Outside temperature over a day can be modeled as a sinusoidal function. Suppose
ID: 2920108 • Letter: O
Question
Outside temperature over a day can be modeled as a sinusoidal function. Suppose you know the high temperature of 87 degrees occurs at 4 PM and the average temperature for the day is 70 degrees. Find the temperature, to the nearest degree, at 5 AM.Outside temperature over a day can be modeled as a sinusoidal function. Suppose you know the high temperature of 87 degrees occurs at 4 PM and the average temperature for the day is 70 degrees. Find the temperature, to the nearest degree, at 5 AM.
Outside temperature over a day can be modeled as a sinusoidal function. Suppose you know the high temperature of 87 degrees occurs at 4 PM and the average temperature for the day is 70 degrees. Find the temperature, to the nearest degree, at 5 AM.
Explanation / Answer
comparing with D(t)= Asin(B(t+C)) +k
high temperature for the day is 87 degrees and the average temperature for the day is 70 degrees
k=70
A=87-70
A=17
period =24 hours
2/B =24
=>B= /12
D(t)= 17sin(( /12)(t+C)) +70
4 pm is 16 hours from midnight , high temperature of 87 degrees occurs at 4 PM
D(16)=87
=> 17sin(( /12)(16+C)) +70=87
=> 17sin(( /12)(16+C)) =17
=>sin(( /12)(16+C)) =1
=>(( /12)(16+C)) =/2
=>(16+C) =6
=>C=-10
so function is D(t)= (17*sin(( /12)*(t-10))) +70 or you can use D(t)= 17sin(( /12)(t+14)) +70
5 pm is 17 hours from midnight
temperature at 5PM =D(17)
temperature at 5PM = (17*sin(( /12)*(17-10))) +70
temperature at 5PM = (17*sin(7 /12)) +70
temperature at 5PM = 86.42
temperature at 5PM = 86 degrees
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