The depth of water in a tank oscillates every 5 hours. The smallest depth is 0.5, and that the largest depth is 7.5. Find a formula for the depth as a function of time t measured in hours for which the depth at time t=0 is the largest depth.

- anonymous

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- anonymous

Sorry, I have to log off now. Good luck.

- anonymous

@zepdrix Can you help? :)

- zepdrix

|dw:1443488320174:dw|So we've got our low and our high point.

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## More answers

- anonymous

So the amplitude is 7.5-.5/2 which is 3.5?

- zepdrix

|dw:1443488404092:dw|We start at our highest point.
And you've calculated the amplitude correctly, ok good!

- zepdrix

Where is the middle of our function going to be located?

- anonymous

Not sure?

- zepdrix

|dw:1443488549410:dw|

- anonymous

3.75?

- zepdrix

3.5 + 0.5
Hmm that doesn't sound right :d

- anonymous

Ohhhh oops I see now, 4

- zepdrix

|dw:1443488682271:dw|Ok great.
And it completes one full oscillation in 5 hours.
So the shape should look something like this, ya?

- anonymous

Beautiful!

- zepdrix

So should we model this curve with `sine` or `cosine`?
What do you think? :)
One will be a bit easier than the other.
You have to think about your starting point.

- anonymous

This is where I always have a problem. How exactly do you decide whether to use sine or cosine?

- zepdrix

|dw:1443488906789:dw|

- anonymous

Does sine always cross at (0,0)?

- zepdrix

|dw:1443489123175:dw|So which one seems more appropriate? :)

- anonymous

cosine!

- zepdrix

Cosine, ok good.

- zepdrix

Sine always crosses through the middle point at the start.
It won't necessarily be (0,0), especially if we have any kind of vertical shift.

- anonymous

Ok! Is the period 2pi/5 or did I do that incorrectly?

- zepdrix

Example: \(\large\rm f(x)=\sin(x)+1\)|dw:1443489226950:dw|Yes, normally this curve would go through (0,0) to start, but it's been shift up by 1.

- anonymous

Oh ok that makes so much sense

- zepdrix

The period is 5 hours.
The letter b that we're going to stick into our formula:\[\large\rm f(x)=A \sin(bx)+d\]Is given by \(\large\rm b=\frac{2\pi}{period}\).
So if that's what you meant, then yes.
Our b is going to be 2pi/5.

- anonymous

yep that's what I meant, great!

- zepdrix

Woops cosine :) not sine, my bad.
Let's stick in what we know so far, amplitude and our b,\[\large\rm f(x)=3.5 \cos\left(\frac{2\pi}{5}x\right)+d\]

- anonymous

How do we find D?

- zepdrix

Ummm depends how you like to think of it I guess...
I like to think of it as a vertical shift of the middle line.

- zepdrix

normally that middle line is the x-axis.
But now it's up at 4, ya?

- anonymous

yep

- anonymous

so is 4 d?

- zepdrix

sounds right!\[\large\rm f(x)=3.5 \cos\left(\frac{2\pi}{5}x\right)+4\]
And to be a little more accurate, it's shifting the entire function up by 4.
You could calculate that shift from any point on the curve.
I think it's just easier to work from the middle.

- zepdrix

yay team \c:/

- anonymous

YAY! Thanks so much, you're the bomb!!

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