Forumet - Matte 4, trigonometriska funktioner hjälp!

Matte 4, trigonometriska funktioner hjälp!

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The number of sunspots on the solar surface is strongly correlated with the ejection of charged particle from the sun, causing northern lights seen on Earth. The number of sunspots shows a periodic behavior (the sunspot cycle) and may be described by a simplified mathematical model in the following way

N(t)= 50cos(0.571t)+60

where t is the time in years.

•Also determine the maximum rate at which the number of sunspots changes.

Det här är mitt tal, och jag förstår ju att jag ska derivera funktionen och hitta maximipunkten för derivatan (-28.55sin(0.571t) = 0) men vet inte riktigt hur jag ska fortsätta härifrån. Någon som kan förklara? 

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Mathematical model


 

N(t)=number of sunspots

 

N(t)=50∙cos(0,571∙t)+60


 

Determine the period of the sunspot cycle


 

0,571∙(t+period)=0,571∙t+2∙π

 

0,571∙t+0,571∙period=0,571∙t+2∙π

 

0,571∙period=2∙π

 

period=(2/0,571)∙π

 

period≈11,0 (years)


 

Determine the maximum number of sunspots


 

-1≤cos(t)≤1

 

-1≤cos(0,571∙t)≤1

 

maximum(cos(0,571∙t))=1


 

maximum(N(t))=maximum(50∙cos(0,571∙t)+60)

 

maximum(N(t))=maximum(50∙cos(0,571∙t))+60

 

maximum(N(t))=50∙maximum(cos(0,571∙t))+60

 

maximum(N(t))=50∙1+60

 

maximum(N(t))=50+60

 

maximum(N(t))=110


 

Determine the minimum number of sunspots


 

-1≤cos(t)≤1

 

-1≤cos(0,571∙t)≤1

 

minimum(cos(0,571∙t))=-1


 

minimum(N(t))=minimum(50∙cos(0,571∙t)+60)

 

minimum(N(t))=minimum(50∙cos(0,571∙t))+60

 

minimum(N(t))=50∙minimum(cos(0,571∙t))+60

 

minimum(N(t))=50∙(-1)+60

 

minimum(N(t))=-50+60

 

minimum(N(t))=10


 

Determine the maximum rate at which the number of sunspots changes


 

N'(t)=rate at which the number of sunspots changes


 

N'(t)=50∙0,571∙(-sin(0,571∙t))+60∙0

 

N'(t)=-28,55∙sin(0,571∙t)+0

 

N'(t)=-28,55∙sin(0,571∙t)


 

-1≤sin(t)≤1

 

-1≤sin(0,571∙t)≤1

 

-1≤-sin(0,571∙t)≤1

 

maximum(-sin(0,571∙t))=1


 

maximum(N'(t))=maximum(-28,55∙sin(0,571∙t))

 

maximum(N'(t))=28,55∙maximum(-sin(0,571∙t))

 

maximum(N'(t))=28,55∙1

 

maximum(N'(t))=28,55

 

maximum(N'(t))≈29 (number per year)