# Sinusoidal Function Help?

• December 19th 2009, 03:06 PM
Jaydee
Sinusoidal Function Help?
...
• December 19th 2009, 04:50 PM
Prove It
Quote:

Originally Posted by Jaydee
Hello all!

My previous thread was closed; I had a bit of a confusion on one of the forum's rules, which is highly embarrassing. I apologize, but I hope I can be cut some slack as it was my very first post here!
Anyways, I am studying for an upcoming grade 11 Pre-Calc exam, and I have a few questions giving me some trouble. Here is one of them, which I know for sure I am doing incorrectly:

1. The number of cups of coffee sold weekly is periodic over the course of a year. The maximum weekly sales of 12,000 cups usually occur around the end of January (week 4). The minimum weekly sales of 2,000 cups usually occur at the end of July (week 30).
The weekly sales can be modeled by a sinusoidal function.
a) Write an equation to model the weekly sales of coffee.
(I attempted and got something like: y = -5000sin[5(x-4)] + 7000 but I’m pretty certain that’s quite off!)

If someone could help me or walk me through this question, that would be much appreciated. I won't lie, this semester I've been pretty much math-stupid, but if someone explains this question to me I will more than likely grasp it with very little difficulty. My teacher hasn't been doing a very good job at... well... teaching. 60% is the class average and we're fast approaching our last unit! : S

JD

If the maximum is 12,000 and the minimum is 2000, what do you think the median value must be?

Once you have that, what do you think the amplitude must be?
• December 19th 2009, 04:53 PM
Jaydee
Well, 12,000 + 2000 / 2 = 7,000, yes?
Therefore from 7000 it is 5000 to 12000, and 5000 to 2000. So the amplitude is 5000.
I have already figured that out correctly, I believe....
• December 19th 2009, 04:56 PM
Prove It
Quote:

Originally Posted by Jaydee
Well, 12,000 + 2000 / 2 = 7,000, yes?
Therefore from 7000 it is 5000 to 12000, and 5000 to 2000. So the amplitude is 5000.
I have already figured that out correctly, I believe....

Correct.

So for the function of the form $y = a\sin{(bx - c)} + d$, you already have $a = 5000$ and $d = 7000$.

What do you think the period has to be?
• December 19th 2009, 05:02 PM
Jaydee
The period is over the course of a year. There are about 52 weeks in a year.

Would I say that my period is 52? Or is it 360/52? I apologize in advance, it has been a while since sinusoidal functions.

I'm leaning more towards 52 being written in the equation, however.
• December 19th 2009, 09:31 PM
Prove It
Yes, the period is 52.

Remember that the period is given by $\frac{2\pi}{b}$.

So now solve $\frac{2\pi}{b} = 52$ for $b$.

Also, how much of a horizontal translation has there been? Remember that a regular sine function begins at the origin.
• December 20th 2009, 10:19 AM
Jaydee
360/52 = 6.9

Horizontal translations are something I have always been pretty dreadful at, and something I need a lot of practice with. I will hazard a guess, however...

If the max starts at week 4, and the min at week 30, and a sine function always starts at the origin, then there has been a horizontal shift 30 to the right?
So x - 30 ?
• December 20th 2009, 02:29 PM
Prove It
No.

First, if you have trigonometric FUNCTIONS or GRAPHS, they are always measured in RADIANS.

So solve the equation $\frac{2\pi}{b} = 52$ for $b$.

You should find that $b = \frac{\pi}{26}$.

So now you have

$y = 5000\sin{\left(\frac{\pi x}{26} - c\right)} + 7000$.

You also have some points that you know lie on the graph, namely $(x, y) = (4, 12 000)$ and $(30, 2000)$.

So, let $x$ and $y$ equal some of these values.

$12000 = 5000\sin{\left(\frac{4\pi}{26} - c\right)} + 7000$

$12000 = 5000\sin{\left(\frac{2\pi}{13} - c\right)} + 7000$

$5000 = 5000\sin{\left(\frac{2\pi}{13} - c\right)}$

$1 = \sin{\left(\frac{2\pi}{13} - c\right)}$

$\frac{2\pi}{13} - c = \arcsin{1}$

$\frac{2\pi}{13} - c = \frac{\pi}{2}$

$c = \frac{2\pi}{13} - \frac{\pi}{2}$

$c = - \frac{9\pi}{26}$.

So this means there has been a horizontal translation of $\frac{9\pi}{26}$ units to the left.

And now, finally, your function is

$y = 5000\sin{\left(\frac{\pi x}{26} + \frac{9\pi}{26}\right)} + 7000$.
• December 21st 2009, 01:22 PM
Jaydee
My apologies. We haven't been taught radians yet--that's grade 12 here in Canada--though I understand your equations. However we will not be asked to use radians on any sort of test situation, so I would like to make sure I am getting the same answer in the 'correct' format for my course.

pt. (4, 12000)
12000 = 5000sin [6.9(4) - d] + 7000
5000 = 5000sin (27.6 - d)
1 = sin (27.6 - d)
arcsin (1) = 27.6 - d
90 - 27.6 = d
d = 62.4

Therefore the equation is ...

y = 5000sin (6.9x - 62.4) + 7000 ?
• December 22nd 2009, 03:18 PM
Prove It
Looks good to me. However, I would be asking your teachers why you are in Year 12 and have not been taught Radian Measure, when it should have been taught in Year 10. Very sloppy of them...