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- Dec 19th 2009, 03:06 PMJaydeeSinusoidal Function Help?
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- Dec 19th 2009, 04:50 PMProve It
- Dec 19th 2009, 04:53 PMJaydee
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.... - Dec 19th 2009, 04:56 PMProve It
- Dec 19th 2009, 05:02 PMJaydee
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. - Dec 19th 2009, 09:31 PMProve It
Yes, the period is 52.

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

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

Also, how much of a horizontal translation has there been? Remember that a regular sine function begins at the origin. - Dec 20th 2009, 10:19 AMJaydee
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 ? - Dec 20th 2009, 02:29 PMProve It
No.

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

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

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

So now you have

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

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

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

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

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

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

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

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

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

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

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

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

And now, finally, your function is

$\displaystyle y = 5000\sin{\left(\frac{\pi x}{26} + \frac{9\pi}{26}\right)} + 7000$. - Dec 21st 2009, 01:22 PMJaydee
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.

So in non-radian-speak, we have...

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 ? - Dec 22nd 2009, 03:18 PMProve 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...