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2.6: Implicitna diferenciacija


V prejšnjih razdelkih smo se naučili najti izpeljanko ( frac {dy} {dx} ) ali (y ^ prime ), če je podano (y ) izrecno kot funkcija (x ). To pomeni, da če za neko funkcijo (f ) vemo (y = f (x) ), lahko najdemo (y ^ prime ). Na primer, glede na (y = 3x ^ 2-7 ) lahko zlahka najdemo (y ^ prime = 6x ). (Tu izrecno navajamo, kako sta povezani povezavi (x ) in (y ). Če poznamo (x ), lahko neposredno najdemo (y ).)

Včasih razmerje med (y ) in (x ) ni eksplicitno; prej je implicitno. Na primer, morda vemo, da je (x ^ 2-y = 4 ). Ta enakost opredeljuje razmerje med (x ) in (y ); če poznamo (x ), bi lahko ugotovili (y ). Ali lahko še vedno najdemo (y ^ prime )? V tem primeru zagotovo; rešujemo, da (y ) dobi (y = x ^ 2-4 ) (zato zdaj izrecno vemo (y )) in nato diferenciramo, da dobimo (y ^ prime = 2x ).

Včasih implicitno razmerje med (x ) in (y ) je zapleteno. Recimo, da imamo ( sin (y) + y ^ 3 = 6-x ^ 3 ). Graf te implicitne funkcije je podan na sliki 2.19. V tem primeru za (y ) ni povsem mogoče rešiti elementarnih funkcij. Presenetljivo pa je, da (y ^ prime ) še vedno lahko najdemo s postopkom, znanim kot implicitna diferenciacija.

Implicitna diferenciacija je tehnika, ki temelji na verižnem pravilu in se uporablja za iskanje izpeljave, kadar je razmerje med spremenljivkami dano implicitno in ne eksplicitno (rešeno za eno spremenljivko v smislu druge).

Začnemo s pregledom pravila verige. Naj sta (f ) in (g ) funkciji (x ). Potem [ frac {d} {dx} Big (f (g (x)) Big) = f ^ prime (g (x)) cdot g '(x). ] Denimo, da ( y = g (x) ). Zgoraj lahko napišemo kot [ frac {d} {dx} Big (f (y)) Big) = f ^ prime (y)) cdot y ^ prime, quad text {or} quad frac {d} {dx} Big (f (y)) Big) = f ^ prime (y) cdot frac {dy} {dx}. label {2.1} tag {2.1} ] Te enačbe so videti nenavadno; ključni koncept, ki se ga je treba tukaj naučiti, je, da lahko najdemo (y ^ prime ), tudi če ne vemo natančno, kako se (y ) in (x ) povežeta.

Ta postopek prikazujemo v naslednjem primeru.

Primer 67: Uporaba implicitne diferenciacije

Poiščite (y ^ prime ), če je ( sin (y) + y ^ 3 = 6-x ^ 3 ).

Rešitev

Začnemo z izpeljavo obeh strani (s čimer ohranjamo enakost.) Imamo:

[ frac {d} {dx} Velika ( sin (y) + y ^ 3 Velika) = frac {d} {dx} Velika (6-x ^ 3 Velika). ]

Desna stran je enostavna; vrne (- 3x ^ 2 ).

Leva stran zahteva večjo pozornost. Vzamemo izpeljani izraz - po - termin. Z uporabo tehnike, ki izhaja iz enačbe 2.1 zgoraj, lahko vidimo, da je [ frac {d} {dx} Big ( sin y Big) = cos y cdot y ^ prime. ]

Isti postopek uporabimo za izraz (y ^ 3 ).

[ frac {d} {dx} Veliki (y ^ 3 Veliki) = frac {d} {dx} Veliki ((y) ^ 3 Veliki) = 3 (y) ^ 2 cdot y ^ prime. ]

Sestavili smo to skupaj z desno stranjo

[ cos (y) y ^ prime + 3y ^ 2y ^ prime = -3x ^ 2. ]

Zdaj rešite za (y ^ prime ).

[ začeti {poravnati *} cos (y) y ^ prime + 3y ^ 2y ^ prime & = -3x ^ 2. velika ( cos y + 3y ^ 2 velika) y ^ prime & = -3x ^ 2 y ^ prime & = frac {-3x ^ 2} { cos y + 3y ^ 2} konec {poravnaj *} ]

Ta enačba za (y ^ prime ) se verjetno zdi nenavadna, saj vsebuje izraza (x ) in (y ). Kako naj se uporablja? To bomo obravnavali naprej.

Implicitne funkcije je na splošno težje obravnavati kot eksplicitne funkcije. Z eksplicitno funkcijo, ki ima vrednost (x ), imamo eksplicitno formulo za izračun ustrezne vrednosti (y ). Pri implicitni funkciji je pogosto treba najti vrednosti (x ) in (y ) ob istem času ki izpolnjujejo enačbo. Veliko lažje je dokazati, da določena točka ustreza enačbi, kot pa dejansko najti takšno točko.

Na primer, lahko enostavno potrdimo, da točka (( sqrt [3] {6}, 0) ) leži na grafu implicitne funkcije ( sin y + y ^ 3 = 6-x ^ 3 ). Če priklopite (0 ) za (y ), vidimo, da je leva stran (0 ). Če nastavimo (x = sqrt [3] 6 ), vidimo, da je tudi desna stran (0 ); enačba je izpolnjena. Naslednji primer najde enačbo tangente na to funkcijo na tej točki.

Primer 68: Uporaba implicitne diferenciacije za iskanje tangente

Poiščite enačbo tangente premice na krivuljo implicitno definirane funkcije ( sin y + y ^ 3 = 6-x ^ 3 ) v točki (( sqrt [3] 6,0) ).

Rešitev

V primeru 67 smo ugotovili, da [y ^ prime = frac {-3x ^ 2} { cos y + 3y ^ 2}. ] Najdemo naklon tangente v točki (( sqrt [ 3] 6,0) ) z namestitvijo ( sqrt [3] 6 ) za (x ) in (0 ) za (y ). Tako imamo v točki (( sqrt [3] 6,0) ) naklon kot [y ^ prime = frac {-3 ( sqrt [3] {6}) ^ 2} { cos 0 + 3 cdot0 ^ 2} = frac {-3 sqrt [3] {36}} {1} približno -9,91. ]

Zato je enačba tangente na implicitno definirano funkcijo ( sin y + y ^ 3 = 6-x ^ 3 ) v točki (( sqrt [3] {6}, 0) ) [y = -3 sqrt [3] {36} (x- sqrt [3] {6}) + 0 približno -9,91x + 18. ] Krivulja in ta tangentna črta sta prikazani na sliki 2.20.

To predlaga splošno metodo za implicitno razlikovanje. V spodnjih korakih predpostavimo, da je (y ) funkcija (x ).

  1. V enačbi vzemimo izpeljanko vsakega člana. Pogoje (x ) obravnavajte kot običajno. Pri jemanju izpeljank izrazov (y ) veljajo običajna pravila, le da moramo zaradi pravila verige vsak izraz pomnožiti z (y ^ prime ).
  2. Poiščite vse izraze (y ^ prime ) na eni strani znaka enačbe, preostale izraze pa postavite na drugo stran.
  3. Izštevaj faktor (y ^ prime ); reši za (y ^ prime ) z deljenjem.

Praktična opomba: Pri ročnem delu je lahko koristno, če namesto (y ^ prime ) uporabite simbol ( frac {dy} {dx} ), saj ga lahko zlahka zamenjate za (y ) ali (y ^ 1 ).

Primer 69: Uporaba implicitne diferenciacije

Glede na implicitno definirano funkcijo (y ^ 3 + x ^ 2y ^ 4 = 1 + 2x ) poiščite (y ^ prime ).

Rešitev

Implicitne izpeljanke bomo uporabljali po pogojih. Izvedek (y ^ 3 ) je (3y ^ 2y ^ prime ).

Drugi izraz, (x ^ 2y ^ 4 ), je malce zapleten. Zahteva pravilo izdelka, saj je produkt dveh funkcij (x ): (x ^ 2 ) in (y ^ 4 ). Njegova izpeljanka je (x ^ 2 (4y ^ 3y ^ prime) + 2xy ^ 4 ). Prvi del tega izraza zahteva (y ^ prime ), ker jemljemo izpeljanko izraza (y ). Drugi del tega ne zahteva, ker jemljemo izpeljanko (x ^ 2 ).

Izpeljanka desne strani je enostavno ugotoviti, da je (2 ). Skupaj dobimo:

[3y ^ 2y ^ prime + 4x ^ 2y ^ 3y ^ prime + 2xy ^ 4 = 2. ]

Premaknite izraze tako, da bo leva stran sestavljena samo iz izrazov (y ^ prime ), desna pa iz vseh ostalih izrazov:

[3y ^ 2y ^ prime + 4x ^ 2y ^ 3y ^ prime = 2-2xy ^ 4. ]

Faktor out (y ^ prime ) z leve strani in rešite, da dobite

[y ^ prime = frac {2-2xy ^ 4} {3y ^ 2 + 4x ^ 2y ^ 3}. ]

Da potrdimo veljavnost našega dela, poiščimo enačbo tangente na to funkcijo v točki. Lahko je potrditi, da točka ((0,1) ) leži na grafu te funkcije. Na tej točki je (y ^ prime = 2/3 ). Enačba tangente je torej (y = 2/3 (x-0) +1 ). Funkcija in njena tangentna črta sta prikazana na sliki 2.21.

Opazite, kako je naša funkcija videti precej drugače kot druge funkcije, ki smo jih videli. Za enega ne uspe preizkus navpične črte. Takšne funkcije so pomembne na številnih področjih matematike, zato je pomembno tudi razvijanje orodij za spopadanje z njimi.

Primer 70: Uporaba implicitne diferenciacije

Glede na implicitno definirano funkcijo ( sin (x ^ 2y ^ 2) + y ^ 3 = x + y ), poiščite (y ^ prime ).

Rešitev

Če ločimo izraz po pojmu, največ težav najdemo v prvem izrazu. Zahteva tako verigo kot pravila za izdelke.

[ začeti {poravnati *} frac {d} {dx} Veliki ( sin (x ^ 2y ^ 2) Veliki) & = cos (x ^ 2y ^ 2) cdot frac {d} { dx} Big (x ^ 2y ^ 2 Big) & = cos (x ^ 2y ^ 2) cdot big (x ^ 2 (2yy ^ prime) + 2xy ^ 2 big) & = 2 (x ^ 2yy ^ prime + xy ^ 2) cos (x ^ 2y ^ 2). end {poravnaj *} ]

Izvode drugih izrazov prepuščamo bralcu. Po prevzemu izpeljank obeh strani imamo

[2 (x ^ 2yy ^ prime + xy ^ 2) cos (x ^ 2y ^ 2) + 3y ^ 2y ^ prime = 1 + y ^ prime. ]

Zdaj moramo biti previdni, da pravilno rešujemo za (y ^ prime ), zlasti zaradi izdelka na levi. Najbolje je izdelek pomnožiti. Če to naredimo, pridemo

[2x ^ 2y cos (x ^ 2y ^ 2) y ^ prime + 2xy ^ 2 cos (x ^ 2y ^ 2) + 3y ^ 2y ^ prime = 1 + y ^ prime. ]

Od tu se lahko varno premikamo po izrazih in dobimo naslednje:

[2x ^ 2y cos (x ^ 2y ^ 2) y ^ prime + 3y ^ 2y ^ prime - y ^ prime = 1 - 2xy ^ 2 cos (x ^ 2y ^ 2). ]

Potem lahko rešimo, da (y ^ prime ) dobi

[y ^ prime = frac {1 - 2xy ^ 2 cos (x ^ 2y ^ 2)} {2x ^ 2y cos (x ^ 2y ^ 2) + 3y ^ 2-1}. ]

Graf te implicitne funkcije je podan na sliki 2.22. Preprosto je preveriti, ali točke ((0,0) ), ((0,1) ) in ((0, -1) ) ležijo na grafu. Nagibe tangente na vsaki od teh točk lahko najdemo z uporabo naše formule za (y ^ prime ).

Pri ((0,0) ) je naklon (- 1 ).

Pri ((0,1) ) je naklon (1/2 ).

Pri ((0, -1) ) je naklon tudi (1/2 ).

Tangentne črte so bile dodane v graf funkcije na sliki 2.23.

Kar nekaj "znanih" krivulj ima enačbe, ki so podane implicitno. Z implicitno diferenciacijo lahko poiščemo naklon na različnih točkah teh krivulj. Dve takšni krivulji bomo raziskali v naslednjih primerih.

Primer 71: Iskanje pobočij tangentnih črt na krog

Poiščite naklon tangente na krog (x ^ 2 + y ^ 2 = 1 ) v točki ((1/2, sqrt {3} / 2) ).

Rešitev

Če vzamemo izpeljanke, dobimo (2x + 2yy ^ prime = 0 ). Rešitev za (y ^ prime ) daje: [y ^ prime = frac {-x} {y}. ]

To je pametna formula. Spomnimo se, da bo naklon črte skozi izhodišče in točko ((x, y) ) na krožnici (y / x ). Ugotovili smo, da je naklon tangencialne črte na krog v tej točki v nasprotju z recipročno vrednostjo (y / x ), in sicer (- x / y ). Zato sta ti dve premici vedno pravokotni.

V točki ((1/2, sqrt {3} / 2) ) imamo naklon tangente kot

[y ^ prime = frac {-1/2} { sqrt {3} / 2} = frac {-1} { sqrt {3}} približno -0,577. ]

Graf kroga in njegove tangente na ((1/2, sqrt {3} / 2) ) je podan na sliki 2.24, skupaj s tanko črtkano črto od začetka, ki je pravokotna na tangentno črto. (Izkazalo se je, da vse normalne črte kroga gredo skozi središče kroga.)

Ta odsek je pokazal, kako najti izpeljave implicitno definiranih funkcij, katerih grafi vključujejo široko paleto zanimivih in nenavadnih oblik. Implicitna diferenciacija se lahko uporablja tudi za nadaljnje razumevanje "običajne" diferenciacije.

Ena luknja v našem trenutnem razumevanju izpeljank je ta: kaj je odvod funkcije kvadratnega korena? To je [ frac {d} {dx} big ( sqrt {x} big) = frac {d} {dx} big (x ^ {1/2} big) = text { ?} ]

Aludiramo na možno rešitev, saj lahko funkcijo kvadratnega korena zapišemo kot funkcijo moči z racionalno (ali delno) močjo. Nato smo v skušnjavi, da uporabimo pravilo napajanja in dobimo [ frac {d} {dx} big (x ^ {1/2} big) = frac12x ^ {- 1/2} = frac {1} {2 sqrt {x}}. ]

Težava pri tem je, da je bilo pravilo napajanja sprva določeno samo za pozitivne celoštevilčne moči, (n> 0 ). Čeprav tega takrat nismo utemeljili, se pravilo moči praviloma dokazuje z uporabo nečesa, imenovanega binomski izrek, ki obravnava samo pozitivna cela števila. Pravilo količnika nam je omogočilo, da razširimo pravilo moči na negativne celoštevilčne moči. Implicitna diferenciacija nam omogoča razširitev pravila moči na racionalne moči, kot je prikazano spodaj.

Naj bo (y = x ^ {m / n} ), kjer sta (m ) in (n ) celi števili brez skupnih faktorjev (torej (m = 2 ) in (n = 5 ) je v redu, vendar (m = 2 ) in (n = 4 ) ni). To eksplicitno funkcijo lahko implicitno prepišemo kot (y ^ n = x ^ m ). Zdaj uporabite implicitno diferenciacijo.

[ začeti {poravnati *} y & = x ^ {m / n} y ^ n & = x ^ m frac {d} {dx} velika (y ^ n velika) & = frak {d} {dx} velik (x ^ m velik) n cdot y ^ {n-1} cdot y ^ prime & = m cdot x ^ {m-1} y ^ prime & = frac {m} {n} frac {x ^ {m-1}} {y ^ {n-1}} quad text {(zdaj nadomestni (x ^ {m / n} ) za (y ))} & = frac {m} {n} frac {x ^ {m-1}} {(x ^ {m / n}) ^ {n-1}} quad text {(uporabi veliko algebre)} & = frac {m} nx ^ {(mn) / n} & = frac {m} nx ^ {m / n -1}. end {poravnaj *} ]

Zgornja izpeljava je ključ do dokaza o razširitvi pravila moči na racionalne moči. Z uporabo omejitev lahko to še enkrat razširimo na vključitev vse moči, vključno z iracionalnimi (tudi transcendentalnimi!) silami, ki podaja naslednji izrek.

Teorem 21: Pravilo moči za diferenciacijo

Naj bo (f (x) = x ^ n ), kjer je (n neq 0 ) realno število. Potem je (f ) diferenciabilna funkcija in (f ^ prime (x) = n cdot x ^ {n-1} ).

Ta izrek nam omogoča, da rečemo, da je izpeljanka (x ^ pi ) ( pi x ^ { pi -1} ).

To končno različico pravila o moči zdaj uporabljamo v naslednjem primeru, drugi preiskavi "slavne" krivulje.

Primer 72: Uporaba pravila napajanja

Poiščite naklon (x ^ {2/3} + y ^ {2/3} = 8 ) v točki ((8,8) ).

Rešitev

To je še posebej zanimiva krivulja, imenovana astroid. To je oblika, ki jo izrisuje točka na robu kroga, ki se valja znotraj večjega kroga, kot je prikazano na sliki 2.25.

Da bi našli naklon astroida v točki ((8,8) ), implicitno vzamemo izpeljanko.

[ začeti {poravnati *} frac23x ^ {- 1/3} + frac23y ^ {- 1/3} y ^ prime & = 0 frac23y ^ {- 1/3} y ^ prime & = - frac23x ^ {- 1/3} y ^ prime & = - frac {x ^ {- 1/3}} {y ^ {- 1/3}} y ^ prime & = - frac {y ^ {1/3}} {x ^ {1/3}} = - sqrt [3] { frac {y} x}. end {poravnaj *} ]

Če priključimo (x = 8 ) in (y = 8 ), dobimo naklon (- 1 ). Astroid s tangentno črto pri ((8,8) ) je prikazan na sliki 2.26.

Implicitna diferenciacija in drugi izpeljani finančni instrument

Za iskanje derivatov višjega reda lahko uporabimo implicitno diferenciacijo. Teoretično je to preprosto: najprej poiščite ( frac {dy} {dx} ), nato vzemite njegovo izpeljanko glede na (x ). V praksi ni težko, pogosto pa zahteva malo algebre. To dokažemo na primeru.

Primer 73: Iskanje druge izpeljanke

Glede na (x ^ 2 + y ^ 2 = 1 ) poiščite ( frac {d ^ 2y} {dx ^ 2} = y ^ { prime prime} ).

Rešitev

Ugotovili smo, da (y ^ prime = frac {dy} {dx} = -x / y ) v primeru 71. Za iskanje (y ^ { prime prime} ) uporabimo implicitno diferenciacijo za (y ^ prime ).

[ začeti {poravnati *} y ^ { prime prime} & = frac {d} {dx} velika (y ^ prime velika) & = frac {d} {dx} levo (- frac xy desno) qquad text {(Zdaj uporabite pravilo količnika.)} & = - frac {y (1) - x (y ^ prime)} {y ^ 2} end {poravnaj *} ]

zamenjaj (y ^ prime ) z (- x / y ):

[ začeti {poravnati *} & = - frac {yx (-x / y)} {y ^ 2} & = - frac {y + x ^ 2 / y} {y ^ 2}. konec {poravnaj *} ]

Čeprav to ni posebej preprost izraz, je uporaben. Vidimo lahko, da (y ^ { prime prime}> 0 ), ko (y <0 ) in (y ^ { prime prime}} <0 ), ko (y> 0 ). V oddelku 3.4 bomo videli, kako je to povezano z obliko grafa.

Logaritmična diferenciacija

Razmislite o funkciji (y = x ^ x ); prikazano je na sliki 2.27. Dobro je definiran za (x> 0 ) in morda bi nas zanimalo, ali najdemo enačbe tangent in premic, ki so normalne na njegov graf. Kako vzamemo njegovo izpeljanko?

Funkcija ni močnostna funkcija: ima "moč" (x ) in ni konstanta. Ni eksponentna funkcija: ima "osnovo" (x ) in ne konstanta .

Tehnika diferenciacije, znana kot logaritemska diferenciacija tukaj postane koristno. Osnovno načelo je to: vzemite naravni dnevnik obeh strani enačbe (y = f (x) ), nato uporabite implicitno diferenciacijo, da poiščete (y ^ prime ). To dokažemo v naslednjem primeru.

Primer 74: Uporaba logaritemske diferenciacije

Glede na (y = x ^ x ) z logaritemsko diferenciacijo poiščite (y ^ prime ).

Rešitev

Kot je predlagano zgoraj, začnemo tako, da vzamemo naravni dnevnik obeh strani in nato uporabimo implicitno diferenciacijo.

[ začeti {poravnati *} y & = x ^ x ln (y) & = ln (x ^ x) text {(uporabi pravilo logaritma)} ln (y) & = x ln x text {(zdaj uporabite implicitno razlikovanje)} frac {d} {dx} Big ( ln (y) Big) & = frac {d} {dx} Big (x ln x Veliki) frac {y ^ prime} {y} & = ln x + x cdot frac1x frac {y ^ prime} {y} & = ln x + 1 y ^ prime & = y big ( ln x + 1 big) text {(nadomestek (y = x ^ x ))} y ^ prime & = x ^ x big ( ln x +1 velika). end {poravnaj *} ]

Če želite »preizkusiti« svoj odgovor, ga uporabimo za iskanje enačbe tangente na (x = 1,5 ). Točka na grafu, skozi katero mora potekati tangentna črta, je ((1,5, 1,5 ^ {1,5} ) približno (1,5, 1,837) ). Z enačbo za (y ^ prime ) najdemo naklon kot

[y ^ prime = 1,5 ^ {1,5} velik ( ln 1,5 + 1 velik) približno 1,837 (1,405) približno 2,582. ]

Tako je enačba tangente (y = 1,6833 (x-1,5) +1,837 ). Slika 2.28 grafov (y = x ^ x ) skupaj s to tangentno črto.

Implicitna diferenciacija se izkaže za koristno, saj nam omogoča, da najdemo trenutne hitrosti sprememb različnih funkcij. Zlasti je razširil pravilo moči na racionalne eksponente, ki smo ga nato razširili na vsa realna števila. V naslednjem poglavju bo uporabljena implicitna diferenciacija za iskanje izpeljank iz inverzno funkcije, kot je (y = sin ^ {- 1} x ).


2.6: Implicitna diferenciacija

BRONX SKUPNOSTNI ŠOL

od mestna univerza v New Yorku

ODDELEK ZA MATEMATIKO IN RAČUNALNIŠTVO

VSEBINA: MTH 31 - Analitična geometrija in račun I (4 kreditne točke / 6 ur na teden)

PREDPOGOJ: MTH 30 ali enakovreden in po potrebi ENG 2 in RDL 2

BESEDILO: Račun (8. izdaja) Jamesa Stewarta, Cengage Learning. ISBN 978-1285740621

Študenti, ki ne potrebujejo MTH 33, lahko uporabljajo

Račun z eno spremenljivko (8. izdaja) Jamesa Stewarta, Cengage Learning ISBN 978-1305266636

Ta tečaj je Pathways Core B (Matematični in kvantitativni razlogi) Tečaj:
Tečaj na tem področju mora izpolnjevati vse naslednje učne rezultate. Študent bo:

a) Iz kvantitativnih predstav, kot so formule, grafi ali tabele, si razlagajte in črpajte ustrezne sklepe.

b) Uporabite algebrske, numerične, grafične ali statistične metode za natančne zaključke in reševanje matematičnih problemov.

c) V ustrezni matematični obliki predstavite kvantitativne probleme, izražene v naravnem jeziku.

d) Učinkovito sporočite kvantitativno analizo ali rešitve matematičnih problemov v pisni ali ustni obliki.

e) Ocenite rešitve težav glede smiselnosti z različnimi sredstvi, vključno z utemeljeno oceno.

f) Uporabi matematične metode za probleme na drugih področjih študija.

Rezultati učenja na tečaju (prispevali so učni rezultati na poti)

Po uspešnem zaključku tega predmeta bo študent sposoben:

1. Ocenite meje na vrednosti in v neskončnosti z uporabo mejnih zakonov in izrek teorema (a, b, c, e)

2. Razlikovanje algebrskih in trigonometričnih funkcij, vključno z uporabo pravil o omejitvah za izdelke, količnike in verige ter implicitno diferenciacijo (a, b)

3. Uporabite diferenciacijo za izračun trenutnih stopenj sprememb in tangente (c, d, e, f)

4. Izračunajte maksimume in minimume funkcij z uporabo računa za reševanje optimizacijskih problemov & # 8232izbiranje v aplikacijah in na drugih področjih študija & # 8232 (b, c, d, e, f)

5. Oblikujte in rešite povezane težave s stopnjami (b, c, d, f)

6. Uporabite metode računanja za skiciranje krivulje (a, b, e)

7. Protidiferencialne algebrske in trigonometrične funkcije (a, b)

8. Približni integrali po Riemannovih vsotah (b, d, e)

9. Ocenite osnovne integrale, vključno z uporabo substitucije in Temeljnega teorema računa (b, d, e)

10. Izračunajte določene integrale geometrijsko ali z uporabo računa za določitev površin, zaprtih z krivuljami (a, b, c, d, f)

ODDELEK TEMA PREDLAGANE VADBE

Poglavje 1: Funkcije in omejitve

1.4 Tangens in hitrostni problemi 49/1, 3, 5, 7

1.5 Meja funkcije 59 / 1-5, 12-14, 17, 23-28

1.6 Izračun mej z uporabo mejnih zakonov 70/1, 3-23 liho

1.8 Neprekinjenost 91/3, 7, 9, 15-21 liho, 25, 33, 37, 39, 41, 44, 45,

Pregled 96 / 1-11 nepar, 17, 23, 27, 29

2.1 Izvedeni finančni instrumenti 113/1, 3, 7, 21–31 nepar, 39–47 nepar, 53, 57, 59

2.2 Izvedeni finančni instrument kot funkcija 125/1, 3, 4, 7, 19, 20, 21, 25-33 liho, 39-51 neparno

2.3 Diferenciacijske formule 140 / 1-43 neparne, 51, 53, 69, 77

2.4 Izpeljanke trigonometričnih funkcij 150 / 1-17 nepar, 25, 29, 39-49 nepar

2.5 Pravilo verige 158 / 1-45 liho, 47, 51, 55, 69, 71

2.6 Implicitna diferenciacija 166 / 1-19 liho, 25, 27, 31, 35, 43, 45

2.7 Stopnje sprememb naravnih in 178 / 1-9 neparne, 15, 18

2.8 Povezane cene 185/1, 3, 9, 10, 11, 13-33 liho

2.9 Linearne aproksimacije in diferenciali 192/1, 3, 5, 7-25 liho, 31

Pregled 196/ 3, 5, 11, 13-37, 45, 51, 59, 61, 75, 77, 79, 82

Poglavje 3: Uporaba diferenciacije

3.1 Največje in najnižje vrednosti 211/3, 5, 15-27 liho, 29-55 liho

3.2 Izrek srednje vrednosti 219/1, 11, 13, 17, 21

3.3 Kako izvedeni finančni instrumenti vplivajo na obliko grafa 227/1, 5, 7, 8, 9-17 liho, 33-41 liho

3.4 Omejitve pri neskončnih vodoravnih asimptotah 241/3, lihe 9-29, 37, 41

3.5 Povzetek skiciranja krivulj 250 / 1-35 nepar

3.7 Težave z optimizacijo 256/3, 5, 7, 11, 17, 21, 27, 31

3.8 Newtonova metoda 276/5, 7, 13-19 neparna, 29

3.9 Antiderivativi 282 / 1-41 neparni, 43, 45, 47

Pregled 286 / 1-27 liho, 38, 41, 46, 49, 55, 57

4.1 Območja in razdalja 303/1, 3, 5, 13, 15, 21, 25

4.2 Določen integral 316/3, 5, 9, 17, 21-25 nepar, 31, 33, 37

4.3 Temeljni izrek računa 327/3, 7-35 nepar, 45, 51, 53

4.4 Nedoločeni integrali in teorem neto sprememb 336 / 1-11 nepar, 19-41 nepar, 55, 57


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Predogled paketov tečajev

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Predogled paketov tečajev

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  • Poglavje DT: Diagnostični testi
    • DT.A: Algebra (46)
    • DT.B: Analitična geometrija (13)
    • DT.C: Funkcije (20)
    • DT.D: Trigonometrija (14)
    • QP.1: Opredelitev in predstavitev funkcij (13)
    • QP.2: Delo s predstavitvami funkcij (14)
    • QP.3: Oznaka funkcije (13)
    • QP.4: Domena in obseg funkcije (12)
    • QP.5: Reševanje linearnih enačb (14)
    • QP.6: Linearne funkcije (15)
    • QP.7: Parabole (13)
    • QP.8: Faktoring kvadratnih enačb in iskanje x-preseki kvadratne funkcije (12)
    • QP.9: Polinomi (17)
    • QP.10: Več o faktoring polinomah (12)
    • QP.11: Iskanje korenin (14)
    • QP.12: Delitev polinoma (14)
    • QP.13: Racionalne funkcije (19)
    • QP.14: Root funkcije (15)
    • QP.15: Racionalizacija števca ali imenovalca (11)
    • QP.16: Eksponentne funkcije (13)
    • QP.17: Logaritmične funkcije (15)
    • QP.18: Trigonometrične funkcije in enotni krog (15)
    • QP.19: Grafi trigonometričnih funkcij (15)
    • QP.20: Trigonometrične identitete (18)
    • QP.21: Posebne funkcije (12)
    • QP.22: Algebrske kombinacije funkcij (14)
    • QP.23: Sestava funkcij (13)
    • QP.24: Pretvorbe funkcij (12)
    • QP.25: Inverzne funkcije (17)
    • 1.1: Štirje načini predstavitve funkcije (113)
    • 1.2: Matematični modeli: Katalog bistvenih funkcij (51)
    • 1.3: Nove funkcije iz starih funkcij (108)
    • 1.4: Tangens in hitrostni problemi (26)
    • 1.5: Omejitev funkcije (82)
    • 1.6: Izračun mej z uporabo mejnih zakonov (99)
    • 1.7: Natančna opredelitev meje (50)
    • 1.8: Neprekinjenost (77)
    • 1: Preverjanje koncepta
    • 1: Resnično napačen kviz (27)
    • 1: Pregledne vaje
    • 1: Načela reševanja problemov (11)
    • 1: Dodatne težave
    • 1: Vprašanja ob pravem času
    • 2.1: Izvedeni finančni instrumenti in stopnje sprememb (104)
    • 2.2: Izvedeni finančni instrument kot funkcija (101)
    • 2.3: Diferenciacijske formule (177)
    • 2.4: Izpeljanke trigonometričnih funkcij (96)
    • 2.5: Pravilo verige (126)
    • 2.6: Implicitna diferenciacija (97)
    • 2.7: Stopnje sprememb v naravoslovnih in družbenih vedah (62)
    • 2.8: Sorodne cene (79)
    • 2.9: Linearne aproksimacije in diferenciali (69)
    • 2: Preverjanje koncepta
    • 2: Resnično napačen kviz (15)
    • 2: Pregledne vaje (1)
    • 2: Težave plus (11)
    • 2: Dodatne težave
    • 2: Vprašanja ob pravem času
    • 3.1: Najvišje in najnižje vrednosti (125)
    • 3.2: Teorem o srednji vrednosti (53)
    • 3.3: Kaj nam izvedeni finančni instrumenti povedo o obliki grafa (110)
    • 3.4: Omejitve pri vodoravnih asimptotah neskončnosti (85)
    • 3.5: Povzetek skiciranja krivulj (83)
    • 3.6: Grafiranje z računi in tehnologijo (36)
    • 3.7: Težave z optimizacijo (103)
    • 3.8: Newtonova metoda (77)
    • 3.9: Antiderivati ​​(126)
    • 3: Preverjanje koncepta
    • 3: Resnično napačen kviz (20)
    • 3: Pregledne vaje
    • 3: Težave plus (7)
    • 3: Dodatne težave
    • 3: Vprašanja ob pravem času
    • 4.1: Težave s področjem in razdaljo (55)
    • 4.2: Določen integral (125)
    • 4.3: Temeljni izrek računa (136)
    • 4.4: Nedoločeni integrali in teorem neto sprememb (102)
    • 4.5: Pravilo o zamenjavi (133)
    • 4: Preverjanje koncepta
    • 4: Resnično napačen kviz (16)
    • 4: Pregledne vaje
    • 4: Težave plus (6)
    • 4: Dodatne težave
    • 4: Vprašanja ob pravem času
    • 5.1: Območja med krivuljami (69)
    • 5.2: Obseg (99)
    • 5.3: Obseg valjastih lupin (64)
    • 5.4: Delo (45)
    • 5.5: Povprečna vrednost funkcije (36)
    • 5: Preverjanje koncepta
    • 5: Resnično napačen kviz
    • 5: Pregledne vaje
    • 5: Težave plus (3)
    • 5: Dodatne težave
    • 5: Vprašanja ob pravem času
    • 6.1: Inverzne funkcije (36)
    • 6.2: Eksponentne funkcije in njihovi izpeljanki (81)
    • 6,2 *: Naravna logaritmična funkcija (41)
    • 6.3: Logaritemske funkcije (40)
    • 6.3 *: Naravna eksponentna funkcija (62)
    • 6.4: Izvedeni logaritemske funkcije (89)
    • 6.4 *: Splošne logaritemske in eksponentne funkcije (37)
    • 6.5: Eksponentna rast in propad (36)
    • 6.6: Inverzne trigonometrične funkcije (38)
    • 6.7: Hiperbolične funkcije (57)
    • 6.8: Nedoločene oblike in pravilo bolnišnice (96)
    • 6: Preverjanje koncepta
    • 6: Resnično napačen kviz (7)
    • 6: Pregledne vaje
    • 6: Težave plus (1)
    • 6: Dodatne težave
    • 6: Vprašanja ob pravem času
    • 7.1: Integracija po delih (93)
    • 7.2: Trigonometrični integrali (89)
    • 7.3: Trigonometrična zamenjava (57)
    • 7.4: Integracija racionalnih funkcij po delnih ulomkih (84)
    • 7.5: Strategija za integracijo (87)
    • 7.6: Integracija z uporabo tabel in tehnologije (62)
    • 7.7: Približna integracija (67)
    • 7.8: Neprimerni integrali (105)
    • 7: Preverjanje koncepta
    • 7: Resnično napačen kviz (13)
    • 7: Pregledne vaje
    • 7: Težave plus (3)
    • 7: Dodatne težave (33)
    • 7: Vprašanja ob pravem času
    • 8.1: Dolžina loka (48)
    • 8.2: Območje revolucionarne površine (44)
    • 8.3: Aplikacije za fiziko in inženirstvo (53)
    • 8.4: Aplikacije za ekonomijo in biologijo (29)
    • 8.5: Verjetnost (33)
    • 8: Preverjanje koncepta
    • 8: Resnično napačen kviz
    • 8: Pregledne vaje
    • 8: Težave plus (2)
    • 8: Dodatne težave (11)
    • 8: Vprašanja ob pravem času
    • 9.1: Modeliranje z diferencialnimi enačbami (29)
    • 9.2: Smerna polja in Eulerjeva metoda (42)
    • 9.3: Ločljive enačbe (60)
    • 9.4: Modeli za rast prebivalstva (35)
    • 9.5: Linearne enačbe (50)
    • 9.6: Sistemi plenilca in plena (23)
    • 9: Preverjanje koncepta
    • 9: Resnično napačen kviz (7)
    • 9: Pregledne vaje
    • 9: Težave plus (2)
    • 9: Dodatne težave (5)
    • 9: Vprašanja ob pravem času
    • 10.1: Krivulje, opredeljene s parametričnimi enačbami (50)
    • 10.2: Račun s parametričnimi krivuljami (71)
    • 10.3: Polarne koordinate (75)
    • 10.4: Račun v polarnih koordinatah (60)
    • 10.5: Stožčasti odseki (74)
    • 10.6: Stožčasti odseki v polarnih koordinatah (34)
    • 10: Preverjanje koncepta
    • 10: Resnično napačen kviz (10)
    • 10: Pregledne vaje
    • 10: Težave plus (1)
    • 10: Dodatne težave (33)
    • 10: Vprašanja ob pravem času
    • 11.1: Zaporedja (86)
    • 11.2: Serija (97)
    • 11.3: Integralni test in ocene vsot (51)
    • 11.4: Primerjalni testi (53)
    • 11.5: Izmenične serije in absolutna konvergenca (59)
    • 11.6: Razmerje in koreninski testi (46)
    • 11.7: Strategija za testiranje serije (50)
    • 11.8: Power Series (56)
    • 11.9: Predstavitve funkcij kot potencialne serije (46)
    • 11.10: Taylor in Maclaurin Series (79)
    • 11.11: Uporabe Taylorjevih polinoma (45)
    • 11: Preverjanje koncepta
    • 11: Resnično napačen kviz (20)
    • 11: Pregledne vaje
    • 11: Težave plus (2)
    • 11: Dodatne težave (15)
    • 11: Vprašanja ob pravem času
    • 12.1: Tridimenzionalni koordinatni sistemi (51)
    • 12.2: Vektorji (51)
    • 12.3: Dot Product (66)
    • 12.4: Navzkrižni izdelek (57)
    • 12.5: Enačbe daljic in ravnin (76)
    • 12.6: Cilindri in kvadrične površine (69)
    • 12: Preverjanje koncepta
    • 12: Resnično napačen kviz (22)
    • 12: Pregledne vaje
    • 12: Težave plus (2)
    • 12: Dodatne težave (12)
    • 12: Vprašanja ob pravem času
    • 13.1: Vektorske funkcije in vesoljske krivulje (48)
    • 13.2: Izvedeni in integrali vektorskih funkcij (55)
    • 13.3: Dolžina in ukrivljenost loka (58)
    • 13.4: Gibanje v vesolju: hitrost in pospešek (47)
    • 13: Preverjanje koncepta
    • 13: Resnično napačen kviz (14)
    • 13: Pregledne vaje
    • 13: Težave plus (1)
    • 13: Dodatne težave (5)
    • 13: Vprašanja ob pravem času
    • 14.1: Funkcije več spremenljivk (75)
    • 14.2: Omejitve in kontinuiteta (50)
    • 14.3: Delni izvedeni finančni instrumenti (94)
    • 14.4: Tangentne ravnine in linearna aproksimacija (45)
    • 14.5: Pravilo verige (65)
    • 14.6: Usmerjeni izvedeni finančni instrumenti in vektor gradienta (78)
    • 14.7: Najvišje in najnižje vrednosti (65)
    • 14.8: Lagrangeovi multiplikatorji (54)
    • 14: Preverjanje koncepta
    • 14: Resnično napačen kviz (12)
    • 14: Pregledne vaje
    • 14: Težave plus (3)
    • 14: Dodatne težave (19)
    • 14: Vprašanja ob pravem času
    • 15.1: Dvojni integrali nad pravokotniki (78)
    • 15.2: Dvojni integrali nad splošnimi regijami (81)
    • 15.3: Dvojni integrali v polarnih koordinatah (52)
    • 15.4: Uporaba dvojnih integralov (48)
    • 15.5: Površina (23)
    • 15.6: Trojni integrali (65)
    • 15.7: Trojni integrali v cilindričnih koordinatah (44)
    • 15.8: Trojni integrali v sferičnih koordinatah (64)
    • 15.9: Sprememba spremenljivk v več integralih (39)
    • 15: Preverjanje koncepta
    • 15: Resnično napačen kviz (9)
    • 15: Pregledne vaje
    • 15: Težave plus (2)
    • 15: Dodatne težave (4)
    • 15: Vprašanja ob pravem času
    • 16.1: Vektorska polja (41)
    • 16.2: Linijski integrali (61)
    • 16.3: Temeljni izrek za linijske integrale (50)
    • 16.4: Greenov izrek (44)
    • 16.5: Curl in divergence (53)
    • 16.6: Parametrične površine in njihova območja (74)
    • 16.7: Površinski integrali (62)
    • 16.8: Stokesov izrek (34)
    • 16.9: Teorem o divergenci (41)
    • 16.10: Povzetek
    • 16: Preverjanje koncepta
    • 16: Resnično napačen kviz (13)
    • 16: Pregledne vaje
    • 16: Težave plus (2)
    • 16: Dodatne težave (9)
    • 16: Vprašanja ob pravem času
    • 17.1: Linearne enačbe drugega reda
    • 17.2: Nehomogene linearne enačbe
    • 17.3: Uporaba diferencialnih enačb drugega reda
    • 17.4: Serijske rešitve
    • 17: Preverjanje koncepta
    • 17: Resnično napačen kviz
    • 17: Pregledne vaje
    • 17: Dodatne težave
    • 17: Vprašanja ob pravem času
    • A.A: Številke, neenakosti in absolutne vrednosti (71)
    • A.B: Koordinatna geometrija in črte (61)
    • A.C .: Grafi enačb druge stopnje (40)
    • A.D .: Trigonometrija (83)
    • A.E: Oznaka Sigma (49)
    • A.F: Dokazi teorem
    • A.G: Odgovori na lihe vaje

    Stewart Calculus postavlja temelje za študente v STEM s poudarkom na reševanju problemov in predstavitvijo konceptov z neprimerljivo jasnostjo in natančnostjo. Izbrali in mentorirali James Stewart, Daniel Clegg in Saleem Watson nadaljujejo njegovo zapuščino.

    Spoznavanje novih avtorjev Stewartovega računa

    Približuje se nova izdaja Stewartovega računa

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    • Paketi tečajev s pripravljenimi nalogami so strokovnjaki s področja predmetov izdelali posebej za ta učbenik, da vam prihranijo čas in jih je mogoče enostavno prilagoditi vašim učnim ciljem.
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    Pi and Phi in the Great Pyramid of Egypt

    Another interesting relationship between Pi and Phi is related to the geometry of the Great Pyramid of Giza. This relationship connects dimensions of the Great Pyramid to both Pi and Phi, but it is not known with certainty whether this was an intentional aspect of its design, whether its design was based on Pi or Phi but not both, or whether it is a simple coincidence. It relates to the fact that 4 divided by square root of phi is almost exactly equal to Pi:

    The square root of Phi (1.6180339887…) = 1.2720196495…

    4 divided by 1.2720196495… = 3.14460551103…

    The difference of these two numbers is less than a 10th of a percent.

    See the Phi, Pi and the Great Pyramid page for more details.


    We derive explicit and new implicit finite-difference formulae for derivatives of arbitrary order with any order of accuracy by the plane wave theory where the finite-difference coefficients are obtained from the Taylor series expansion. The implicit finite-difference formulae are derived from fractional expansion of derivatives which form tridiagonal matrix equations. Our results demonstrate that the accuracy of a (2N + 2)th-order implicit formula is nearly equivalent to that of a (6N + 2)th-order explicit formula for the first-order derivative, and (2N + 2)th-order implicit formula is nearly equivalent to (4N + 2)th-order explicit formula for the second-order derivative. In general, an implicit method is computationally more expensive than an explicit method, due to the requirement of solving large matrix equations. However, the new implicit method only involves solving tridiagonal matrix equations, which is fairly inexpensive. Furthermore, taking advantage of the fact that many repeated calculations of derivatives are performed by the same difference formula, several parts can be precomputed resulting in a fast algorithm. We further demonstrate that a (2N + 2)th-order implicit formulation requires nearly the same memory and computation as a (2N + 4)th-order explicit formulation but attains the accuracy achieved by a (6N + 2)th-order explicit formulation for the first-order derivative and that of a (4N + 2)th-order explicit method for the second-order derivative when additional cost of visiting arrays is not considered. This means that a high-order explicit method may be replaced by an implicit method of the same order resulting in a much improved performance. Our analysis of efficiency and numerical modelling results for acoustic and elastic wave propagation validates the effectiveness and practicality of the implicit finite-difference method.

    Many scientific and engineering problems involve numerically solving partial differential equations. A variety of differ- ence techniques, such as the finite-difference (FD), the pseudospectral (PS) and the finite-element (FM) methods, have been developed. However, because of the straightforward implementation, requiring small memory and computation time, the FD is the most popular and has been widely utilized in seismic modelling (Kelly et al 1976, Dablain 1986, Virieux 1986, Igel et al 1995, Etgen and O'Brien 2007, Bansal and Sen 2008) and migration (Claerbout 1985, Larner and Beasley 1987, Li 1991, Ristow and Ruhl 1994, Zhang et al 2000, Fei and Liner 2008). In order to improve the efficiency, the accuracy and the stability of a FDM in numerical modelling, several variants of FD methods have been developed.

    A conventional FD needs a fixed number of grid points per wavelength in any one layer. In fact, a coarse mesh can be used in the high-velocity layer for its large wavelength. Therefore, variable grid schemes are proposed to significantly reduce computational time and memory requirements. When there is an abrupt transition from the adaptive region of a coarse mesh to a much finer mesh, large amplitude artificial reflections from the adaptive zone may occur. To overcome these problems, the scheme of interpolating the field variables in the adaptive region has been developed (Wang and Schuster 1996, Hayashi and Burns 1999). Other advanced methods include the PS-2 method for regular grids (Zahradník and Priolo 1995) and a simple case of so-called rectangular irregular grids. This method is developed further and applied to both nonplanar topography and internal discontinuities (Opršal and Zahradník 1999).

    To save computational cost, besides optimizing mesh sizes according to the local parameters, different temporal sampling in different parts of the numerical grid is introduced by Falk et al ( 1996). Since the method is restricted to ratios of time steps between the different domains of 2n, a new method is proposed to handle any positive integer ratio and does not depend on the time-step ratio by Tessmer ( 2000).

    To improve the modelling accuracy of the first-order elastic and viscoelastic wave equations, staggered-grid FD schemes that involve defining different components of one physical parameter at different staggered points are usually applied to compute the derivatives in the equations (Virieux 1986, Levander 1988, Robertsson et al 1994, Graves 1996). However, boundary conditions of the elastic wave field at a free surface, i.e. the high-contrast discontinuity between vacuum and rock, must be defined in the FD algorithm (Robertsson 1996, Graves 1996, Opršal and Zahradník 1999, Saenger and Bohlen 2004). To avoid this problem, a rotated staggered-grid technique (Gold et al 1997, Saenger et al 2000) is presented where high contrast discontinuities can be incorporated without using explicit boundary conditions and without averaging elastic moduli. A velocity–stress rotated staggered-grid algorithm is used to simulate seismic waves in an elastic and viscoelastic model with 3D topography of the free surface (Saenger and Bohlen 2004). The accuracy for modelling Rayleigh waves utilizing the standard staggered-grid and the rotated staggered-grid method is also investigated by Bohlen and Saenger ( 2006).

    A conventional method uses FD operators with low-order accuracy to calculate space derivatives therefore, it needs small processor memory and less computation time, but leads to low accuracy results. High-order FD schemes are developed to improve the accuracy of the conventional finite-difference method. The FD scheme with any order accuracy has been derived for the first-order derivatives and used to solve the wave equations (Dablain 1986, Fornberg 1987, Crase 1990, Visbal and Gaitonde 2001, Hestholm 2007). The FD coefficients are determined by the Taylor series expansion (Dablain 1986) or by an optimization (Fornberg 1987). Using the Taylor series expansion also, the FD method with any even-order accuracy is presented for any order derivatives (Liu et al 1998) and utilized to simulate wave propagation in two-phase anisotropic media (Liu and Wei 2008).

    However, most of these methods make use of the explicit finite-difference method (EFDM). Some development on the implicit finite-difference method (IFDM) has also been reported in the literature. To yield good modelling results, implicit finite-difference formulae are skilfully derived for the elastic wave equation (Emerman et al 1982). These formulae express the value of a variable at some point at a future time in terms of the value of the variable at that point and at its neighbouring points at present time, past times and future times. An IFDM for time derivatives has also been implemented in seismic migration algorithms (Ristow and Ruhl 1997, Shan 2007, Zhang and Zhang 2007).

    Here, we focus on the space derivative calculation by a FDM. In our formulation, an EFDM directly calculates the derivative value at some point in terms of the function values at that point and at its neighbouring points. However, an IFDM expresses the derivative value at some point in terms of the function values at that point and at its neighbouring points and the derivative values at its neighbouring points. For example, a compact finite-difference method (CFDM) is one such IFDM (Lele 1992). In areas other than geophysics and seismology, several variants of the IFDM have been widely studied (Ekaterinaris 1999, Meitz and Fasel 2000, Lee and Seo 2002, Nihei and Ishii 2003). Zhang and Chen ( 2006) proposed a new numerical method, named the traction image method, to accurately and efficiently implement the traction-free boundary conditions in finite-difference simulation in the presence of surface topography. In this method, the physical traction-free boundary conditions provide a constraint on the derivatives of the velocity components along the free surface, which leads to a solution to calculate the derivative of the velocity components by a compact scheme.

    In this paper, we derive both explicit and implicit finite-difference formulae with even-order accuracy for any order derivative. Further, we develop a practical IFDM and demonstrate its efficiency and applicability with some numerical results.


    4.5 Substitution Method

    From the Fundamental Theorem, we see that differentiation and integration are as inverse process to each other. If we reverse the rule of differentiation, we will get a method to integrate a function.

    Theorem 4.7 (The Substitution Rule) If (u=g(x)) is a differentiable function whose range is an interval (I) and (f) is continuous on (I) , then [ int f(g(x))g'(x)mathrmx=int f(u)mathrmu. ]

    For a definite integral, we also need to substitute limits of integration when applying the substitution rule.

    Theorem 4.8 (The Substitution Rule for Definite Integral) If (g'(x)) is continuous on ([a, b]) and (f(x)) is continuous on the range of (u=g(x)) , ([g(a), g(b)]) , then [ int_a^bf(g(x))g'(x)mathrmx=int_^f(u)mathrmu. ]

    Applying the chain rule to functions with symmetries will simplify the calculation.

    Proposition 4.1 Suppose that (f) is a continuous function on ([-a, a]) .

    1. If (f) is an odd function, that is (f(-x)=-f(x)) , then [displaystyle int_<-a>^af(x)mathrmx =0.]
    2. If (f) is an even function, that is (f(-x)=f(x)) , then [displaystyle int_<-a>^af(x)mathrmx =2int_0^af(x)mathrmx.]

    Exercise 4.31 Evaluate the following integral. [ displaystyle int xsqrtmathrm x. ]

    Opomba: You may also try the substitution (u=sqrt) .

    Exercise 4.32 Evaluate the following integral. [ displaystyle int frac>mathrm x. ]

    Here, I would like to show you another way. Let (u=sqrt<1-x^3>) . Then (u^2=1-x^3) and (mathrmx=frac<2umathrmu><-3x^2>) . Therefore, [ egin int frac>mathrm x=&int fracfrac<2umathrmu><-3x^2> =& -frac23int mathrmu =&-frac23u+C =&-frac23sqrt<1-x^3>+C end ]

    Exercise 4.33 Evaluate the following integral. [ displaystyle int(sin hetacos^2 heta)mathrm heta. ]

    Let (u=cos heta) . Then (mathrm heta=fracu><-sin theta>) . Therefore, [ egin int (sin hetacos^2 heta)mathrm heta=&int((sin heta) u^2) fracu><-sin heta> =& -int u^2 mathrmu =&-frac13u^3+C =&-frac13cos^2 heta+C end ]

    Exercise 4.34 Evaluate the following integral. [ displaystyle intsec^2 t an tmathrm t. ]

    Let (u= an t) . Then (mathrmt=fracu>) . Therefore, [ egin intsec^2 t an tmathrm t=&int umathrmu =&frac12u^2+C =&frac12 an^2t+C end ]

    Exercise 4.35 Evaluate the following integral. [ displaystyle int sqrt[3] <2x+1>mathrm x. ]

    Exercise 4.36 Evaluate the following integral. [ displaystyle int (3x-2)^5 mathrm t. ]

    One way is to use the binomial formula the expand the integrand and then integrate.

    Here we use a substitution to make the calculation easier.

    Exercise 4.37 Evaluate the following integral. [ displaystyle int_1^4 frac)>>mathrm x. ]

    Opomba: To avoid the mistake of forgetting substitute the integral limits, it will be better to find the indefinite integral first.

    Exercise 4.38 Evaluate the following integral. [ displaystyle int_1^2 fracmathrm x. ]

    Exercise 4.39 Evaluate the following integral. [ displaystyle int_0^ sin(2 heta)cos(2 heta)mathrm heta. ]

    There are different ways to find this integral.

    One way is to let (u=sin(2 heta)) . Then (mathrm u=2cos(2 heta)mathrm heta) and [ sin(2 heta)cos(2 heta)mathrm heta=frac12umathrmu ] Therefore, [ egin int_0^ sin(2 heta)cos(2 heta)mathrm heta =&int_^frac12umathrmu =&frac12 int_<0>^ <1>umathrmu =&frac14 u^2|_<0>^<1>=frac14. konec ]

    Exercise 4.40 Evaluate the following integral. [ displaystyle int_<-2>^<2>(x^3+xsec^2x)mathrm x. ]

    Note that the function (f(x)=x^3+xsec^2x) is an odd function, that is (f(-x)=-f(x)) . If we let (u=-x) ,then (mathrmx=-mathrmu) and
    [ egin int_<-2>^<2>(x^3+xsec^2x)mathrm x=&-int_<2>^<-2>((-u)^3+(-u)sec^2(-u))mathrm u =&-int_<-2>^<2>(u^3+usec^2u)mathrm u =&-int_<-2>^<2>(x^3+xsec^2x)mathrm x end ] Add to both sides the definite integral, we get [ 2int_<-2>^<2>(x^3+xsec^2x)mathrm x=0. ] Therefore, [ int_<-2>^<2>(x^3+xsec^2x)mathrm x=0. ]


    2.6: Implicit Differentiation

    This page will be updated as the semester progresses.

    Appendix D (Trigonometry) Completed notes
    Section 1.1 (Vectors) Completed notes
    Section 1.2 (The dot product) Completed notes
    Section 1.3 (Vector functions) Completed notes
    Section 2.2 (The limit of a function) Completed notes
    Section 2.3 (Calculating limits using the limit laws) Completed notes
    Section 2.5 (Continuity) Completed notes
    Section 2.6 (Limits at infinity horizontal asymptotes) Completed notes
    Section 2.7 (Tangents, velocities, and the other rates of change) Completed notes
    Section 3.1 (Derivatives) Completed notes
    Section 3.2 (Differentiation formulas) Partially completed notes
    Section 3.4 (Derivatives of trigonometric functions) Completed notes
    Review for Test 1

    Section 3.5 (The chain rule) Completed notes
    Section 3.6 (Implicit differentiation) Completed notes
    Section 3.7 (Derivatives of vector functions) Completed notes
    Section 3.8 (Higher derivatives) Completed notes
    Section 3.9 (Slopes and tangents to parametric curves) Completed notes
    Section 3.10 (Related rates) Completed notes
    Section 3.11 (Differentials linear and quadratic approximations) Completed notes
    Section 4.1 (Exponential functions and their derivatives) Completed notes
    Section 4.2 (Inverse functions) Completed notes
    Section 4.3 (Logarithmic functions) Completed notes
    Review for Test 2

    Section 4.4 (Derivatives of logarithmic functions) Completed notes
    Section 4.5 (Exponential growth and decay) Completed notes
    Section 4.6 (Inverse trigonometric functions) Completed notes
    Section 4.8 (L'Hospital's Rule) Completed notes
    Section 5.1 (What does f' say about f?) Completed notes
    Section 5.2 (Maximum and minimum values) Completed notes
    Section 5.3 (Derivatives and the shapes of curves) Completed notes
    Section 5.5 (Applied maximum and minimum problems) Completed notes
    Section 5.7 (Antiderivatives) Completed notes
    Section 6.1 (Sigma notation) Completed notes
    Section 6.2 (Area) Completed notes
    Section 6.3 (Definite integral) Completed notes
    Section 6.4 (The fundamental theorem of calculus) Completed notes


    3 THE PURPOSE OF FREE WILL

    The prevalent endorsement of the belief in free will raises an important fundamental question – Why would anyone believe in free will? If one believes in free will – then what is free will meant for?

    One group of scholars views free will beliefs as a mechanism that allows the self to pursue self-enhancing desired states and goals and seek own wants and needs (Hume, 1748 Edwards, 1754 ). Put more simply – free will is only worth having if it enables the individual to get what she or he wants (Dennett, 2003 ).

    A second view often referred to as the “action-control perspective” argues that the concept of free will has evolved to allow the self to coexist with others in society as to override inherent immediate biological urges that mainly focus on the self (Kant, 1797/1967 ) thus allowing for prospection, long-term planning, action control, and coordination with others in society (Baumeister, 2005 , 2008a ). The belief in free will could have possibly evolved so that people would be able to a deal with a world of increasingly complicated choices and complex societal interactions that require coordination and inhibition of self (Baumeister, 2008a Laurene, Rakos, Tisak, Robichaud, & Horvath, 2011 Rakos et al., 2008 ).

    The close conceptual relationship that free will holds with moral responsibility supports the view that free will is a notion embedded in societal considerations. The concept of free will may be regarded by societies and religions as a solution to the predicament of laypersons that associate determinism with inevitability, reduced accountability, and thus lower action control over socially undesirable behaviors. Based on the idea of free will as a social tool, the belief that a person could make different free choices in a given situation is considered essential to legal, moral, and political judgments (Juth & Lorentzon, 2010 Searle, 2007 ). More broadly, society often regards it appropriate to adjust legal and moral judgments based on the assessment of whether a wrongdoer acted out of his or her own free will (Greene & Cohen, 2004 Roskies, 2006 ). In order to legally hold a person accountable and bring a person to trial, it is now commonly expected that it be proven that the person could have done otherwise, meaning that there were no external influences coercing the person to act in this way (e.g., having a gun to the person's head) or that the person did not merely act out of uncontrollable urges (e.g., temporary insanity Burns & Bechara, 2007 ). Similarly, a contract between two people is only considered valid if the two sides have entered the contract out of their own free will, meaning that both sides were free from any coercion (Cohen, 1933 ).

    A developmental perspective argues free will to be rooted in the perception people experience in their everyday choices while growing up – even if such a perception is illusory, serving as a self-indicator regarding the ability to execute and increasing one's motivation to enter difficult choice situations (Bandura, 2006 Rakos, 2004 Wegner, 2004 ). Nichols ( 2004 ) showed that children between the ages of three and five typically endorse free will and reject determinism by making the claim that a person in a given scenario could have chosen to act differently, much more so than a physical object could have. Nichols goes on further to argue that the perception of having free will in kids is innate rather than learned – that freedom of an agent is inferred by native evidence to form the belief that humans are different than objects in their ability to act otherwise. Other studies have extended these findings by demonstrating that not only do kids at the age of five perceive people to have the capacity to choose more freely than objects do but that they also clearly distinguish between free and un-free actions by the same human agent (Chernyak, Kushnir, & Wellman, 2010 Kushnir, Wellman, & Chernyak, 2009 ).

    To summarize, the role of free will in people's beliefs could be the pursuit of own goals and desires or in the evolutionary role of free will as overcoming self to allow people to coexist with others in society. This belief could also be rooted in an innate intuitive perception developed by people while growing up to self-motivate when faced with making choices.


    2.6: Implicit Differentiation

    MATH 180: ELEMENTS OF CALCULUS I
    University of New Mexico, Spring 2015

    Inštruktor: Janet Vassilev
    E-mail:
    [email protected]
    Webpage:
    http://math.unm.edu/

    jvassil
    Phone: 277-2214
    Office: SMLC 324
    Office Hours:
    Monday and Friday 1-1:50
    , Wednesday 10-10:50 or by appointment.

    Prerequisite: Grade of C or better in MATH 121

    Textbook: APPLIED CALCULUS for the Managerial, Life and Social Sciences, Ninth Edition, Brooks / Cole 2014, S. T. Tan
    Hardcover for both Math 180 and Math 181: ISBN-10: 1-133-60771-3 ISBN-13: 978-1-133-60771-7
    UNM Custom Edition Softcover for Math 180 only: ISBN-10: 1-305-02608-X ISBN-13: 978-1-305-02608-7

    Calculator: Calculators will not be allowed on any of the exams. A scientific calculator may be necessary for some homework assignments or quizzes.

    Homework: Your daily homework is your most important effort in this course. It is imperative that you do all of the assigned problems, especially the hard ones, because this is how you actually learn the material. Expect 2-3 hours of homework for every hour of class meeting time (on average 6-9 hours per week). Keep all of your homework together in a folder so that if you are having trouble in the course, you can bring it with you when you go to see me or get tutoring.

    Kvizi: There will be weekly quizzes. The quiz problems will be very similar to the homework problems, if not the same. Most of the quizzes will be in-class and announced, but occasionally there may be a pop quiz. No make-up quizzes will be given, even if you have an excused absence. The two lowest quiz scores will be dropped at the end of the semester.

    Exams: There will be three in-class exams, 100 points each. You have to show all your work and use proper mathematical notation to receive full credit. A correct answer without work will receive no more than 1 point. I do not give make up exams. If you are ill or have some form of excused absence, you must contact me on or before the day of the exam in order to have your final grade calculated without this test. I typically replace the score for this exam by your average on the final.

    Končni izpit: The final exam, comprehensive and worth 200 points, will be held on Monday, May 4 th between 7:30 am and 9:30 am. The location of the exam will be announced near the end of the semester.

    Important Note: Notes of any kind, 3x5 cards, books, cell phones, computers, headphones etc. are not allowed on any tests, including the Final Exam.

    Ocenjevanje : To get full credit on graded work, students must address all mathematical components presented by the problem, showing all steps and calculations. The use of proper notation, well-structured procedures, and legibility will be taken into account when assigning points. Your grade will be determined based on your performance on the following:

    Grading: The grades will for the most part be assigned as follows: 90% to 100% = A 80% to 89% = B 70% to 79% = C below 70% D or F
    However, there may be a slight curve. If so, the curve will be announce for each test. There will be no extra credit. Students who withdraw after week 3 will receive the grade W. No W’s will be given to students who have not withdrawn.

    Communication: Please check your UNM e-mail regularly or make sure to forward your e-mail from that address to an account that you check at least daily. I may send you important information and updates to your UNM e-mail address. If you e-mail me, include your full name and that you are a student in my 180 class.

    Attendance: Attendance is mandatory. A student with three or more unexcused absences may be dropped from the course. Tardiness or early departure may be regarded as absence. It is the student’s responsibility to withdraw from the course if he/she stops attending. A failing grade of “F” will be assigned if the student stops attending and does not withdraw.

    Student Behavior: According to the Code of Conduct as stated in the Policies and Regulations for UNM, student activities that interfere with the rights of others to pursue their education or to conduct their University duties and responsibilities will lead to disciplinary action. This includes any activities that are disruptive to the class and any acts of academic dishonesty. Students are expected to behave in a courteous and respectful manner toward the instructor and their fellow students. Students should turn off their cell-phones before the beginning of each class, and be prepared to remain seated the entire class. Students may be dropped from a class for inappropriate behavior.

    Students with Disabilities: We accommodate students with documented disabilities. During the first two weeks of the semester, those students should inform the instructor of their particular needs.

    Help: If you are struggling, seek help immediately. In addition to your instructor's office hours, there is extra help available at:
    - The Calculus Tutoring Table, staffed by instructors every day, 3 rd floor DSH near the elevator
    - CAPS: Center for Academic Program Support, 3rd floor Zimmerman Library, 277-4560

    - MEP Engineering Annex, room 210, or call the study group at 277-8795
    - CATS: Counseling and Therapy Services, Student Health Center, 277-4537 (for test anxiety, etc.)


    Class 11th Maths - Video Tutorials in Hindi

    We warmly welcome you to our online courses. Maths is a subject that most of us don't like because it is very difficult to understand isn't it? What if learning maths becomes easy?

    Yes, I am here to help you to do maths in an easy way. I will make sure that you learn the formulas and easy tricks to solve all your maths assignments and prepare perfectly for your examinations.

    I will make sure that the dear students score great marks in their examinations and leave everyone shockingly amazed.

    In this course, you will be receiving

    1. Ebooks
    2. Live sessions
    3. Video tutorials that will be present in Hindi so that it becomes easy for you to understand
    4. Online course study materials
    5. Tips and tricks to learn the formulas
    6. Solve question papers
    7. Mock tests
    8. Doubt clearing sessions

    In this course, you will be learning about

    1. Complex numbers and every chapter related to it
    2. Quadrant arguments
    3. The principal value of an argument
    4. Cube root of unity
    5. Properties of modulus
    6. Properties of conjugate
    7. DE Moivre's theorem
    8. Questions on Nth root
    9. Questions based on complex numbers ( part 1 to part 16 )
    10. Binomial coefficient
    11. Odd terms sum And questions on it
    12. Sequences and series
    13. Arithmetic progression and questions on it
    14. Sum of arithmetic progression and questions on it
    15. Sum of n terms of G.P
    16. Chain rule
    17. Existence of limit
    18. Limits using trigonometric identities
    19. L' Hospital's rule
    20. Introduction to differentiation
    21. Parametric equations and many more chapters will be covered from the syllabus.

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