Samband och förändring: Skillnad mellan sidversioner

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Rad 82: Rad 82:


=== Öka din förståelse av räta linjen med Geogebra ===
=== Öka din förståelse av räta linjen med Geogebra ===
Denna GGB är amerikansk och k och m heter


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Versionen från 28 november 2012 kl. 23.08

Kapitel 4 handlar om Samband och förändring och består av 14 delar.

4.1 Procent

Procentbegreppet och tre problemtyper, 174-178

måndag

Promille och ppm, 178-180

tisdag

Procentenheter, 181-183

Detta gör vi på tisdagen i vecka 45.

Mikael Bondestam om skillnaden mellan procent och procentenheter.

Förändringsfaktor, 184-188

ons

Index, 189-191

fre

Kul grej: Bråk, decimal procent i GeoGebra

Ränta, 192-195

måndag

Genomgång diagnos 8

Uppg5

Börja med att som repetition göra uppgift 5 från Diagnos 8. Gör det på det krångliga sättet (elevlösning) och jämför med hur enklet det blir med hjälp av förändringsfaktorn.

Kolla: Wolfram Alpha är enastående på uppgift 10.

Genomgång:

Exempel 1 på sid 193 i boken Det lönar sig att lösa uppgiften i Excel.

4.2 Funktionsbegreppet

Vad är en funktion? 196-200

tis

Följande GeoGebrafil har jag gjort själv. Den visar tre sätt att rita parabler.

genom att mata in ordet parabel samt ange tre punkter 
genom att angen linje och en punkt (styrlinje och brännpunkt
genom att ange funktionen (inklusive start och stoppvärden = definitionsmängd)

Observera att de tre punkterna I, G, H på den blå parabeln motsvarar lösningen på Exempel 1 i boken sidan 197.

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Definitionsmängd och värdemängd, 201-203

tis

GeoGebra: jag instruerar och eleverna prövar att rita en trinangel med omskriven cirkel.

4.3 Linjära funktioner

Repetition: Förra gången stiftade vi bekantskap med en parabel som naturligtvis låter sig ritas i GeoGebra. En av kurvorna är precis den som kommer ur bokens Exempel 1 på sidan 197.

Man kan naturligtvis rita kurvan i Wolfram Alpha oxå. Det är bara att högerlicka på uttrycket i GeoGebra och kopiera till inmatningsfältet. Wolfram Alpha finns förresten som en Gadget till er som har Vistra eller 7:an.

Här kommer en grafisk lösning till exempel 2 på sidan 206 (GeoGebra):

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Öka din förståelse av räta linjen med Geogebra

Denna GGB är amerikansk och k och m heter

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Sidorna 204-208

onsdag

Klurig läxa

Tristan och Isolde

4.4 Proportionalitet

Direkt efter diagnosen

Titta på denna länk

Sen har jag gjoret en busenkel GeoGebra om räta linjens ekvation (linjära funktioner). Den är gjord i tre steg. titta i konstruktionsprotokollet. Visa/ konstruktionsprotokoll.

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Denna GGB är min egen och naturligtvis fri enligt CC som allt på denna sida.--hakan 21 november 2011 kl. 08.29 (UTC)

Övningar

Här är en som är enkel:http://geogebratube.org/student/m23347

http://geogebratube.org/student/m23346

Här är en bra men den ser inte snygg ut i Mac-Kan fixas till.


De bör bäddas in i sidan.

Direkt proportionalitet, 209-212

fre

Direkt proportionalitet är å ena sida enklare än räta linjen. Det är ett specialfall när m = 0. Det betyder att linjen går genom origo.

Å andra sidan dyker proportionaliteten upp i en mängd sammanhang i exempelvis fysiken. Här kommer ett sträcka-tid-diagram (st-diagram). Det är teoriavsnittet i boken sid 209.

Teori

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Exempel 1

Kommer snart

Exempel 2, sid 210

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Fler proportionaliteter, 213-215

måndag

Genomgång av Veckodiagnosen

Vi går igenom uppgift 3 och 5 från Diagnos 9. Trean kommer nedan men femman var enbart på tavlan.

Uppgift 3 löd så här:

3. Ulla lånar 180 000 för att köpa en bil. Lånet är med rak amortering på sex år och räntan är 5,6 %. Hur mycket måste Ulla betala varje månad?

Detta kan bli en mycket jobbig uppgift om man ska ge ett svar för varje månad. Det är ju 72 månader på sex år. Här får man själv göra några avgränsningar av uppgiften så att den blir rimlig.

Till att börja med kan man ju visa att man förstår att rak amortering innebär att beloppet delas upp i lika stora delar per månad.

180 000 / 6 = 30 000 kr per år
30 000 /12 = 2 500 per månad i amortering

Till detta kommer en ränta på det kvarvarande beloppet. Räntan kommer därför att sjunka månad för månad.

Här kan det räcka med att visa vad räntan blir för två eller tre månade, exempelvis efter en månad, 12 månader och 24 månader.

Excel

Om man vill kan man göra en kalkyl i Excel över lånekostnaden månad för månad.

Algebraisk lösning av uppgift 3

180 000 kr ==> Amortering 2500 per månad
ränta 5.6 % ==> förändringsfaktorn 1.056
månad     lån [tKr]        räntekostnad          att betala
 1        180              180*1.056             2500+180*1.056
 2        177.5            177.5*1.056           2500+177.5*1.056 
 3        175              175*1.056             2500+175*1.056 
 ..
 n        180-2500(n-1)    180-2500(n-1)*1.056   2500+(180-2500(n-1))*1.056

Månadskostnaden för månad nummer n är alltså 2500+(180-2500(n-1))*1.056

Intro - Önskebrunnen på Tom Tits

Beräkning av djupet

s = at2/2

Vi hjälps åt med att ta

Mätning av djupet

En annan metod är att ta ett måttband och mäta djupet. Det visade att brunnen var ungefär 5 m djup.

Teori - proportionaliteter

5x2 - funktionen ovan ser ut så här i GGB. Observera att funktionen kan beskrivas med hjälp av ord, värdetabell, graf och funktion.:

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På sidan 214 finns en rosa ruta med sex olinjära funktioner. Skriv in dem i inmatningsfältet i GGB och undersök hur de ser ut.

4.5 Potensfunktioner

Sidorna, 216-217

tisdag

4.6 Exponentialfunktioner

Sidorna, 218-222

ons

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4.7 Mer om grafiska lösningar

Sidorna 223-230

onsdag

Teori

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Vektorer

Vi skulle behöva repetera vektorer helt kort.

Öva matte

Khans Academy

Repetition: Övning på räta linjens ekvation:

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Den kommer från: GeoGebra.org och är naturligtvis fri att använda enligt CC.