I need to create and equation that approximates my data for this sinusidal equation , how would I create one? I know it will have to begin with y=

jjennylove Apr 25, 2019

#1**+1 **

No clue, Jenny....did your instructor point you to any technology that might be used to do this???.....

CPhill Apr 25, 2019

#2**+2 **

Well actually this is a sample , I was unsure how they found out the period was 12 , why it was pi/6 , as well as why they changed 25 to negatitive in order to be that equation from the table above.

jjennylove
Apr 25, 2019

#3**0 **

sorry for the confusion , hopefully i explained it better where im confused

jjennylove
Apr 25, 2019

#5**+2 **

The period is 12 because there are 12 months in a year.

The lowest temp is 40 degrees in January

The highest is 89 in july (6 months later)

The average of these is (40+89)/2 = 129/2 = 64.5

89-64.5 = 24.5

y=-24.3cos(nx)

12=2pi/n

n=12/(2pi) = 6/pi

\(y= -24.5 cos \frac{\pi x}{6}+64.5 \qquad\text{The angle is in radians}\\\)

Here is the graph

Oh dear this is meant to be a sinusoidal graph.

The sine curve is just this with a phase shift of a quarter of a period (3 months)

\(y= -24.5 cos \frac{\pi x}{6}+64.5 \qquad\text{The angle is in radians}\\ y= -24.5 sin \frac{\pi }{6}(x+3)+64.5 \\ or\\ y= 24.5 sin \frac{\pi }{6}(x-3)+64.5 \)

Here are the graphs

Melody
Apr 26, 2019

#6**+1 **

im not quite sure where you got the 64.5? Your equation looks a bit different from the example.

jjennylove
Apr 26, 2019

#7**+1 **

The middle of summer is the top of the sine curve - that is july 89 degrees

The middle of winter is the bottom of the curve -that is Jan 40 degrees

The middle of the curve is the average of these. (89+40)/2 = 129/2 = 64.5

Melody
Apr 26, 2019

#8**0 **

I posted another question on y profile called " Pleas explain how they got these values" ,I am very confused.

jjennylove
Apr 28, 2019