# Can anyone help?

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Draw a scatter diagram of the data for one period. Find the sinusoidal function of the form y=A sin(ωx-ϕ)+β that fits the data. Show your work.

Draw a scatter diagram of the data for one period. Find the sinusoidal function of the form y=A sin(ωx-ϕ)+β that fits the data. Show your work.

##### 3 Answers

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Here's a rough method...

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Here's one naive and slightly long-winded approach, which may or may not be the sort of thing the posers of the question had in mind...

Given:

We want to match this to:

#f(x) = A sin(omega x - phi) + beta#

where

The natural period of

So

Note that

#(6.21, 3.91)# ,#(7.02, 4.36)# ,#(7.84, 5.31)# ,

#(8.19, 6.21)# ,#(8.06, 7.02)# ,#(7.41, 7.84)# ,

#(6.30, 8.19)# ,#(5.21, 8.06)# ,#(4.28, 7.41)# ,

#(3.91, 6.30)# ,#(4.36, 5.21)# ,#(5.31, 4.28)#

The centre of the circle is

#m=1/12(3.91+4.36+5.31+6.21+7.02+7.84+8.19+8.06+7.41+6.30+5.21+4.28) = 6.175#

If all of the

The distance between

#sqrt((x_2-x_1)^2+(y_2-y_1)^2)#

Putting

#2.27# ,#2.00# ,#1.88# ,#2.02# ,#2.07# ,#2.07# ,#2.02# ,#2.12# ,#2.26# ,#2.27# ,#2.06# ,#2.08#

with mean:

So far, we have determined

Notice that

So we might as well approximate the minimum of the sinusoid at

So our formula is:

#f(x) = A sin(omega x - phi) + beta = 2.09 sin(pi/6 x - (2pi)/3) + 6.175#

which results in monthly values:

#4.09# ,#4.37# ,#5.13# ,#6.17# ,#7.22# ,#7.98# ,

#8.27# ,#7.98# ,#7.22# ,#6.17# ,#5.13# ,#4.37#

graph{(y-(2.09 sin(pi x / 6-2pi/3) + 6.175))((x-1)^2+(y-3.91)^2-0.03)((x-2)^2+(y-4.36)^2-0.03)((x-3)^2+(y-5.31)^2-0.03)((x-4)^2+(y-6.21)^2-0.03)((x-5)^2+(y-7.02)^2-0.03)((x-6)^2+(y-7.84)^2-0.03)((x-7)^2+(y-8.19)^2-0.03)((x-8)^2+(y-8.06)^2-0.03)((x-9)^2+(y-7.41)^2-0.03)((x-10)^2+(y-6.30)^2-0.03)((x-11)^2+(y-5.21)^2-0.03)((x-12)^2+(y-4.28)^2-0.03) = 0 [-0.85, 19.15, 1.56, 11.56]}

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Describe your changes (optional) 200

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This looks more complicated than it is. Break it down into the basic components. You are given the equation form (GOOD - often we have to figure that out on our own!) that we want to fit the data to. Nevermind the complicated form, *basically* , it is a sine function, modified by an amplitude (A) and an offset (B). Everything inside the sin() doesn't matter, as whatever they are must reduce to some value from which the sine is calculated.

Thus, we really just need to "fit" the equation:

The only observed value it the rainfall, so the only "x" metric is the month. We should end up with a cyclical curve showing the expected rainfall as a function of the month number, 1-12.

A sine function only has values between

Similarly, the data range is 3.91 - 8.19, or

NOW we have the basic equation:

Given a value of y and a month number (1-12) calculate the multipliers

For May this would be:

For August the same process gives:

Set these equal (

Omega is then

The final equation is thus:

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Describe your changes (optional) 200

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See below.

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Considering

Solving this system of nonlinear equations with a numerical method like Newton-Raphson's we get

Attached a plot solving the fitting

NOTE:

The term

Describe your changes (optional) 200