Question

The following data represents the normal monthly precipitation for a certain city.( I will post the table)
Draw a scatter diagram of the data for one period. Find the sinusoidal function of the form (I will post) that fits the data.

Answer 1

1st pic is the tale
2nd pic is for "Find the sinusoidal function of the form___________ that fits the data"

Answer 2

i mean table srry

Answer 3

so they want you to find a regression trig equation? or do they want you to interpolate this?

Answer 4

i have no idea. i think basically they want the sinusoidal function

Answer 5

thats a tough one

Answer 6

well here's the scatter plot of the data

Answer 7

so the good news is that we can see where it peaks, but the bad news is that there isn't enough data to find out where it bottoms out (or where it repeats)

Answer 8

my guess is that that point at (1,3.91) is the lowest point, but that's just a wild guess really

Answer 9

if this is the case though, you can find the following

A = (max - min)/2

B = (max+min)/2

w = 2pi/T where T is the period

phi can be found by plugging in one point and solving for it

Answer 10

this won't be perfect, but it's the closest you can get (i think)

Answer 11

so what would the answer be?

Answer 12

@jim_thompson5910

Answer 13

what's the max y value?

Answer 14

8.19

Answer 15

what about the min y value

Answer 16

3.91

Answer 17

so max - min = ???

Answer 18

4.28

Answer 19

@jim_thompson5910

Answer 20

now cut that in half

Answer 21

to find the amplitude

Answer 22

2.14

Answer 23

so A = 2.14

Answer 24

Now add the max and min, then divide by 2. This will give you the value B

Answer 25

B=8.005

Answer 26

hmm i'm getting 6.05

Answer 27

how are you getting that

Answer 28

opps my mistake ur right. 6.05

Answer 29

as for w, you need to find the period (basically the time length it takes to repeat)

Answer 30

how would i find the period?

Answer 31

well this is where the guess work comes in, but it looks like it starts to repeat at x = 12

Answer 32

oops i mean it starts to dip back down at x = 12, agreed?

Answer 33

seems like it would be hitting the x axis at 16

Answer 34

well i guess it repeats, there so this means

T = 12

since the length from x = 1 to x = 12 is 12

Answer 35

so

w = 2pi/T

w = 2pi/12

w = pi/6

Answer 36

so 2.14sin(pi/6 x-(phi))+6.05

how do u figure out phi?

Answer 37

finally

y = A*sin(wx - phi) + B

y = 2.14*sin((pi/6)x - phi) + 6.05

6.21 = 2.14*sin((pi/6)(4) - phi) + 6.05

6.21 = 2.14*sin(2pi/3 - phi) + 6.05

6.21 - 6.05 = 2.14*sin(2pi/3 - phi)

0.16 = 2.14*sin(2pi/3 - phi)

0.16/2.14 = sin(2pi/3 - phi)

0.07476635 = sin(2pi/3 - phi)

sin(2pi/3 - phi) = 0.07476635

2pi/3 - phi = arcsin(0.07476635)

Can you take it from here?

Answer 38

im confused.. :/ what are you trying to do?

Answer 39

I'm solving for phi

Answer 40

think of phi as a variable

Answer 41

okay so having this, 2pi/3 - phi = arcsin(0.07476635) how would i solve for phi?

Answer 42

2pi/3 - phi = arcsin(0.07476635)

2pi/3 - phi = 4.28779745 ... use the "arcsine" feature on your calculator

- phi = 4.28779745 - 2pi/3

- phi = 2.19340235

phi = -2.19340235

Answer 43

okay so 2.14sin(pi/6 x-(-2.19))+6.05 is the answer?

Answer 44

Yes, it's an approximation though (see graph attached) because it doesn't go through every point. This is known as an regression equation (or regression curve).

Answer 45

so y=2.14sin(pi/6 x-(-2.19))+6.05 i have to put the y= part right?

Answer 46

@jim_thompson5910

Answer 47

yes you can also state it as

y=2.14sin(pi/6 x+2.19)+6.05

Answer 48

thank u so much for your time! U are a big help!

Answer 49

i'm glad