Samband och förändring: Skillnad mellan sidversioner
Hakan (diskussion | bidrag) |
Hakan (diskussion | bidrag) |
||
Rad 79: | Rad 79: | ||
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<br /> | |||
=== En övning på räta linjen med Geogebra === | |||
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{{clear}} | |||
=== Sidorna 204-208 === | === Sidorna 204-208 === |
Versionen från 20 november 2012 kl. 08.35
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
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.
- Lutning på GeoGebra.se
- taxifärd från Geogebrainstitutet
- linjär funktion med glidare från Geogebrainstitutet
Här kommer en grafisk lösning till exempel 2 på sidan 206 (GeoGebra):
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En övning på räta linjen med Geogebra
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Sidorna 204-208
onsdag
Klurig läxa
4.4 Proportionalitet
Direkt efter diagnosen
Titta på denna länk
- Uträkning till sista uppgiften på Diagnos 9 i Wolfram Alpha.
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)
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
Repetition: Övning på räta linjens ekvation:
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Den kommer från: GeoGebra.org och är naturligtvis fri att använda enligt CC.