help-28

[queen dome]

erstmal a

kapieren tue ich beide net

[leaden ermine]

hast du schon mal was von integration gehört

@queen dome

[queen dome]

jaaa a hab ich denke mal kapiert

die stamfunktion von -x^2 ist 1/3x^3 oder?

[leaden ermine]

ja und von 2ax?

[queen dome]

ax^2?

:,)

[leaden ermine]

ja

sehr gut

und was sind die grenzen

[queen dome]

Die Intervall grenze ist 0-2a?

omgggggggggggggggggggggg

[leaden ermine]

ja 0 bis 2a

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ich bin im kopf die ganze zeit bei der Fläche zwischen 2 graphen und hab vergessen das wenn man zwischen graph und x-Achse den Inhalt will nur die Nullstellen nutzt

wow

mb

[leaden ermine]

kein problem

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und bei b?

[leaden ermine]

ja

kannst du in deinen worten sagen was die wollen

[queen dome]

Anhand der Quader Seiten die richtigen werte für den Hochpunkt (a/a^2) finden? mithilfe von dem flächeninhalt aus aufgabe a)?

[leaden ermine]

du meinst quadrat

[queen dome]

ya

ja

ja

[leaden ermine]

ja also es geht darum a auszurechnen

[queen dome]

ja?

[leaden ermine]

die information die gegeben ist, dass die parabel die selbe fläche hat wie das quadrat

[queen dome]

ja also die fläche aus a nehmen und wurzel ziehen?

[leaden ermine]

außerdem weißt du welche höhe das quadrat hat

[queen dome]

hab ich?

a^2?

[leaden ermine]

nicht wurzel ziehen

ja genau

was ist dann auch die breite von dem quadrat

[queen dome]

es ist ein quadrat

jede seite ist a^2 lang

oder nicht?

[leaden ermine]

ja

was ist dann die fläche von dem quadrat

[queen dome]

2a^2?

[leaden ermine]

nein

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ok :,)

[leaden ermine]

ja genau

mein discord spackt

also die höhe und breite ist a²

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alles güd

[leaden ermine]

die fläche von einem rechteck ist höhe × breite

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a^2 mal a^2

[leaden ermine]

ja!

was ergibt das

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a^4?

[leaden ermine]

ja

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ahhhhhhhhhhhhhhhhhh

[leaden ermine]

und das ist laut aufgabe gleich was

gleich welcher fläche

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unser Integrall

[leaden ermine]

ja und was war der wert

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hab ich noch nicht :,D

[leaden ermine]

doch

das ist das aus der a)

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jaa aber a hab ich noch net fertig ausgerechnet nur aufgestellt

:,)

[leaden ermine]

ja aber der wert steht eh schon in der a)😂

[queen dome]

ohh

[leaden ermine]

du sollst nur nachweisen und du hast ja das integral richtig aufgestellt

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aso ja

upsi

[leaden ermine]

kannst du später ausrechnen

jetzt bei der b) musst du einfach das gleichsetzen

[queen dome]

ja und F(x) ist ja -1/3x^3+2a^2

[leaden ermine]

Fläche Quadrat = Fläche Parabel

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a^4= 4/3a^3

[leaden ermine]

ax^2*

[queen dome]

jajajajaaa

[leaden ermine]

(-1/3)x^3 + ax^2

[queen dome]

jaja

die klammern darf ich net vergessen

gut

darf ich einfach geteilt durch a^3 rechnen und dann streichen?

oder geht das nicht?

[leaden ermine]

wo bist du gerade

du springst die ganze zeit rum

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sryy

a^4= 4/3a^3

[leaden ermine]

oh ja super

ok ja aber warum darfst du das?

[queen dome]

also darf ich da geteilt durch a^3 oder net?

[maiden folio]

j ammple

[leaden ermine]

was weißt du über a

[maiden folio]

kylian mbape

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das a a ist? und wir auf beiden seiten a stehen haben?

:,)

[leaden ermine]

a ist echt größer 1

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stimmt

mist

[leaden ermine]

das heißt du kannst durch a teilen ohne probleme weil ist eh niemals 0

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3/4 ist es aber nicht

[leaden ermine]

ne

a^4 = (4/3)a^3 wenn du durch a^3 teilst was bleibt jeweils übrig

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a= 3/4

[leaden ermine]

wieso 3/4

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ohhhhhhhhhhhhhhhhh

4/3

mann

ich bin so mannnnnnnnnnnnnn

[leaden ermine]

mit der 4/3 ist nix passiert

ja

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4/3

mannn ich bin immer ich drehe zahlen vergesse das - mit aus der aufgabe

mannn

würdest du mit mir noch mehr aufgaben in die richtung machen wollen?

[leaden ermine]

kann ich machen

[queen dome]

dankeeee wirklich

diese hab ich noch

ich hatte mit a angefangen mein ich

aber bei e funktionen tue ich mich immer schwer

[leaden ermine]

ironisch weil e funktionen sind sehr einfach zu ableiten

[queen dome]

jaaaaaaaaaaaaaaa

aber ich kp

ich bin in mathe immer blonder als blond und mach die blödesten fehler

so ansich kapiere ich diese Themen aber gleichzeitig auch null

[leaden ermine]

fühl ich

also zuerst einmal weißt du was ein wende punkt ist

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ofc

zweite ableitung gleich 0

[leaden ermine]

wo kommen welche ungefähr in frage im bild?

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und dritte ungleich 0

[leaden ermine]

ja ich meinte eher geometrisch haha

aber ja

[queen dome]

ungefähr bei 1.1 und bei -3?

[leaden ermine]

das ist ein punkt wo die funktion ihr krümmungsverhalten ändert

1.1 ist nicht ganz richtig

[queen dome]

1?

[leaden ermine]

-3 akzeptier ich

ne das ist ja ein tiefpunkt

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asoooooo jajaja 0

[leaden ermine]

ja da ungefähr

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hab auf die y-achse geachtet xd

mbmb

[leaden ermine]

so warum habe ich das gefragt

weil es ist nach dem punkt gefragt mit negativer steigung

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da wir "eine negative" steigung estimmen sollen

[leaden ermine]

ja

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alsoo die bei 0?

[leaden ermine]

und ich will dass du so eine idee bekommst was dich erwarten wird

[queen dome]

weil da unsere steigung - ist?

[leaden ermine]

negative nicht +

[queen dome]

jajaja

mann

[leaden ermine]

ok jetzt machst du folgendes

[queen dome]

okok?

[leaden ermine]

.

[leaden ermine]

.

Also ableiten

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x ableiten bei mir ist immer so eine sache :,)

[leaden ermine]

du brauchst hier eine regel

[queen dome]

Diese Kettenregel

oder Punktregel?

[leaden ermine]

produkt regel meinste

[queen dome]

die mit u mal v´ + u´mal v

?

[leaden ermine]

ja genau

[queen dome]

jaja

[leaden ermine]

die beiden brauchst du

also hier leitest du e^x aber (x-a)^2 nicht + e^x nicht abgeleitet aber (x-a)^2 schon

[queen dome]

ja??

ich denke

ja?

[leaden ermine]

ja

[queen dome]

lass mich gucken ob ich das auch so hab

[leaden ermine]

das ist nur der ansatz

[queen dome]

ich das da

hab

[leaden ermine]

hmm okay ja geht auch

wenn du es ausmultiplizierst würdest du keine kettenregel brauchen

aber ich empfehle das nicht

wenn du (x-a)^100 hast dann willst du lieber die kettenregel verwenden

[queen dome]

bei uns haben wir das immer so gemacht und das dauert immer so langggggg

[queen dome]

okidoki

[leaden ermine]

ok also deine ableitungen sehen top aus

[queen dome]

ja aber dann darauf kettenregel udn wieder zusammenfassen

dabei mache ich immer fehler

[queen dome]

genau

so ist das so viel schöner aufgeschrieben wow

[leaden ermine]

ok gut jetzt einfach einsetzen und am besten auch faktorisieren

[queen dome]

jaaa???

[leaden ermine]

,, f'_a(x) = e^x\cdot(x-a)^2+e^x(2x-2a) = e^x\cdot(x-a)^2+2e^x(x-a)

[queen dome]

unser 0?

[leaden ermine]

0?

[queen dome]

alles gutt :,)

[leaden ermine]

hier kannst du e^x und (x-a) ausklammern

ist das klar?

[queen dome]

wie? was?

jaa?

aso ja

[leaden ermine]

ok gut

[queen dome]

jaja

jajajaj alles gut

[leaden ermine]

das ist günstig wenn du es so aufschreibst weill wenn du es gleich 0 setzt kannst du die nullstellen instant ablesen

[queen dome]

jaa?

bestimmt

[leaden ermine]

ok jetzt brauchst du die zweite ableitung

[queen dome]

genau

[leaden ermine]

jetzt haben wir hier 3 faktoren

das ist bisschen ekelhaft deswegen mache ich folgendes

,, f'a(x) = e^x(x-a)(x-a+2) = \underbrace{e^x}{=u} \dotproduct \underbrace{((x-a)^2+2(x-a))}_{=v}

[queen dome]

okkkkkkkkkk

jajajjaaa

[queen dome]

2 raus und die beiden hinteren dadurch verbinden

[leaden ermine]

das (x-a)^2+2(x-a) kannst du auch eigentlich zsmfassen

x^2-2ax+a^2+2x-2a

[queen dome]

jaa

und das ableiten

die a falles weg? oder?

weil wir nach x ableiten?

[leaden ermine]

ja

[queen dome]

also 2x+2?

und dann noch halt e^x

kettenregel?

[leaden ermine]

,, f'a(x)=\underbrace{e^x}{=u} \dotproduct \underbrace{(x^2+(2-2a)x-2a+a^2)}_{=v}

[queen dome]

ich das da?

hab

[leaden ermine]

moment

[queen dome]

ja alles gut

[leaden ermine]

(x-a)^2 + 2(x-a) = x^2-2ax+a^2 + 2x-2a

da können wir die x terme fassen

x^2+(2x-2ax)-2a+a^2

x^2+(2-2a)x-2a+a^2

[queen dome]

kein sein ja?

[leaden ermine]

wir machen das um das leben einfacher zumachen

[queen dome]

okidoki

[leaden ermine]

jetzt wieder u' und v' finden

[queen dome]

u` bleibt wie u

[leaden ermine]

jaa

[queen dome]

2x+ 2-2a

[leaden ermine]

klasse

[queen dome]

jaa

wiw

woww

oha

ich bete ich bekomme das morgen auch hin

[queen dome]

wow

kurzfassen

[leaden ermine]

yep lass e^x klammern

[queen dome]

es geht theoretisch noch kleiner oder?

[leaden ermine]

,, f_a''(x) = e^x\left(x^2+x(4-2a)+(a^2-4a+2)\right) \stackrel{!}{=} 0

[queen dome]

ja genau

[leaden ermine]

so in der art

jetzt setzt du das gleich 0

[queen dome]

genau

[leaden ermine]

du musst also die quadratische gleichung lösen

[queen dome]

e^x=0

[leaden ermine]

jaa

ist das überhaupt möglich

[queen dome]

und dann alles in der klammer

nope

[leaden ermine]

richtig

[queen dome]

blitzz ⚡

das da hin mageln

[leaden ermine]

,, x^2+x(4-2a)+(a^2-4a+2) = 0 \text{ wobei } \begin{cases} a &= 1 \ b&=(4-2a) \ c&=(a^2-4a+2) \end{cases}

[queen dome]

dayyyyyyyyyyyyyyyyyyum

[leaden ermine]

wenn du mitternachtsformel verwendest

[queen dome]

mitternachtsformel haben wir nie gemacht

:)

nur pq :,)

[leaden ermine]

was dann

opfer

[queen dome]

ich weiß

und sowas sitzt im mathe Lk peinlich von mir ikik

[leaden ermine]

der nachteil ist bei pq muss der koeffizient von x^2 immer 1 sein

viele vergessen das :)

[queen dome]

jep

davor darf nichts stehen

daher wenn geteil durch die zahl

[leaden ermine]

ok also versuchs mal und dann sag wast du hast

[queen dome]

bei c jetzt?

[leaden ermine]

c?

[queen dome]

b ist ja 2?

[leaden ermine]

was für n b

[queen dome]

was?

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hier bei c

[leaden ermine]

die lsg von der quadratischengleichung

[queen dome]

und hier b

huhhhh

[leaden ermine]

die stehen rechts

[queen dome]

ja

[leaden ermine]

a ist ja eine zahl hier

konstante

[queen dome]

und bei c= pq formel?

[leaden ermine]

achso stimmt

[queen dome]

und b= 2

[leaden ermine]

b war p und und q c denke ich

[queen dome]

oder soo

p= 2

[leaden ermine]

wieso ist p = 2?

[queen dome]

a ist 1

[leaden ermine]

achsoooo

hab das voll übersehen 😭

[queen dome]

und 4-2mal1=b

ist 2

[leaden ermine]

dann hätten wir von anfang an alles mit a = 1 machen können bruh

[queen dome]

alles güd

stimmt

das

hab ich nicht beachtet

[leaden ermine]

weil so kann man sich leichter verrechnen

[queen dome]

indeed

[leaden ermine]

ok ja klar

p = 2 und q = -1

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1-4+2

[queen dome]

jaa

oder

ja

[leaden ermine]

ja

ich habe nur geguckt ob alles soweit passt rechnerisch

[queen dome]

alles güd

du multifunktionstalet einfach 2 leuten gleichzeitig helfen :,)

[leaden ermine]

das kein problem

manchmal ists schwieriger nur einer person zu helfen :,)

[queen dome]

WOW

okokokokok du genie

wow

an sich denke ich bekomme ich das morgen hin hast du noch iwi Tipps für Integrale, Funktionsscharen und Trassierung?

weil ich halt so null iwi plan davon hob

[leaden ermine]

noch nie von trassierung gehört haha

[queen dome]

ohh

soo

stetigkeit

und

differenzierbar

knickfrei

so zeugs

[leaden ermine]

ah ok ich seh

ich nur ein komisches wort

ja und

ja

also das sollte kein rpoblem sein für dich

also tipps hmm

[queen dome]

hoffe ich

[leaden ermine]

du solltest integrieren und ableiten drauf haben

aufgabe richtig lesen und so viele informationen daraus gewinnen wie möglich

[queen dome]

mein lehrer plediert dazu aufgaben die Man kann unrechenbar zu machen

[queen dome]

jaa dabei hab ich mega probleme xd

[leaden ermine]

ja deswegen haben wir auch lehrermangel

[gritty flax]

Gut tipps

[queen dome]

bei mir an der Schule gehts

aber der lehrer kapiert nicht das wir zu dumm für ihn sind

integrieren einfache sachen ja

aber wenn da aufeinmal ein bruch oder eine wurzel kommt

bin ich raus

[leaden ermine]

wieso

,, \sqrt[n]{x^m} = x^{m/n}

[queen dome]

da schaltet mein hirn aus

ja das weiß ich

[leaden ermine]

bei brüchen wenn sie kompliziert sind einfach quotientenregel

[queen dome]

?

kp was das ist :D

[gritty flax]

Und power rule funktionniert mit die quotienten power

[queen dome]

ok? :,)

[leaden ermine]

yakubros ist ein auslandsschüler

[gritty flax]

x^n => x^(n+1) / (n+1)

[queen dome]

die mathe power

ihr seid Mathe Mann

[leaden ermine]

ich bin nicht so gut

[queen dome]

"nenenenee nulllllllllllllll"

[gritty flax]

||Die Sache ist, ich habe nie den Wortschatz für Mathematik gelernt kekw||

[queen dome]

jaa darum bin ich überglücklich hier auf deutsche zu treffen

🫶

[leaden ermine]

yaku der matrose ist franzose

[queen dome]

könne wir noch was zu Funktionsscharen machen? weil kp was das macht

ich weiß nur das wir da 2 x a dinga haben

und das daher alles mit dem graphen möglich ist iwi

ich guck eben in mein buch ob ich aufgaben finde wenn das ok geht

wenn das ok geht bei euch

[gritty flax]

Funktionsscharen handelt sich dabei einfach um eine Möglichkeit, durch die Variation eines einzelnen Parameters mehrere Funktionen zu erhalten
(Sorry maybe some grammaire issue)

[queen dome]

alles gut

[gritty flax]

Ist das abitur ?

[queen dome]

jep

12.2

[gritty flax]

Ich seh

[queen dome]

Mein english geht fit

besser als Mathe

Die sachen oben waren auch Abi aufgaben aus den vorheriegen jahren

[leaden ermine]

damn

kindergarten

[queen dome]

Mit dir ja

allein weiß net

würdest du mit mir noch ein paar sachen machen wollen oder nicht?

Ich hätte noch mehr aufgaben

:,)

[gritty flax]

Beste Vorbereitung

[queen dome]

joaaa

hoffe ich

[leaden ermine]

könnte machen bis 23

[queen dome]

gerne dankee dir sooooooooo dolll

🫶

Nr 17?

die unterste?

Ich vermute das meine Klausur hierdrauf aufgebaut wird auf solchen Aufgaben

aber verstehen tue ich nichts

oder neh lieber 15 denke ich

Nr 15 oder Nr 13

[leaden ermine]

du entscheidest

[queen dome]

Lieber 13

sachaufgaben liegen mir schwer

[leaden ermine]

ok es geht darum iwas mit einer exponentialfkt zu modellieren

wie habt ihr exponentialfkt definiert

[queen dome]

gute frage

[leaden ermine]

wie sieht so eine aus

[queen dome]

weiß nuch noch exponentielles wachstum

a mal e ^k mal x

[leaden ermine]

ok

gibt auch ab^x

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aber ob es das ist was dort angesprochen wird ka

[queen dome]

ja das kenne ich auch

[leaden ermine]

aber man kann mit beides das modellieren

was wäre für dich besser denkst du

also eig ist es gleich

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mit k

[leaden ermine]

weil

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mit ln und so

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kp :D

[leaden ermine]

also was ich sagen will ist e^k und b ist beides das gleiche

[queen dome]

ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

aber ich denke mit k mag ich lieber

[leaden ermine]

beispiel wenn b = e ist dann wäre k = 1 einfach

b = e^1

[queen dome]

ahhhhhhhhhhhhhh

[leaden ermine]

ok gut!

[queen dome]

okok

dann lets go

[leaden ermine]

lets go lesen

[queen dome]

Null stelle bei 4000

:)

[leaden ermine]

ok glücklicherweise hast du eine tabelle

[queen dome]

ja

[leaden ermine]

von 4000 zu 3000 ist ungefährt eine reduktion von 80 % weil 0.8(4000) ~ 3000

[queen dome]

ohhhhhhhhhhhhhhh

[leaden ermine]

eigentlich 3200 aber ja so in etwas passts

[queen dome]

okok

unser a ist 4000?

[leaden ermine]

hast du vorher nullstelle gesagt?

[queen dome]

jein?

[leaden ermine]

das geht nicht : |

[queen dome]

wenns stimmt ja wenn nicht nein

aua ok

[leaden ermine]

es ist der y-achsenabschnitt

[queen dome]

neinnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnn

ahhhhhhhhhhh

ok

aber a ist immernoch 4000

bestimmt :,)

Lightningg

was

[leaden ermine]

ach komm

aber mein punkt ist du weißt ja von vorher e^x wird niemals null

das gleiche gilt auch für beliebige von ae^(kx)

[queen dome]

oh

ok

ja

[leaden ermine]

ok

aber ja eine Bedingung ist B(0) = 4000

was ist dann a?

[queen dome]

4000

da alles ^0 = 0 ist

[leaden ermine]

jaa

4000e^(kx) ok gut

[queen dome]

-k herausfinden mit ln

punkt einsetzen

aber erst

[leaden ermine]

moment

wir müssen heraus finden wie viel nach 1 meter da ist

mir ist auch aufgefallen die tabelle ist wahrscheinlich für c)

also nach einem meter sind 80 % gesunken

[queen dome]

nnenenenneneneneeeee

NEEEEEEEEEEEEEEEEEEEEEE

NEINNN mannnnnnnnnnnnnnn

[leaden ermine]

80 % des Wertes an der Oberfläche

[queen dome]

warum macht mathe das

warum

[leaden ermine]

chill du schaffst

du musst dich nur fokusieren

oberfläche waren es 4000

[queen dome]

also 4000 = 100%

[leaden ermine]

nach 1 meter sind 80 % davon

[queen dome]

wir wollen 80% da dies = 1m ist

okok

[leaden ermine]

wie viel sind 80 % von 4000

[queen dome]

Ich hab aufgepasst

3200

[leaden ermine]

gg

also B(1) = 3200 muss gelten

als zweite Bedingung

und jetzt lös für k

[queen dome]

aber passt a = 4000 noch?

okidoki

geteilt durch 4000

?

[leaden ermine]

ja

[queen dome]

und dann ln

ln

[leaden ermine]

ja

[queen dome]

k = ln(0.8)

[leaden ermine]

ops hab x = 1 vergessen

egal du checkst

[queen dome]

alles güd

jaja das habe ich in der letzten klausur gehabt

[leaden ermine]

jetzt guck mal

[queen dome]

8 Punkte :,)

[queen dome]

okidoki

Ich gucke

[leaden ermine]

,, 4000e^{\ln(0.8)x} = 4000(e^{\ln(0.8)})^x = 4000(0.8)^x \text{ also } b = 0.8 = e^{\ln(0.8)} = e^k

[queen dome]

ja

jajajaj

[leaden ermine]

mit beiden formeln kannst du das problem modellieren

[queen dome]

icher verstehe

[leaden ermine]

mit e kann man schöner ableiten

ok perfekt

[queen dome]

ist er

er denkt wir schecken tun wir aber nicht

[leaden ermine]

aber egal jetzt

[queen dome]

ist notirt

notiert

[leaden ermine]

wenn wir schon dabei sind

ln(ab) = ln(a)+ln(b) und ln(a/b) = ln(a)-ln(b)

kann beim ableiten nützlich sein

bsp

[queen dome]

ohhhhhhhhhhhhh

[leaden ermine]

viel einfacher wa

[queen dome]

ja

jajajjaa

ja

[leaden ermine]

als unnötig produkt regel

ok gut

[queen dome]

notiert

mega danke

[leaden ermine]

ok

gut

[queen dome]

👍

diese Differenzierbarkeit bekomme ich denke ich hin

und Integrale bete ich auch

Funktionsscharen guck ich mir morgen früh nochmal an

das klappt save 🥲

[leaden ermine]

Funktionenscharen was gibts da zu wissen?

[queen dome]

gute frage

[leaden ermine]

weil in abhängikeit von a kriegst du unterschiedliche funktionen

die aber sich bissel ähneln

[queen dome]

jaja das weiß ich und so mit (a>0)

und so

[leaden ermine]

ja das amchen so lehrer die faul sind

die sagen so bestimmen sie für a = 0 oder a = 1 dies und das

[queen dome]

wir müssen von denen so Wendepunkte und Extrempunkte berechnen

[leaden ermine]

ja und da komtms aufs ableiten an

[queen dome]

ja in etwa

[leaden ermine]

und was die bedingungen waren

[queen dome]

jaa bindingungen jaja

[leaden ermine]

ein tipp bei maxima und minima

[queen dome]

ja?

[leaden ermine]

wenn du ein interval gegeben hast, vergiss nicht die randpunkte abzuchecken

[queen dome]

jaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

jajajja

okokok

[leaden ermine]

ok alles klar

ich finde du bist eigentlich fit

[queen dome]

hoffe ich :,)

[leaden ermine]

also du checkst schnell und viel

und kannst auch

[queen dome]

normalerweise weiß mein gehirn speicher mehr als ich denke

[leaden ermine]

versuch einfach morgen ruhig und entspannt zu bleiben

und vorallem nicht mit den leuten vor der klausur sein die filme schieben

das lenkt dich nur ab

[queen dome]

jajaja 👍

dankeeeeeeeeeee dirrrrrrrrrrrrrr 🫶

soooooooo megaaaaaaaa viel

[leaden ermine]

kein problem

viel glück

[queen dome]

soll ich dich auf dem laufenden halten oder net?

[leaden ermine]

kannst du machen :D

[queen dome]

dankeeeeeeeeeeeeeeee ich nehme all das glück was ich bekomme

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[lucid mulch]

An observer is located 400 feet from a building. The observer notices a kite flying at a 29° angle of elevation from his line of sight. How high is the kite flying over the building? You must show all work and calculations to receive full credit.

[tight mortar]

isn’t this the same as the other question u asked? just with different numbers

it’s the same work and answer just with the numbers shifted

[lucid mulch]

Yeah, I know. But, I got the answer 193.92 ft, and I think I miscalculated.

Oh nvm, I got it actually now

[torn jolt]

Show your calculations

[lucid mulch]

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[soft idol]

I need help with problem 57

[astral sinew]

what have you tried

[soft idol]

taking out d on the bottom

thats all i can think to do

[glacial stream]

I mean that and maybe get the surd to numerator

[weak bolt]

7 = 4 + 3

[glacial stream]

I don't think there's much else to do

[glacial stream]

$$\sqrt[7]{x} = (x)^{\frac{1}{7}}$$

[soft idol]

it says this is the answer

[glossy valve]

HitenTandon

[glacial stream]

Yea so just get the surd into the numerator

[soft idol]

whats surd

[weak bolt]

$\sqrt[4]{d^4 \times d^3}$

[glossy valve]

HitenTandon

╰ 𝕃 𝕌 ℂ 𝕀 𝔽 𝔼 ℝ ╮

[glacial stream]

Oh you meant to do that yea, he has already done that

from his own admission

[soft idol]

this is just what i did

[glossy valve]

HitenTandon

[soft idol]

shit heah

because you cant have a radical on bottom right

[glacial stream]

Yea

[soft idol]

do i multiply the d 4root d^3

or just the radical

[glacial stream]

You multiply a number which removes the radical from the denominator

if you multiply $\sqrt[4]{d^3}$ on both, you'll get $d^2\sqrt[4]{d^2}$, right?

[glossy valve]

HitenTandon

[glacial stream]

And that's not what you want

So you'll multiply it with $\sqrt[4]{d}$ on both numerator and denominator instead

[glossy valve]

HitenTandon

[glacial stream]

.

[soft idol]

shit yeah because you want it to come oht

making it d4 and not d3

ok

i have the top right

but not thebottom

it says its d2

but i got d5

[glossy valve]

HitenTandon

[soft idol]

i just figures

its d4 then 4root d4

chill bro

[steel solar]

But what’s $\sqrt[4]{d^4}$?

[glossy valve]

;(

[soft idol]

just d?

[steel solar]

Yeah.

So what’s d times d?

[soft idol]

i already have the d^4 tho

[steel solar]

Where?

[glacial stream]

That's not $d^4$ that's the 4 from the $\sqrt[4]{d^3}$

[glossy valve]

HitenTandon

[steel solar]

?!

Notation abuse?

[soft idol]

oh shit

getting caught in my sloppy work again

[glacial stream]

😭

[soft idol]

thanks

[glacial stream]

Alr, if your query has been solved, please use .close to close the thread, also, please don't create multiple threads for your problem

[soft idol]

thats fine buddy

just think its outrageous im getting ignored while other people are getting helped lmao

[steel solar]

?!

[glacial stream]

Sometimes it can take time to reply, but Helpers do come, but I was there on your thread when it was closed

I was legit reading your question as you closed it

[soft idol]

dang lol

[glacial stream]

We can only help so many peoplpe at once

Please wait your turn

[soft idol]

once i sent mine before theirs?

[glacial stream]

It can still happen

It's not about when you sent it, sometimes it may happen that some helper who's helping other people doesn't even know how to help with your query

[soft idol]

k buddy

.close

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[steep hazel]

can someone help me with polar curves?

[steep hazel]

heres the question:

finding the area under the cruves

[steel solar]

Okay.

What have you tried?

[steep hazel]

i did integral from 0 to pi/2

[steel solar]

You also know the area for a polar curve is given by $\int_{\alpha}^{\beta}\frac12r^2d\theta$ for $\alpha\le\theta\le\beta$, right?

[steep hazel]

of (1/2)(1+costheta)

[glossy valve]

;(

[steel solar]

Square (1+cos(theta)).

[steep hazel]

sorry yeah i squared it

wheni put it in a calculator i get 2.178

but my answer key says 1.885

[steel solar]

Hmm.

,w 1/2 times integral from 0 to pi/2 of (1+cos(x))^2

[glossy valve]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by exiled_hype

[steel solar]

Looks about right.

[steep hazel]

so is the answer key wrong then?

[steel solar]

,w plot r=1+cos(theta)

[glossy valve]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by exiled_hype

[steel solar]

Yes.

[steep hazel]

well more specifcially, it says (3pi/8) +(sqrt2)/2

if that helps

[steel solar]

Uh, ok.

You wanna work out the integral?

Just notice that $\cos^2(x)=\frac12(1+\cos(2x))$.

[glossy valve]

;(

[steep hazel]

why am i noticing that

oh

okay

thanks i got it

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[half oyster]

how can I find d/du?

I'm 99% sure it's chain rule but like I'm kinda bad at it

I think chain rule u go from outer function to inner, so how would that work?

or should I do product rule twice?

[slender bane]

holy heck, alright, I think you go product&chain crazy here

do you know how to do product rule with three fuctions?

[half oyster]

nah

should I just solve the first 2 first?

x^9 * (x+2)^3

[slender bane]

alright so if you have y=f(x)h(x)g(x)

[half oyster]

then whatever that gives run it back with the other part?

[slender bane]

I can tell you the rules cause it is easily memerable, it is proved with that as well.

[slender bane]

y'=f(x)h(x)g'(x)+f(x)h'(x)g(x)+f'(x)h(x)g(x)

[half oyster]

oh so it's the same u just add another part

[slender bane]

yeah

[half oyster]

so what's the chain rule then?

[slender bane]

first of all you gotta do chain to the overall function right?

[half oyster]

1/u * u' I think

[slender bane]

and then when you do the inside, you gotta split it into three function, each function needs to be chained as well

[slender bane]

yes that is correct

[half oyster]

okay so y' = 1/[f(x)h(x)g(x)] * [f(x)h(x)g'(x)+f(x)h'(x)g(x)+f'(x)h(x)g(x)]

[slender bane]

well except for x^9, that derivitive is easily see-able

[slender bane]

absolutely correct

[half oyster]

okay bet

makes sense, thanks so much

[slender bane]

np

[half oyster]

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[errant plank]

yo wsg

[modest obsidian]

state your problem or close this channel please @errant plank

[onyx glen]

!redir

[full forum]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[modest obsidian]

can you manually close this channel seems like he wont use it

[onyx glen]

i will let it time out or let him speak

but i am not closing it until there's good cause to

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[serene maple]

Hi

[serene maple]

is x⁰ = 1ˣ true?

[void nova]

Yes (as long as x ≠ 0)

[serene maple]

Ok thanks

[rapid rain]

yeah x = 0 is up for debate (0^0)

best not dabble into it if you're not ready to stand your ground

xd

[serene maple]

ok thanks

One more thing for this are my answers correct
(i) Trapezium
(ii) AB = 𝑥+4
(iii) CD = 3(𝑥+4)
(iv) 𝑥=9.125
(v) AB = 13.125 cm, CD = 39.375 cm

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@serene maple Has your question been resolved?

[knotty grail]

all correct

[serene maple]

ok thank you

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[solemn sand]

Like dy/dx is eqn of slope of tangent to the curve which is will slope of tangent at different points as we put different values of x that lie on that curve, similarly is d²y/dx² smth that also gives slope or anything curve related?

[umbral gale]

acceleration

uh with respect to time but

second derivative of a function tells you about the concavity of a function

[hybrid zephyr]

Physics?

[narrow path]

position derivative wrt time is velocity

than you get the second derivative and its acceleration

[hybrid zephyr]

He mentioned if it was time....

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[cold anvil]

I cant manage to get the answr

Can someone tell me what i did wrong?

[acoustic forum]

You did +24 when it's -24

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[cold anvil]

Tqtq

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[finite marlin]

Hello,

[plush mortar]

hi

[finite marlin]

I have a problem about the calculus of a determinant in python, can anyone help me in one hour please ?

[plush mortar]

do you mean calculation of a detemrinant

?

[glacial stream]

!da2a

[full forum]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[finite marlin]

Yes

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[obtuse valley]

help please

step 1 i need to take derivative?

[lime ether]

there is no question here

[obtuse valley]

no thats cheatsheet

the absolute extrema

[gritty rose]

can you write a complete question what you want to know

[obtuse valley]

alright i found cheat sheet online

step 1 to find absolute extrema

suppose find critical point f’(x) right but they put f(x)

[lime ether]

what

[gritty rose]

https://tutorial.math.lamar.edu/classes/calcI/CriticalPoints.aspx

suppose find critical point f’(x)
that's incorrect

[obtuse valley]

oh ok

so in cheatsheet incorrect?

[gritty rose]

no

suppose find critical point f’(x)
is incorrect

the cheatsheet doesn't say that

[obtuse valley]

i think i got it

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[opaque locust]

for part D

[opaque locust]

how come we dont need to do a +5

for the total distance

[boreal slate]

for which part?

oh part d

you shouldnt need a +5

but it seems like the particle will change directions and you need to account for that

[opaque locust]

okay

also

for part c

am i doing something wrong with the calculator

cause i keep getting 0 even though the answer is supposed to be like 1.02 or something

[boreal slate]

you have to be careful with getting derivatives

what you are differentiating is a constant

what you need to do is notate it like

[opaque locust]

?

[boreal slate]

$\frac{d}{dt} \vert_{t=1} v_Q(t)$

[glossy valve]

LocalLunatic

[opaque locust]

tyy

[boreal slate]

yeah like that

[opaque locust]

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[rapid solar]

Can someone tell me why In this question the domain is only the first quadrant but we take the second quadrant as well? I asked my teacher in class and he said because the domain has no relation with quadrants?? Is that true?

[zenith pewter]

0<x<π/2
0<2x<π

[rapid solar]

thank you!!!

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[mortal marten]

im getting 3 integer values for a after defining the mod and then using wavy curve but the q says there has to be a minimum of 4

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[cold sail]

do u know if i can assume the flow is at rest at height delta h

[umbral dome]

typically the velocity at a free surface is negligible, so yes

[cold sail]

in such qus tho why are we sometimes given the cross section then

if the qu is very solvable without using the cross section

[umbral dome]

well it's solvable given the assumption that the cross section of the pond is much larger than the cross section of the pipe (which is reasonable in this case)

[cold sail]

like in here isnt this solvable without using the cross section and just assuming the flow to be at rest at the height h_0

or is there somethign im missing here

[umbral dome]

well that's only the cross section of the pipe itself

which wouldn't pertain to bernoulli's equation since we would also need the cross section of the tank to find a velocity, but might be relevant to finding the volumetric flow through the pipe

[cold sail]

cant u just directly use bernoulli's and say something like p_atm / p + g(h_0 + h_p) = u^2_0/2 + p_atm/g + g(0)

and then rearrange for u_0 to get the speed?

[umbral dome]

you're correct that you don't need the cross section of the pipe it for part a, it might be useful for later parts

[cold sail]

aight i shall give this qu a go then

[full forum]

@cold sail Has your question been resolved?

[cold sail]

im stuck on part a 💀

it says to write the speed as a function of h but using p_atm / p + g(h_0 + h_p) = u^2_0/2 + p_atm/g + g(0) i get that the speed is constant

what am i doing wrong here

[umbral dome]

shouldn't you be using the height h rather than the initial height h_0?

[cold sail]

but

the flow isnt at rest at height h i swear

[cold sail]

even if that's the case when i use bernoulli's shouldnt it give me a non constant answer then? unless im applying it wrong?

[umbral dome]

well if you only use the initial height then you are assuming that it stays at that water level forever, in which case the height would have to be constant

[cold sail]

oh ye

cz bernoulli's equation only holds along streamlines

but as the height drops then so will the streamline

implying that point is no longer on the streamline

[umbral dome]

you want to apply bernoulli's to the point at wherever currently happens to be the water surface

[cold sail]

@umbral dome is it the smae idea in this qu then

we dont assume the water to stay at a constant level of delta h

[umbral dome]

well for that one i think you are meant to assume that the pond is so large that the volume doesn't drain significantly

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[quartz remnant]

i need help seeing a vision for where to go with this

[thick hedge]

maybe write the numerator as cos(a-b)-cos(a+b)

and see where that takes you

[quartz remnant]

ok

[thick hedge]

If that doesn't work, brute force it

[quartz remnant]

what does that mean

[willow sun]

tan2x=2tanx/1-tan²x

And try subbing in tan²x from the first eq

[quartz remnant]

i tried that but didnt know where to go after

[paper rain]

is that x-a and x-b?

[quartz remnant]

ye

im not getting anywhere

@thick hedge yea idk what to do

[thick hedge]

Hmm, one minute

[quartz remnant]

kk

[thick hedge]

$tan^2(x) = \frac{tan(x) - tan(a)}{ 1+ tan(x) tan(a)} \frac{ tan(x) - tan(b)}{ 1 + tan(x) tan(b)}$

[glossy valve]

What a wonderful world !

[raw dirge]

you almost definitely need to simply and use sum to product

do you know the sum to product formulae?

[sacred cloak]

expand the denominator and divide both sides by cosacosb I think

[quartz remnant]

ugh i hate that stuff

but yes

i know it

ok idk who to listen to lmao

[sacred cloak]

$\frac{2sin(a)sin(b)}{sin(a)cos(b)+ sin(b)cos(a)} * \frac{cos(a)cos(b)}{cos(a)cos(b)}$

[sacred cloak]

Im just testing stuff out so don't listen to me 💀

[quartz remnant]

lmao alr

[glossy valve]

rawr

[sacred cloak]

$= \frac{tan(a)tan(b)}{tan(a) + tan(b)}$

[glossy valve]

rawr

[sacred cloak]

Idk what to do with this

but you can maybe compare it to the double angle formula for tan2x, it seems similar

[quartz remnant]

ok

[raw dirge]

how many lines of working are u expected for this

[quartz remnant]

dk

i was just given that screenshot

[full forum]

@quartz remnant Has your question been resolved?

[inland moth]

uh not related to answering this question at all but there are commands for sin cos and tan

$\sin{x}$

[glossy valve]

Shioshi

[full forum]

@quartz remnant Has your question been resolved?

[languid orchid]

e

eee

[glossy valve]

@glacial stream

[quartz remnant]

i am

ty

i think im making progress

[glacial stream]

Cool

[quartz remnant]

.close

[full forum]

Channel closed

Closed by @quartz remnant

Use .reopen if this was a mistake.

[full forum]

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[sturdy fern]

help with ci and cii

[bright bronze]

Z1OZ2 is given to be equilateral triangle

[sturdy fern]

yes

[bright bronze]

so, that gives you two hints

OZ1 = OZ2

[sturdy fern]

60 degress anbd same side?

[bright bronze]

and the angle Z1OZ2 = 60

[bright bronze]

yea

[sturdy fern]

but what does it mean to express z1 in z2?

[bright bronze]

You need a single equation that relates z1 and z2 based on the given information

[sturdy fern]

im not sure what that would be

[bright bronze]

You are given the complex numbers in the form Ae^theta

now you know the relationship between the angles and the magnitudes

so you can relate r1 and r2

[sturdy fern]

same?

[bright bronze]

and similarly theta and alpha angles

[sturdy fern]

r1 and r2

[vapid notch]

the argument of z1 is theta+60

[bright bronze]

yea

[vapid notch]

and the modulus of both are equal

i.e r1=r2

[bright bronze]

maybe you can let the op think this by themselves