rachelmaddie
Full Member
- Joined
- Aug 30, 2019
- Messages
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Hi. Can someone please check my work for this problem?
a
Amplitude = maximum - minimum/2 = (77-40)/2 = 18.5
b
Vertical shift = maximum +minimum/2. (77 + 40)/2 = 58.5
c
The period of the function is (data range/# of cycles), which in this case 24 hrs/1 = 24 hrs
d
We know that t =0 at 5 pm, so we want to find a formula for the temperature as a function of time f(t) where f(0) = the temperature at 5 pm (77 degrees)
So now we need to calculate the actual value of our phase shift, which is 25%*period = 25%*24 = 6
So our period = 24 = 2*pi/B
We start with:
24 = 2*pi/B
Multiply both sides by B:
24*B = (2*pi/B)*B
Simplify - the B's on the right side cancel out:
24*B = 2*pi
Divide both sides by 24:
(24*B)/24 = (2*pi)/24
Simplify again - the 24s on the left side cancel out:
B = pi/12
To write a sinusoidal function that models this temp. variation use the formula:
y = A sin(B(x + C)) + D
where Amplitude is A
Period is 2pi/B
Phase shift is C (positive to the left)
Vertical shift is D
Substituting this into formula we get:
y = 18.5*sin((pi/12)*(t + 6)) + 58.5
e
Find the value of t at 10 AM, replace t in the equation we found with that value, calculate the value of the equation with that t, and then compare the result to the recorded temperature at 10 AM
Since t = 0 at 5 pm, then t = -1 at 4 pm, -2 at 3 pm, t = -7 at 10 am.
Replace t with -7 in the equation we found for part d and calculate the value of the equation.
y = 18.5*sin((pi/12)*(-7 + 6)) + 58.5
y = 53.7118476656
The model’s temperature at 10AM is 52 degrees.
In comparison, the numbers are close but not identical, which is to be expected because one is a real measurement and the other is from a model.
a
Amplitude = maximum - minimum/2 = (77-40)/2 = 18.5
b
Vertical shift = maximum +minimum/2. (77 + 40)/2 = 58.5
c
The period of the function is (data range/# of cycles), which in this case 24 hrs/1 = 24 hrs
d
We know that t =0 at 5 pm, so we want to find a formula for the temperature as a function of time f(t) where f(0) = the temperature at 5 pm (77 degrees)
So now we need to calculate the actual value of our phase shift, which is 25%*period = 25%*24 = 6
So our period = 24 = 2*pi/B
We start with:
24 = 2*pi/B
Multiply both sides by B:
24*B = (2*pi/B)*B
Simplify - the B's on the right side cancel out:
24*B = 2*pi
Divide both sides by 24:
(24*B)/24 = (2*pi)/24
Simplify again - the 24s on the left side cancel out:
B = pi/12
To write a sinusoidal function that models this temp. variation use the formula:
y = A sin(B(x + C)) + D
where Amplitude is A
Period is 2pi/B
Phase shift is C (positive to the left)
Vertical shift is D
Substituting this into formula we get:
y = 18.5*sin((pi/12)*(t + 6)) + 58.5
e
Find the value of t at 10 AM, replace t in the equation we found with that value, calculate the value of the equation with that t, and then compare the result to the recorded temperature at 10 AM
Since t = 0 at 5 pm, then t = -1 at 4 pm, -2 at 3 pm, t = -7 at 10 am.
Replace t with -7 in the equation we found for part d and calculate the value of the equation.
y = 18.5*sin((pi/12)*(-7 + 6)) + 58.5
y = 53.7118476656
The model’s temperature at 10AM is 52 degrees.
In comparison, the numbers are close but not identical, which is to be expected because one is a real measurement and the other is from a model.