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.

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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.

Mathematics
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Sorry, I have to log off now. Good luck.
@zepdrix Can you help? :)
|dw:1443488320174:dw|So we've got our low and our high point.

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So the amplitude is 7.5-.5/2 which is 3.5?
|dw:1443488404092:dw|We start at our highest point. And you've calculated the amplitude correctly, ok good!
Where is the middle of our function going to be located?
Not sure?
|dw:1443488549410:dw|
3.75?
3.5 + 0.5 Hmm that doesn't sound right :d
Ohhhh oops I see now, 4
|dw:1443488682271:dw|Ok great. And it completes one full oscillation in 5 hours. So the shape should look something like this, ya?
Beautiful!
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.
This is where I always have a problem. How exactly do you decide whether to use sine or cosine?
|dw:1443488906789:dw|
Does sine always cross at (0,0)?
|dw:1443489123175:dw|So which one seems more appropriate? :)
cosine!
Cosine, ok good.
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.
Ok! Is the period 2pi/5 or did I do that incorrectly?
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.
Oh ok that makes so much sense
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.
yep that's what I meant, great!
Woops cosine :) not sine, my bad. Let's stick in what we know so far, amplitude and our b,\[\large\rm f(x)=3.5 \cos\left(\frac{2\pi}{5}x\right)+d\]
How do we find D?
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.
normally that middle line is the x-axis. But now it's up at 4, ya?
yep
so is 4 d?
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.
yay team \c:/
YAY! Thanks so much, you're the bomb!!

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