Question

I have done all that i can for this math project, however there is a few questions that confuse me. Can someone please help me, This is 15% of my final exam???

We are given the data for the monthly temperatures for the city of Rock Hill in 2012.
Jan=(43) Fed=(47) Mar=(54) Apr=(63) May=(70) June=(78) July=(81) Aug=(80) Sep=(73) Oct=(63) Nov=(54) Dec=(45)

The parts i don't get are.....
3.) Find a trigonometric model which closely approximates the data. Use x for the month and y for the average temperature. Tip: y=a sin (bx+c)+d

Answer 1

Find the Average value. This should serve as "d".

Find the minimum and maximum values. With a little luck, they will be about the same distance from the average value. This will be quite convenient as it will give you a clue about "a". It won't give it to you directly, but almost.

It might be reasonable to believe the period is 12. What would that say about "b"?

At this point, I'm not sure we need "c".

Answer 2

im confused. once we plug that in do we slove?

Answer 3

You must understand the structure.

y=a sin (bx+c)+d

d is the average value.
a is about the amplitude.
b has to do with the period.
c has to do with phase shift, which we may need.

Answer 4

I understand that part

Answer 5

Well, then get to it.

What's the average?

Answer 6

of all the temperatrues together or the month and the average?

Answer 7

?? There are 12 values. What is the average of the 12 values?

Answer 8

62.6

Answer 9

One thing. You basic sine function is this

y = sin(x)

This is the same as your monster sine function

y = a*sin(bx + c) + d

Right now, if we pick a = b = 1 and c = d = 0, we have y = sin(x). This is NOT a very good representation of the data.

With the calculation of the average, we have

y = a*sin(bx + c) + 62.2

Again, with a = b = 1 and c = 0, we have y = sin(x) + 62.2

This is a better representation of the data.

Now, let's find the min and max.

Answer 10

ok would the max be 81 and the min be 43?

Answer 11

Right. How far is each of those from the average?

Answer 12

19.6 for the lowest and 18.4 for the highest

Answer 13

Does that fill you with warm, fuzzy feelings about the sine function being an adequate representation of the data? If they were EXACTLY the same, I would be all excited. What do you think? Close enough to "the same" to proceed?

Answer 14

i think so, so its y=19.6sin(12x+c)+62.6 ?

Answer 15

Don't get ahead of yourself. One piece at a time.

Is 19.6 a good choice? It's a good choice for the maximum, but not that great a choice for the minimum. I used (81-43)/2 = 19. This is awesome for neither max nor min, but is a good compromise for both. Agreed?

If so, we can move on to the period.

Answer 16

sorry, i get excited. i agree.

Answer 17

All right, we now have \(y = 19\cdot sin(x) + 62.6\) and this still has b = 1 and c = 0.

Okay, now we need a period of 12. The secret to this is the existing period of the sine function which is \(2\pi\). What do we need to make the period 12 instead of \(2\pi\)? You already know it has something to do with 'b', but what exactly is it?

Answer 18

2pi/b? =pi/6?

Answer 19

Perfect. \(12 = \dfrac{2\pi}{b} \implies b = \dfrac{\pi}{6}\). Now we have \(y = 19\cdot \sin\left(\dfrac{\pi}{6}x\right) + 62.6\).

Are you drawing each of these so we can see how close were getting?

Answer 20

yes sir/maam

Answer 21

Good. What do we have left? Is the minimum of our present function in January?

Answer 22

yes

Answer 23

?? Not the minimum of the data. Where is the minimum of the new function? Looks to me like it's in September. That will never do.

Answer 24

oh i though it was the min of the data

Answer 25

Agreed that the function minimum is in September?

Answer 26

yes

Answer 27

Okay, how do we move it four months? We need it to be in January. We need "c"!! Remember how I said I wasn't sure we needed "c"? Well, now I'm sure. We need it.

What value for 'x' will move it up four months?

Answer 28

Just for the record, I'm going to guess that you will not select the correct value, but it is totally not your fault. I'll explain after you come up with a first attempt.

Answer 29

would it be 10?

Answer 30

?? I was expecting 8 or 4. We want to move the minimum four months to the future or eight months back. 10?

10 is actually not too bad, but only because of some weird coincidences.

The real values are -2.1 or +4.2. Try them out and see which you like better. I'll explain it after you are convinced.

Answer 31

Im sorry i'm a litle confused

Answer 32

Try it out: \(y = 19\cdot\sin\left(\dfrac{\pi}{6}x - 2.1\right) + 62.6\).

See how good it is. The problem statement said we could "closely [approximate] the data". Does this function do that??

Answer 33

it looks like it.

Answer 34

Okay, then what's up with that -2.1? We wanted to move 4 and we used 2.1? How can that possible makes sense?

You should not like the answer AT ALL. The problem statement is written badly.

This:

y=a sin (bx+c)+d

would more helpfully be written as

y=a sin (b[x+c])+d

That makes ALL the difference.

We could use 4 if the problem statement had presented it correctly.

\(\dfrac{\pi}{6}(x - 4) = \dfrac{\pi}{6}x - 2.094395\) and there is our crazy -2.1.

If it were me, I might present the final answer as \(y = 19\cdot\sin\left(\dfrac{\pi}{6}(x - 4)\right) + 62.6\). This will be wonderfully correct. It will show the direct relationship with the four months we had to move the minimum value. We will also emphasize that the original problem statement was not the most useful form we could have used from the beginning.

What say you? Am I making any sense?

Answer 35

yes

Answer 36

so is thats the answer?

Answer 37

Is it a good representation of the data? If so, then yes.

It is an exciting piece of work. Rather amazing, I think, that we can do that sort of thing with just a little effort.

Answer 38

yes sir/maam. Could you help me with another part?

Answer 39

Don't you sleep?