are : Note tht this is often Cl II topi. The three si formuls re, L s SA p ys (rotte out -is) SA p s (rotte out y-is) where s is epenent upon the form of the funtion eing worke with s follows. y s + if y f, s + y if f y, y y y () () s + t if f t, y g t, t t t r if, s r + q r f q q q With surfe re you my hve to sustitute in for the or y epening on your hoie of s to mth the ifferentil in the s. With prmetri n polr you will lwys nee to sustitute. Improper Integrl An improper integrl is n integrl with one or more infinite limits n/or isontinuous integrns. Integrl is lle onvergent if the limit eists n hs finite vlue n ivergent if the limit oesn t eist or hs infinite vlue. This is typilly Cl II topi. Infinite Limit. lim t f f tæ. lim - tæ- t. f f + f provie BOTH integrls re onvergent. - - Disontinuous Integrn f f t. Disont. t : f lim f +. Disont. t : lim -. Disontinuity t tæ t f f < < : f ( ) f ( ) f ( ) + provie oth re onvergent. Comprison Test for Improper Integrls : If f g 0 on [, ) then,. If f onv. then g ( ) onv.. If g ( ) ivg. then f Useful ft : If > 0 then For given integrl f ( ) p onverges if p > n iverges for p. tæ ivg. Approimting Definite Integrls - n n (must e even for Simpson s Rule) efine D n n ivie [, ] into n suintervls [ 0, ], [, ],, [, n n ] * * * Mipoint Rule : f ªD Èf + f + + f ( n) - with 0 L, n n then, * Î i, i- i D f ª Èf 0 + f ++ f + + f n- + f n Î L D f ª Èf 0 + 4f + f + + f - + 4f - + f Î L is mipoint [ ] Trpezoi Rule : Simpson s Rule : n n n Visit for omplete set of Clulus notes. 005 Pul Dwkins

13 Clulus Chet Sheet Derivtives Definition n Nottion f ( + h) f If y f then the erivtive is efine to e f lim. h 0 h If y f then ll of the following re equivlent nottions for the erivtive. f y f y f Df If y f then,. m f is the slope of the tngent line to y f t n the eqution of the tngent line t is y f + f. given y If f ( ) n. ( f ) f If y f ll of the following re equivlent nottions for erivtive evlute t. f y f y Df Interprettion of the Derivtive f is the instntneous rte of. hnge of f ( ) t.. If f ( ) is the position of n ojet t time then f is the veloity of the ojet t. Bsi Properties n Formuls g re ifferentile funtions (the erivtive eists), n n re ny rel numers,. ( f ± g) f ± g. fg f g+ fg Prout Rule f f g fg 4. Quotient Rule g g ( ) ( sin) os ( os) sin ( tn) se ( se) setn n n n Power Rule ( f ( g )) f g g This is the Chin Rule 5. ( ) 0 Common Derivtives ( s) sot ( ot) s ( sin ) ( os ) ( tn ) ( ) ln ( ) ( e ) e ( ln ), > 0 ( ln ), 0 ( log ), > 0 ln Clulus Chet Sheet Chin Rule Vrints The hin rule pplie to some speifi funtions. n n. ( f ) n f f 5. ( os f ) f sin f f f. ( e ) f e 6. ( tn f ) f se f f. ( ln f ) 7. se [ f ] f se f tn f f f 4. ( sin f ) f os f 8. ( tn f ) + f [ ] [ ] Higher Orer Derivtives The Seon Derivtive is enote s The n th Derivtive is enote s ( ) f ( f f n is efine s ) n n f f n is efine s n f ( f ( )), i.e. the erivtive of the ( n ) ( n ) f ( f ), i.e. the erivtive of first erivtive, f. the (n-) st n erivtive, f. Impliit Differentition + y sin y + y y here, so prouts/quotients of n y 9y Fin y if e. Rememer will use the prout/quotient rule n erivtives of y will use the hin rule. The trik is to ifferentite s norml n every time you ifferentite y you tk on y (from the hin rule). After ifferentiting solve for y. e ( y ) y yy ( y) y e y y e e os + 9y y + y + yy y y + y 9y 9y e 9 os y 9 os y y y Critil Points 9 9 is ritil point of f. f 0 or. f Inresing/Deresing Conve Up/Conve Down oesn t eist. provie either Inresing/Deresing. If f > 0 for ll in n intervl I then f ( ) is inresing on the intervl I. < for ll in n intervl I then. If f 0 f ( ) is eresing on the intervl I. for ll in n intervl I then. If f 0 f ( ) is onstnt on the intervl I. e y y e 9y 9y 9 os ( y) Conve Up/Conve Down. If f > 0 for ll in n intervl I then f ( ) is onve up on the intervl I. < for ll in n intervl I then. If f 0 f ( ) is onve own on the intervl I. Infletion Points is infletion point of f onvity hnges t. if the Visit for omplete set of Clulus notes. 005 Pul Dwkins Visit for omplete set of Clulus notes. 005 Pul Dwkins

14 Asolute Etrem. is n solute mimum of f ( ) if f f for ll in the omin. is n solute minimum of f ( ). if f f for ll in the omin. Fermt s Theorem f hs reltive (or lol) etrem t If, then is ritil point of Clulus Chet Sheet f. Etreme Vlue Theorem f is ontinuous on the lose intervl If [, ] then there eist numers n so tht,.,,. f ( ) is the s. m. in [, ],. f ( ) is the s. min. in [, ]. Fining Asolute Etrem To fin the solute etrem of the ontinuous, use the funtion f ( ) on the intervl [ ] following proess.. Fin ll ritil points of f ( ) in [, ].. Evlute f ( ) t ll points foun in Step.. Evlute f ( ) n f ( ). 4. Ientify the s. m. (lrgest funtion vlue) n the s. min.(smllest funtion vlue) from the evlutions in Steps &. Etrem Reltive (lol) Etrem. is reltive (or lol) mimum of f f for ll ner. f ( ) if. is reltive (or lol) minimum of f f for ll ner. f ( ) if st Derivtive Test If is ritil point of. rel. m. of f ( ) if f 0 f then is > to the left of n f < 0 to the right of.. rel. min. of f ( ) if f 0 < to the left of n f > 0to the right of. is. not reltive etrem of f ( ) if f the sme sign on oth sies of. n Derivtive Test If Men Vlue Theorem, is ritil point of f ( ) suh tht f 0 then. is reltive mimum of f ( ) if f < 0.. is reltive minimum of f ( ) if f > 0.. my e reltive mimum, reltive minimum, or neither if f 0. Fining Reltive Etrem n/or Clssify Critil Points f.. Fin ll ritil points of. Use the st erivtive test or the n erivtive test on eh ritil point. If f ( ) is ontinuous on the lose intervl [ ] n ifferentile on the open intervl (, ) f f then there is numer < < suh tht f. Newton s Metho If n is the n th guess for the root/solution of f ( ) 0 then (n+) st guess is provie f ( n ) eists. f f n+ n ( n ) ( ) n Clulus Chet Sheet Relte Rtes Sketh piture n ientify known/unknown quntities. Write own eqution relting quntities n ifferentite with respet to t using impliit ifferentition (i.e. on erivtive every time you ifferentite funtion of t). Plug in known quntities n solve for the unknown quntity. E. A 5 foot ler is resting ginst wll. The ottom is initilly 0 ft wy n is eing pushe towrs the wll t ft/se. How fst 4 is the top moving fter se? is negtive euse is eresing. Using Pythgoren Theorem n ifferentiting, + y 5 + yy 0 After se we hve 0 7n so y Plug in n solve for y. 7 7( 4 ) + 76 y 0 y ft/se E. Two people re 50 ft prt when one strts wlking north. The ngleθ hnges t 0.0 r/min. At wht rte is the istne etween them hnging when θ 0.5 r? We hve θ 0.0 r/min. n wnt to fin. We n use vrious trig fns ut esiest is, seθ seθ tnθθ We knowθ 0.5 so plug in θ n solve. se( 0.5) tn( 0.5)( 0.0) ft/se Rememer to hve lultor in rins! Optimiztion Sketh piture if neee, write own eqution to e optimize n onstrint. Solve onstrint for one of the two vriles n plug into first eqution. Fin ritil points of eqution in rnge of vriles n verify tht they re min/m s neee. E. We re enlosing retngulr fiel with E. Determine point(s) on y + tht re 500 ft of fene mteril n one sie of the losest to (0,). fiel is uiling. Determine imensions tht will mimize the enlose re. Mimize A y sujet to onstrint of + y 500. Solve onstrint for n plug into re. A y( 500 y) 500 y 500y y Differentite n fin ritil point(s). A 500 4y y 5 By n eriv. test this is rel. m. n so is the nswer we re fter. Finlly, fin The imensions re then Minimize f ( 0) ( y ) + n the onstrint is y +. Solve onstrint for n plug into the funtion. y f y + y + y y y+ Differentite n fin ritil point(s). f y y By the n erivtive test this is rel. min. n so ll we nee to o is fin vlue(s). ± The points re then (, ) n (, ) Visit for omplete set of Clulus notes. 005 Pul Dwkins Visit for omplete set of Clulus notes. 005 Pul Dwkins